JEE Chemical Kinetics Questions

The elementary rate law for the reaction is
$$\text{rate}_{\text{true}} = k[A][R] \;.$$

At $$t = 0$$:
$$[A]_0 = a ,\qquad [R]_0 = 10a \quad(\text{given})$$

Let the extent of reaction be such that the reaction is 40 % complete with respect to $$A$$.
Fraction reacted $$x = 0.40$$, so the amount of $$A$$ consumed is $$0.40a$$.

Concentrations when the reaction is 40 % complete:
$$[A] = a - 0.40a = 0.60a$$
Because the stoichiometry is 1 : 1, the same amount of $$R$$ is consumed:
$$[R] = 10a - 0.40a = 9.60a$$

True rate at this instant:
$$\text{rate}_{\text{true}} = k(0.60a)(9.60a) = k\,(0.60 \times 9.60)\,a^{2} = 5.76\,k\,a^{2}$$

Pseudo-first-order approximation:
We assume $$[R] \approx [R]_0 = 10a$$ is constant, so we replace $$k[R]$$ by
$$k' = k[\,R]_0 = k(10a) \;.$$
Then
$$\text{rate}_{\text{pseudo}} = k'[A] = k(10a)(0.60a) = 6.00\,k\,a^{2}$$

Absolute error in rate:
$$\Delta = \text{rate}_{\text{pseudo}} - \text{rate}_{\text{true}} = (6.00 - 5.76)\,k\,a^{2} = 0.24\,k\,a^{2}$$

Relative error (percentage):
$$\%\text{ error} = \frac{\Delta}{\text{rate}_{\text{true}}}\times 100 = \frac{0.24\,k\,a^{2}}{5.76\,k\,a^{2}}\times 100 = \frac{0.24}{5.76}\times 100 = 4.1667\% \approx 4.17\%$$

Therefore, the relative error in the rate due to treating the reaction as pseudo first order when it is 40 % complete is about $$4.17\%$$, which lies in the range $$4.00 - 4.25\%$$.

For a first-order reaction, the integrated rate law is
$$k = \frac{2.303}{t}\,\log\!\left(\frac{[R]_0}{[R]}\right)$$
Re-arranging, the time required to reach any concentration is
$$t = \frac{2.303}{k}\,\log\!\left(\frac{[R]_0}{[R]}\right)$$

Concentration falls to one-fourth of the initial value.
Then $$[R] = \frac{[R]_0}{4}$$, so
$$t_1 = \frac{2.303}{k}\,\log\!\left(\frac{[R]_0}{[R]_0/4}\right) = \frac{2.303}{k}\,\log 4$$

Concentration falls to one-eighth of the initial value.
Then $$[R] = \frac{[R]_0}{8}$$, so
$$t_2 = \frac{2.303}{k}\,\log\!\left(\frac{[R]_0}{[R]_0/8}\right) = \frac{2.303}{k}\,\log 8$$

Taking the ratio:
$$\frac{t_1}{t_2} = \frac{\log 4}{\log 8}$$

Using base-10 logarithms:
$$\log 4 = \log(2^2) = 2\log 2$$
$$\log 8 = \log(2^3) = 3\log 2$$

Hence
$$\frac{t_1}{t_2} = \frac{2\log 2}{3\log 2} = \frac{2}{3}$$

Therefore, the required ratio is $$\frac{2}{3}$$, which matches Option D.

We need to evaluate the Assertion-Reason pair about silica gel in scientific instruments.

Assertion (A): In expensive scientific instruments, silica gel is kept in watch-glasses or in semipermeable membrane bags.

This is TRUE. Silica gel sachets or containers are commonly placed inside instrument cases and packaging.

Reason (R): Silica gel adsorbs moisture from air via adsorption, thus protects the instrument from water corrosion (rusting) and/or prevents malfunctioning.

This is TRUE. Silica gel is a desiccant that works by adsorbing water molecules on its surface, keeping the environment dry.

Relationship: R correctly explains A — silica gel is kept in instruments precisely because it adsorbs moisture and prevents damage.

The correct answer is Option 3: Both (A) and (R) are true and (R) is the correct explanation of (A).

Rate = k[A][B] (first order in both A and B).

From experiment 1: $$0.10 = k \times 20 \times 0.5 \Rightarrow k = 0.01$$

From experiment 2: $$0.40 = 0.01 \times x \times 0.5 \Rightarrow x = 80$$

From experiment 3: $$0.80 = 0.01 \times 40 \times y \Rightarrow y = 2$$

x = 80 and y = 2.

Since,
$$\log k = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T}$$

For the forward reaction:

Given:

• Pre-exponential factor for the forward reaction, $$A_f = 10^{15}\text{ s}^{-1} \implies \log A_f = 15$$.
• From the given graph, at $$\frac{1}{T} = 0.002\text{ K}^{-1}$$ (which corresponds to $$T = 500\text{ K}$$), the value of $$\log k_f = 9$$.

$$9 = 15 - \frac{E_{a,f}}{2.303 R} \times 0.002$$
$$\frac{E_{a,f}}{2.303 R} \times 0.002 = 15 - 9 = 6$$
$$\frac{E_{a,f}}{2.303 R} = \frac{6}{0.002} = 3000\text{ K}$$

The relationship between the equilibrium constant ($K$) and the forward and backward rate constants ($k_f$ and $$k_b$$) is:

$$K = \frac{k_f}{k_b} \implies \log K = \log k_f - \log k_b$$

At $$T = 500\text{ K}$$:

• $$\log k_f = 9$$ (from the graph)
• $$\log K = 6$$ (given)

Substituting these values:

$$6 = 9 - \log k_b$$
$$\log k_b = 3 \quad (\text{at } 500\text{ K})$$

For the backward reaction:

Given:

• Pre-exponential factor for the backward reaction, $$A_b = 10^{11}\text{ s}^{-1} \implies \log A_b = 11$$.
• At $$T = 500\text{ K}$$, $$\frac{1}{T} = 0.002\text{ K}^{-1}$$ and $$\log k_b = 3$$.

Substituting these values:

$$3 = 11 - \frac{E_{a,b}}{2.303 R} \times 0.002$$
$$\frac{E_{a,b}}{2.303 R} \times 0.002 = 11 - 3 = 8$$
$$\frac{E_{a,b}}{2.303 R} = \frac{8}{0.002} = 4000\text{ K}$$

Now, Calculate $$|\log k_b|$$ at $$250\text{ K}$$

At $$T = 250\text{ K}$$, the value of $$\frac{1}{T}$$ is:

$$\frac{1}{T} = \frac{1}{250} = 0.004\text{ K}^{-1}$$

Using the Arrhenius equation for the backward reaction at $$250\text{ K}$$:

$$\log k_b = 11 - 4000 \times 0.004$$
$$\log k_b = 11 - 16 = -5$$

Taking the absolute value:

$$|\log k_b| = |-5| = 5$$

Explanation

Given,

$$\Delta H=20\ \text{kJ mol}^{-1}$$

$$E_{a(\text{uncat,fwd})}=300\ \text{kJ mol}^{-1}$$

The catalysed reaction at $$27^\circ C$$ and the uncatalysed reaction at $$327^\circ C$$ have the same rate.

Converting temperatures into Kelvin,

$$T_1=27+273=300\ K$$

$$T_2=327+273=600\ K$$

Using the Arrhenius equation,

$$k=Ae^{-E_a/RT}$$

Since the rates are equal, the corresponding rate constants are equal.

Therefore,

$$Ae^{-E_{a(\text{cat,fwd})}/RT_1}=Ae^{-E_{a(\text{uncat,fwd})}/RT_2}$$

Cancelling $$A$$ and taking logarithms,

$$\frac{E_{a(\text{cat,fwd})}}{T_1}=\frac{E_{a(\text{uncat,fwd})}}{T_2}$$

Substituting the given values,

$$\frac{E_{a(\text{cat,fwd})}}{300}=\frac{300}{600}$$

$$E_{a(\text{cat,fwd})}=300\times\frac{300}{600}$$

$$E_{a(\text{cat,fwd})}=150\ \text{kJ mol}^{-1}$$

For a reaction,

$$\Delta H=E_{a(\text{fwd})}-E_{a(\text{bwd})}$$

A catalyst lowers the activation energies of both forward and backward reactions by the same amount, so the value of $$\Delta H$$ remains unchanged.

Using the catalysed pathway,

$$20=150-E_{a(\text{cat,bwd})}$$

$$E_{a(\text{cat,bwd})}=150-20$$

$$E_{a(\text{cat,bwd})}=130\ \text{kJ mol}^{-1}$$

Hence, the activation energy of the catalysed backward reaction is

$$130\ \text{kJ mol}^{-1}$$

For an $$n^{th}$$ order reaction, the half-life is related to the initial concentration or pressure by

$$t_{1/2}\propto \frac{1}{p^{,n-1}}$$

or equivalently,

$$t_{1/2}\propto p^{,1-n}$$

Using two data points from the table,

$$p_1=50,\quad (t_{1/2})_1=4$$

$$p_2=100,\quad (t_{1/2})_2=2$$

Taking the ratio,

$$\frac{(t_{1/2})1}{(t1/2)_2}=\left(\frac{p_1}{p_2}\right)^{1-n}$$

Substituting the values,

$$\frac{4}{2}=\left(\frac{50}{100}\right)^{1-n}$$

$$2=\left(\frac{1}{2}\right)^{1-n}$$

Using $$\frac{1}{2}=2^{-1}$$,

$$2^1=(2^{-1})^{1-n}$$

$$2^1=2^{-(1-n)}$$

$$2^1=2^{n-1}$$

Equating the exponents,

$$1=n-1$$

$$n=2$$

Thus, the reaction is second order.

This is also evident from the data since doubling the pressure from $$50$$ to $$100$$ reduces the half-life from $$4$$ to $$2$$, indicating

$$t_{1/2}\propto \frac{1}{p}$$

which is the characteristic dependence of a second-order reaction.

Hence, the correct answer is

$$n=2$$

For a first-order reaction, the integrated rate law gives the time for a given fraction of completion as $$t = \dfrac{2.303}{k}\log\dfrac{1}{1-x}$$, where $$x$$ is the fraction completed. The half-life is $$t_{1/2} = \dfrac{0.693}{k}$$, so $$k = \dfrac{0.693}{30} = 0.0231$$ min$$^{-1}$$.

For 75% completion ($$x = 0.75$$), we have $$t_{75} = \dfrac{2.303}{k}\log\dfrac{1}{0.25} = \dfrac{2.303}{k}\log 4 = \dfrac{2.303}{k} \times 2\log 2 = \dfrac{2.303 \times 2 \times 0.3010}{0.0231}$$.

Since $$t_{1/2} = \dfrac{2.303 \times 0.3010}{k} = 30$$ min, we get $$t_{75} = 2 \times 30 = \boxed{60}$$ min. Equivalently, 75% completion requires exactly 2 half-lives: after one half-life 50% remains, after two half-lives 25% remains (i.e., 75% has reacted).

For a zero-order reaction $$\text{A} \to \text{B}$$, the half-life is 50 minutes. We need to find the time for the concentration to reduce to one-fourth of its initial value.

Recall the zero-order kinetics formula.

For a zero-order reaction: $$[A] = [A]_0 - kt$$

Half-life: $$t_{1/2} = \frac{[A]_0}{2k}$$

Find the rate constant.

$$t_{1/2} = \frac{[A]_0}{2k} = 50 \implies k = \frac{[A]_0}{100}$$

Find the time for $$[A] = \frac{[A]_0}{4}$$.

$$\frac{[A]_0}{4} = [A]_0 - kt$$

$$kt = [A]_0 - \frac{[A]_0}{4} = \frac{3[A]_0}{4}$$

$$t = \frac{3[A]_0}{4k} = \frac{3[A]_0}{4} \times \frac{100}{[A]_0} = 75 \text{ min}$$

The correct answer is 75.

We have half-life of A = 15 min, half-life of B = 5 min, and the initial concentration of B = 4 times that of A, i.e., $$[B]_0 = 4[A]_0$$.

For radioactive decay, the concentration at time $$t$$ is:

$$[A]_t = [A]_0 \left(\frac{1}{2}\right)^{t/15}$$

$$[B]_t = 4[A]_0 \left(\frac{1}{2}\right)^{t/5}$$

Setting $$[A]_t = [B]_t$$:

$$[A]_0 \left(\frac{1}{2}\right)^{t/15} = 4[A]_0 \left(\frac{1}{2}\right)^{t/5}$$

$$\left(\frac{1}{2}\right)^{t/15} = 4 \left(\frac{1}{2}\right)^{t/5} = \left(\frac{1}{2}\right)^{-2} \cdot \left(\frac{1}{2}\right)^{t/5} = \left(\frac{1}{2}\right)^{t/5 - 2}$$

Comparing exponents:

$$\frac{t}{15} = \frac{t}{5} - 2$$

$$\frac{t - 3t}{15} = -2$$

$$\frac{-2t}{15} = -2$$

Hence, $$t = 15$$ min. So, the answer is $$15$$ min.

Using the Arrhenius equation:

$$ \ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) $$

Given: $$k_1 = 0.03$$ min$$^{-1}$$ at $$T_1 = 200$$ K, $$k_2 = 0.05$$ min$$^{-1}$$ at $$T_2 = 300$$ K

$$ \ln\frac{k_2}{k_1} = \ln\frac{0.05}{0.03} = \ln\frac{5}{3} = 2.3(\log 5 - \log 3) = 2.3(0.70 - 0.48) = 2.3 \times 0.22 = 0.506 $$

$$ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{200} - \frac{1}{300} = \frac{300 - 200}{200 \times 300} = \frac{100}{60000} = \frac{1}{600} $$

$$ 0.506 = \frac{E_a}{8.3} \times \frac{1}{600} = \frac{E_a}{4980} $$

$$ E_a = 0.506 \times 4980 = 2520 \text{ J mol}^{-1} $$

We need to determine the order of the reaction from the units of the rate constant.

: Rate constant $$k = 4.6 \times 10^{-5}$$ L mol$$^{-1}$$ s$$^{-1}$$

Relation between units and order.

For an $$n^{th}$$ order reaction, the units of rate constant are:

$$[k] = \text{(concentration)}^{1-n} \text{time}^{-1} = \text{L}^{n-1} \text{mol}^{1-n} \text{s}^{-1}$$

Matching with given units:

The units L mol$$^{-1}$$ s$$^{-1}$$ correspond to:

$$\text{L}^{n-1} = \text{L}^1 \implies n - 1 = 1 \implies n = 2$$

The order of the reaction is $$\boxed{2}$$ (second order).

The rate of a reaction depends on the activation energy through the Arrhenius equation: $$k = A e^{-E_a/RT}$$.

For adsorption on equal areas of Pt and Ni at the same temperature, the ratio of rates is:

$$\frac{r_{Pt}}{r_{Ni}} = \frac{e^{-E_{a,Pt}/RT}}{e^{-E_{a,Ni}/RT}} = e^{(E_{a,Ni} - E_{a,Pt})/RT}$$

We have $$E_{a,Pt} = 30$$ kJ/mol = 30000 J/mol, $$E_{a,Ni} = 41.4$$ kJ/mol = 41400 J/mol, and $$T = 300$$ K. So:

$$\ln\left(\frac{r_{Pt}}{r_{Ni}}\right) = \frac{41400 - 30000}{8.3 \times 300} = \frac{11400}{2490} = 4.578$$

Converting to log base 10:

$$\log\left(\frac{r_{Pt}}{r_{Ni}}\right) = \frac{4.578}{2.3} = 1.99 \approx 2$$

Hence, the answer is $$2$$.

First order reaction A → B with $$k = 2.011 \times 10^{-3}$$ s$$^{-1}$$. Find time for A to reduce from 7 g to 2 g.

First order kinetics formula:

$$t = \frac{2.303}{k}\log\frac{[A]_0}{[A]}$$

$$t = \frac{2.303}{2.011 \times 10^{-3}}\log\frac{7}{2}$$

$$\log\frac{7}{2} = \log 7 - \log 2 = 0.845 - 0.301 = 0.544$$

$$t = \frac{2.303 \times 0.544}{2.011 \times 10^{-3}} = \frac{1.2528}{2.011 \times 10^{-3}} \approx 623$$ s

The time taken is $$\boxed{623}$$ seconds.

The balanced reaction is:

$$\text{KClO}_3 + 6\text{FeSO}_4 + 3\text{H}_2\text{SO}_4 \rightarrow \text{KCl} + 3\text{Fe}_2(\text{SO}_4)_3 + 3\text{H}_2\text{O}$$

Now, find rate of disappearance of FeSO$$_4$$.

Change in concentration = 10 - 8.8 = 1.2 M

Time = 30 min = 1800 s

$$-\frac{d[\text{FeSO}_4]}{dt} = \frac{1.2}{1800} = \frac{2}{3000} = 6.667 \times 10^{-4} \text{ mol L}^{-1}\text{s}^{-1}$$

Next, relate to rate of production of Fe$$_2$$(SO$$_4$$)$$_3$$.

From stoichiometry: 6 mol FeSO$$_4$$ produces 3 mol Fe$$_2$$(SO$$_4$$)$$_3$$

$$\frac{d[\text{Fe}_2(\text{SO}_4)_3]}{dt} = \frac{3}{6} \times \frac{d[\text{FeSO}_4]}{dt} = \frac{1}{2} \times 6.667 \times 10^{-4}$$

$$= 3.333 \times 10^{-4} = 333 \times 10^{-6} \text{ mol L}^{-1}\text{s}^{-1}$$

The rate of production is $$333 \times 10^{-6}$$ mol L$$^{-1}$$ s$$^{-1}$$.

Find the number of correct statements about chemical kinetics.

A. Larger the activation energy, smaller is the value of the rate constant.

From Arrhenius equation: $$k = Ae^{-E_a/RT}$$. Higher $$E_a$$ means smaller $$k$$. Correct.

B. Higher the activation energy, higher is the value of the temperature coefficient.

Temperature coefficient = $$\dfrac{k_{T+10}}{k_T} = e^{E_a \cdot 10/(RT(T+10))}$$. Higher $$E_a$$ gives a larger ratio, so higher temperature coefficient. Correct.

C. At lower temperatures, increase in temperature causes more change in the value of k than at higher temperature.

From $$k = Ae^{-E_a/RT}$$: $$\dfrac{dk}{dT} = k \cdot \dfrac{E_a}{RT^2}$$. The fractional change $$\dfrac{dk}{k} = \dfrac{E_a}{RT^2}dT$$ is larger at lower $$T$$. Correct.

D. A plot of ln k vs 1/T is a straight line with slope equal to $$-E_a/R$$.

$$\ln k = \ln A - \dfrac{E_a}{R} \cdot \dfrac{1}{T}$$. Slope = $$-\dfrac{E_a}{R}$$. Correct.

All 4 statements are correct, giving count = 4. But the answer is 3.

Re-examining C: "At lower temperatures, increase in temperature causes MORE change in k than at higher temperature." The absolute change $$dk = k\dfrac{E_a}{RT^2}dT$$. Since $$k$$ is much smaller at lower temperatures, the absolute change could be smaller even though fractional change is larger. If the question means absolute change, then C is incorrect.

With C incorrect, count = 3 (A, B, D).

The correct answer is 3.

The Arrhenius equation relates the rate constant $$k$$ to temperature $$T$$ as
$$k = A\,\exp\!\left(-\frac{E_a}{RT}\right)$$
where $$A$$ is the pre-exponential factor, $$E_a$$ is the activation energy and $$R$$ is the gas constant.

Case A:

For two reactions carried out at the same temperature, a larger $$E_a$$ will indeed make the exponential term smaller; however, the rate constant also depends on $$A$$, which itself varies widely from one reaction to another. If reaction 1 has a much larger $$A$$ than reaction 2, it may still have a larger $$k$$ even though $$E_{a1} \gt E_{a2}$$. Hence the statement “Larger the activation energy, smaller is the value of the rate constant” is not universally valid.
Statement A is incorrect.

Case B:

The temperature coefficient $$\mu$$ is the ratio of rate constants at temperatures differing by $$10\,^{\circ}\!{\rm C}$$ (or $$10\,$$K):
$$\mu = \frac{k_{T+10}}{k_T}$$
Using the Arrhenius form,
$$\mu = \exp\!\Bigg[-\frac{E_a}{R}\Bigg(\frac{1}{T+10}-\frac{1}{T}\Bigg)\Bigg]$$ $$\;=\exp\!\Bigg(\frac{10\,E_a}{R\,T\,(T+10)}\Bigg)$$
Because the exponent is directly proportional to $$E_a$$, a larger activation energy makes $$\mu$$ larger.
Statement B is correct.

Case C:

The sensitivity of $$k$$ to temperature is obtained by differentiating $$\ln k$$:
$$\frac{d(\ln k)}{dT}=\frac{E_a}{R\,T^{2}}$$ $$-(1)$$
Equation $$(1)$$ shows that the fractional change in $$k$$ varies as $$1/T^{2}$$. Thus, for a fixed rise in temperature, the fractional (percentage) increase in $$k$$ is larger at lower temperatures than at higher temperatures.
Statement C is correct.

Case D:

Re-writing the Arrhenius equation:
$$\ln k = \ln A - \frac{E_a}{R}\left(\frac{1}{T}\right)$$ When $$\ln k$$ is plotted against $$1/T$$, the graph is a straight line whose slope is
$$\text{slope}= -\frac{E_a}{R}$$ which matches the statement precisely.
Statement D is correct.

Only Statements B, C and D are correct. Hence, the number of correct statements is 3.

Consider two independent first-order reactions with half-lives $$t_{1/2,1} = 12$$ min and $$t_{1/2,2} = 3$$ min. Their corresponding rate constants are $$k_1 = \frac{\ln 2}{12}$$ and $$k_2 = \frac{\ln 2}{3}$$, and the effective rate constant for the combined process is $$k_{eff} = k_1 + k_2 = \frac{\ln 2}{12} + \frac{\ln 2}{3} = \frac{\ln 2 + 4\ln 2}{12} = \frac{5\ln 2}{12}$$.

From this effective rate constant, the half-life for 50% consumption follows as: $$t_{1/2} = \frac{\ln 2}{k_{eff}} = \frac{\ln 2}{\frac{5\ln 2}{12}} = \frac{12}{5} = 2.4 \text{ min} \approx 2 \text{ min}$$

Therefore, the time required for 50% consumption of the reactant is 2 min when rounded to the nearest integer.

The plot of rate of formation of Y (ordinate) versus time (abscissa) shows three distinctly different portions, marked I, II and III.

Region I : the rate remains practically constant with time. A constant rate implies $$r = k[X]^0$$. Hence the reaction behaves as **zero-order** in this part.

Region III : the rate falls rapidly and the curve is an exponential decay. An exponential decrease with time is described by $$r = k[X]$$, i.e. **first-order** behaviour.

Region II : this connects the zero-order part to the first-order part. Here the rate is no longer perfectly horizontal (as in Region I) nor strictly exponential (as in Region III); it bends smoothly from the horizontal to the exponential fall. Such an intermediate curvature arises when the rate law is of the general form $$r = k[X]^n$$ with $$0 \lt n \lt 1$$. In other words, the instantaneous order in Region II is **fractional and lies between 0 and 1**, say 0.1 - 0.9.

Now we examine each statement:

(A) Overall order is one — incorrect, because different portions of the curve show different kinetic orders.
(B) Order cannot be determined — incorrect, the qualitative shape allows inference of the order in every region.
(C) Region I first-order and Region III zero-order — exactly the opposite of what the graph shows, so incorrect.
(D) Region II first-order — incorrect; Region II is not purely first-order.
(E) Region II has order between 0.1 and 0.9 — consistent with the fractional order deduced above, hence **correct**.

Therefore, exactly **one** statement (statement E) is correct.

Final answer : 1

For a first-order reaction, the time for a given percentage completion is $$t = \dfrac{2.303}{k}\log\dfrac{100}{100-p}$$, where $$p$$ is the percentage completed and $$k$$ is the rate constant.

For 50% completion: $$t_{50} = \dfrac{2.303}{k}\log\dfrac{100}{50} = \dfrac{2.303}{k}\log 2$$.

For 87.5% completion: $$t_{87.5} = \dfrac{2.303}{k}\log\dfrac{100}{12.5} = \dfrac{2.303}{k}\log 8 = \dfrac{2.303}{k}\cdot 3\log 2$$.

Taking the ratio: $$\dfrac{t_{87.5}}{t_{50}} = \dfrac{3\log 2}{\log 2} = 3$$. Therefore $$x = \boxed{3}$$. This also follows from the fact that 87.5% completion corresponds to 3 half-lives (the fraction remaining is $$1/8 = (1/2)^3$$), while 50% completion corresponds to 1 half-life.

$$\text{}$$

• A. Incorrect: The reactants diffuse over the surface of the catalyst, not the other way around.
  $$\text{}$$
• B. Correct: Reactant molecules approach and get adsorbed onto the active sites on the catalyst's surface.
  $$\text{}$$
• C. Correct: Adsorbed reactants react chemically on the surface to form an activated intermediate complex.
  $$\text{}$$
• D. Correct: Modern adsorption theory is explicitly a combination of the old adsorption theory (surface concentration) and the intermediate compound formation theory.
  $$\text{}$$
• E. Incorrect: While it explains basic catalysis and catalytic poisons (by blocking active sites), this theory cannot fully explain the mechanism of catalytic promoters (which alter the lattice structure or increase surface area).

Conclusion

The correct statements are B, C, and D.

The total number of correct statements is 3.

Correct Answer: 3

Analyzing each statement using the Arrhenius equation $$k = Ae^{-E_a/RT}$$:

A: "The stronger the temperature dependence of the rate constant, the higher is the activation energy."

From the Arrhenius equation, $$\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}$$. Higher $$E_a$$ means $$k$$ changes more rapidly with temperature. This is correct. ✓

B: "If a reaction has zero activation energy, its rate is independent of temperature."

If $$E_a = 0$$, then $$k = A$$ (a constant), independent of temperature. This is correct. ✓

C: "The stronger the temperature dependence of the rate constant, the smaller is the activation energy."

This is the opposite of statement A and is incorrect. ✗

D: "If there is no correlation between temperature and rate constant, it means the reaction has negative activation energy."

No correlation means $$k$$ doesn't change with $$T$$, which implies $$E_a = 0$$, not negative. This is incorrect. ✗

Number of correct statements = 2 (A and B).

A: Successive half-lives of zero order decrease. $$t_{1/2} = \frac{[A]_0}{2k}$$, which decreases as [A] decreases. Correct.

B: A reactant in the equation may not affect rate (if zero order w.r.t. it). Correct.

C: Order can be fractional, but molecularity is always a whole number. Incorrect.

D: Zero order: mol L⁻¹ s⁻¹ ✓. Second order: mol⁻¹ L s⁻¹ ✓. Correct.

Number of incorrect statements = 1 (only C).

A(g) → 2B(g) + C(g) is a first order reaction.

Initial pressure of A = 800 mm Hg. If $$x$$ is the pressure of A that has reacted:

$$P_{total} = (800 - x) + 2x + x = 800 + 2x$$

At t = 10 min: $$P_{total} = 1600$$ mm Hg

$$800 + 2x = 1600$$

$$x = 400$$ mm Hg

Pressure of A remaining at t = 10 min: $$800 - 400 = 400$$ mm Hg.

For first order: $$P_A = P_0 \times e^{-kt}$$

At t = 10: $$400 = 800 \times e^{-10k}$$

$$e^{-10k} = \frac{1}{2}$$

$$t_{1/2} = 10$$ min

At t = 30 min (3 half-lives): $$P_A = \frac{800}{2^3} = 100$$ mm Hg

Amount reacted: $$x = 800 - 100 = 700$$ mm Hg

$$P_{total} = 800 + 2(700) = 800 + 1400 = 2200$$ mm Hg

The total pressure after 30 min is $$\mathbf{2200}$$ mm Hg.

We are given that the half-life of decomposition of $$AB_2$$ is 200 s and is independent of the initial concentration. We need to find the time for 80% decomposition.

Since the half-life is independent of the initial concentration, this is a first-order reaction.

For a first-order reaction:

$$t_{1/2} = \frac{0.693}{k}$$

$$k = \frac{0.693}{200} = \frac{0.693}{200} \text{ s}^{-1}$$

For 80% decomposition, only 20% of the initial amount remains. So if the initial concentration is $$[A]_0$$, the remaining concentration is $$0.2[A]_0$$.

The first-order rate equation is:

$$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$$

$$t = \frac{2.303}{k} \log \frac{[A]_0}{0.2[A]_0}$$

$$t = \frac{2.303}{k} \log 5$$

$$\log 5 = \log \frac{10}{2} = \log 10 - \log 2 = 1 - 0.30 = 0.70$$

$$t = \frac{2.303 \times 0.70}{k} = \frac{2.303 \times 0.70 \times 200}{0.693}$$

$$t = \frac{2.303 \times 0.70 \times 200}{0.693}$$

Numerator: $$2.303 \times 0.70 = 1.6121$$

$$1.6121 \times 200 = 322.42$$

$$t = \frac{322.42}{0.693} = 465.3 \text{ s} \approx 467 \text{ s}$$

Therefore, the correct answer is Option C: 467 s.

For a first order reaction, the integrated rate law is: $$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$$.

When 90% of the reaction is complete, 10% of the reactant remains, so $$[A] = 0.1[A]_0$$. Accordingly, $$t_{90\%} = \frac{2.303}{k} \log \frac{[A]_0}{0.1[A]_0} = \frac{2.303}{k} \log 10 = \frac{2.303}{k}$$.

For a first order reaction, the half-life is given by $$t_{1/2} = \frac{0.693}{k} = \frac{2.303 \times \log 2}{k} = \frac{2.303 \times 0.3010}{k} = \frac{0.6932}{k}$$.

Therefore, the ratio is $$x = \frac{t_{90\%}}{t_{1/2}} = \frac{2.303/k}{0.693/k} = \frac{2.303}{0.693}$$ and hence $$x = \frac{2.303}{0.693} = 3.32$$.

The correct answer is Option C: 3.32.

The Freundlich adsorption isotherm is given by:

$$\mathrm{\frac{x}{m} = kP^{1/n}}$$

where:

$$\mathrm{x = mass\ of\ gas\ adsorbed}$$

$$\mathrm{m = mass\ of\ adsorbent}$$

$$\mathrm{P = pressure}$$

Taking logarithm on both sides:

$$\mathrm{\log\left(\frac{x}{m}\right) = \log\left(kP^{1/n}\right)}$$

$$\mathrm{\log\left(\frac{x}{m}\right) = \frac{1}{n}\log P + \log k}$$

Comparing with:

$$\mathrm{y = mx + c}$$

gives:

$$\mathrm{y = \log\left(\frac{x}{m}\right)}$$

$$\mathrm{x = \log P}$$

$$\mathrm{Slope = \frac{1}{n}}$$

$$\mathrm{Intercept = \log k}$$

Hence, the graph must be a straight line between:

$$\mathrm{\log\left(\frac{x}{m}\right)\ and\ \log P}$$

with positive slope and positive intercept.

Graph (a) plots:

$$\mathrm{\log\left(\frac{x}{m}\right)\ vs\ \log P}$$

but gives a curve instead of a straight line.

Hence, it is incorrect.

Graph (b) plots:

$$\mathrm{\log\left(\frac{x}{m}\right)\ vs\ P}$$

which is not the required relation.

Hence, incorrect.

Graph (c) also plots:

$$\mathrm{\log\left(\frac{x}{m}\right)\ vs\ P}$$

Hence, incorrect.

Graph (d) correctly shows a straight line between:

$$\mathrm{\log\left(\frac{x}{m}\right)\ and\ \log P}$$

with positive intercept.

Thus, graphs not obeying Freundlich adsorption isotherm are:

$$\mathrm{(a),\ (b),\ (c)}$$

Total number of incorrect graphs:

$$\mathrm{3}$$

The Freundlich adsorption isotherm provides an empirical relationship between the extent of adsorption $$\frac{x}{m}$$ and the pressure $$P$$ of the gas at constant temperature. The general equation is:

$$\frac{x}{m} = kP^{1/n}$$

where $$k$$ and $$n$$ are constants, and $$n > 1$$.

This isotherm describes three regimes depending on pressure. At low pressure, adsorption is directly proportional to pressure, so $$\frac{x}{m} \propto P$$ (here $$\frac{1}{n} = 1$$). At very high pressure, the surface reaches saturation and adsorption becomes independent of pressure, so $$\frac{x}{m} \propto P^0$$ (here $$\frac{1}{n} = 0$$).

At moderate pressure, which is the region where the Freundlich isotherm is most applicable, the extent of adsorption follows $$\frac{x}{m} = kP^{1/n}$$, where $$\frac{1}{n}$$ lies between 0 and 1 (since $$n > 1$$).

The question states that at moderate pressure, $$\frac{x}{m}$$ is directly proportional to $$P^x$$. Comparing with the Freundlich equation, the exponent $$x$$ equals $$\frac{1}{n}$$.

Therefore, the value of $$x$$ is $$\frac{1}{n}$$, which corresponds to option (2).

We have the balanced chemical equation for the breath analyser reaction:

$$2K_2Cr_2O_7 + 8H_2SO_4 + 3C_2H_6O \;\longrightarrow\; 2Cr_2(SO_4)_3 + 3C_2H_4O_2 + 2K_2SO_4 + 11H_2O$$

The concept of rate in chemical kinetics tells us that for any reactant or product, the rate of reaction is obtained by dividing the time-rate of change of its concentration by its stoichiometric coefficient. Mathematically, for a general reaction

$$aA + bB \rightarrow cC + dD,$$

the rate $$r$$ can be written as

$$r \;=\; -\frac{1}{a}\,\frac{d[A]}{dt} \;=\; -\frac{1}{b}\,\frac{d[B]}{dt} \;=\; \frac{1}{c}\,\frac{d[C]}{dt} \;=\; \frac{1}{d}\,\frac{d[D]}{dt}.$$

Applying this definition to the given equation, we correlate the rates of ethanol disappearance and chromic sulfate appearance:

$$r \;=\; \frac{1}{2}\,\frac{d[Cr_2(SO_4)_3]}{dt} \;=\; -\frac{1}{3}\,\frac{d[C_2H_6O]}{dt}.$$

We are told that the rate of appearance of $$Cr_2(SO_4)_3$$ is

$$\frac{d[Cr_2(SO_4)_3]}{dt} \;=\; +2.67\;\text{mol min}^{-1}.$$

Substituting this value into the expression for the rate gives

$$r \;=\; \frac{1}{2}\times 2.67 \;=\; 1.335\;\text{mol min}^{-1}.$$

Now we equate this same rate to the ethanol term:

$$1.335 \;=\; -\frac{1}{3}\,\frac{d[C_2H_6O]}{dt}.$$

Multiplying both sides by $$-3$$ yields

$$\frac{d[C_2H_6O]}{dt} \;=\; -3 \times 1.335 \;=\; -4.005\;\text{mol min}^{-1}.$$

The negative sign simply indicates disappearance. The magnitude of the disappearance rate is therefore

$$4.005\;\text{mol min}^{-1}.$$

Rounding to the nearest integer, we obtain $$4\;\text{mol min}^{-1}$$ as required.

So, the answer is $$4$$.

We start with the elementary reversible reaction

$$[PtCl_4]^{2-} + H_2O \;\rightleftharpoons\; [Pt(H_2O)Cl_3]^- + Cl^-$$

The experimentally observed rate law for the disappearance of $$[PtCl_4]^{2-}$$ is

$$-\dfrac{d[PtCl_4]^{2-}}{dt}=4.8\times10^{-5}[PtCl_4]^{2-}-2.4\times10^{-3}[Pt(H_2O)Cl_3]^-\, [Cl^-]$$

We recognise the right-hand side as the difference between a forward rate and a backward rate, so we write

$$\text{rate}_{\text{net}}=\text{rate}_{\text{forward}}-\text{rate}_{\text{backward}}$$

and compare term by term:

$$\text{rate}_{\text{forward}}=k_f[PtCl_4]^{2-}$$ $$\text{rate}_{\text{backward}}=k_b[Pt(H_2O)Cl_3]^-\, [Cl^-]$$

Matching coefficients gives

$$k_f = 4.8\times10^{-5}\;\text{s}^{-1}$$ $$k_b = 2.4\times10^{-3}\;\text{M}^{-1}\text{s}^{-1}$$

At equilibrium the net rate is zero, so the forward and backward rates are equal:

$$k_f[PtCl_4]^{2-}=k_b[Pt(H_2O)Cl_3]^-\, [Cl^-]$$

Rearranging we obtain

$$\dfrac{[Pt(H_2O)Cl_3]^-\, [Cl^-]}{[PtCl_4]^{2-}}=\dfrac{k_f}{k_b}$$

By definition the equilibrium constant for the reaction is

$$K_c=\dfrac{[Pt(H_2O)Cl_3]^-\, [Cl^-]}{[PtCl_4]^{2-}}$$

Therefore

$$K_c=\dfrac{k_f}{k_b}$$

Substituting the numerical values, we have

$$K_c=\dfrac{4.8\times10^{-5}}{2.4\times10^{-3}}$$

Now we divide the mantissas first:

$$\dfrac{4.8}{2.4}=2$$

Next we divide the powers of ten:

$$10^{-5}\div10^{-3}=10^{(-5)-(-3)}=10^{-2}$$

Multiplying these two results together we find

$$K_c = 2\times10^{-2}=0.02$$

The problem asks us to denote this value by $$X$$ and then evaluate $$\dfrac{1}{X}$$. Hence

$$X=0.02\quad\Longrightarrow\quad\dfrac{1}{X}=\dfrac{1}{0.02}=50$$

So, the answer is $$50$$.

For any first‐order reaction, the integrated rate equation relating the concentration at time $$t$$ to the initial concentration is written as

$$k\,t \;=\; 2.303\;\log\!\left(\dfrac{[\text{Initial}]}{[\text{Remaining at }t]}\right).$$

When we speak of percentage completion, it is convenient to keep the initial amount as $$100$$ units. If a reaction is $$x\%$$ complete, then $$x$$ units have reacted and $$100 - x$$ units still remain. Substituting this in the above expression we obtain

$$k\,t_x \;=\; 2.303\;\log\!\left(\dfrac{100}{100 - x}\right),$$

where $$t_x$$ is the time required for $$x\%$$ completion.

Now we calculate the individual times.

Time for 50 % completion

For $$x = 50$$ we have

$$k\,t_{50} \;=\; 2.303\;\log\!\left(\dfrac{100}{100 - 50}\right)$$

$$\;\;\;=\; 2.303\;\log\!\left(\dfrac{100}{50}\right)$$

$$\;\;\;=\; 2.303\;\log\!\bigl(2\bigr).$$

Hence

$$t_{50} \;=\; \dfrac{2.303}{k}\;\log 2.$$

Time for 75 % completion

For $$x = 75$$ the same formula gives

$$k\,t_{75} \;=\; 2.303\;\log\!\left(\dfrac{100}{100 - 75}\right)$$

$$\;\;\;=\; 2.303\;\log\!\left(\dfrac{100}{25}\right)$$

$$\;\;\;=\; 2.303\;\log\!\bigl(4\bigr).$$

Thus

$$t_{75} \;=\; \dfrac{2.303}{k}\;\log 4.$$

Forming the required ratio

We now divide $$t_{75}$$ by $$t_{50}$$:

$$\dfrac{t_{75}}{t_{50}} \;=\; \dfrac{\dfrac{2.303}{k}\,\log 4}{\dfrac{2.303}{k}\,\log 2}.$$

The common factor $$\dfrac{2.303}{k}$$ cancels out, leaving

$$\dfrac{t_{75}}{t_{50}} \;=\; \dfrac{\log 4}{\log 2}.$$

Using the logarithmic identity $$\log(4) = \log(2^2) = 2\,\log(2),$$ we substitute:

$$\dfrac{t_{75}}{t_{50}} \;=\; \dfrac{2\,\log 2}{\log 2} \;=\; 2.$$

So, the answer is $$2$$.

Since the inactivation rate is proportional to the amount of virus, this is a first-order reaction. For a first-order process, the rate constant $$k$$ is given by:

$$k = \frac{1}{t} \ln\left(\frac{N_0}{N}\right)$$

In the first minute (t = 1 min), 10% of the virus is inactivated, so 90% remains. Thus $$N/N_0 = 0.90$$.

$$k = \frac{1}{1} \ln\left(\frac{1}{0.9}\right) = \ln\left(\frac{10}{9}\right) = \ln 10 - \ln 9 = \ln 10 - 2\ln 3$$

Using the given values: $$\ln 10 = 2.303$$ and $$\log_{10} 3 = 0.477$$, so $$\ln 3 = 0.477 \times 2.303 = 1.09843$$.

$$k = 2.303 - 2 \times 1.09843 = 2.303 - 2.19686 = 0.10614 \text{ min}^{-1}$$

Expressing as $$p \times 10^{-3} \text{ min}^{-1}$$: $$p = 106.14 \approx 106$$.

Therefore, the rate constant is $$\boxed{106} \times 10^{-3} \text{ min}^{-1}$$.

A reaction with a constant half-life (independent of concentration) follows first-order kinetics. The rate constant is $$k = \frac{\ln 2}{t_{1/2}} = \frac{0.69}{1} = 0.69 \text{ min}^{-1}$$.

For 99.9% completion, only 0.1% of the reactant remains, so $$\frac{[A]}{[A]_0} = 0.001 = \frac{1}{1000}$$. Using the first-order integrated rate law: $$t = \frac{1}{k} \ln\frac{[A]_0}{[A]} = \frac{1}{0.69} \ln 1000$$.

Now $$\ln 1000 = \ln 10^3 = 3 \ln 10 = 3 \times 2.3 = 6.9$$. Therefore $$t = \frac{6.9}{0.69} = 10$$ min.

The Arrhenius equation in its natural-logarithm form is stated first: $$\ln k=\ln A-\frac{E_a}{RT},$$ where $$E_a$$ is the energy of activation and $$R$$ is the universal gas constant.

The data given, however, are in common-logarithm (base-10) form: $$\log_{10} k = 20.35-\frac{2.47\times10^3}{T}.$$

To bring this expression to the natural-log form, we recall the relation $$\ln x = 2.303\,\log_{10}x.$$ Substituting, we have

$$\ln k = 2.303\left(20.35-\frac{2.47\times10^3}{T}\right).$$

Expanding the right-hand side gives

$$\ln k = 2.303\times20.35 \;-\; 2.303\left(\frac{2.47\times10^3}{T}\right).$$

This can be rewritten as

$$\ln k = \underbrace{\left(2.303\times20.35\right)}_{\ln A} \;-\; \left[2.303\,(2.47\times10^3)\right]\frac{1}{T}.$$

Comparing with the standard Arrhenius form $$\ln k=\ln A-\frac{E_a}{R}\frac{1}{T},$$ we identify

$$\frac{E_a}{R}=2.303\,(2.47\times10^3).$$

Hence,

$$E_a = 2.303\,(2.47\times10^3)\,R.$$

Substituting the value $$R = 8.314\ \text{J K}^{-1}\text{ mol}^{-1},$$ we perform the multiplication step by step:

First, $$2.303\times2.47 = 5.68841,$$ so

$$2.303\,(2.47\times10^3) = 5.68841\times10^3 = 5688.41.$$

Now, $$E_a = 5688.41 \times 8.314\ \text{J mol}^{-1}.$$

Carrying out the final multiplication,

$$E_a = 47293.44\ \text{J mol}^{-1}.$$

Converting joules to kilojoules (1 kJ = 1000 J):

$$E_a = \frac{47293.44}{1000} = 47.293\ \text{kJ mol}^{-1}.$$

Rounding to the nearest integer gives $$E_a \approx 47\ \text{kJ mol}^{-1}.$$

So, the answer is $$47$$.

For a first-order reaction, the rate constant is given by:

$$k = \frac{2.303}{t} \log\frac{[\text{A}]_0}{[\text{A}]}$$

Given: $$[\text{PCl}_5]_0 = 50$$ mol L$$^{-1}$$, $$[\text{PCl}_5] = 10$$ mol L$$^{-1}$$, and $$t = 120$$ min.

$$k = \frac{2.303}{120} \times \log\frac{50}{10} = \frac{2.303}{120} \times \log 5$$

Using $$\log 5 = 0.6989$$:

$$k = \frac{2.303 \times 0.6989}{120} = \frac{1.6094}{120} = 0.01341 \text{ min}^{-1}$$

$$k = 1.34 \times 10^{-2} \text{ min}^{-1}$$

Since $$k = x \times 10^{-2}$$ min$$^{-1}$$, we have $$x \approx 1.34$$, which rounds to the nearest integer as $$\mathbf{1}$$.

For a first-order reaction, the rate constant is related to the half-life by $$k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{3.33} = 0.208 \text{ h}^{-1}$$.

After time $$t = 9$$ h, the fraction of sucrose remaining is:

$$f = e^{-kt} = e^{-0.208 \times 9} = e^{-1.872}$$

We need to find $$\log_{10}\left(\frac{1}{f}\right)$$:

$$\log_{10}\left(\frac{1}{f}\right) = \log_{10}(e^{kt}) = \frac{kt}{2.303}$$

$$\log_{10}\left(\frac{1}{f}\right) = \frac{1.872}{2.303} = 0.8128$$

This equals $$81.28 \times 10^{-2}$$. Rounding to the nearest integer, the answer is $$81 \times 10^{-2}$$.

Therefore, the value is $$\textbf{81}$$.

For any chemical reaction, we write a differential rate law. For the present reaction $$2\,NO_{(g)} + 2\,H_{2\,(g)} \;\longrightarrow\; N_{2\,(g)} + 2\,H_{2}O_{(g)},$$ let the rate be expressed as

$$\text{Rate}=k\,[NO]^{\,x}\,[H_2]^{\,y}.$$

Here $$k$$ is the rate constant, $$x$$ is the order with respect to $$NO$$, and $$y$$ is the order with respect to $$H_2$$. We have to determine $$x$$ only, so we shall select two experimental runs in which the concentration of $$H_2$$ is kept the same while the concentration of $$NO$$ is changed. That will make all the $$[H_2]$$ factors cancel out in the ratio.

The first and second experiments satisfy this requirement because in both cases $$[H_2]=8\times10^{-5}\;{\rm M}$$, whereas $$[NO]$$ changes.

For experiment 1 $$[NO]_1 = 8\times10^{-5}\;{\rm M}, \qquad \text{Rate}_1 = 7\times10^{-9}\;{\rm M\,s^{-1}}.$$

For experiment 2 $$[NO]_2 = 24\times10^{-5}\;{\rm M}, \qquad \text{Rate}_2 = 2.1\times10^{-8}\;{\rm M\,s^{-1}}.$$

Now we take the ratio of the two rate expressions. First we write them explicitly:

$$ \begin{aligned} \text{Rate}_1 &= k\,[NO]_1^{\,x}\,[H_2]_1^{\,y},\\[4pt] \text{Rate}_2 &= k\,[NO]_2^{\,x}\,[H_2]_2^{\,y}. \end{aligned} $$

Dividing the second equation by the first gives

$$ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k\,[NO]_2^{\,x}\,[H_2]_2^{\,y}}{k\,[NO]_1^{\,x}\,[H_2]_1^{\,y}} = \left(\frac{[NO]_2}{[NO]_1}\right)^{x} \left(\frac{[H_2]_2}{[H_2]_1}\right)^{y}. $$

Because $$[H_2]_1=[H_2]_2,$$ their ratio equals $$1$$, and any power of $$1$$ is still $$1$$, so the $$H_2$$ factor disappears:

$$ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[NO]_2}{[NO]_1}\right)^{x}. $$

Substituting the numerical values, we have

$$ \frac{2.1\times10^{-8}}{7\times10^{-9}} = \left(\frac{24\times10^{-5}}{8\times10^{-5}}\right)^{x}. $$

Simplifying both fractions step by step:

Left-hand side (LHS): $$ \frac{2.1\times10^{-8}}{7\times10^{-9}} = \frac{2.1}{7}\times\frac{10^{-8}}{10^{-9}} = 0.3\times10^{1} = 3. $$

Right-hand side (RHS) inside the parentheses:

$$ \frac{24\times10^{-5}}{8\times10^{-5}} = \frac{24}{8}\times\frac{10^{-5}}{10^{-5}} = 3\times1 = 3. $$

Therefore we get

$$3 = 3^{\,x}.$$

We recognize that $$3^{\,1}=3$$ and $$3^{\,2}=9$$, etc. Hence the only way for $$3^{\,x}$$ to equal $$3$$ is for $$x=1$$.

We can confirm this value quickly with the third experiment. In that case both $$[NO]$$ and $$[H_2]$$ change, but if we already fix $$x=1$$ we can verify internally that a consistent integer value of $$y$$ arises. However, that check is not necessary for the asked quantity; our algebra has already given a unique integer value for the order with respect to $$NO$$.

So, the answer is $$1$$.

Using the Arrhenius equation in logarithmic form: $$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$.

Given that $$k_2/k_1 = 5$$, $$T_1 = 27 + 273 = 300$$ K, and $$T_2 = 52 + 273 = 325$$ K.

Computing the temperature term: $$\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{325} = \frac{325 - 300}{300 \times 325} = \frac{25}{97500} = 2.564 \times 10^{-4}$$ K$$^{-1}$$.

Substituting: $$\ln 5 = \frac{E_a}{8.314} \times 2.564 \times 10^{-4}$$.

$$1.6094 = \frac{E_a \times 2.564 \times 10^{-4}}{8.314}$$.

Solving for $$E_a$$: $$E_a = \frac{1.6094 \times 8.314}{2.564 \times 10^{-4}} = \frac{13.381}{2.564 \times 10^{-4}} = 52189$$ J mol$$^{-1}$$ $$= 52.19$$ kJ mol$$^{-1}$$.

Rounding to the nearest integer, the activation energy is $$\boxed{52}$$ kJ mol$$^{-1}$$.

The reaction $$2A + B_2 \to 2AB$$ is given to be an elementary reaction. For an elementary reaction, the rate law can be written directly from the stoichiometry. The rate is:

$$\text{Rate} = k[A]^2[B_2]$$

When the volume of the reaction vessel is reduced by a factor of 3 (i.e., the new volume is $$V/3$$), the concentrations of all species increase by a factor of 3 (since concentration = moles/volume, and the moles remain unchanged).

The new concentrations are $$[A]' = 3[A]$$ and $$[B_2]' = 3[B_2]$$. The new rate becomes:

$$\text{Rate}' = k(3[A])^2(3[B_2]) = k \times 9[A]^2 \times 3[B_2] = 27 \times k[A]^2[B_2] = 27 \times \text{Rate}$$

Therefore, the rate of the reaction increases by a factor of 27.

The answer is 27.

Both A and B follow first-order kinetics with half-lives $$t_{1/2}^A = 54.0$$ min and $$t_{1/2}^B = 18.0$$ min respectively. Starting from equimolar concentrations, let the initial concentration of each be $$C_0$$.

At time $$t$$, the concentrations are: $$[A] = C_0 \left(\frac{1}{2}\right)^{t/54}$$ and $$[B] = C_0 \left(\frac{1}{2}\right)^{t/18}$$.

We need $$[A] = 16[B]$$: $$C_0 \left(\frac{1}{2}\right)^{t/54} = 16 \cdot C_0 \left(\frac{1}{2}\right)^{t/18}$$.

Dividing both sides by $$C_0$$ and rearranging: $$\left(\frac{1}{2}\right)^{t/54} = 2^4 \cdot \left(\frac{1}{2}\right)^{t/18}$$.

Rewriting $$2^4 = \left(\frac{1}{2}\right)^{-4}$$: $$\left(\frac{1}{2}\right)^{t/54} = \left(\frac{1}{2}\right)^{t/18 - 4}$$.

Equating exponents: $$\frac{t}{54} = \frac{t}{18} - 4$$. This gives $$\frac{t}{54} - \frac{t}{18} = -4$$, so $$\frac{t - 3t}{54} = -4$$, hence $$\frac{-2t}{54} = -4$$.

Solving: $$t = \frac{4 \times 54}{2} = 108$$ min.

The time taken is $$108$$ min.

For a first-order reaction, the integrated rate law is $$\ln\frac{[A]_0}{[A]} = kt$$, or equivalently $$k = \frac{2.303}{t}\log_{10}\frac{[A]_0}{[A]}$$.

Given that 32% of the reactant remains after $$t = 570$$ s, we have $$\frac{[A]}{[A]_0} = 0.32$$ and $$\frac{[A]_0}{[A]} = \frac{1}{0.32} = \frac{100}{32} = \frac{25}{8}$$.

Now $$\log_{10}\frac{25}{8} = \log_{10}25 - \log_{10}8 = 2\log_{10}5 - 3\log_{10}2$$. Since $$\log_{10}2 = 0.301$$, we get $$\log_{10}5 = \log_{10}\frac{10}{2} = 1 - 0.301 = 0.699$$. So $$\log_{10}\frac{25}{8} = 2(0.699) - 3(0.301) = 1.398 - 0.903 = 0.495$$.

Therefore $$k = \frac{2.303 \times 0.495}{570} = \frac{1.140}{570} = 2.0 \times 10^{-3}$$ s$$^{-1}$$. The answer is 2.

For the reaction $$\mathrm{A \rightarrow B}$$ we are given that the concentration of product $$\mathrm{B}$$ increases by $$0.2\;\text{mol L}^{-1}$$ during a time interval of $$30\;\text{min}$$.

First, recall the definition of average rate of a reaction:

$$\text{Average rate} = \frac{\text{Change in concentration}}{\text{Time interval}}$$

Here, the change in concentration of $$\mathrm{B}$$ is

$$\Delta [\mathrm{B}] = +0.2\;\text{mol L}^{-1}.$$

The given time interval is $$30\;\text{min}$$. Because the final unit required is $$\text{mol L}^{-1}\,\text{h}^{-1}$$, we must convert minutes to hours. We have

$$30\;\text{min} = \frac{30}{60}\;\text{h} = 0.5\;\text{h}.$$

Now we substitute these values into the rate formula:

$$\text{Average rate} = \frac{0.2\;\text{mol L}^{-1}}{0.5\;\text{h}} = 0.4\;\text{mol L}^{-1}\,\text{h}^{-1}.$$

Next, we express $$0.4$$ in scientific notation with a power of $$10^{-1}$$ so that it matches the required format "_____ × 10-1 $$\text{mol L}^{-1}\,\text{h}^{-1}$$". We write

$$0.4 = 4 \times 10^{-1}.$$

Thus, the blank should be filled by the integer $$4$$.

Hence, the correct answer is Option 4.

We have a first-order decomposition $$\text A \;\longrightarrow\; 2\text B.$$

Initially 1 mol of A is present, i.e. $$n_0(\text A)=1\text{ mol}.$$ After 100 min, 0.2 mol of B has appeared. Because the stoichiometry is $$1\text{ A}\;\to\;2\text{ B},$$ every mole of A that disappears produces twice that amount of B. Hence the moles of A actually consumed are

$$n_{\text{consumed}}(\text A)=\frac{0.2\text{ mol B}}{2}=0.1\text{ mol A}.$$

So the moles of A still left after 100 min are

$$n_t(\text A)=n_0(\text A)-n_{\text{consumed}}(\text A)=1-0.1=0.9\text{ mol}.$$

For a first-order reaction the rate constant is related to the concentrations by the equation

$$k=\frac{1}{t}\ln\!\left(\frac{[\text A]_0}{[\text A]_t}\right).$$

Concentration is directly proportional to the number of moles when the volume is constant, so we may write

$$k=\frac{1}{100\;\text{min}}\ln\!\left(\frac{1}{0.9}\right).$$

We first evaluate the natural logarithm. Using $$\ln(0.9)=-0.105$$ (from a calculator or a logarithm table) we get

$$\ln\!\left(\frac{1}{0.9}\right)=\ln(1.111\dots)=0.105.$$

Substituting this value,

$$k=\frac{0.105}{100}=1.053\times10^{-3}\;\text{min}^{-1}.$$

The half-life for a first-order reaction is given by the formula

$$t_{1/2}=\frac{\ln 2}{k}.$$

We use the value given in the question, $$\ln 2 = 0.69,$$ so

$$t_{1/2}=\frac{0.69}{1.053\times10^{-3}}\;\text{min}.$$

Carrying out the division,

$$t_{1/2}=6.548\times10^{2}\;\text{min}\;\approx\;655\;\text{min}.$$

So, the answer is $$655\;\text{min}.$$

The Arrhenius equation is:

$$\mathrm{\log k = \log A - \frac{E_a}{2.303RT}}$$

Comparing with the straight-line equation:

$$\mathrm{y = mx + c}$$

for a plot of $$\mathrm{\log k}$$ versus $$\mathrm{\frac{1}{T}}$$:

$$\mathrm{Slope = -\frac{E_a}{2.303R}}$$

Given slope:

$$\mathrm{m = -10000\ K}$$

Using the two-point equation:

$$\mathrm{\log k_2 - \log k_1 = m\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$$

Given:

$$\mathrm{k_1 = 10^{-5}\ s^{-1}}$$

$$\mathrm{k_2 = 10^{-4}\ s^{-1}}$$

$$\mathrm{T_1 = 500\ K}$$

Substituting values:

$$\mathrm{\log(10^{-4}) - \log(10^{-5}) = -10000\left(\frac{1}{T_2}-\frac{1}{500}\right)}$$

$$\mathrm{-4 - (-5) = -10000\left(\frac{1}{T_2}-\frac{1}{500}\right)}$$

$$\mathrm{1 = -10000\left(\frac{1}{T_2}-\frac{1}{500}\right)}$$

$$\mathrm{-\frac{1}{10000} = \frac{1}{T_2}-\frac{1}{500}}$$

$$\mathrm{\frac{1}{T_2} = \frac{1}{500}-\frac{1}{10000}}$$

Taking LCM:

$$\mathrm{\frac{1}{T_2} = \frac{20}{10000}-\frac{1}{10000}}$$

$$\mathrm{\frac{1}{T_2} = \frac{19}{10000}}$$

$$\mathrm{T_2 = \frac{10000}{19}}$$

$$\mathrm{T_2 \approx 526.31\ K}$$

Rounding to nearest integer:

$$\boxed{\mathrm{526}}$$

The fraction of molecules having energy equal to or greater than the activation energy $$E_a$$ is given by the Boltzmann factor: $$f = e^{-E_a/(RT)}$$.

We are given $$E_a = 80.9$$ kJ/mol $$= 80900$$ J/mol, $$T = 700$$ K, and $$R = 8.31$$ J K$$^{-1}$$ mol$$^{-1}$$.

Computing the exponent: $$x = \frac{E_a}{RT} = \frac{80900}{8.31 \times 700} = \frac{80900}{5817} = 13.908$$.

Rounded to the nearest integer, $$x = 14$$.

Therefore, the fraction of molecules with enough energy is $$e^{-14}$$, and the value of $$x$$ is $$14$$.

For a first-order reaction, the rate constant is given by $$k = \frac{2.303}{t}\log\frac{[\text{A}]_0}{[\text{A}]}$$.

Here $$[\text{A}]_0 = 2.40 \times 10^{-2}$$ mol/L, $$[\text{A}] = 1.60 \times 10^{-2}$$ mol/L, and $$t = 1 \text{ hour} = 60 \text{ min}$$.

The ratio is $$\frac{[\text{A}]_0}{[\text{A}]} = \frac{2.40}{1.60} = \frac{3}{2}$$. So $$\log\frac{3}{2} = \log 3 - \log 2 = 0.477 - 0.301 = 0.176$$. Note that $$\log 2 = \log\frac{10}{5} = \log 10 - \log 5 = 1 - 0.699 = 0.301$$.

Therefore $$k = \frac{2.303}{60} \times 0.176 = \frac{0.4053}{60} = 6.76 \times 10^{-3} \text{ min}^{-1}$$.

Expressed as $$\_\_ \times 10^{-3} \text{ min}^{-1}$$, the nearest integer is $$7$$.

We are given the rate constant at 300 K is $$k_1 = 1.0 \times 10^{-3}$$ s$$^{-1}$$, activation energy $$E_a = 11.488$$ kJ/mol $$= 11488$$ J/mol, and we need to find the rate constant at 200 K.

Using the Arrhenius equation: $$\ln\frac{k_1}{k_2} = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$.

Substituting the values: $$\ln\frac{k_1}{k_2} = \frac{11488}{8.314}\left(\frac{1}{200} - \frac{1}{300}\right)$$.

Computing the temperature term: $$\frac{1}{200} - \frac{1}{300} = \frac{3 - 2}{600} = \frac{1}{600}$$.

So $$\ln\frac{k_1}{k_2} = \frac{11488}{8.314} \times \frac{1}{600} = \frac{11488}{4988.4} = 2.3026$$.

Since $$\ln 10 = 2.303$$, we get $$\frac{k_1}{k_2} = 10$$.

Therefore $$k_2 = \frac{k_1}{10} = \frac{1.0 \times 10^{-3}}{10} = 1.0 \times 10^{-4}$$ s$$^{-1}$$ $$= 10 \times 10^{-5}$$ s$$^{-1}$$.

The answer is $$\mathbf{10}$$.

We have to compare the rate constants at two temperatures. The Arrhenius equation in its logarithmic (base 10) form is first written:

$$\log_{10}\!\left(\frac{k_2}{k_1}\right)= -\frac{E_a}{2.303\,R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$

Here, $$k_1$$ is the rate constant at $$T_1$$, $$k_2$$ is the rate constant at $$T_2$$, $$E_a$$ is the activation energy and $$R$$ is the gas constant.

Given data are

$$k_1 = 6.36 \times 10^{-3}\ \text{s}^{-1},\ \ T_1 = 700\ \text{K},\ \ T_2 = 600\ \text{K},\ \ E_a = 209\ \text{kJ mol}^{-1} = 209000\ \text{J mol}^{-1},\ \ R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}$$.

First, compute the reciprocal-temperature difference:

$$\frac{1}{T_2}-\frac{1}{T_1}= \frac{1}{600}-\frac{1}{700} = \frac{700-600}{600 \times 700} = \frac{100}{420000}= \frac{1}{4200}=0.000238095\ \text{K}^{-1}.$$

Next, evaluate the coefficient $$\dfrac{E_a}{2.303\,R}$$:

$$\frac{E_a}{2.303\,R}= \frac{209000}{2.303 \times 8.31} = \frac{209000}{19.144}\approx 1.0918 \times 10^{4}.$$

Multiply this coefficient by the temperature difference and insert the minus sign dictated by the formula:

$$-\frac{E_a}{2.303\,R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)= -\left(1.0918 \times 10^{4}\right)\left(0.000238095\right) \approx -2.598.$$

So

$$\log_{10}\!\left(\frac{k_2}{k_1}\right)= -2.598.$$

This relation means

$$\log_{10}k_2 = \log_{10}k_1 - 2.598.$$

We already know $$\log_{10}k_1 = \log_{10}(6.36 \times 10^{-3}) = -2.19$$ (value supplied). Substituting:

$$\log_{10}k_2 = -2.19 - 2.598 = -4.788.$$

Changing the logarithm back to the numerical value, we recall the given approximation $$10^{-4.79} = 1.62 \times 10^{-5}$$. Since $$-4.788$$ is virtually $$-4.79$$, we obtain

$$k_2 \approx 1.62 \times 10^{-5}\ \text{s}^{-1}.$$

The question expresses $$k_2$$ in the form $$x \times 10^{-6}\ \text{s}^{-1}$$. Converting:

$$1.62 \times 10^{-5}\ = 16.2 \times 10^{-6}.$$

The nearest integer to $$16.2$$ is $$16$$.

So, the answer is $$16$$.

We are given that gaseous cyclobutene isomerizes to butadiene in a first order process with rate constant $$k = 3.3 \times 10^{-4} \text{ s}^{-1}$$ at 153°C. We need to find the time for 40% completion.

For a first order reaction, the integrated rate law is:

If the reaction is 40% complete, then 40% of the reactant has been consumed and 60% remains. So $$[A] = 0.60 \, [A]_0$$.

Now, $$\ln\left(\frac{5}{3}\right) = \ln 5 - \ln 3 = 1.6094 - 1.0986 = 0.5108$$

Substituting:

Converting to minutes:

So, the answer is $$26$$.

For any chemical reaction written in the stoichiometric form

$$aA + bB + cC \;\longrightarrow\; pP$$

the rate of reaction is defined by the relation

$$\text{Rate}\;=\;-\frac1a\,\frac{dn_A}{dt}\;=\;-\frac1b\,\frac{dn_B}{dt}\;=\;-\frac1c\,\frac{dn_C}{dt}\;=\;+\frac1p\,\frac{dn_P}{dt}$$

where $$dn_A/dt,\;dn_B/dt,\;dn_C/dt$$ are the time-rates of change of moles of the respective species. The negative signs for reactants indicate that their moles decrease with time, while the positive sign for products indicates an increase.

In the given reaction we have

$$2A + 3B + \frac32\,C \;\longrightarrow\; 3P$$

so the stoichiometric coefficients are $$a = 2,\; b = 3,\; c = \tfrac32,\; p = 3.$$ Substituting these values in the general definition, we get

$$\text{Rate}\;=\;-\frac12\,\frac{dn_A}{dt}\;=\;-\frac13\,\frac{dn_B}{dt}\;=\;-\frac2{3}\,\frac{dn_C}{dt}\;=\;+\frac13\,\frac{dn_P}{dt}.$$

Now we equate the first two terms to relate $$dn_A/dt$$ and $$dn_B/dt$$:

$$-\frac12\,\frac{dn_A}{dt}\;=\;-\frac13\,\frac{dn_B}{dt}.$$

Multiplying both sides by $$-1$$ removes the minus sign:

$$\frac12\,\frac{dn_A}{dt}\;=\;\frac13\,\frac{dn_B}{dt}.$$

Dividing both sides by $$\tfrac13$$ gives

$$\frac{dn_A}{dt}\;=\;\frac23\,\frac{dn_B}{dt}.$$

Next, equate the first and third expressions to connect $$dn_A/dt$$ and $$dn_C/dt$$:

$$-\frac12\,\frac{dn_A}{dt}\;=\;-\frac23\,\frac{dn_C}{dt}.$$

Again, cancelling the minus sign yields

$$\frac12\,\frac{dn_A}{dt}\;=\;\frac23\,\frac{dn_C}{dt}.$$

Multiplying both sides by $$2$$ gives

$$\frac{dn_A}{dt}\;=\;\frac43\,\frac{dn_C}{dt}.$$

Collecting the two results, we have

$$\frac{dn_A}{dt}\;=\;\frac23\,\frac{dn_B}{dt}\;=\;\frac43\,\frac{dn_C}{dt}.$$

This set of equalities is exactly what is written in Option C.

Hence, the correct answer is Option C.

For a first-order reaction, the half-life $$t_{1/2}$$ is related to the rate constant $$k$$ by the formula

$$t_{1/2}=\frac{0.693}{k}\,.$$

We have the following data:

For compound A, $$t_{1/2(A)} = 300\,\text{s}$$, so

$$k_A = \frac{0.693}{t_{1/2(A)}} = \frac{0.693}{300} = 0.00231\ \text{s}^{-1}.$$

For compound B, $$t_{1/2(B)} = 180\,\text{s}$$, so

$$k_B = \frac{0.693}{t_{1/2(B)}} = \frac{0.693}{180} = 0.00385\ \text{s}^{-1}.$$

Let the initial concentrations of A and B be equal and denoted by $$[A]_0 = [B]_0$$. For first-order kinetics, the concentration at time $$t$$ is

$$[A] = [A]_0 e^{-k_A t}, \qquad [B] = [B]_0 e^{-k_B t}\,.$$

We are asked to find the time $$t$$ when the concentration of A becomes four times that of B, i.e.

$$[A] = 4[B].$$

Substituting the exponential expressions, we get

$$[A]_0 e^{-k_A t} = 4\,[B]_0 e^{-k_B t}.$$

Because $$[A]_0 = [B]_0$$, these cancel, leaving

$$e^{-k_A t} = 4\,e^{-k_B t}.$$

Dividing both sides by $$e^{-k_B t}$$ gives

$$e^{\,(-k_A + k_B)\,t} = 4.$$

Taking natural logarithms on both sides,

$$( -k_A + k_B )\,t = \ln 4.$$

Since $$\ln 4 = 2\ln 2 = 2 \times 0.693 = 1.386$$, we have

$$t = \frac{1.386}{k_B - k_A}.$$

Now substitute the calculated rate constants:

$$k_B - k_A = 0.00385 - 0.00231 = 0.00154\ \text{s}^{-1},$$

so

$$t = \frac{1.386}{0.00154} = 900\ \text{s}.$$

Hence, the correct answer is Option B.

We have the forward reaction

$$2\,H_2(g)+2\,NO(g)\;\longrightarrow\;N_2(g)+2\,H_2O(g)$$

and its experimentally observed rate law

$$\text{rate}_{f}=k_f[NO]^2[H_2].$$

To obtain a suitable rate law for the reverse reaction we make use of the fact that, at equilibrium, the forward rate equals the reverse rate. First we write a general form for the backward rate:

$$\text{rate}_{b}=k_b\,[N_2]^{\,x}[H_2O]^{\,y}[H_2]^{\,z}[NO]^{\,w},$$

where the exponents $$x,y,z,w$$ are to be chosen so that (i) equality of the two rates at equilibrium gives a constant ratio $$k_f/k_b$$ and (ii) no term on the right is superfluous. Because only $$N_2$$ and $$H_2O$$ appear as products, we take $$w=0$$; hence

$$\text{rate}_{b}=k_b\,[N_2]^{\,x}[H_2O]^{\,y}[H_2]^{\,z}.$$

At equilibrium we must have

$$k_f[NO]^2[H_2]=k_b\,[N_2]^{\,x}[H_2O]^{\,y}[H_2]^{\,z}. \qquad (1)$$

Meanwhile, the equilibrium‐constant expression for the reaction is

$$K_c=\frac{[N_2][H_2O]^2}{[NO]^2[H_2]^2}. \qquad (2)$$

We now substitute the right side of (2) into (1) so that every concentration on the right of (1) can be replaced by a power of those on the left:

From (2) we have

$$[NO]^2[H_2]^2=\frac{[N_2][H_2O]^2}{K_c},$$

and therefore

$$[NO]^2[H_2]=\frac{[N_2][H_2O]^2}{K_c[H_2]}. \qquad (3)$$

Substituting (3) into (1) gives

$$k_f\left(\frac{[N_2][H_2O]^2}{K_c[H_2]}\right)=k_b\,[N_2]^{\,x}[H_2O]^{\,y}[H_2]^{\,z}.$$

Canceling common factors $$ [N_2] $$ and $$ [H_2O]^2 $$ on both sides yields

$$\frac{k_f}{K_c[H_2]}=k_b\,[N_2]^{\,x-1}[H_2O]^{\,y-2}[H_2]^{\,z}.$$

The left side is a constant (because $$k_f$$ and $$K_c$$ are constants), so the concentration factors on the right must together reduce to unity. This is possible only when

$$x-1=0,\quad y-2=0,\quad z+1=0.$$

Thus

$$x=1,\; y=2,\; z=-1.$$

Substituting these values back into the general form of the reverse rate law gives

$$\text{rate}_{b}=k_b\;\frac{[N_2][H_2O]^2}{[H_2]}\;.$$

Therefore, the observed rate expression for the reverse reaction is

$$\text{rate}_{b}=k_b\,[N_2][H_2O]^2/[H_2].$$

This corresponds to Option D.

Hence, the correct answer is Option D.

For any chemical reaction of the type

$$2A + B \longrightarrow C + D$$

the empirical rate law is written as

$$\text{rate}=k\,[A]^m\,[B]^n$$

where $$k$$ is the rate constant and $$m,n$$ are the orders of the reaction with respect to $$A$$ and $$B$$ respectively. Our first goal is to determine $$m$$ and $$n$$ from the experimental data.

We start by comparing Experiments I and II. In these two runs the concentration of $$A$$ is kept the same while that of $$B$$ is doubled:

$$[A]_{\text I}=0.1,\;[A]_{\text{II}}=0.1$$

$$[B]_{\text I}=0.1,\;[B]_{\text{II}}=0.2=2\times0.1$$

The corresponding initial rates are

$$r_{\text I}=6.00\times10^{-3},\;r_{\text{II}}=2.40\times10^{-2}=4\times6.00\times10^{-3}$$

So the rate increases by a factor of $$4$$ when $$[B]$$ is doubled. Using the rate law:

$$\frac{r_{\text{II}}}{r_{\text I}}=\frac{k\,[A]^m_{\text{II}}\,[B]^n_{\text{II}}}{k\,[A]^m_{\text I}\,[B]^n_{\text I}} =\frac{[A]^m_{\text{II}}}{[A]^m_{\text I}}\; \frac{[B]^n_{\text{II}}}{[B]^n_{\text I}} =(1)\left(\frac{0.2}{0.1}\right)^n=2^n$$

The left-hand side equals $$4$$, so

$$2^n=4\;\Longrightarrow\;n=2$$

Next we compare Experiments I and III, where $$[B]$$ is kept constant and $$[A]$$ is doubled:

$$[A]_{\text I}=0.1,\;[A]_{\text{III}}=0.2=2\times0.1$$

$$[B]_{\text I}=0.1,\;[B]_{\text{III}}=0.1$$

The rates are

$$r_{\text I}=6.00\times10^{-3},\;r_{\text{III}}=1.20\times10^{-2}=2\times6.00\times10^{-3}$$

Thus

$$\frac{r_{\text{III}}}{r_{\text I}}=\left(\frac{0.2}{0.1}\right)^m=2^m=2$$

which immediately gives

$$m=1$$

Hence the complete rate law is

$$\boxed{\text{rate}=k\,[A]^1\,[B]^2}$$

To find the numerical value of $$k$$ we substitute data from any experiment, say Experiment I:

$$6.00\times10^{-3}=k\,(0.1)\,(0.1)^2=k\,(0.1)(0.01)=k\,(0.001)$$

$$\Rightarrow\;k=\frac{6.00\times10^{-3}}{1.00\times10^{-3}}=6.0$$

Now we use this rate constant to determine the missing concentrations.

Experiment IV

$$\text{rate}=7.20\times10^{-2},\;[B]=0.2,\;k=6.0,\;[A]=X$$

Substituting in the rate law:

$$7.20\times10^{-2}=6.0\;(X)\;(0.2)^2=6.0\;X\;(0.04)=0.24\,X$$

$$\Rightarrow\;X=\frac{7.20\times10^{-2}}{0.24}=0.30$$

Experiment V

$$\text{rate}=2.88\times10^{-1},\;[A]=0.3,\;k=6.0,\;[B]=Y$$

Substituting:

$$2.88\times10^{-1}=6.0\;(0.3)\;Y^2=1.8\,Y^2$$

$$\Rightarrow\;Y^2=\frac{2.88\times10^{-1}}{1.8}=0.16$$

$$\Rightarrow\;Y=\sqrt{0.16}=0.40$$

Thus we obtain

$$X=0.3,\qquad Y=0.4$$

Hence, the correct answer is Option C.

We begin by recalling that the order of a reaction is defined exclusively from the experimentally obtained rate law, while a reaction mechanism tells us the actual sequence of elementary steps by which reactants are converted into products. The two ideas are related, but they are not the same; a given overall order may correspond to either a single elementary step or to many elementary steps, depending upon the kinetic details.

For an elementary (single-step) reaction the rate law can be written directly from the molecularity. If the step is $$\text{A} + \text{B} \longrightarrow \text{products},$$ then, by the law of mass action, the rate is $$\text{rate} = k[\text{A}][\text{B}],$$ which is second order overall. However, if the experimentally determined rate law of the overall process turns out to be second order, this does not compel us to accept that the reaction is elementary; the same rate law can arise after combining several elementary steps and applying the steady-state or pre-equilibrium approximations. Thus:

$$\text{“second-order”} \;\centernot\;\Rightarrow\;\text{“single step”}.$$

Next, when we meet a zero-order reaction, the experimental rate law is

$$\text{rate} = k[\text{A}]^0 = k,$$

which is independent of the concentration of the reactant. A constant rate that does not slow down as the reactants are consumed seldom originates from just one bimolecular or unimolecular collision. Instead, zero-order kinetics most commonly appear (i) in surface-catalysed processes where all active sites are saturated, or (ii) in photochemical reactions where the rate is controlled by a constant photon flux. In such cases several elementary events are involved: adsorption, surface reaction, desorption, or excitation and relaxation steps. Because at least two distinct molecular events are necessary, a zero-order overall rate law is essentially a consequence of a multi-step mechanism.

For a first-order reaction the empirical rate law is $$\text{rate} = k[\text{A}].$$ Many gas-phase isomerisations and certain radioactive decays are indeed single-step processes and first order, but the same first-order law can just as well arise from consecutive reactions in which the first step is slow and the later steps are fast. Therefore “first order” does not guarantee “single step”.

Lastly, declaring that “a zero-order reaction is a single step reaction” contradicts the discussion above, because a single elementary step that involves one or more reactant molecules must have an order equal to its molecularity (never zero).

Let us check each option one by one:

Option A: “A second order reaction is always a multistep reaction.” We have seen that a single elementary bimolecular collision already gives a second-order law, so the word “always” makes the statement false.

Option B: “A zero order reaction is a multi-step reaction.” Because zero-order kinetics usually stem from surface saturation or similar situations that necessarily involve more than one elementary step, this statement is true.

Option C: “A first order reaction is always a single step reaction.” Counter-examples exist (e.g. slow first step followed by rapid ones), so the statement is false.

Option D: “A zero order reaction is a single step reaction.” This is the converse of option B and is false for the same reasons.

Only option B survives scrutiny.

Hence, the correct answer is Option 2.

For any reaction, the Arrhenius equation relates the rate constant $$k$$ to the activation energy $$E_a$$ as

$$k = A\,e^{-E_a/RT}$$

where $$A$$ is the pre-exponential factor, $$R$$ is the gas constant, and $$T$$ is the absolute temperature. We are told that in the presence of the enzyme the reaction is $$10^6$$ times faster, that is

$$\dfrac{k_{\text{enzyme}}}{k_{\text{no enzyme}}}=10^6.$$

Using the Arrhenius form for each rate constant, we have

$$\dfrac{A\,e^{-E_{a,\text{enzyme}}/RT}}{A\,e^{-E_{a,\text{no enzyme}}/RT}} = 10^6.$$

The pre-exponential factor $$A$$ cancels, giving

$$e^{-\dfrac{E_{a,\text{enzyme}}}{RT}} \, e^{+\dfrac{E_{a,\text{no enzyme}}}{RT}} = 10^6.$$

Combining the exponents,

$$e^{-\dfrac{E_{a,\text{enzyme}} - E_{a,\text{no enzyme}}}{RT}} = 10^6.$$

Taking natural logarithms on both sides, we get

$$-\dfrac{E_{a,\text{enzyme}} - E_{a,\text{no enzyme}}}{RT} = \ln(10^6).$$

Now $$\ln(10^6) = 6\ln(10) = 6(2.303).$$ Substituting this value,

$$-\dfrac{E_{a,\text{enzyme}} - E_{a,\text{no enzyme}}}{RT} = 6(2.303).$$

Multiplying both sides by $$-RT$$ gives the change in activation energy:

$$E_{a,\text{enzyme}} - E_{a,\text{no enzyme}} = -6(2.303)RT.$$

Thus the activation energy decreases by $$6(2.303)RT$$ when the enzyme is added.

Hence, the correct answer is Option A.

We are told that the milk “splits” (curdles) exactly when the population of Lactobacillus acidophilus becomes double. A doubling time in a first-order growth process is related to the rate constant by the well-known relation

$$t_{\text{double}}=\dfrac{\ln 2}{k}\;.$$

So, if the doubling (splitting) time is known, the corresponding rate constant is

$$k=\dfrac{\ln 2}{t_{\text{double}}}\;.$$

The experiment gives

• at $$T_1 = 300\ \text{K}$$, splitting time $$t_1 = 60\ \text{min}$$,

• at $$T_2 = 400\ \text{K}$$, splitting time $$t_2 = 40\ \text{min}\;.$$

Using the formula above, the two rate constants are therefore

$$k_1=\dfrac{\ln 2}{t_1}= \dfrac{\ln 2}{60}\;,$$ $$k_2=\dfrac{\ln 2}{t_2}= \dfrac{\ln 2}{40}\;.$$

We next apply the Arrhenius equation, stated as

$$k = A\,e^{-E_a/(RT)},$$

where $$E_a$$ is the activation energy, $$R$$ the gas constant and $$A$$ the pre-exponential factor. Taking natural logarithms of both sides gives

$$\ln k = \ln A - \dfrac{E_a}{R}\,\dfrac{1}{T}\;.$$

For two temperatures $$T_1$$ and $$T_2$$ this yields

$$\ln\dfrac{k_2}{k_1}= -\dfrac{E_a}{R}\Bigl(\dfrac{1}{T_2}-\dfrac{1}{T_1}\Bigr) = \dfrac{E_a}{R}\Bigl(\dfrac{1}{T_1}-\dfrac{1}{T_2}\Bigr).$$

We already have the ratio of rate constants:

$$\dfrac{k_2}{k_1}= \dfrac{\dfrac{\ln 2}{40}}{\dfrac{\ln 2}{60}} = \dfrac{60}{40}=1.5\;,$$ so that $$\ln\dfrac{k_2}{k_1}= \ln(1.5)\approx 0.4\;.$$ (The numerical hint $$\ln(2/3)=0.4$$ implies $$\ln(1.5)= -\ln(2/3)\approx 0.4$$.)

Now we evaluate the temperature term:

$$\dfrac{1}{T_1}-\dfrac{1}{T_2}= \dfrac{1}{300}-\dfrac{1}{400} =\dfrac{400-300}{300\times 400} =\dfrac{100}{120000} =\dfrac{1}{1200} = 0.0008333\ \text{K}^{-1}\;.$$

Substituting these two results into the Arrhenius relation gives

$$0.4=\dfrac{E_a}{R}\,(0.0008333).$$

Solving for $$E_a$$:

$$E_a = \dfrac{0.4 \times R}{0.0008333} = \dfrac{0.4 \times 8.3\ \text{J mol}^{-1}\text{K}^{-1}}{0.0008333}.$$ Straight multiplication gives $$0.4 \times 8.3 = 3.32\ \text{J mol}^{-1},$$ and finally $$E_a = \dfrac{3.32}{0.0008333}\ \text{J mol}^{-1} \approx 3.98\times10^{3}\ \text{J mol}^{-1}.$$ Converting joules to kilojoules, $$E_a \approx 3.98\ \text{kJ mol}^{-1}.$$

Hence, the correct answer is Option C.

The radioactive decay of any nuclide is governed by the exponential law

$$N \;=\; N_0\,e^{-\lambda t},$$

where $$N_0$$ is the initial number of nuclei, $$N$$ is the number left after time $$t$$, and $$\lambda$$ is the decay constant.

It is often more convenient to write this law using the half-life $$T_{1/2}$$. The relationship between the decay constant and the half-life is

$$\lambda \;=\; \frac{\ln 2}{T_{1/2}}.$$

Eliminating $$\lambda$$ from the first equation gives the form that directly involves the half-life:

$$N \;=\; N_0\Bigl(\tfrac12\Bigr)^{t/T_{1/2}}.$$

In the present problem we are told that the half-life of $$^{90}\text{Sr}$$ is $$T_{1/2}=6.93\text{ yr}$$. A newborn baby absorbs an initial mass of $$1\,\mu\text{g}$$ of this isotope. We are asked how long it will take for this quantity to drop by 90 %, that is, for only 10 % of the original amount to remain.

Mathematically, “drop by 90 %’’ means

$$\frac{N}{N_0}=0.10.$$

Substituting this fraction and the given half-life into the decay equation, we have

$$0.10 \;=\;\Bigl(\tfrac12\Bigr)^{t/6.93}.$$

To solve for $$t$$ we take the natural logarithm of both sides:

$$\ln(0.10) \;=\; \ln\!\Bigl[\Bigl(\tfrac12\Bigr)^{t/6.93}\Bigr].$$

Using the property $$\ln(a^b)=b\,\ln a$$, the right-hand side simplifies:

$$\ln(0.10) \;=\; \frac{t}{6.93}\,\ln\!\Bigl(\tfrac12\Bigr).$$

Now we isolate $$t$$:

$$t \;=\; 6.93 \times \frac{\ln(0.10)}{\ln(1/2)}.$$

Both logarithms are negative, so their ratio is positive. We calculate them explicitly:

$$\ln(0.10) = -2.302585093,\qquad \ln\!\Bigl(\tfrac12\Bigr) = -0.693147181.$$

Dividing gives

$$\frac{\ln(0.10)}{\ln(1/2)} = \frac{-2.302585093}{-0.693147181} = 3.321928094.$$

Multiplying by 6.93 yr, we find

$$t = 6.93 \times 3.321928094 = 23.02296\text{ yr}.$$

Rounding appropriately, the required time is

$$t \;\approx\; 23.03\text{ years}.$$

So, the answer is $$23.03\text{ years}$$.

We know that for any chemical reaction the temperature-dependence of the rate constant is expressed by the Arrhenius equation

$$k = A \, e^{-E_a/RT}$$

where $$k$$ is the rate constant, $$A$$ is the pre-exponential factor, $$E_a$$ is the activation energy, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature in kelvin.

Because the order of the reaction does not change with temperature, the rate is directly proportional to the rate constant. Thus the ratio of the two measured rates is the same as the ratio of the two corresponding rate constants. We are told that the rate decreases by a factor of $$3.555$$ when the temperature is lowered from $$40^{\circ}\text C$$ to $$30^{\circ}\text C$$, so we can write

$$\dfrac{k_{30}}{k_{40}}=\dfrac{1}{3.555}$$

Using the logarithmic form of the Arrhenius equation for two temperatures,

$$\ln\!\left(\dfrac{k_{30}}{k_{40}}\right) = -\dfrac{E_a}{R}\left(\dfrac{1}{T_{30}}-\dfrac{1}{T_{40}}\right)$$

First we convert the two Celsius temperatures into kelvin:

$$T_{40}=40^{\circ}\text C + 273 = 313\ \text K$$

$$T_{30}=30^{\circ}\text C + 273 = 303\ \text K$$

Now we substitute the known quantities into the logarithmic equation. On the left-hand side we have

$$\ln\!\left(\dfrac{k_{30}}{k_{40}}\right)=\ln\!\left(\dfrac{1}{3.555}\right)=\ln(1)-\ln(3.555)= -\ln(3.555)$$

Evaluating the natural logarithm,

$$\ln(3.555)\approx 1.269$$

so that

$$\ln\!\left(\dfrac{k_{30}}{k_{40}}\right)\approx -1.269$$

On the right-hand side we have the temperature factor,

$$\dfrac{1}{T_{30}}-\dfrac{1}{T_{40}}=\dfrac{1}{303}-\dfrac{1}{313}$$

Finding a common denominator,

$$\dfrac{1}{303}-\dfrac{1}{313}=\dfrac{313-303}{303\times313}=\dfrac{10}{94839}\ \text{K}^{-1}$$

Numerically,

$$\dfrac{10}{94839}\approx 1.0546\times10^{-4}\ \text{K}^{-1}$$

Now we insert these values into the logarithmic Arrhenius expression:

$$-1.269 = -\dfrac{E_a}{R}\,(1.0546\times10^{-4})$$

Because both sides carry a negative sign, they cancel, giving

$$1.269 = \dfrac{E_a}{R}\,(1.0546\times10^{-4})$$

We isolate $$E_a$$ by multiplying both sides by $$R$$ and dividing by $$1.0546\times10^{-4}$$:

$$E_a = \dfrac{1.269\,R}{1.0546\times10^{-4}}$$

Taking $$R = 8.314\ \text{J mol}^{-1}\text K^{-1}$$, we have

$$E_a = \dfrac{1.269 \times 8.314}{1.0546\times10^{-4}}\ \text{J mol}^{-1}$$

Multiplying in the numerator,

$$1.269 \times 8.314 \approx 10.55$$

and then dividing,

$$E_a = \dfrac{10.55}{1.0546\times10^{-4}}\ \text{J mol}^{-1} \approx 1.000 \times 10^{5}\ \text{J mol}^{-1}$$

Converting joules to kilojoules (by dividing by $$1000$$),

$$E_a \approx 100\ \text{kJ mol}^{-1}$$

So, the answer is $$100$$.

For a first-order reaction we use the integrated rate-law written with common (base 10) logarithms:

$$k \;=\; \frac{2.303}{t}\,\log\!\left(\frac{[A]_0}{[A]}\right)$$

Here $$k$$ is the rate constant, $$t$$ is the elapsed time, $$[A]_0$$ is the initial concentration and $$[A]$$ is the concentration remaining after time $$t$$.

We are first told that 75 % of the reactant is consumed in 90 minutes. 75 % consumed means 25 % left, so

$$\frac{[A]}{[A]_0}=0.25 \quad\Longrightarrow\quad \frac{[A]_0}{[A]}=\frac{1}{0.25}=4.$$

Substituting these numbers into the formula gives

$$k = \frac{2.303}{90}\,\log 4.$$

Using the given value $$\log 2 = 0.30$$ we obtain $$\log 4 = 2\,\log 2 = 2 \times 0.30 = 0.60$$, so

$$k = \frac{2.303}{90} \times 0.60.$$

Now we are asked to find the time required for 60 % of the reaction to be completed. 60 % consumed means 40 % left, hence

$$\frac{[A]}{[A]_0}=0.40 \quad\Longrightarrow\quad \frac{[A]_0}{[A]}=\frac{1}{0.40}=2.5.$$

Applying the same first-order expression for this new situation, we write

$$k = \frac{2.303}{t_{60\%}}\;\log 2.5,$$

where $$t_{60\%}$$ is the unknown time for 60 % completion. Solving for $$t_{60\%}$$ gives

$$t_{60\%} = \frac{2.303}{k}\,\log 2.5.$$

We already have $$k = \dfrac{2.303}{90} \times 0.60$$, so substituting this value of $$k$$ into the expression for $$t_{60\%}$$ yields

$$t_{60\%} = \frac{2.303}{\dfrac{2.303}{90}\times 0.60}\;\log 2.5.$$

The factor $$2.303$$ cancels out from numerator and denominator, leaving

$$t_{60\%} = \frac{90}{0.60}\,\log 2.5.$$

Using the given value $$\log 2.5 = 0.40$$, we have

$$t_{60\%} = \frac{90}{0.60} \times 0.40.$$

Carrying out the multiplication and division step by step:

$$\frac{90}{0.60} = 150,$$

so

$$t_{60\%} = 150 \times 0.40 = 60.$$

Therefore, approximately 60 minutes are required for 60 % completion of the same first-order reaction.

So, the answer is $$60 \text{ minutes}$$.

We are told that when the temperature is raised from 27 °C to 42 °C, the number of molecules having energy greater than the threshold (activation) energy becomes five times. In kinetic theory, the fraction of molecules possessing energy equal to or greater than the activation energy $$E_a$$ at temperature $$T$$ is proportional to the Boltzmann factor $$e^{-E_a/RT}$$. Therefore, for two temperatures $$T_1$$ and $$T_2$$ we can write

$$\frac{\text{Number of molecules at }T_2}{\text{Number of molecules at }T_1} =\frac{e^{-E_a/RT_2}}{e^{-E_a/RT_1}} = e^{-E_a/RT_2 + E_a/RT_1} = e^{E_a\left(\dfrac{1}{RT_1}-\dfrac{1}{RT_2}\right)}.$$

According to the question this ratio equals 5, so

$$5 = e^{E_a\left(\dfrac{1}{RT_1}-\dfrac{1}{RT_2}\right)}.$$

Taking natural logarithm on both sides,

$$\ln 5 = E_a\left(\dfrac{1}{RT_1}-\dfrac{1}{RT_2}\right).$$

Now we insert the numerical data. First convert the given Celsius temperatures to Kelvin:

$$T_1 = 27^\circ\text{C} + 273 = 300\ \text{K},$$

$$T_2 = 42^\circ\text{C} + 273 = 315\ \text{K}.$$

Compute the term in parentheses:

$$\frac{1}{T_1}-\frac{1}{T_2} =\frac{1}{300}-\frac{1}{315} =\frac{315-300}{300 \times 315} =\frac{15}{94500} =\frac{1}{6300}.$$

Substituting $$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$$ and $$\ln 5 = 1.6094$$ into the logarithmic equation, we get

$$1.6094 = E_a \left(\frac{1}{8.314}\right)\left(\frac{1}{6300}\right).$$

Rearranging for $$E_a$$,

$$E_a = \frac{1.6094 \times 8.314}{\dfrac{1}{6300}} = 1.6094 \times 8.314 \times 6300.$$

First multiply $$1.6094$$ and $$8.314$$:

$$1.6094 \times 8.314 = 13.3805516.$$

Now multiply this result by 6300:

$$13.3805516 \times 6300 = 84\,297.47508.$$

Rounding to the nearest whole number (since the data are given to four significant figures),

$$E_a \approx 84\,297\ \text{J mol}^{-1}.$$

Hence, the correct answer is Option C.

We assume that the rate follows the usual rate law expression $$r = k[A]^m[B]^n$$, where $$m$$ is the order with respect to $$A$$ and $$n$$ is the order with respect to $$B$$. The symbol $$k$$ is the rate constant.

For the initial experiment we let the initial concentrations be $$[A]=a$$ and $$[B]=b$$. The initial rate is given to be

$$r_1 = k\,a^{\,m}\,b^{\,n} = 0.3\;\text{mol L}^{-1}\text{s}^{-1}.$$

Now both $$[A]$$ and $$[B]$$ are doubled, so they become $$2a$$ and $$2b$$ respectively. The new rate therefore is

$$r_2 = k\,(2a)^{\,m}\,(2b)^{\,n}.$$

Using the laws of exponents, we write

$$r_2 = k\,2^{\,m}a^{\,m}\,2^{\,n}b^{\,n} = 2^{\,m+n}\,k\,a^{\,m}b^{\,n} = 2^{\,m+n}\,r_1.$$

We are told that the numerical value of the rate increases from $$0.3$$ to $$2.4$$, so

$$\dfrac{r_2}{r_1}=\dfrac{2.4}{0.3}=8.$$

Comparing with the algebraic factor we obtained, we have

$$2^{\,m+n}=8.$$

Since $$8=2^{3}$$, we immediately deduce

$$m+n = 3.$$

Thus, the total order of the reaction is three, but we still need individual orders $$m$$ and $$n$$.

Next, only $$[A]$$ is doubled while $$[B]$$ is kept unchanged. The concentration of $$A$$ becomes $$2a$$ and that of $$B$$ stays $$b$$. The resulting rate is

$$r_3 = k\,(2a)^{\,m}\,b^{\,n} = 2^{\,m}\,k\,a^{\,m}b^{\,n}=2^{\,m}\,r_1.$$

The problem states that the rate rises from $$0.3$$ to $$0.6$$, so

$$\dfrac{r_3}{r_1}=\dfrac{0.6}{0.3}=2.$$

Equating the two factors gives

$$2^{\,m}=2 \quad\Longrightarrow\quad m = 1.$$

Substituting $$m = 1$$ into the earlier relation $$m+n = 3$$, we obtain

$$1 + n = 3 \quad\Longrightarrow\quad n = 2.$$

Therefore, the order with respect to $$A$$ is $$1$$ and the order with respect to $$B$$ is $$2$$. The only statement among the options that reflects this outcome is “Order of the reaction with respect to B is 2.”

Hence, the correct answer is Option A.

We have the reversible elementary reaction $$A_2 \;\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}\; 2A.$$

For any elementary step, the rate law is directly proportional to the product of the concentrations of the reactant molecules raised to the power of their stoichiometric coefficients. Therefore:

1. Forward step $$A_2 \longrightarrow 2A$$ has rate

$$r_{\text{f}} = k_1[A_2].$$

2. Backward step $$2A \longrightarrow A_2$$ has rate

$$r_{\text{b}} = k_{-1}[A]^2$$

because two molecules of $$A$$ collide simultaneously in this elementary process.

Next, we translate these molecular events into the rate of change of concentration of $$A$$. The forward step creates two moles of $$A$$ per molecular event, while the backward step consumes two moles of $$A$$ per molecular event. Hence:

$$\frac{d[A]}{dt} = (+2)\,r_{\text{f}} \;-\; (2)\,r_{\text{b}}.$$

Substituting $$r_{\text{f}} = k_1[A_2]$$ and $$r_{\text{b}} = k_{-1}[A]^2$$ we obtain

$$\frac{d[A]}{dt} = 2\,k_1[A_2] \;-\; 2\,k_{-1}[A]^2.$$

So the required expression matches Option A.

Hence, the correct answer is Option A.

We have a straight-line plot of $$\ln k$$ versus $$1/T$$ for the reaction. For any Arrhenius plot, the governing relation is first written as

$$k \;=\; A\,e^{-\dfrac{E_a}{RT}}.$$

Taking natural logarithms on both sides gives the linear form

$$\ln k \;=\; \ln A \;-\;\dfrac{E_a}{R}\;\left(\dfrac1T\right).$$

In the form $$y = mx + c$$, the ordinate is $$y = \ln k$$, the abscissa is $$x = 1/T$$, and the slope is

$$m = -\dfrac{E_a}{R}.$$

From the figure (reading two well-separated points and calculating), the slope of the straight line is found to be approximately

$$m \;=\; -4.6 \times 10^{3}\;{\rm K}.$$

Stating this numerically,

$$-\dfrac{E_a}{R} = -4.6 \times 10^{3}\;{\rm K}, \qquad\text{so}\qquad \dfrac{E_a}{R} = 4.6 \times 10^{3}\;{\rm K}.$$

Now we wish to compare rate constants at two temperatures. Using the Arrhenius form for two temperatures $$T_1$$ and $$T_2$$, the standard manipulation is

$$\ln\!\left(\dfrac{k_2}{k_1}\right) \;=\; -\dfrac{E_a}{R}\;\Bigl(\dfrac1{T_2} - \dfrac1{T_1}\Bigr).$$

First, list the known data:

$$T_1 = 400\;{\rm K},\qquad k_1 = 10^{-5}\;{\rm s^{-1}},$$

$$T_2 = 500\;{\rm K},\qquad k_2 = \text{?}$$

Calculate the difference in reciprocal temperatures:

$$\dfrac1{T_2} - \dfrac1{T_1} \;=\; \dfrac1{500} \;-\; \dfrac1{400}.$$

Writing both denominators explicitly,

$$\dfrac1{500} = 2.0 \times 10^{-3}\;{\rm K^{-1}},\qquad \dfrac1{400} = 2.5 \times 10^{-3}\;{\rm K^{-1}}.$$

So,

$$\dfrac1{T_2} - \dfrac1{T_1} = (2.0 - 2.5)\times10^{-3} = -0.5 \times 10^{-3} \;=\; -5.0 \times 10^{-4}\;{\rm K^{-1}}.$$

Substituting slope information $$E_a / R = 4.6 \times 10^{3}\;{\rm K}$$ into the two-temperature equation gives

$$\ln\!\left(\dfrac{k_2}{k_1}\right) = -\bigl(4.6 \times 10^{3}\bigr)\;\Bigl(-5.0 \times 10^{-4}\Bigr).$$

Multiplying the two negatives yields a positive result:

$$\ln\!\left(\dfrac{k_2}{k_1}\right) = (4.6 \times 10^{3})\,(5.0 \times 10^{-4}) = 2.3.$$

Recognising that $$\ln 10 \approx 2.303$$, we can rewrite

$$\ln\!\left(\dfrac{k_2}{k_1}\right) \;\approx\; \ln 10.$$

Therefore,

$$\dfrac{k_2}{k_1} \;\approx\; 10.$$

Finally, multiplying by the known $$k_1$$ value,

$$k_2 \;=\; 10\,k_1 \;=\; 10 \times 10^{-5}\;{\rm s^{-1}} \;=\; 10^{-4}\;{\rm s^{-1}}.$$

Hence, the correct answer is Option A.

For the consecutive first-order reaction scheme $$A \xrightarrow{k_1} B \xrightarrow{k_2} C$$ we first recall the basic kinetic expressions for each elementary step.

The rate of formation of $$B$$ from $$A$$ is given by the first-order rate law $$\text{Rate of formation of }B = k_1[A]$$ because the step $$A \to B$$ is first order in $$A$$ with rate constant $$k_1$$.

Simultaneously, $$B$$ is being consumed in the next first-order step $$B \to C$$, so its rate of disappearance is $$\text{Rate of consumption of }B = k_2[B]$$, where $$k_2$$ is the corresponding rate constant.

The net rate of change of the concentration of $$B$$ is therefore the difference between its formation and its consumption:

$$\frac{d[B]}{dt}=k_1[A]-k_2[B].$$

The problem states that the net rate of formation of $$B$$ is set to zero. In mathematical form, this is the steady-state approximation:

$$\frac{d[B]}{dt}=0.$$

Substituting this condition into the net rate expression gives

$$0 = k_1[A] - k_2[B].$$

We now isolate $$[B]$$ algebraically. First, add $$k_2[B]$$ to both sides:

$$k_2[B] = k_1[A].$$

Next, divide both sides by $$k_2$$ to solve for $$[B]$$:

$$[B] = \frac{k_1}{k_2}[A].$$

This final expression matches option C in the list provided.

Hence, the correct answer is Option 3.

We are told that the reaction $$2X \to B$$ follows zeroth-order kinetics. For any zeroth-order reaction the rate law is stated first:

$$\text{Rate}=k\;,$$

where $$k$$ is the zeroth-order rate constant having the units $$\text{mol L}^{-1}\text{h}^{-1}$$ (or $$\text{M h}^{-1}$$).

The integrated form of the zeroth-order rate law, obtained by integrating $$d[X]/dt=-k$$, is

$$[X]=[X]_0-k\,t$$

where $$[X]_0$$ is the initial concentration and $$[X]$$ is the concentration after time $$t$$.

We also need the relationship between the half-life and the rate constant for a zeroth-order reaction. The half-life $$t_{1/2}$$ is defined as the time when $$[X]$$ becomes $$[X]_0/2$$. Substituting $$[X]=[X]_0/2$$ and $$t=t_{1/2}$$ into the integrated law:

$$\frac{[X]_0}{2}=[X]_0-k\,t_{1/2}$$

Simplifying, we obtain the well-known zeroth-order half-life formula:

$$t_{1/2}=\frac{[X]_0}{2k}$$

Now we use the data from the first experiment, where $$[X]_0=0.2\ \text{M}$$ and $$t_{1/2}=6\ \text{h}$$, to determine $$k$$. Rearranging the formula just written gives

$$k=\frac{[X]_0}{2\,t_{1/2}}$$

Substituting the numerical values:

$$k=\frac{0.2\ \text{M}}{2 \times 6\ \text{h}}=\frac{0.2}{12}\ \text{M h}^{-1}=0.016666\ldots\ \text{M h}^{-1}$$

So, $$k=0.0167\ \text{M h}^{-1}$$ when rounded to four significant figures.

Next we analyse the second situation, where the initial concentration is $$[X]_0=0.5\ \text{M}$$ and we wish to know the time $$t$$ taken for the concentration to fall to $$[X]=0.2\ \text{M}$$. We again start from the integrated rate law

$$[X]=[X]_0-k\,t$$

and solve for $$t$$:

$$t=\frac{[X]_0-[X]}{k}$$

Substituting the known quantities:

$$t=\frac{0.5\ \text{M}-0.2\ \text{M}}{0.016666\ldots\ \text{M h}^{-1}} =\frac{0.3\ \text{M}}{0.016666\ldots\ \text{M h}^{-1}}$$

Carrying out the division gives

$$t=18\ \text{h}$$

Therefore, the time required for the concentration of $$X$$ to decrease from $$0.5\ \text{M}$$ to $$0.2\ \text{M}$$ is 18 hours.

Hence, the correct answer is Option C.

We begin by recalling that the rate equation for a zero-order reaction is given by the formula $$\text{Rate}=k$$, where $$k$$ is the zero-order rate constant. For such reactions the integrated form is

$$[A]=[A]_0-k\,t,$$

where $$[A]_0$$ is the initial amount (or concentration) of the substance, $$[A]$$ is the amount remaining after time $$t$$, and $$k$$ has units of amount per time. The presence of units $$\mu\text{g\,year}^{-1}$$ for $$k$$ confirms that the decomposition of X follows zero-order kinetics.

We are told that the initial mass of X is $$[A]_0=5\;\mu\text{g}$$ and the mass after decomposition is $$[A]=2.5\;\mu\text{g}$$. The rate constant is $$k=0.05\;\mu\text{g\,year}^{-1}$$. Substituting these values into the integrated equation, we get

$$2.5 = 5 - 0.05\,t.$$

Now we isolate $$t$$ step by step. First subtract $$5$$ from both sides:

$$2.5 - 5 = -0.05\,t.$$

This simplifies to

$$-2.5 = -0.05\,t.$$

Next divide both sides by $$-0.05$$ to solve for $$t$$:

$$t = \dfrac{-2.5}{-0.05}.$$

Carrying out the division in the numerator and denominator, we have

$$t = \dfrac{2.5}{0.05}.$$

Since $$2.5 \div 0.05 = 50$$, we arrive at

$$t = 50\;\text{years}.$$

Thus, 50 years are required for the decomposition of 5 $$\mu\text{g}$$ of X to 2.5 $$\mu\text{g}$$.

Hence, the correct answer is Option D.

For any reaction of the general type $$aA+bB\rightarrow \text{products}$$ we write the empirical rate law as

$$\text{Rate}=k[A]^m[B]^n$$

where $$k$$ is the rate constant and $$m,n$$ are the (experimental) orders with respect to the reactants A and B, respectively. Our task is to determine the values of $$m$$ and $$n$$ from the given initial-rate data.

First, label the three experiments for easy reference:

Experiment 1: $$[A]_1=0.05\ \text{mol L}^{-1},\quad [B]_1=0.05\ \text{mol L}^{-1},\quad R_1=0.045\ \text{mol L}^{-1}\text{s}^{-1}$$
Experiment 2: $$[A]_2=0.10\ \text{mol L}^{-1},\quad [B]_2=0.05\ \text{mol L}^{-1},\quad R_2=0.090\ \text{mol L}^{-1}\text{s}^{-1}$$
Experiment 3: $$[A]_3=0.20\ \text{mol L}^{-1},\quad [B]_3=0.10\ \text{mol L}^{-1},\quad R_3=0.72\ \text{mol L}^{-1}\text{s}^{-1}$$

We now compare pairs of experiments so that only one concentration changes at a time.

Step 1: Find $$m$$ by keeping $$[B]$$ constant.

Between Experiments 1 and 2, $$[B]$$ is the same (0.05 mol L−1), so the rate ratio depends only on $$[A]$$:

$$\frac{R_2}{R_1}= \frac{k[A]_2^m[B]_2^n}{k[A]_1^m[B]_1^n} =\left(\frac{[A]_2}{[A]_1}\right)^m\left(\frac{[B]_2}{[B]_1}\right)^n =\left(\frac{[A]_2}{[A]_1}\right)^m(1)^n =\left(\frac{0.10}{0.05}\right)^m.$$

The numerical ratio of rates is $$\dfrac{0.090}{0.045}=2$$, so

$$2=\left(\frac{0.10}{0.05}\right)^m=2^m.$$

Hence $$m=1$$.

Step 2: Find $$n$$ using $$m=1$$.

Now compare Experiments 2 and 3; both $$[A]$$ and $$[B]$$ double. Using the rate law with $$m=1$$, we have

$$\frac{R_3}{R_2}= \frac{k[A]_3^1[B]_3^n}{k[A]_2^1[B]_2^n} =\left(\frac{[A]_3}{[A]_2}\right)^1\left(\frac{[B]_3}{[B]_2}\right)^n =\left(\frac{0.20}{0.10}\right)\left(\frac{0.10}{0.05}\right)^n =2\cdot 2^n.$$

The experimental rate ratio is $$\dfrac{0.72}{0.090}=8$$, so

$$8=2\cdot 2^n \quad\Longrightarrow\quad 4=2^n \quad\Longrightarrow\quad n=2.$$

Step 3: Write the complete rate law.

With $$m=1$$ and $$n=2$$, the empirical rate expression becomes

$$\text{Rate}=k[A]^1[B]^2=k[A][B]^2.$$

Step 4: (Optional) Verify $$k$$ is constant.

Using Experiment 1 data:

$$k=\frac{R_1}{[A]_1[B]_1^2} =\frac{0.045}{(0.05)(0.05)^2} =\frac{0.045}{0.000125}=360.$$

Using Experiment 2:

$$k=\frac{0.090}{(0.10)(0.05)^2} =\frac{0.090}{0.00025}=360,$$

confirming that $$k$$ is indeed constant and our orders are correct.

Therefore, the reaction follows the rate law $$\boxed{\text{Rate}=k[A][B]^2}$$, which corresponds to Option D.

Hence, the correct answer is Option D.

We begin with the Arrhenius equation for the temperature-dependence of the rate constant:

$$k=A\,e^{-E_a/(RT)}$$

Taking natural logarithms on both sides, we use the property $$\ln(e^x)=x$$ to obtain

$$\ln k=\ln A-\dfrac{E_a}{RT}.$$

The question says that a graph of $$\ln k$$ (vertical axis) is drawn against $$\dfrac{1}{RT}$$ (horizontal axis). So, let us write the above equation explicitly in the straight-line form $$y=mx+c$$ with

$$y=\ln k,\qquad x=\dfrac{1}{RT}.$$

Substituting the chosen $$x$$ into the Arrhenius expression, we have

$$\ln k=\ln A-E_a\left(\dfrac{1}{RT}\right).$$

Comparing this with $$y=mx+c$$ we identify

$$m=-E_a,$$

because the coefficient of $$x$$ (which is $$1/(RT)$$) is $$-E_a$$, while the intercept $$c$$ is $$\ln A$$.

According to the statement of the problem, the experimentally obtained straight line has a gradient (slope) equal to $$-y$$ unit. Thus,

$$m=-y.$$

But we have already shown that $$m=-E_a$$, so equating the two expressions for the slope gives

$$-E_a=-y\quad\Longrightarrow\quad E_a=y.$$

Therefore, the energy required to activate the reactant, $$E_a,$$ is simply $$y$$ in magnitude and carries the same unit mentioned for the gradient.

Hence, the correct answer is Option B.

We have the balanced chemical equation

$$2\,\text{N}_2\text{O}_5(g) \;\longrightarrow\; 4\,\text{NO}_2(g) \;+\; \text{O}_2(g).$$

For any reaction written as $$aA \;\longrightarrow\; bB,$$ the rate is defined by the general relation

$$\text{Rate} \;=\; -\dfrac{1}{a}\,\dfrac{\Delta[A]}{\Delta t} \;=\; \dfrac{1}{b}\,\dfrac{\Delta[B]}{\Delta t}.$$

Here, the stoichiometric coefficients are $$a = 2$$ for $$\text{N}_2\text{O}_5$$ and $$b = 4$$ for $$\text{NO}_2.$$

The initial concentration of $$\text{N}_2\text{O}_5$$ is given as $$[\text{N}_2\text{O}_5]_0 = 3.00\ \text{mol L}^{-1}.$$ After a time interval of $$\Delta t = 30\ \text{min},$$ its concentration becomes $$[\text{N}_2\text{O}_5]_t = 2.75\ \text{mol L}^{-1}.$$

So, the change in concentration of $$\text{N}_2\text{O}_5$$ is

$$\Delta[\text{N}_2\text{O}_5] \;=\; [\text{N}_2\text{O}_5]_t - [\text{N}_2\text{O}_5]_0 \;=\; 2.75 - 3.00 \;=\; -0.25\ \text{mol L}^{-1}.$$

The negative sign simply shows a decrease; for the rate we use its magnitude:

$$|\Delta[\text{N}_2\text{O}_5]| = 0.25\ \text{mol L}^{-1}.$$

Using the definition of rate for the disappearance of $$\text{N}_2\text{O}_5,$$ we write

$$\text{Rate}_{\text{disappearance of } \text{N}_2\text{O}_5} = -\dfrac{1}{2}\,\dfrac{\Delta[\text{N}_2\text{O}_5]}{\Delta t}.$$

Substituting the numerical values,

$$\text{Rate}_{\text{disappearance of } \text{N}_2\text{O}_5} = -\dfrac{1}{2}\,\dfrac{-0.25\ \text{mol L}^{-1}}{30\ \text{min}} = \dfrac{0.25}{2 \times 30}\ \text{mol L}^{-1}\text{min}^{-1} = \dfrac{0.25}{60}\ \text{mol L}^{-1}\text{min}^{-1} = 0.004167\ \text{mol L}^{-1}\text{min}^{-1}.$$

Now we relate this rate to the rate of formation of $$\text{NO}_2.$$ From the stoichiometry, every $$2$$ moles of $$\text{N}_2\text{O}_5$$ that decompose give $$4$$ moles of $$\text{NO}_2.$$ Therefore,

$$\dfrac{1}{4}\,\dfrac{\Delta[\text{NO}_2]}{\Delta t} = -\dfrac{1}{2}\,\dfrac{\Delta[\text{N}_2\text{O}_5]}{\Delta t}.$$

So,

$$\dfrac{\Delta[\text{NO}_2]}{\Delta t} = 2\;\Bigl(-\dfrac{\Delta[\text{N}_2\text{O}_5]}{\Delta t}\Bigr).$$

The factor $$2$$ arises because $$4/2 = 2.$$ Hence, the rate of formation of $$\text{NO}_2$$ is twice the rate of disappearance of $$\text{N}_2\text{O}_5.$$ Substituting,

$$\text{Rate}_{\text{formation of } \text{NO}_2} = 2 \times 0.004167\ \text{mol L}^{-1}\text{min}^{-1} = 0.008334\ \text{mol L}^{-1}\text{min}^{-1} \times 2 = 0.01667\ \text{mol L}^{-1}\text{min}^{-1}.$$

Expressing in scientific notation,

$$\text{Rate}_{\text{formation of } \text{NO}_2} = 1.667 \times 10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}.$$

This numerical value corresponds to Option A.

Hence, the correct answer is Option A.

lnk = -Ea/Rt

First, we recall the Arrhenius equation in its logarithmic two-temperature form:

$$\ln\!\left(\dfrac{k_2}{k_1}\right)=\dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$$

Here $$k_1$$ and $$k_2$$ are the rate constants at absolute temperatures $$T_1$$ and $$T_2$$, $$E_a$$ is the activation energy, and $$R$$ is the gas constant.

We have the given data:

$$k_1 = 2.5 \times 10^{-4}\ \text{dm}^3\ \text{mol}^{-1}\ \text{s}^{-1}$$ at $$327^{\circ}\text{C}$$, $$k_2 = 1.0\ \text{dm}^3\ \text{mol}^{-1}\ \text{s}^{-1}$$ at $$527^{\circ}\text{C}$$.

Temperatures must be converted to Kelvin:

$$T_1 = 327 + 273 = 600\ \text{K}$$ $$T_2 = 527 + 273 = 800\ \text{K}$$

Next we form the ratio of the rate constants:

$$\dfrac{k_2}{k_1}= \dfrac{1.0}{2.5 \times 10^{-4}} = \dfrac{1.0}{0.00025}=4000$$

Taking the natural logarithm,

$$\ln\!\left(\dfrac{k_2}{k_1}\right)=\ln(4000)$$ $$= \ln(4\times10^3)=\ln4+\ln10^3$$ $$=1.3863+6.9078=8.2941$$

Now we evaluate the temperature term:

$$\dfrac{1}{T_1}-\dfrac{1}{T_2}= \dfrac{1}{600}-\dfrac{1}{800}$$ $$=0.0016667-0.00125$$ $$=0.00041667\ \text{K}^{-1}$$

Substituting these numerical values and $$R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$$ into the Arrhenius expression:

$$8.2941 = \dfrac{E_a}{8.314}\,(0.00041667)$$

First multiply $$R$$ and the reciprocal-temperature difference:

$$8.314 \times 0.00041667 = 0.0034657$$

Then multiply both sides of the equation by this product to isolate $$E_a$$:

$$E_a = \dfrac{8.2941}{0.00041667}\times 8.314$$ $$= 68.985\ \text{J mol}^{-1}\times \dfrac{1}{0.00041667}$$ $$= 68.985 \div 0.00041667$$ $$\approx 1.6556 \times 10^{5}\ \text{J mol}^{-1}$$

Converting joules to kilojoules,

$$E_a = 165.6\ \text{kJ mol}^{-1} \approx 166\ \text{kJ mol}^{-1}$$

Hence, the correct answer is Option A.

We start with the general stoichiometric equation for the reaction

$$xA \;\longrightarrow\; yB$$

The instantaneous rate of disappearance of the reactant A is written as $$-\dfrac{d[A]}{dt}$$ and the rate of formation of the product B as $$\dfrac{d[B]}{dt}$$. For any elementary reaction, these rates are related to the common rate $$r$$ by the stoichiometric coefficients, according to the formula

$$r \;=\; -\dfrac{1}{x}\,\dfrac{d[A]}{dt} \;=\; \dfrac{1}{y}\,\dfrac{d[B]}{dt}.$$

Rearranging the first and last members, we obtain a direct relation between the two measurable rates:

$$-\dfrac{d[A]}{dt} \;=\; \dfrac{x}{y}\,\dfrac{d[B]}{dt}.$$

Now we take the common logarithm (base 10) of both sides. Using the rule $$\log_{10}(mn)=\log_{10}m+\log_{10}n$$, we get

$$\log_{10}\!\left(-\dfrac{d[A]}{dt}\right) \;=\; \log_{10}\!\left(\dfrac{x}{y}\,\dfrac{d[B]}{dt}\right) \;=\; \log_{10}\!\left(\dfrac{d[B]}{dt}\right) +\log_{10}\!\left(\dfrac{x}{y}\right).$$

The question itself gives us the following empirical relation:

$$\log_{10}\!\left(-\dfrac{dA}{dt}\right) \;=\; \log_{10}\!\left(\dfrac{dB}{dt}\right) +0.3010.$$

By simple comparison of the two expressions, we identify

$$\log_{10}\!\left(\dfrac{x}{y}\right)=0.3010.$$

We recall the fundamental logarithmic value $$\log_{10}2=0.3010.$$ Therefore

$$\dfrac{x}{y}=2.$$

This means the stoichiometric coefficient of A is exactly twice that of B. The simplest integral choice satisfying this ratio is

$$x=2,\qquad y=1,$$

so the reaction must be

$$2A \;\longrightarrow\; B.$$

Now we inspect each option to see which chemical conversion fits the form $$2A \to B$$:

A. $$\mathrm{C_2H_4}\;(A) \;\longrightarrow\; \mathrm{C_4H_8}\;(B)$$ can be written as $$2\,\mathrm{C_2H_4} \to \mathrm{C_4H_8}.$$ Here two ethene molecules dimerise to give one molecule of butene, exactly matching the required stoichiometry.

B. $$3\,\mathrm{C_2H_2} \to \mathrm{C_6H_6}$$ involves a 3 : 1 ratio, not 2 : 1.

C. The isomerisation $$\mathrm{n\!-\!C_4H_{10}} \to \mathrm{iso\!-\!C_4H_{10}}$$ is a 1 : 1 process.

D. The dissociation $$\mathrm{N_2O_4} \to 2\,\mathrm{NO_2}$$ has the inverse 1 : 2 ratio.

Only option A satisfies the deduced stoichiometric requirement $$2A \to B$$.

Hence, the correct answer is Option A.

We begin by writing the general rate law for the reaction

$$2A + B \longrightarrow \text{product}$$

In its most general form, the rate can be expressed as

$$\text{Rate}=k[A]^m[B]^n$$

where $$m$$ and $$n$$ are the orders of the reaction with respect to $$A$$ and $$B$$ respectively, and $$k$$ is the rate constant.

We now compare the experiments two at a time so that one concentration changes while the other remains (essentially) the same, allowing us to isolate each order.

Determining the order with respect to $$B$$

In Experiments I and II the concentration of $$A$$ is held constant at $$0.10\ \text{mol L}^{-1}$$ while $$[B]$$ changes from $$0.20$$ to $$0.25\ \text{mol L}^{-1}$$. The corresponding rates are equal:

$$\frac{\text{Rate}_{\text{II}}}{\text{Rate}_{\text{I}}} =\frac{6.93\times10^{-3}}{6.93\times10^{-3}}=1$$

Using the rate‐law expression, we have

$$\frac{k[0.10]^m[0.25]^n}{k[0.10]^m[0.20]^n}=1$$

$$\Rightarrow \left(\frac{0.25}{0.20}\right)^n=1$$

$$\Rightarrow \left(\frac{0.25}{0.20}\right)^n=\left(1.25\right)^n=1$$

The only power of $$1.25$$ that gives $$1$$ is $$n=0$$. Thus,

$$n=0$$    and   the reaction is zero order in $$B$$.

Determining the order with respect to $$A$$

Because the rate is independent of $$B$$, we can use any two experiments to find $$m$$. Choosing Experiments I and III:

Experiment I: $$[A]=0.10,\ [B]=0.20,\ \text{Rate}=6.93\times10^{-3}$$

Experiment III: $$[A]=0.20,\ [B]=0.30,\ \text{Rate}=1.386\times10^{-2}$$

The ratio of rates is

$$\frac{\text{Rate}_{\text{III}}}{\text{Rate}_{\text{I}}} =\frac{1.386\times10^{-2}}{6.93\times10^{-3}}=2$$

Because $$n=0$$, the $$[B]$$ terms drop out, giving

$$\frac{k[0.20]^m}{k[0.10]^m}=2$$

$$\Rightarrow \left(\frac{0.20}{0.10}\right)^m=2$$

$$\Rightarrow (2)^m=2$$

$$\Rightarrow m=1$$

Hence the reaction is first order in $$A$$.

Writing the specific rate law

With $$m=1$$ and $$n=0$$, the rate law simplifies to

$$\text{Rate}=k[A]$$

Calculating the rate constant $$k$$

Using the data from Experiment I (any experiment will do):

$$k=\frac{\text{Rate}}{[A]} =\frac{6.93\times10^{-3}\ \text{mol L}^{-1}\text{min}^{-1}}{0.10\ \text{mol L}^{-1}} =6.93\times10^{-2}\ \text{min}^{-1}$$

Finding the time to consume half of $$A$$

Because the reaction is first order in $$A$$, we recall the integrated first‐order law

$$\ln\!\left(\frac{[A]_0}{[A]}\right)=kt$$

For the half‐life, we set $$[A]=\dfrac{[A]_0}{2}$$, so

$$\ln\!\left(\frac{[A]_0}{[A]_0/2}\right)=\ln 2=kt_{1/2}$$

Therefore the half‐life is

$$t_{1/2}=\frac{\ln 2}{k}$$

Substituting $$k=6.93\times10^{-2}\ \text{min}^{-1}$$:

$$t_{1/2}=\frac{0.693}{6.93\times10^{-2}} =\frac{0.693}{0.0693} =10\ \text{minutes}$$

Hence, the correct answer is Option B.

First, we recall the definition of catalysis. A catalyst is a substance that changes the rate of a chemical reaction but itself remains chemically unchanged at the end. When the catalyst and the reactants are present in different physical phases, the phenomenon is called heterogeneous catalysis. Thus, in heterogeneous catalysis we always have at least two phases: one for the catalyst and another for the reacting species.

Now, let us analyze each option and compare the physical phases involved.

Option A - Combustion of Coal. Coal is a solid, and it reacts with oxygen gas to give products such as $$\text{CO}_2$$, $$\text{CO}$$, water vapour, etc. There is no catalyst employed in the ordinary burning of coal. The reaction proceeds simply because of high temperature once ignition has occurred. Since a catalyst is altogether absent, the question of heterogeneity or homogeneity of catalysis does not arise. Therefore this process is not an example of heterogeneous catalysis.

Option B - Hydrogenation of Vegetable Oils. The industrial hydrogenation of unsaturated vegetable oils is carried out in the presence of finely divided nickel, which is a solid. The reactants, hydrogen gas and liquid oil, together constitute different phases from the solid nickel catalyst. Hence we have a solid catalyst and fluid reactants, i.e. $$\text{solid} \;|\; \text{liquid+gas}$$, clearly satisfying the definition of heterogeneous catalysis. Therefore this is indeed an example of heterogeneous catalysis.

Option C - Ostwald’s Process. In the Ostwald process for the manufacture of nitric acid, ammonia gas, $$\text{NH}_3(g)$$, is oxidised in the presence of a platinum-rhodium gauze catalyst, which is a solid. So the catalyst phase is solid while the reactant phase is gaseous, written symbolically as $$\text{solid} \;|\; \text{gas}$$. Because the phases are different, this is a classic case of heterogeneous catalysis.

Option D - Haber’s Process. The Haber process for synthesising ammonia involves nitrogen gas and hydrogen gas reacting over a finely divided iron catalyst that is promoted by $$\text{K}_2\text{O}$$ and $$\text{Al}_2\text{O}_3$$. The iron catalyst is solid whereas both reactants are gaseous, again giving $$\text{solid} \;|\; \text{gas}$$ and fulfilling the criterion for heterogeneous catalysis.

Summarising the above discussion, we see that Options B, C, and D all employ a solid catalyst with reactants in gaseous or liquid phases, making them examples of heterogeneous catalysis. Option A, on the other hand, involves no catalyst at all and therefore cannot belong to the heterogeneous catalysis category.

Hence, the correct answer is Option A.

We are given that 50% of the reaction occurs in 100 seconds, and 75% of the reaction occurs in 200 seconds.

Let the initial concentration be $$a$$. Then:

After 100 s: concentration remaining = $$\frac{a}{2}$$

After 200 s: concentration remaining = $$a - 0.75a = \frac{a}{4}$$

Notice that during the first 100 seconds, the concentration drops from $$a$$ to $$\frac{a}{2}$$ (i.e., it halves). During the next 100 seconds (from 100 s to 200 s), the concentration drops from $$\frac{a}{2}$$ to $$\frac{a}{4}$$ (it halves again).

So the half-life is the same in both intervals: $$t_{1/2} = 100$$ s in each case.

Now recall the key property of a first-order reaction: the half-life of a first-order reaction is independent of the initial concentration. The formula is:

$$t_{1/2} = \frac{0.693}{k}$$

Since the half-life remains constant at 100 s regardless of the concentration (it was 100 s when starting from $$a$$, and again 100 s when starting from $$\frac{a}{2}$$), this confirms that the reaction is first order.

Verification with the integrated rate law:

For a first-order reaction: $$\ln\frac{[A]_0}{[A]_t} = kt$$

At $$t = 100$$ s: $$\ln\frac{a}{a/2} = k \times 100$$

$$\ln 2 = 100k$$

$$k = \frac{\ln 2}{100} = \frac{0.693}{100} = 6.93 \times 10^{-3}$$ s$$^{-1}$$

At $$t = 200$$ s: $$\ln\frac{a}{a/4} = k \times 200$$

$$\ln 4 = 200k$$

$$k = \frac{\ln 4}{200} = \frac{2 \ln 2}{200} = \frac{\ln 2}{100} = 6.93 \times 10^{-3}$$ s$$^{-1}$$

The rate constant $$k$$ is the same in both cases, confirming first-order kinetics.

Why not zero order? For a zero-order reaction, $$t_{1/2} = \frac{a}{2k}$$, which depends on the initial concentration. If we start with concentration $$\frac{a}{2}$$, the new half-life would be $$\frac{a}{4k} = \frac{t_{1/2}}{2} = 50$$ s, not 100 s. So the half-life would change, which contradicts the data.

Why not second order? For a second-order reaction, $$t_{1/2} = \frac{1}{ka}$$. Starting from $$\frac{a}{2}$$, the half-life would become $$\frac{1}{k \cdot a/2} = \frac{2}{ka} = 2 \times 100 = 200$$ s, not 100 s. Again, the half-life changes, which contradicts the data.

Therefore, the reaction is first order.

The correct answer is Option D.

We start by recalling that physical adsorption, also called physisorption, is governed by weak van der Waals forces. Because these forces are non-specific and act at appreciable distances, several characteristic trends follow automatically.

First, according to Le-Chatelier’s principle, an increase in external pressure $$P$$ drives the equilibrium $$\text{Gas (adsorbate)} \rightleftharpoons \text{Adsorbed layer}$$ to the right. Hence, higher $$P$$ results in a larger amount of adsorbed gas. So the statement “Higher the pressure, more the adsorption” is definitely a property of physical adsorption.

Next, adsorption takes place at the surface of the solid adsorbent. The greater the available surface area $$A_s$$, the larger the number of sites on which gas molecules can settle. Therefore, the sentence “Greater the surface area, more the adsorption” is again fully consistent with physisorption.

Further, the process is exothermic. For an exothermic process, lowering the temperature $$T$$ shifts the equilibrium to give more product—in this case, more adsorbed species. Thus “Lower the temperature, more the adsorption” also matches the well-known behaviour of physical adsorption.

Now we analyse the last statement: “Unilayer adsorption occurs.” Because the binding forces are weak, molecules that are already adsorbed do not block further molecules from sticking on top of them; instead, additional layers can pile up one after another. Hence physisorption generally produces multilayer, not monolayer, coverage. Monolayer (unilayer) formation is characteristic of chemisorption, where strong chemical bonds confine adsorption to the first layer only.

Therefore, the sentence about unilayer adsorption does not represent a property of physical adsorption while the other three statements do.

Hence, the correct answer is Option D.

For the first-order reaction $$A \rightarrow P$$, we first recall the two fundamental relations that connect rate constant $$k$$, half-life $$t_{1/2}$$ and concentration:

1. The half-life formula for a first-order reaction is stated as $$t_{1/2}=\frac{0.693}{k}.$$

2. The integrated rate law is $$\ln\!\left(\frac{[A]_0}{[A]}\right)=kt,$$ where $$[A]_0$$ is the initial concentration and $$[A]$$ is the concentration at time $$t$$.

We are given the half-life $$t_{1/2}=10\ \text{days}$$. Using the half-life formula, we substitute and solve for $$k$$:

$$10=\frac{0.693}{k} \quad\Rightarrow\quad k=\frac{0.693}{10}=0.0693\ \text{day}^{-1}.$$

Next, “$$\dfrac14$$-th conversion” means that one-quarter of the reactant has reacted, so three-quarters remains. Mathematically,

$$\frac{[A]}{[A]_0}=\frac34 \quad\Longrightarrow\quad \frac{[A]_0}{[A]}=\frac43.$$

We now insert this ratio into the integrated rate law:

$$\ln\!\left(\frac43\right)=kt.$$

Substituting the value of $$k$$ obtained above gives

$$t=\frac{\ln\!\left(\tfrac43\right)}{0.0693}.$$

To evaluate the natural logarithm, we expand it using the property $$\ln\!\left(\tfrac43\right)=\ln4-\ln3.$$ Since $$\ln2=0.693,$$ we first find $$\ln4$$:

$$\ln4=\ln(2^2)=2\ln2=2(0.693)=1.386.$$

The question supplies $$\ln3=1.1,$$ hence

$$\ln\!\left(\tfrac43\right)=1.386-1.1=0.286.$$

We now divide by the rate constant:

$$t=\frac{0.286}{0.0693}\ \text{days}\approx4.13\ \text{days}.$$

Rounding suitably, the time required is about $$4.1\ \text{days}.$$

Hence, the correct answer is Option C.

We have a single-reactant gas-phase decomposition. For such a reaction the differential rate law can be written in terms of pressure as

where $$P$$ is the instantaneous pressure, $$k$$ is the rate constant and $$n$$ is the order that we have to find.

The initial pressure is given as

The problem supplies the rate at two different extents of reaction, so we first convert the percentages reacted into the corresponding instantaneous pressures.

When 5 % has reacted, the fraction of acetaldehyde that is still present is $$1-0.05=0.95$$. Hence

At this stage the rate is

When 33 % has reacted, the unreacted fraction is $$1-0.33=0.67$$. Therefore

and the rate is

Because the rate law is $$r=k\,P^{\,n}$$ at any moment, we can write for the two sets of conditions

Dividing the first equation by the second eliminates the unknown rate constant $$k$$:

Substituting the numerical values,

The pressure ratio is

So we have

To isolate $$n$$, we take natural logarithms of both sides (any logarithm base would work):

Using the numerical logarithms $$\ln 2.00 = 0.6931$$ and $$\ln 1.418 \approx 0.3500$$, we obtain

The exponent is essentially 2, which means the reaction is second order with respect to acetaldehyde.

Hence, the correct answer is Option B.

The behaviour of many adsorption systems at low pressure is described by the empirical Freundlich adsorption isotherm. Its mathematical form is stated first, because we will need it for the derivation:

$$\frac{x}{m}=k\,P^{\;1/n}$$

Here $$\frac{x}{m}$$ is the amount of gas adsorbed per unit mass of the adsorbent, $$P$$ is the equilibrium pressure of the gas, while $$k$$ and $$n>1$$ are empirical constants characteristic of the particular adsorbate-adsorbent pair and the temperature.

Now we are told that a graph of $$\log\left(\dfrac{x}{m}\right)$$ versus $$\log P$$ is a straight line. To see the equation of that straight line, we take the common (base-10) logarithm of both sides of the Freundlich equation. Using the property of logarithms $$\log(a\,b)=\log a+\log b$$ and $$\log(a^{b})=b\log a$$, we proceed step by step:

First take the logarithm of the left side:

$$\log\left(\frac{x}{m}\right)=\log\left(k\,P^{1/n}\right)$$

Next separate the product inside the logarithm:

$$\log\left(k\,P^{1/n}\right)=\log k+\log\left(P^{1/n}\right)$$

Now apply the power rule of logarithms to the second term:

$$\log\left(P^{1/n}\right)=\frac{1}{n}\,\log P$$

Substituting this back we get:

$$\log\left(\frac{x}{m}\right)=\log k+\frac{1}{n}\,\log P$$

We can compare this equation with the standard equation of a straight line, which in the form $$y=mx+c$$ has slope $$m$$ and intercept $$c$$. Identifying the variables:

$$y=\log\left(\dfrac{x}{m}\right),\quad x=\log P,$$

we see that the coefficient of $$\log P$$—namely $$\dfrac{1}{n}$$—is the slope of the straight line.

Hence, the slope of the plot of $$\log\left(\dfrac{x}{m}\right)$$ versus $$\log P$$ is $$\dfrac{1}{n}$$.

Hence, the correct answer is Option B.

We begin with the balanced gaseous decomposition

$$2\,N_2O_5(g) \;\longrightarrow\; 4\,NO_2(g)+O_2(g)$$

The vessel initially contains only $$N_2O_5$$ at a pressure of $$P_0 = 50\;{\rm mm\;Hg}$$. Let the fraction of the original $$N_2O_5$$ that has decomposed after any time $$t$$ be denoted by $$\alpha$$ (so $$0\le \alpha \le 1$$).

Because pressure is directly proportional to the number of moles at constant volume and temperature, we translate the stoichiometry into partial pressures:

• Initial moles (or mole‐pressure units) of $$N_2O_5$$  $$= 1$$ (reference basis).
• When the fraction $$\alpha$$ has reacted, the mole distribution is

$$\begin{aligned} N_2O_5 &: 1-\alpha\\[4pt] NO_2 &: 2\alpha \quad\;(\text{because }2N_2O_5\to4NO_2\Rightarrow1\to2)\\[4pt] O_2 &: \dfrac{\alpha}{2}\quad (\text{because }2N_2O_5\to O_2\Rightarrow1\to\frac12) \end{aligned}$$

Hence the total number of mole‐units, and therefore the total pressure, become

$$P = P_0\big[(1-\alpha)+2\alpha+\tfrac{\alpha}{2}\big] = P_0\big[1+1.5\,\alpha\big].$$

Data for 50 minutes. After 50 min the pressure has risen from 50 mm Hg to 87.5 mm Hg. Substituting these numbers,

$$87.5 = 50\,(1+1.5\,\alpha_1).$$

Dividing both sides by 50,

$$1+1.5\,\alpha_1 = 1.75 \quad\Longrightarrow\quad 1.5\,\alpha_1 = 0.75 \quad\Longrightarrow\quad \alpha_1 = 0.50.$$

Thus, exactly 50 % of the $$N_2O_5$$ has decomposed in 50 minutes.

Finding the first-order rate constant. For a first-order reaction we have the integrated rate law

$$k t = -\ln(1-\alpha),$$

where $$\alpha$$ is the fraction decomposed in time $$t$$.

Using $$\alpha_1 = 0.50$$ at $$t_1 = 50\;{\rm min}$$,

$$k = \frac{-\ln(1-\alpha_1)}{t_1} = \frac{-\ln(1-0.50)}{50} = \frac{\ln 2}{50}\;{\rm min^{-1}}.$$

Fraction decomposed in 100 minutes. For $$t_2 = 100\;{\rm min}$$,

$$k t_2 = \frac{\ln 2}{50}\times 100 = 2\,\ln 2 = \ln 4.$$

Therefore,

$$1-\alpha_2 = e^{-k t_2}=e^{-\ln 4}= \frac{1}{4}\quad\Longrightarrow\quad \alpha_2 = 1-\frac14 = \frac34 = 0.75.$$

Total pressure after 100 minutes. Substituting $$\alpha_2 = 0.75$$ into $$P = P_0(1+1.5\alpha)$$,

$$P_{100} = 50\Bigl(1 + 1.5 \times 0.75\Bigr) = 50\bigl(1 + 1.125\bigr) = 50 \times 2.125 = 106.25\;{\rm mm\;Hg}.$$

The calculated pressure, 106.25 mm Hg, matches Option B.

Hence, the correct answer is Option B.

We begin with the Arrhenius equation, which gives the temperature-dependence of a rate constant:

$$k = A\,e^{-\dfrac{E_a}{RT}}$$

Here $$k$$ is the rate constant, $$A$$ is the pre-exponential factor, $$E_a$$ is the activation energy, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

For the two reactions we are told that the pre-exponential factors are identical, so we can write

$$k_1 = A\,e^{-\dfrac{E_{a1}}{RT}} \qquad\text{and}\qquad k_2 = A\,e^{-\dfrac{E_{a2}}{RT}}$$

The activation energy of reaction $$A_1$$ is greater than that of reaction $$A_2$$ by $$10\ \text{kJ mol}^{-1}$$. Converting to joules, we have

$$E_{a1} = E_{a2} + 10\ \text{kJ mol}^{-1} = E_{a2} + 10000\ \text{J mol}^{-1}$$

We want the natural logarithm of the ratio of the two rate constants:

$$\ln\frac{k_2}{k_1}$$

Substituting the expressions for $$k_1$$ and $$k_2$$ gives

$$\ln\frac{k_2}{k_1} = \ln\!\left(\frac{A\,e^{-\dfrac{E_{a2}}{RT}}}{A\,e^{-\dfrac{E_{a1}}{RT}}}\right)$$

The factors $$A$$ cancel, and using the property $$\ln\left(\dfrac{e^x}{e^y}\right)=x-y$$ we obtain

$$\ln\frac{k_2}{k_1} = -\frac{E_{a2}}{RT} -\!\left(-\frac{E_{a1}}{RT}\right) = \frac{E_{a1}-E_{a2}}{RT}$$

But $$E_{a1}-E_{a2}=10000\ \text{J mol}^{-1}$$, so

$$\ln\frac{k_2}{k_1} = \frac{10000\ \text{J mol}^{-1}}{R\,T}$$

Now we substitute $$R = 8.314\ \text{J mol}^{-1}\ \text{K}^{-1}$$ and $$T = 300\ \text{K}$$:

$$\ln\frac{k_2}{k_1} = \frac{10000}{8.314 \times 300}$$

First calculate the denominator:

$$8.314 \times 300 = 2494.2$$

Then perform the division:

$$\ln\frac{k_2}{k_1} = \frac{10000}{2494.2} \approx 4.01$$

The value is extremely close to $$4$$, and among the given options $$4$$ is the only matching choice.

Hence, the correct answer is Option C.

For any elementary reaction the Arrhenius equation gives the temperature dependence of the rate constant:

$$k = A\,e^{-E_a/RT}$$

where $$k$$ is the rate constant, $$A$$ is the frequency factor, $$E_a$$ is the activation energy, $$R$$ is the gas constant and $$T$$ is the absolute temperature in kelvin.

Taking two temperatures $$T_1$$ and $$T_2$$ for the same reaction, the ratio of the two rate constants is

$$\frac{k_2}{k_1}=e^{-E_a/R\,(1/T_2-1/T_1)} \;.$$

This may be rewritten, bringing the negative sign inside, as

$$\frac{k_2}{k_1}=e^{E_a/R\,(1/T_1-1/T_2)} \;.$$

We are told that for reaction A the rate doubles when the temperature is raised from $$T_1 = 300\ {\rm K}$$ to $$T_2 = 310\ {\rm K}$$, so $$k_2/k_1 = 2$$. Substituting these numbers, we have

$$2 = e^{E_{a,A}/R \,(1/300 - 1/310)} \;.$$

Taking natural logarithm on both sides,

$$\ln 2 = \frac{E_{a,A}}{R}\bigl(\, \frac{1}{300} - \frac{1}{310}\,\bigr).$$

Calculating the bracketed term first,

$$\frac{1}{300} = 0.0033333\quad{\rm K^{-1}},$$ $$\frac{1}{310} = 0.0032258\quad{\rm K^{-1}},$$

so

$$\frac{1}{300}-\frac{1}{310}=0.0033333-0.0032258 = 0.0001075\ {\rm K^{-1}}.$$

Now inserting this value,

$$\ln 2 = 0.693147 = \frac{E_{a,A}}{R}\times 0.0001075.$$

Hence

$$\frac{E_{a,A}}{R}= \frac{0.693147}{0.0001075}\approx 6.45\times 10^{3}\ {\rm K}.$$

For reaction B the problem states that its activation energy is twice that of reaction A, therefore

$$E_{a,B}=2E_{a,A}\quad\Longrightarrow\quad\frac{E_{a,B}}{R}=2\left(\frac{E_{a,A}}{R}\right)\approx 2\times6.45\times10^{3}=1.289\times10^{4}\ {\rm K}.$$

Let reaction B start at $$T_1 = 300\ {\rm K}$$ and let the temperature be raised to $$T_2 = 300+\Delta T$$. We want the rate to double again, so $$k_2/k_1 = 2$$. Using the same ratio form for reaction B, we write

$$2 = e^{E_{a,B}/R\,(1/300 - 1/T_2)}.$$

Taking the natural logarithm,

$$\ln 2 = \frac{E_{a,B}}{R}\Bigl(\frac{1}{300}-\frac{1}{T_2}\Bigr).$$

Now substitute the numerical value obtained for $$E_{a,B}/R$$:

$$0.693147 = 1.289\times10^{4}\Bigl(\frac{1}{300}-\frac{1}{T_2}\Bigr).$$

Dividing both sides by $$1.289\times10^{4}$$ gives

$$\frac{1}{300}-\frac{1}{T_2}= \frac{0.693147}{1.289\times10^{4}}\approx 5.378\times10^{-5}\ {\rm K^{-1}}.$$

Hence

$$\frac{1}{T_2}= \frac{1}{300}-5.378\times10^{-5} = 0.0033333-0.00005378 = 0.0032795\ {\rm K^{-1}}.$$

Taking the reciprocal to obtain $$T_2$$,

$$T_2 = \frac{1}{0.0032795}\approx 304.92\ {\rm K}.$$

The required rise in temperature is therefore

$$\Delta T = T_2 - 300 = 304.92 - 300 \approx 4.92\ {\rm K}.$$

Hence, the correct answer is Option A.

We start with the Arrhenius equation, which relates the rate constant $$k$$ to the activation energy $$E_a$$:

$$k = A\,e^{-E_a/(RT)}$$

Here $$A$$ is the pre-exponential factor, $$R$$ is the gas constant and $$T$$ is the absolute temperature. Because $$A$$ and $$E_a$$ are assumed to be independent of temperature, we can take the natural logarithm of the ratio of two rate constants at two different temperatures to eliminate $$A$$. Thus, for temperatures $$T_1$$ and $$T_2$$:

$$\ln\!\left(\dfrac{k_2}{k_1}\right)=\dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$$

For the given reaction the temperature rises from $$T_1 = 300\ \text{K}$$ to $$T_2 = 310\ \text{K}$$ and the rate quadruples, so $$k_2/k_1 = 4$$. Substituting these values we get:

$$\ln(4)=\dfrac{E_a}{R}\left(\dfrac{1}{300}-\dfrac{1}{310}\right)$$

We first evaluate the left-hand side. Since $$4 = 2^2$$,

$$\ln(4)=2\ln(2)=2(0.693)=1.386$$

Next we calculate the temperature difference term:

$$\dfrac{1}{300}-\dfrac{1}{310}=\dfrac{310-300}{300\times310}=\dfrac{10}{93000}=\dfrac{1}{9300}\approx 1.07527\times10^{-4}\ \text{K}^{-1}$$

Now we substitute $$R = 8.314\ \text{J mol}^{-1}\ \text{K}^{-1}$$ and rearrange the equation to solve for $$E_a$$:

$$E_a = \dfrac{R\ln(4)}{\dfrac{1}{300}-\dfrac{1}{310}}$$

$$E_a = \dfrac{8.314 \times 1.386}{1.07527\times10^{-4}}$$

First compute the numerator:

$$8.314 \times 1.386 = 11.526$$

Now divide by the denominator:

$$E_a = \dfrac{11.526}{1.07527\times10^{-4}}\ \text{J mol}^{-1}$$

$$E_a \approx 1.07206\times10^{5}\ \text{J mol}^{-1}$$

To express this in kilojoules per mole, we divide by 1000:

$$E_a \approx 107.2\ \text{kJ mol}^{-1}$$

Hence, the correct answer is Option C.

We are told that the decomposition of $$H_2O_2$$ follows first-order kinetics. For any first-order reaction the integrated rate law is stated first:

$$\ln \frac{[A]_0}{[A]} = kt.$$

If we prefer common logarithms, the same relation is written as

$$k = \frac{2.303}{t}\,\log\!\frac{[A]_0}{[A]}.$$

We have an initial concentration $$[A]_0 = 0.5\;{\rm M}$$ that falls to $$[A] = 0.125\;{\rm M}$$ in a time $$t = 50\;{\rm min}.$$ Substituting these numbers gives

$$k = \frac{2.303}{50}\,\log\!\frac{0.5}{0.125}.$$

The ratio inside the logarithm is $$\dfrac{0.5}{0.125}=4,$$ and $$\log 4 = 0.6020.$$ Hence

$$k = \frac{2.303 \times 0.6020}{50}.$$

Multiplying the numerator we obtain $$2.303 \times 0.6020 = 1.386,$$ so

$$k = \frac{1.386}{50} = 0.0277\;{\rm min^{-1}}.$$

For a first-order reaction the instantaneous rate of disappearance of $$H_2O_2$$ is

$$-\frac{d[H_2O_2]}{dt} = k[H_2O_2].$$

Now we are interested in the moment when $$[H_2O_2] = 0.05\;{\rm M}.$$ Putting this value into the expression for the rate, we get

$$-\frac{d[H_2O_2]}{dt} = 0.0277 \times 0.05.$$

Carrying out the multiplication,

$$-\frac{d[H_2O_2]}{dt} = 1.385 \times 10^{-3}\;{\rm mol\;L^{-1}\,min^{-1}}.$$

This is the rate at which $$H_2O_2$$ is being consumed. The balanced chemical equation for the decomposition is

$$2H_2O_2 \rightarrow 2H_2O + O_2.$$

From the stoichiometry we see that two moles of $$H_2O_2$$ produce one mole of $$O_2$$, so the rate of formation of oxygen is one-half of the rate of disappearance of $$H_2O_2$$. Thus,

$$\frac{d[O_2]}{dt} = \frac{1}{2}\left(-\frac{d[H_2O_2]}{dt}\right) = \frac{1}{2}\,(1.385 \times 10^{-3}).$$

Dividing by two,

$$\frac{d[O_2]}{dt} = 6.93 \times 10^{-4}\;{\rm mol\;L^{-1}\,min^{-1}}.$$

Because the options are expressed simply in $$\rm mol\;min^{-1},$$ and the numerical value matches exactly, we choose the corresponding option.

Hence, the correct answer is Option D.

We are told that the experimentally determined rate law for the reaction $$A + B \rightarrow \text{Product}$$ is

$$r = k[A][B]$$

Here $$r$$ is the rate of the reaction, $$[A]$$ and $$[B]$$ are the molar concentrations of the reactants, and $$k$$ is the rate constant. By definition, the rate constant $$k$$ depends only on temperature (and the presence of a catalyst, if any). It does not change when we merely alter the concentrations of the reactants.

Nevertheless, to see this algebraically, let us write the rate twice—once with the original concentrations and once after changing the concentration of $$B$$.

Originally we have

$$[A]_1 = 0.1\ \text{mol L}^{-1}, \qquad [B]_1 = 0.1\ \text{mol L}^{-1}.$$

Substituting these values into the rate law, the initial rate becomes

$$r_1 = k [A]_1 [B]_1 = k(0.1)(0.1) = 0.01k.$$

Now the concentration of $$B$$ is increased to $$[B]_2 = 0.3\ \text{mol L}^{-1}$$ while $$[A]$$ is kept the same, so $$[A]_2 = 0.1\ \text{mol L}^{-1}$$. The new rate is therefore

$$r_2 = k [A]_2 [B]_2 = k(0.1)(0.3) = 0.03k.$$

Dividing the new rate by the original rate gives

$$\frac{r_2}{r_1} = \frac{0.03k}{0.01k} = 3.$$

The factor of 3 tells us that the rate triples, but notice that the factor $$k$$ cancels out in the ratio:

$$\frac{0.03\cancel{k}}{0.01\cancel{k}} = 3.$$

Since $$k$$ cancels, it is clear that changing the concentration of $$B$$ does not and cannot alter the numerical value of the rate constant. The only quantities affected are the rate itself and any directly concentration-dependent terms.

Hence the rate constant remains exactly the same, $$k$$.

Hence, the correct answer is Option D.

In a multi-step reaction mechanism, the overall rate is determined by the rate-determining step (RDS), which is the slowest step in the sequence. For such a mechanism, the effective rate constant of the overall reaction is approximately equal to the rate constant of the RDS.

By comparing the rate constants, we see that $$k_i$$ ($$5.2 \times 10^9$$) is significantly smaller than $$k_{ii}$$ ($$2.6 \times 10^{10}$$). Since step (i) is the slower process, it acts as the rate-determining step.

We begin by examining the Assertion: “Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen.” Atmospheric air indeed contains roughly $$78\%$$ nitrogen as $$N_2$$ and about $$21\%$$ oxygen as $$O_2$$. Under ordinary atmospheric conditions (room temperature, normal pressure) these two gases coexist without undergoing any noticeable chemical reaction to give oxides such as $$NO$$ or $$NO_2$$. This statement is therefore correct.

Next we look at the Reason: “The reaction between nitrogen and oxygen requires high temperature.” The conversion of $$N_2$$ and $$O_2$$ into nitric oxide may be written as

$$N_2(g) + O_2(g) \rightarrow 2\,NO(g)$$

For this reaction to proceed, the very strong triple bond in $$N_2$$ (bond dissociation energy approximately $$945\ \text{kJ mol}^{-1}$$) must be broken. Breaking such a bond demands a large amount of energy, so a high activation energy barrier exists. As a consequence, an appreciable rate of reaction is observed only when the temperature is very high, for example during lightning discharges in the atmosphere or in internal-combustion engines where flame temperatures reach several thousand kelvin. Thus, the Reason is also a correct statement.

Now we must decide whether the Reason actually explains the Assertion. The Assertion states that oxides of nitrogen are not formed under normal atmospheric conditions. The Reason tells us that formation of these oxides needs high temperature. Because the usual atmospheric temperature is far below the required temperature threshold, the reaction hardly proceeds, which directly accounts for the observed lack of oxides. Therefore, the Reason supplies the genuine cause for the Assertion.

Both statements are true, and the Reason is the correct explanation for the Assertion. Hence, the correct answer is Option B.

First, recall the statement of the rate law for an elementary reaction. For any elementary step involving $$m+n+\dots$$ reacting molecules, collision theory tells us that its rate is directly proportional to the product of the concentrations of all those molecules, i.e.

$$\text{Rate}=k[A]^m[B]^n\dots$$

Here $$k$$ is the rate constant. The presence of each extra reactant concentration term reflects the fact that all of those molecules must collide simultaneously in the correct orientation for a successful reaction event.

Now consider what happens when the molecularity, and therefore the order, becomes large, say greater than $$3$$. For four different species the elementary event must involve a simultaneous collision of four particles. According to the kinetic theory of gases, the probability of such a four-body collision occurring in a tiny volume element in a very short time interval is proportional to the product of four separate number densities.

Mathematically, using simple probability arguments, if the probability of finding one specified molecule in that small region is $$p$$, then the probability of locating two independent specified molecules there together is $$p^2$$, for three it is $$p^3$$, and for four it falls to $$p^4$$. Because $$p<1$$, every additional multiplying factor makes the overall probability shrink drastically:

$$p > p^2 > p^3 > p^4 \quad (\text{for }0<p<1).$$

Thus, when more than three reactant molecules have to strike one another at exactly the same instant, the likelihood becomes so vanishingly small that, in practice, such elementary steps are almost never observed. Instead, reactions that appear overall to involve many molecules actually proceed through a sequence of simpler bimolecular or termolecular elementary steps.

Let us now weigh each option against this reasoning:

Option A talks about a loss of active species on collision. While some collisions may indeed be ineffective, that fact does not specifically explain why orders >3 are rare.

Option B states that the probability of the simultaneous collision of all reacting species is very low. This argument matches exactly with the statistical reasoning we have just developed, so it directly addresses the rarity of high-order reactions.

Option C claims that entropy and activation energy necessarily rise when more molecules are involved. Although activation energies can change, that is not the central, universally valid reason for the scarcity of elementary steps of order >3.

Option D refers to an equilibrium shift owing to elastic collisions, which is unrelated to the fundamental kinetics question being asked.

Therefore, only Option B pinpoints the core kinetic explanation: the extremely low probability of so many molecules meeting simultaneously with the correct orientation and energy.

Hence, the correct answer is Option B.

The rate equation given is Rate = k [A] [B]. This means the rate depends on the concentrations of both A and B.

Initially, let the concentration of A be [A] and the concentration of B be [B]. So the initial rate is:

$$ \text{Rate}_1 = k \times [A] \times [B] $$

Now, the concentration of A is kept the same, so it remains [A]. The concentration of B is doubled, so it becomes 2[B].

The new rate is:

$$ \text{Rate}_2 = k \times [A] \times (2[B]) $$

Simplify this expression:

$$ \text{Rate}_2 = k \times [A] \times 2[B] $$

$$ \text{Rate}_2 = 2 \times (k \times [A] \times [B]) $$

Notice that $$ k \times [A] \times [B] $$ is exactly the initial rate $$\text{Rate}_1$$. So we can write:

$$ \text{Rate}_2 = 2 \times \text{Rate}_1 $$

This shows that the new rate is twice the initial rate. Therefore, the rate doubles.

Hence, the correct answer is Option A.

The reaction is $$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$$, which follows first-order kinetics. The initial pressure of $$N_2O_5$$ is 50 mm Hg, and after 30 minutes, the total pressure increases to 87.5 mm Hg. We need to find the total pressure after 60 minutes.

Since the reaction is first-order with respect to $$N_2O_5$$, the rate depends on its concentration. At constant volume and temperature, concentration is proportional to partial pressure, so we can use partial pressures in the first-order rate equation.

Let $$P_0 = 50$$ mm Hg be the initial pressure of $$N_2O_5$$. As the reaction proceeds, $$N_2O_5$$ decomposes, and products are formed. For the reaction $$2N_2O_5 \rightarrow 4NO_2 + O_2$$, when 2 moles of $$N_2O_5$$ decompose, they produce 5 moles of products (4 moles of $$NO_2$$ and 1 mole of $$O_2$$), resulting in a net increase of 3 moles (since 5 moles produced minus 2 moles decomposed).

Define $$y$$ as the pressure decrease corresponding to the decomposition of $$N_2O_5$$ based on the stoichiometry. If the initial pressure of $$N_2O_5$$ is $$P_0$$, then:

• The pressure of $$N_2O_5$$ remaining is $$P_0 - 2y$$.
• The pressure of $$NO_2$$ produced is $$4y$$.
• The pressure of $$O_2$$ produced is $$y$$.

The total pressure $$P_t$$ at any time is:

Solving for $$y$$:

The partial pressure of $$N_2O_5$$ at time $$t$$ is:

For a first-order reaction:

Here, $$[A]_0 = P_0 = 50$$ mm Hg and $$[A] = P_{N_2O_5} = \frac{5P_0 - 2P_t}{3}$$, so:

Simplify the argument:

At $$t = 30$$ min, $$P_t = 87.5$$ mm Hg:

Calculate inside the denominator:

So:

Solving for $$k$$:

Now, find the total pressure $$P$$ at $$t = 60$$ min:

Substitute $$k = \frac{\ln 2}{30}$$:

Since $$\ln a = \ln b$$ implies $$a = b$$:

Solve for $$P$$:

Hence, the pressure after 60 minutes is 106.25 mm Hg, which corresponds to Option A.

So, the answer is 106.25 mm Hg.

We have the gaseous equilibrium $$A(g) \rightleftharpoons B(g)$$ in which the enthalpy change is given as $$\Delta H = -40\; \text{kJ mol}^{-1}$$. A negative value tells us that the forward reaction is exothermic, that is, the products lie lower on the energy scale than the reactants.

For any elementary reaction we picture a potential-energy diagram with a single transition state. Let the reactants be our zero level of energy. Then:

• The energy of the transition state above the reactants is the forward activation energy, so its height is $$E_f$$.

• Because the products are $$\Delta H$$ lower than the reactants, their energy is $$\Delta H$$ (here $$-40\; \text{kJ mol}^{-1}$$).

• The reverse activation energy $$E_b$$ is the difference between the transition-state energy and the energy of the products.

Hence, starting from these definitions, we write

$$E_b = (\text{transition-state energy}) - (\text{product energy}) = E_f - \Delta H.$$

Re-arranging this very useful relation we also get

$$\Delta H = E_f - E_b.$$

Next, the question states that the ratio of the forward to the reverse activation energies is

$$\frac{E_f}{E_b} = \frac{2}{3}.$$

Using the relation $$E_b = E_f - \Delta H$$ and substituting the numerical value $$\Delta H = -40\; \text{kJ mol}^{-1}$$, we obtain

$$E_b = E_f - (-40) = E_f + 40.$$

Now we substitute this expression for $$E_b$$ into the given ratio:

$$\frac{E_f}{E_b} = \frac{E_f}{E_f + 40} = \frac{2}{3}.$$

Cross-multiplying gives every algebraic step clearly:

$$3\,E_f = 2\,(E_f + 40).$$

$$3E_f = 2E_f + 80.$$

$$3E_f - 2E_f = 80.$$

$$E_f = 80\; \text{kJ mol}^{-1}.$$

Once $$E_f$$ is known we return to $$E_b = E_f + 40$$:

$$E_b = 80 + 40 = 120\; \text{kJ mol}^{-1}.$$

Thus the forward and reverse activation energies are

$$E_f = 80\; \text{kJ mol}^{-1}, \qquad E_b = 120\; \text{kJ mol}^{-1}.$$

Comparing with the options provided, these values correspond exactly to Option C.

Hence, the correct answer is Option C.

The half-life period of a first-order reaction is given as 15 minutes. For a first-order reaction, the half-life ($$ t_{1/2} $$) is related to the rate constant ($$ k $$) by the formula:

Substituting the given half-life:

Solving for $$ k $$:

The amount of substance left after time $$ t $$ for a first-order reaction is given by:

where $$ N_0 $$ is the original amount and $$ N $$ is the amount left at time $$ t $$. We need to find the amount after one hour, which is 60 minutes. So, $$ t = 60 $$ minutes.

Substitute $$ k $$ and $$ t $$:

Simplify the exponent:

So the equation becomes:

Using the property $$ e^{c \ln a} = a^c $$, we get:

Therefore:

Alternatively, using the half-life concept: the number of half-lives in 60 minutes is $$ \frac{60}{15} = 4 $$. After each half-life, the amount halves, so after 4 half-lives, the amount is $$ \left( \frac{1}{2} \right)^4 = \frac{1}{16} $$ of the original amount.

Hence, the amount left after one hour is $$ \frac{1}{16} $$ of the original amount. Comparing with the options, Option A corresponds to $$ \frac{1}{16} $$ of the original amount.

Hence, the correct answer is Option A.

For the reaction $$3A + 2B \rightarrow C + D$$, we need to write the differential rate law. The rate of a reaction is defined as the rate of disappearance of reactants or the rate of appearance of products, divided by their respective stoichiometric coefficients. This ensures the rate is consistent regardless of which species we measure.

The stoichiometric coefficients are 3 for A, 2 for B, 1 for C, and 1 for D. Therefore, the rate of the reaction (denoted as rate) can be expressed in multiple ways:

Rate $$ = -\frac{1}{3} \frac{d[A]}{dt} $$ because A is a reactant and its concentration decreases over time.

Rate $$ = -\frac{1}{2} \frac{d[B]}{dt} $$ because B is a reactant and its concentration decreases.

Rate $$ = +\frac{1}{1} \frac{d[C]}{dt} = \frac{d[C]}{dt} $$ because C is a product and its concentration increases.

Rate $$ = +\frac{1}{1} \frac{d[D]}{dt} = \frac{d[D]}{dt} $$ because D is a product and its concentration increases.

All these expressions represent the same rate. The differential rate law relates this rate to the concentrations of the reactants, typically written as $$ k [A]^n [B]^m $$, where $$ k $$ is the rate constant, and $$ n $$ and $$ m $$ are the orders of the reaction with respect to A and B.

We are asked to compare the expressions involving $$ \frac{d[A]}{dt} $$ and $$ \frac{d[C]}{dt} $$. From above:

Rate $$ = -\frac{1}{3} \frac{d[A]}{dt} $$

and

Rate $$ = \frac{d[C]}{dt} $$

Since both equal the rate, we set them equal:

$$ -\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} $$

Additionally, the rate is given by $$ k [A]^n [B]^m $$, so:

$$ -\frac{1}{3} \frac{d[A]}{dt} = k [A]^n [B]^m $$

and

$$ \frac{d[C]}{dt} = k [A]^n [B]^m $$

Therefore, we can write:

$$ -\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k [A]^n [B]^m $$

Now, let's examine the options:

Option A: $$ +\frac{1}{3} \frac{d[A]}{dt} = -\frac{d[C]}{dt} = k [A]^n [B]^m $$

This implies $$ +\frac{1}{3} \frac{d[A]}{dt} = k [A]^n [B]^m $$ and $$ -\frac{d[C]}{dt} = k [A]^n [B]^m $$. However, since A is a reactant, $$ \frac{d[A]}{dt} $$ is negative, so $$ +\frac{1}{3} \frac{d[A]}{dt} $$ would be negative, but $$ -\frac{d[C]}{dt} $$ would be negative (as $$ \frac{d[C]}{dt} $$ is positive), while the rate $$ k [A]^n [B]^m $$ is positive. This does not match our derived equation.

Option B: $$ \frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k [A]^n [B]^m $$

This implies $$ \frac{1}{3} \frac{d[A]}{dt} = k [A]^n [B]^m $$ and $$ \frac{d[C]}{dt} = k [A]^n [B]^m $$. Since A is decreasing, $$ \frac{d[A]}{dt} $$ is negative, so $$ \frac{1}{3} \frac{d[A]}{dt} $$ is negative, but $$ \frac{d[C]}{dt} $$ is positive. A negative quantity cannot equal a positive quantity unless both are zero, which is not generally true. Hence, this is incorrect.

Option C: $$ -\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k [A]^n [B]^m $$

This matches exactly what we derived: $$ -\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k [A]^n [B]^m $$.

Option D: $$ -\frac{d[A]}{dt} = \frac{d[C]}{dt} = k [A]^n [B]^m $$

This implies $$ -\frac{d[A]}{dt} = k [A]^n [B]^m $$ and $$ \frac{d[C]}{dt} = k [A]^n [B]^m $$. From our rate definition, $$ -\frac{d[A]}{dt} = 3 \times \left( -\frac{1}{3} \frac{d[A]}{dt} \right) = 3 \times \text{rate} $$, and $$ \frac{d[C]}{dt} = \text{rate} $$. So $$ -\frac{d[A]}{dt} = 3 \times \frac{d[C]}{dt} $$, meaning $$ -\frac{d[A]}{dt} $$ is three times $$ \frac{d[C]}{dt} $$. The option claims they are equal, which would require $$ 3 \times \frac{d[C]}{dt} = \frac{d[C]}{dt} $$, implying $$ \frac{d[C]}{dt} = 0 $$, which is not true in general. Hence, this is incorrect.

Therefore, the correct option is C.

Hence, the correct answer is Option C.

We have the reaction $$2A + B \rightarrow C + D$$ and we want to obtain a mathematical expression for the initial rate of formation of $$C$$. In chemical kinetics, the general differential rate law for such a reaction is written as a product of the concentrations of the reactants raised to unknown powers (orders of reaction). So we write

$$\text{rate}= \frac{dc}{dt}=k[A]^m[B]^n$$

where $$k$$ is the rate constant at the given temperature, and $$m$$ and $$n$$ are the partial reaction orders with respect to $$A$$ and $$B$$ respectively. Our task is to determine $$m$$ and $$n$$ from the experimental data.

The table of data is:

Experiment 1: $$[A]_0 = 0.1\ \text{M},\quad [B]_0 = 0.1\ \text{M},\quad \text{rate}_1 = 1.2\times10^{-3}$$

Experiment 2: $$[A]_0 = 0.1\ \text{M},\quad [B]_0 = 0.2\ \text{M},\quad \text{rate}_2 = 1.2\times10^{-3}$$

Experiment 3: $$[A]_0 = 0.2\ \text{M},\quad [B]_0 = 0.1\ \text{M},\quad \text{rate}_3 = 2.4\times10^{-3}$$

First we compare Experiments 1 and 2. The concentration of $$A$$ is kept the same, $$[A]=0.1\ \text{M}$$, while $$[B]$$ is doubled from $$0.1\ \text{M}$$ to $$0.2\ \text{M}$$. Mathematically, we write

$$\frac{\text{rate}_2}{\text{rate}_1}= \frac{k[A]^m[B]_2^n}{k[A]^m[B]_1^n}=\left(\frac{[B]_2}{[B]_1}\right)^n.$$

Substituting the numerical values, we have

$$\frac{1.2\times10^{-3}}{1.2\times10^{-3}} = \left(\frac{0.2}{0.1}\right)^n.$$

The left side is $$1$$, and the right side simplifies to $$2^n$$. Hence

$$1 = 2^n \quad\Rightarrow\quad 2^n = 1 \quad\Rightarrow\quad n = 0.$$

So the reaction is zero-order with respect to $$B$$.

Next we compare Experiments 1 and 3. This time $$[B]$$ is kept constant at $$0.1\ \text{M}$$, while $$[A]$$ is doubled from $$0.1\ \text{M}$$ to $$0.2\ \text{M}$$. Using the same procedure,

$$\frac{\text{rate}_3}{\text{rate}_1}= \frac{k[A]_3^m[B]^n}{k[A]_1^m[B]^n}= \left(\frac{[A]_3}{[A]_1}\right)^m.$$

Substituting the values,

$$\frac{2.4\times10^{-3}}{1.2\times10^{-3}} = \left(\frac{0.2}{0.1}\right)^m.$$

The left side is $$2$$, and the right side is $$2^m$$. Therefore

$$2 = 2^m \quad\Rightarrow\quad 2^m = 2 \quad\Rightarrow\quad m = 1.$$

Thus the reaction is first-order with respect to $$A$$.

Combining the two results $$m=1$$ and $$n=0$$, the rate law becomes

$$\text{rate}= \frac{dc}{dt}= k[A]^1[B]^0 = k[A].$$

This matches the expression given in option D.

Hence, the correct answer is Option D.

The reaction is given as: $$2SO_2 + O_2 \rightleftharpoons 2SO_3$$

The rate of reaction can be expressed in terms of any reactant or product using the stoichiometric coefficients. For a general reaction: $$aA + bB \rightarrow cC + dD$$, the rate is defined as: $$\text{rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$$

For this reaction, the stoichiometric coefficients are:

• For $$SO_2$$, the coefficient is 2.
• For $$O_2$$, the coefficient is 1.
• For $$SO_3$$, the coefficient is 2.

Therefore, the rate of reaction can be written as: $$\text{rate} = -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{1}{1} \frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt}$$

We are given that $$\frac{d[O_2]}{dt} = -2.5 \times 10^{-4}$$ mol L$$^{-1}$$ s$$^{-1}$$. Using the expression for $$O_2$$: $$\text{rate} = -\frac{d[O_2]}{dt}$$ because the coefficient of $$O_2$$ is 1.

Substituting the given value: $$\text{rate} = - \left( -2.5 \times 10^{-4} \right) = 2.5 \times 10^{-4}$$ mol L$$^{-1}$$ s$$^{-1}$$.

Now, we need to find the rate in terms of $$[SO_2]$$, which is $$\frac{d[SO_2]}{dt}$$. Using the expression for $$SO_2$$: $$\text{rate} = -\frac{1}{2} \frac{d[SO_2]}{dt}$$.

Rearranging for $$\frac{d[SO_2]}{dt}$$: $$\frac{d[SO_2]}{dt} = -2 \times \text{rate}$$.

Substituting the value of rate: $$\frac{d[SO_2]}{dt} = -2 \times \left( 2.5 \times 10^{-4} \right)$$.

Calculating: $$-2 \times 2.5 \times 10^{-4} = -5.0 \times 10^{-4}$$.

So, $$\frac{d[SO_2]}{dt} = -5.00 \times 10^{-4}$$ mol L$$^{-1}$$ s$$^{-1}$$.

Comparing with the options, this matches option D.

Hence, the correct answer is Option D.

The activation energy ($$E_A$$) can be found using the Arrhenius equation, which relates the rate constant ($$k$$) to temperature ($$T$$) and activation energy. The formula for two different temperatures is:

$$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_A}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

where:

• $$k_1$$ and $$k_2$$ are the rate constants at temperatures $$T_1$$ and $$T_2$$ respectively,
• $$R$$ is the gas constant (8.314 J K⁻¹ mol⁻¹),
• $$E_A$$ is the activation energy in J/mol.

First, convert the temperatures from Celsius to Kelvin:

$$T_1 = 100^\circ \text{C} + 273 = 373 \text{K}$$

$$T_2 = 150^\circ \text{C} + 273 = 423 \text{K}$$

Given:

$$k_1 = 1.3 \times 10^{-4} \text{M}^{-1} \text{s}^{-1}$$

$$k_2 = 1.3 \times 10^{-3} \text{M}^{-1} \text{s}^{-1}$$

Calculate the ratio $$\frac{k_2}{k_1}$$:

$$\frac{k_2}{k_1} = \frac{1.3 \times 10^{-3}}{1.3 \times 10^{-4}} = 10$$

Take the natural logarithm of the ratio:

$$\ln\left(\frac{k_2}{k_1}\right) = \ln(10) \approx 2.302585$$

Now, compute the difference in the reciprocal temperatures:

$$\frac{1}{T_1} = \frac{1}{373} \approx 0.002681$$

$$\frac{1}{T_2} = \frac{1}{423} \approx 0.002364$$

$$\frac{1}{T_1} - \frac{1}{T_2} = 0.002681 - 0.002364 = 0.000317$$

Alternatively, use the exact fraction:

$$\frac{1}{T_1} - \frac{1}{T_2} = \frac{T_2 - T_1}{T_1 T_2} = \frac{423 - 373}{373 \times 423} = \frac{50}{373 \times 423}$$

Calculate $$373 \times 423$$:

$$373 \times 400 = 149200$$

$$373 \times 23 = 373 \times 20 + 373 \times 3 = 7460 + 1119 = 8579$$

$$373 \times 423 = 149200 + 8579 = 157779$$

So,

$$\frac{1}{T_1} - \frac{1}{T_2} = \frac{50}{157779} \approx 0.000317$$

Now rearrange the Arrhenius equation to solve for $$E_A$$:

$$E_A = \frac{R \cdot \ln\left(\frac{k_2}{k_1}\right)}{\frac{1}{T_1} - \frac{1}{T_2}}$$

Substitute the values:

$$E_A = \frac{8.314 \times 2.302585}{0.000317}$$

First, compute the numerator:

$$8.314 \times 2.302585 = 19.14369169$$

Now divide by the denominator:

$$E_A = \frac{19.14369169}{0.000317} \approx 60402.77 \text{J/mol}$$

Alternatively, using the exact fraction:

$$E_A = \frac{8.314 \times 2.302585 \times 157779}{50}$$

First, compute $$8.314 \times 157779 = 1311774.606$$

Then, $$1311774.606 \times 2.302585 \approx 3020890.06$$

Then, divide by 50:

$$\frac{3020890.06}{50} = 60417.8012 \text{J/mol}$$

Both methods give approximately 60400 J/mol. Convert to kJ/mol by dividing by 1000:

$$E_A \approx 60.4 \text{kJ/mol}$$

Rounding to the nearest whole number, $$E_A \approx 60 \text{kJ/mol}$$.

Comparing with the options:

• A. 16
• B. 60
• C. 99
• D. 132

Hence, the correct answer is Option B.

For the reaction $$2N_2O_5 \to 4NO_2 + O_2$$, the rate of reaction can be defined using the stoichiometric coefficients. The rate is given by:

$$ \text{rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} $$

This equation comes from the fact that for every 2 moles of $$N_2O_5$$ consumed, 4 moles of $$NO_2$$ are produced. Therefore, the rate of disappearance of $$N_2O_5$$ and the rate of appearance of $$NO_2$$ are related by the stoichiometry.

Now, the problem gives two expressions:

First, $$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$$.

Second, $$+\frac{d[NO_2]}{dt} = k'[N_2O_5]$$.

From the stoichiometric relationship:

$$ -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} $$

Multiply both sides by 4 to simplify:

$$ 4 \times \left( -\frac{1}{2} \frac{d[N_2O_5]}{dt} \right) = 4 \times \left( \frac{1}{4} \frac{d[NO_2]}{dt} \right) $$

$$ -2 \frac{d[N_2O_5]}{dt} = \frac{d[NO_2]}{dt} $$

Using the first given expression, $$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$$, multiply both sides by 2:

$$ 2 \times \left( -\frac{d[N_2O_5]}{dt} \right) = 2 \times k[N_2O_5] $$

$$ -2 \frac{d[N_2O_5]}{dt} = 2k[N_2O_5] $$

But from earlier, $$-2 \frac{d[N_2O_5]}{dt} = \frac{d[NO_2]}{dt}$$, so substitute:

$$ \frac{d[NO_2]}{dt} = 2k[N_2O_5] $$

The second given expression is $$\frac{d[NO_2]}{dt} = k'[N_2O_5]$$. Therefore:

$$ 2k[N_2O_5] = k'[N_2O_5] $$

Since the concentration of $$N_2O_5$$ is not zero, we can divide both sides by $$[N_2O_5]$$:

$$ 2k = k' $$

This means that $$k'$$ is twice $$k$$, or $$2k = k'$$.

Looking at the options:

A. k = k'

B. 2k = k'

C. k = 2k'

D. k = 4k'

Hence, the correct answer is Option B.

The reaction given is X → Y, which is exothermic. This means the enthalpy change, ΔH, is negative. The problem states the enthalpy of reaction is 135 kJ mol⁻¹, so we write ΔH = -135 kJ mol⁻¹.

The activation energy for the forward reaction (X → Y) is given as Ea(forward) = 150 kJ mol⁻¹.

We need to find the activation energy for the reverse reaction (Y → X), denoted as Ea(reverse).

Recall the relationship between activation energies and enthalpy change for a reaction: ΔH = Ea(forward) - Ea(reverse).

Substitute the known values into this equation:

ΔH = Ea(forward) - Ea(reverse)

-135 = 150 - Ea(reverse)

Now, solve for Ea(reverse). First, isolate the term with Ea(reverse) by moving 150 to the left side:

-135 - 150 = - Ea(reverse)

-285 = - Ea(reverse)

Multiply both sides by -1 to solve for Ea(reverse):

285 = Ea(reverse)

So, Ea(reverse) = 285 kJ mol⁻¹.

Comparing with the options:

A. 280 kJ mol⁻¹

B. 285 kJ mol⁻¹

C. 270 kJ mol⁻¹

D. 15 kJ mol⁻¹

Hence, the correct answer is Option B.

For a zero-order reaction, the rate of reaction is constant and independent of the concentration of the reactant. The integrated rate law for a zero-order reaction is given by:

$$ [A]_t = [A]_0 - kt $$

where:

• $$ [A]_t $$ is the concentration of the reactant at time $$ t $$,
• $$ [A]_0 $$ is the initial concentration,
• $$ k $$ is the rate constant,
• $$ t $$ is the time.

We are given:

• Rate constant $$ k = 2.0 \times 10^{-2} $$ mol L$$^{-1}$$ s$$^{-1}$$,
• Time $$ t = 25 $$ seconds,
• Concentration at time $$ t $$, $$ [A]_t = 0.5 $$ M (which is equivalent to mol L$$^{-1}$$).

We need to find the initial concentration $$ [A]_0 $$.

Substitute the given values into the equation:

$$ 0.5 = [A]_0 - (2.0 \times 10^{-2}) \times 25 $$

First, calculate the value of $$ kt $$:

$$ kt = (2.0 \times 10^{-2}) \times 25 $$

Multiply 2.0 by 25:

$$ 2.0 \times 25 = 50 $$

Now, since $$ 2.0 \times 10^{-2} = 0.02 $$, we have:

$$ kt = 0.02 \times 25 = 0.5 $$

Alternatively, using scientific notation:

$$ 2.0 \times 10^{-2} = 0.02 $$

$$ 0.02 \times 25 = 0.5 $$

So, $$ kt = 0.5 $$ mol L$$^{-1}$$.

Now, substitute back into the equation:

$$ 0.5 = [A]_0 - 0.5 $$

To solve for $$ [A]_0 $$, add 0.5 to both sides:

$$ 0.5 + 0.5 = [A]_0 $$

$$ 1.0 = [A]_0 $$

Therefore, the initial concentration is 1.0 M.

Comparing with the options:

• A. 0.5M
• B. 1.25M
• C. 12.5M
• D. 1.0M

Hence, the correct answer is Option D.

The given reaction is:

$$2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$$

The instantaneous rate of disappearance of $$MnO_4^-$$ is given as $$4.56 \times 10^{-3}$$ Ms$$^{-1}$$. This means $$-\frac{d[MnO_4^-]}{dt} = 4.56 \times 10^{-3}$$ Ms$$^{-1}$$.

To find the rate of appearance of $$I_2$$, which is $$\frac{d[I_2]}{dt}$$, we use the stoichiometric coefficients from the balanced equation. The rate of reaction can be expressed in terms of any reactant or product. For the reaction, the rate $$r$$ is defined as:

$$r = -\frac{1}{2} \frac{d[MnO_4^-]}{dt} = \frac{1}{5} \frac{d[I_2]}{dt}$$

We know $$-\frac{d[MnO_4^-]}{dt} = 4.56 \times 10^{-3}$$ Ms$$^{-1}$$. Substituting this into the expression for the rate:

$$r = -\frac{1}{2} \times (-\frac{d[MnO_4^-]}{dt}) = \frac{1}{2} \times (4.56 \times 10^{-3})$$

Calculate the value:

$$\frac{1}{2} \times 4.56 \times 10^{-3} = 2.28 \times 10^{-3}$$ Ms$$^{-1}$$

So, $$r = 2.28 \times 10^{-3}$$ Ms$$^{-1}$$.

Now, using the rate expression for $$I_2$$:

$$r = \frac{1}{5} \frac{d[I_2]}{dt}$$

Substitute the value of $$r$$:

$$2.28 \times 10^{-3} = \frac{1}{5} \frac{d[I_2]}{dt}$$

To solve for $$\frac{d[I_2]}{dt}$$, multiply both sides by 5:

$$\frac{d[I_2]}{dt} = 5 \times 2.28 \times 10^{-3}$$

Calculate the product:

$$5 \times 2.28 = 11.4$$

So, $$\frac{d[I_2]}{dt} = 11.4 \times 10^{-3} = 1.14 \times 10^{-2}$$ Ms$$^{-1}$$.

Therefore, the rate of appearance of $$I_2$$ is $$1.14 \times 10^{-2}$$ Ms$$^{-1}$$.

Comparing with the options:

A. $$4.56 \times 10^{-4}$$ Ms$$^{-1}$$

B. $$1.14 \times 10^{-2}$$ Ms$$^{-1}$$

C. $$1.14 \times 10^{-3}$$ Ms$$^{-1}$$

D. $$5.7 \times 10^{-3}$$ Ms$$^{-1}$$

Hence, the correct answer is Option B.

A radioactive isotope decays over time, and we are given its half-life is 3 days. This means that every 3 days, the amount of the isotope reduces to half of what it was at the start of that period. The isotope was received after 12 days, and at that time, 3 grams are left. We need to find the initial mass.

The formula for the remaining amount of a radioactive substance after time $$ t $$ is:

$$ N = N_0 \left( \frac{1}{2} \right)^{t / T_{1/2}} $$

where $$ N $$ is the amount left, $$ N_0 $$ is the initial amount, $$ t $$ is the time elapsed, and $$ T_{1/2} $$ is the half-life period.

Here, $$ N = 3 $$ grams, $$ t = 12 $$ days, and $$ T_{1/2} = 3 $$ days. We substitute these values into the formula:

$$ 3 = N_0 \left( \frac{1}{2} \right)^{12 / 3} $$

First, simplify the exponent: $$ 12 / 3 = 4 $$, so the equation becomes:

$$ 3 = N_0 \left( \frac{1}{2} \right)^4 $$

Now, calculate $$ \left( \frac{1}{2} \right)^4 $$:

$$ \left( \frac{1}{2} \right)^4 = \frac{1^4}{2^4} = \frac{1}{16} $$

So the equation is:

$$ 3 = N_0 \times \frac{1}{16} $$

To solve for $$ N_0 $$, multiply both sides by 16:

$$ N_0 = 3 \times 16 $$

$$ N_0 = 48 $$

Therefore, the initial mass of the isotope was 48 grams.

Looking at the options:

A. 12 g

B. 36 g

C. 48 g

D. 24 g

Hence, the correct answer is Option C.

We start with the Arrhenius equation for the temperature dependence of a rate constant:

$$k = A\,e^{-\dfrac{E_a}{RT}}$$

where $$k$$ is the rate constant, $$A$$ is the pre-exponential factor, $$E_a$$ is the activation energy, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

For two different temperatures $$T_1$$ and $$T_2$$ with their corresponding rate constants $$k_1$$ and $$k_2$$, we write the Arrhenius equation twice and then divide:

$$\dfrac{k_2}{k_1}= \dfrac{A\,e^{-\dfrac{E_a}{RT_2}}}{A\,e^{-\dfrac{E_a}{RT_1}}}=e^{-\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)}$$

Taking natural logarithm on both sides gives

$$\ln\!\left(\dfrac{k_2}{k_1}\right)=\dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$$

We are told that the rate doubles when the temperature rises from $$300\,\text{K}$$ to $$310\,\text{K}$$, so

$$\dfrac{k_2}{k_1}=2,\qquad T_1=300\,\text{K},\qquad T_2=310\,\text{K}$$

Substituting these values, we have

$$\ln 2=\dfrac{E_a}{R}\left(\dfrac{1}{300}-\dfrac{1}{310}\right)$$

First we evaluate the temperature difference term:

$$\dfrac{1}{300}-\dfrac{1}{310}= \dfrac{310-300}{300\times310}= \dfrac{10}{93000}= \dfrac{1}{9300}\,\text{K}^{-1}$$

Now we use $$R = 8.314\;\text{J K}^{-1}\text{mol}^{-1}$$ and convert $$\ln 2$$ from the given common logarithm. The problem provides $$\log_{10}2 = 0.301$$, and we know the relation $$\ln x = 2.303\log_{10}x$$. Hence

$$\ln 2 = 2.303 \times 0.301 \approx 0.693$$

Putting everything into the equation:

$$0.693 = \dfrac{E_a}{8.314}\left(\dfrac{1}{9300}\right)$$

Rearranging for $$E_a$$ gives

$$E_a = 0.693 \times 8.314 \times 9300$$

First multiply $$0.693$$ by $$8.314$$:

$$0.693 \times 8.314 \approx 5.762$$

Then multiply by $$9300$$:

$$E_a = 5.762 \times 9300 \,\text{J mol}^{-1} \approx 5.3605 \times 10^{4}\,\text{J mol}^{-1}$$

Finally convert joules to kilojoules (1 kJ = 1000 J):

$$E_a \approx \dfrac{5.3605 \times 10^{4}\,\text{J}}{1000} = 53.6 \,\text{kJ mol}^{-1}$$

Hence, the correct answer is Option C.

The integrated rate law for a first-order reaction $$A \rightarrow$$ products is
$$k = \frac{1}{t}\,\ln\!\left(\frac{[A]_0}{[A]}\right)$$

Initial concentration $$[A]_0 = 0.1\,\text{M}$$, final concentration after $$t = 40\,\text{min}$$ is $$[A] = 0.025\,\text{M}$$. Substituting,

$$k = \frac{1}{40}\,\ln\!\left(\frac{0.1}{0.025}\right) = \frac{1}{40}\,\ln(4)$$

Since $$\ln(4) = 1.3863$$,

$$k = \frac{1.3863}{40} = 0.03466\,\text{min}^{-1}$$

For a first-order reaction, instantaneous rate at any moment is
$$\text{rate} = k\,[A]$$

At the required concentration $$[A] = 0.01\,\text{M}$$,

$$\text{rate} = 0.03466 \times 0.01 = 3.466 \times 10^{-4}\,\text{M min}^{-1}$$

Rounded to three significant figures, $$\text{rate} = 3.47 \times 10^{-4}\,\text{M min}^{-1}$$

Option B which is: $$3.47 \times 10^{-4}$$ M/min

For every irreversible, non‐elementary reaction of the form $$A \rightarrow \text{Products}$$, the half-life $$t_{1/2}$$ depends on the initial concentration $$a_0$$ according to the rate law.

General relation between $$t_{1/2}$$ and $$a_0$$
If the reaction is of overall order $$n$$ (rate $$r = k\,a_0^{\,n}$$), the integrated rate expression gives

• For $$n = 1$$ (first order) : $$t_{1/2} = \dfrac{0.693}{k}$$  — independent of $$a_0$$.
• For $$n \neq 1$$ (zero, second, third, …) :

$$t_{1/2} = \dfrac{2^{\,n-1}-1}{(n-1)\,k\,a_0^{\,n-1}} \qquad -(1)$$

Log-log form
Taking common (base-10) logarithms of $$-(1)$$:

$$\log t_{1/2} = \log\!\left[\dfrac{2^{\,n-1}-1}{(n-1)\,k}\right] \;-\;(n-1)\,\log a_0 \qquad -(2)$$

Equation $$-(2)$$ is of the straight-line form $$y = c + m\,x$$ with
  • ordinate $$y = \log t_{1/2}$$,
  • abscissa $$x = \log a_0$$,
  • slope $$m = -(n-1)$$.

Interpreting the given plot
The figure supplied in the question is a straight line whose slope is +1 (i.e. $$\log t_{1/2}$$ increases by one unit when $$\log a_0$$ increases by one unit).

Setting $$m = +1$$ in $$m = -(n-1)$$ gives
$$+1 = -(n-1) \;\Longrightarrow\; n - 1 = -1 \;\Longrightarrow\; n = 0$$.

Conclusion
The reaction is zero order.

Option B which is: zero

The given rate law is
$$\text{rate}_1 = k[A]^n[B]^m$$

Apply the stated changes in concentrations:
  • $$[A]$$ is doubled  →  $$[A]' = 2[A]$$
  • $$[B]$$ is halved  →  $$[B]' = \tfrac12[B]$$

The new rate becomes
$$\text{rate}_2 = k([A]')^n([B]')^m = k\,(2[A])^n\!\left(\tfrac12[B]\right)^{\!m}$$

Simplify the powers of 2:
$$\text{rate}_2 = k\,2^{\,n}[A]^n \cdot 2^{-\,m}[B]^m = 2^{\,n-m}\,k[A]^n[B]^m = 2^{\,n-m}\,\text{rate}_1$$

Hence the ratio of the new rate to the original rate is
$$\frac{\text{rate}_2}{\text{rate}_1} = 2^{\,n-m}$$

Therefore, the correct choice is:
Option D which is: $$2^{(n-m)}$$

The Freundlich adsorption isotherm is an empirical relation that connects the extent of adsorption with the equilibrium pressure (or concentration) of the adsorbate. It is written as

$$\frac{x}{m}=k\,p^{1/n} \qquad (k \gt 0,\; n \gt 1) \;$$ $$-(1)$$

Here

• $$x$$ = mass of gas adsorbed (at equilibrium)
• $$m$$ = mass of adsorbent
• $$p$$ = equilibrium pressure of the gas
• $$k$$ and $$n$$ are adsorption constants that depend on the nature of the adsorbent-adsorbate pair and temperature.

To transform $$-(1)$$ into the equation of a straight line, take logarithms (base 10 or natural - either gives linearity). Using common logarithms:

$$\log\!\left(\frac{x}{m}\right)=\log k + \frac{1}{n}\,\log p \qquad -(2)$$

Equation $$-(2)$$ is of the form $$y = c + m\,x$$, where

• dependent variable, $$y = \log(x/m)$$,
• independent variable, $$x = \log p$$,
• slope, $$m = \dfrac{1}{n}$$,
• intercept, $$c = \log k$$.

Hence, a plot of $$\log(x/m)$$ on the ordinate versus $$\log p$$ on the abscissa yields a straight line with slope $$1/n$$ and intercept $$\log k$$.

Therefore, the correct choice is:
Option C which is: $$\log x/m$$ vs $$\log p$$

The empirical rule for many reactions states: for every $$10^\circ \text{C}$$ rise in temperature, the rate constant (and hence the rate) becomes twice its previous value.

Given a total temperature rise of $$50^\circ \text{C}$$, we can divide this rise into five successive $$10^\circ \text{C}$$ steps:

Number of steps $$= \frac{50^\circ \text{C}}{10^\circ \text{C}} = 5$$.

If the rate after one step is multiplied by $$2$$, then after $$n$$ identical steps the overall multiplication factor is $$2^n$$.

Thus, for $$n = 5$$ steps:

Overall rate increase factor $$= 2^5 = 32$$.

Hence the rate of the reaction becomes $$32$$ times its original value when the temperature is raised by $$50^\circ \text{C}$$.

Option B which is: $$32$$ times

For a first-order reaction the rate constant $$k$$ is related to the half-life $$t_{1/2}$$ by

$$k = \frac{0.693}{t_{1/2}} \qquad -(1)$$

Given $$t_{1/2} = 6.93\text{ min}$$, substitute in (1):
$$k = \frac{0.693}{6.93} = 0.100\;\text{min}^{-1}$$

Let the initial concentration be $$[A]_0$$ and the concentration after time $$t$$ be $$[A]$$. For a first-order reaction

$$\ln\!\left(\frac{[A]_0}{[A]}\right)=kt \qquad -(2)$$

Completion of $$99\%$$ of the reaction means $$1\%$$ is left, i.e. $$[A] = 0.01[A]_0$$.
Hence $$\dfrac{[A]_0}{[A]} = \dfrac{[A]_0}{0.01[A]_0}=100$$.

Insert this value in (2):
$$\ln(100)=kt$$

Using $$\ln(100)=2\ln(10)=2\times2.303=4.606$$, we get

$$t = \frac{4.606}{k} = \frac{4.606}{0.100}=46.06\text{ min}$$

Therefore, the time required for $$99\%$$ completion is

Option C which is: $$46.06$$ minutes

For a general reaction $$aA \rightarrow bB$$ the rate is defined as
$$\text{rate}= -\frac{1}{a}\frac{d[A]}{dt}=+\frac{1}{b}\frac{d[B]}{dt} \quad -(1)$$

Comparing the given reaction $$\frac{1}{2}A \rightarrow 2B$$ with $$aA \rightarrow bB$$, we have
$$a=\frac{1}{2}, \qquad b=2$$

Substituting these values into equation $$-(1)$$ gives
$$-\frac{1}{\frac{1}{2}}\frac{d[A]}{dt}=+\frac{1}{2}\frac{d[B]}{dt}$$

Simplify the left coefficient: $$\frac{1}{\frac{1}{2}}=2$$, therefore
$$-2\,\frac{d[A]}{dt}= \frac{1}{2}\frac{d[B]}{dt}$$

Rearranging for $$-\dfrac{d[A]}{dt}$$:
$$-\frac{d[A]}{dt} = \frac{1}{4}\frac{d[B]}{dt}$$

Thus the disappearance rate of $$A$$ is one-fourth the appearance rate of $$B$$.

Option B which is: $$-\frac{d[A]}{dt} = \frac{1}{4}\frac{d[B]}{dt}$$

For any reaction, the energy profile looks like:

Reactants $$\xrightarrow{\;E_{a(fwd)}\;}$$ Transition state $$\xrightarrow{\;E_{a(rev)}\;}$$ Products

Hence, the difference between the two activation energies equals the enthalpy change of reaction:

$$\Delta H = E_{a(fwd)} - E_{a(rev)} \; -(1)$$

Given (without catalyst):

$$E_{a(fwd)} = 180\;{\rm kJ\,mol^{-1}}, \qquad E_{a(rev)} = 200\;{\rm kJ\,mol^{-1}}$$

Substituting in (1):

$$\Delta H = 180 - 200 = -20\;{\rm kJ\,mol^{-1}}$$

The negative sign means the forward reaction $$A_2 + B_2 \rightarrow 2AB$$ is exothermic by 20 kJ mol−1.

In the presence of a catalyst, the activation energy of each direction is reduced by 100 kJ mol−1:

$$E_{a(fwd,\,cat)} = 180 - 100 = 80\;{\rm kJ\,mol^{-1}}$$

$$E_{a(rev,\,cat)} = 200 - 100 = 100\;{\rm kJ\,mol^{-1}}$$

Using equation (1) again:

$$\Delta H_{\rm(cat)} = 80 - 100 = -20\;{\rm kJ\,mol^{-1}}$$

Thus the catalyst lowers both activation energies equally, but the difference—and therefore the enthalpy change—remains the same. Only its magnitude is usually reported in such problems.

Therefore, the enthalpy change of the reaction in the presence of the catalyst is $$20\;{\rm kJ\,mol^{-1}}$$.

Option D which is: 20

Step 1: Determine the Reaction Order with respect to $$B$$

• The problem states: "When concentration of $$B$$ alone was doubled, the half-life did not change."
• For a reaction where the half-life ($$t_{1/2}$$) is independent of the initial concentration of that reactant, the reaction is first-order with respect to that specific component.
• Therefore, the order with respect to $B$ ($$y$$) = $$1$$.

Step 2: Determine the Reaction Order with respect to $$A$$

• The problem states: "When the concentration of $A$ alone was doubled, the rate increased by two times."
• Since doubling the concentration ($$2^1$$) causes a direct doubling of the rate ($$2^1$$), the rate is directly proportional to the concentration of $$A$$.
• Therefore, the order with respect to $$A$$ ($$x$$) = $$1$$.

Step 3: Calculate the Overall Order of the Reaction

The rate law can be written as:

$$\text{Rate} = k[A]^1[B]^1$$

$$\text{Overall Order } (n) = x + y = 1 + 1 = 2 \text{ (Second-order reaction)}$$

Step 4: Find the Unit of the Rate Constant ($$k$$)

The general formula for the units of a rate constant is:

$$\text{Unit of } k = (\text{mol L}^{-1})^{1-n} \text{ s}^{-1}$$

Substituting $$n = 2$$ into the formula:

$$\text{Unit of } k = (\text{mol L}^{-1})^{1-2} \text{ s}^{-1} = (\text{mol L}^{-1})^{-1} \text{ s}^{-1} = \text{L mol}^{-1} \text{ s}^{-1}$$

Conclusion:

The reaction is overall second-order, meaning the rate constant is expressed in liters per mole-second.

Answer: Option A — $$\text{L mol}^{-1}\text{s}^{-1}$$

Radioactive decay follows first-order kinetics. The activity $$A$$ at any time $$t$$ is related to the initial activity $$A_0$$ by

$$A = A_0 \left(\frac{1}{2}\right)^{\,\frac{t}{t_{1/2}}}$$

where $$t_{1/2}$$ is the half-life.

Given data:
  • Half-life $$t_{1/2}=30$$ days
  • Initial activity $$A_0 = 10A_{\text{perm}}$$ (ten times the permissible activity)
  • Required final activity $$A = A_{\text{perm}}$$

Set up the decay equation:

$$A_{\text{perm}} = 10A_{\text{perm}}\left(\frac{1}{2}\right)^{\,\frac{t}{30}}$$

Divide both sides by $$A_{\text{perm}}$$ to cancel it:

$$1 = 10\left(\frac{1}{2}\right)^{\,\frac{t}{30}}$$

Rearrange:

$$\left(\frac{1}{2}\right)^{\,\frac{t}{30}} = \frac{1}{10}$$

Take logarithms (base 10 or natural, any base works) on both sides:

$$\frac{t}{30}\,\log\!\left(\frac{1}{2}\right) = \log\!\left(\frac{1}{10}\right)$$

Solve for $$t$$:

$$t = 30 \times \frac{\log\!\left(\frac{1}{10}\right)}{\log\!\left(\frac{1}{2}\right)}$$

Because $$\log\!\left(\frac{1}{10}\right) = -1$$ (in base-10) and $$\log\!\left(\frac{1}{2}\right) = -0.3010$$,

$$t = 30 \times \frac{-1}{-0.3010} \approx 30 \times 3.322 \approx 99.7\ \text{days}$$

Rounding to the nearest whole day, the safe time is approximately $$100$$ days.

Option D which is: 100 days

For any rate-law expression of the form $$r = k\,[\text{reactant}]^{n}$$, the exponent $$n$$ is called the order of the reaction with respect to that reactant.

The problem states that the reaction is second order in carbon monoxide, so $$n = 2$$. Hence the rate law involving the concentration of carbon monoxide is

$$r = k \,[CO]^{2}\quad\ldots(1)$$

Let the initial concentration of CO be $$[CO]_1$$ and the corresponding initial rate be $$r_1$$. From (1),

$$r_1 = k \,[CO]_1^{2} \quad\ldots(2)$$

Now the concentration of CO is doubled, i.e. $$[CO]_2 = 2[CO]_1$$. The new rate $$r_2$$ becomes

$$r_2 = k \,(2[CO]_1)^{2} = k \,4[CO]_1^{2} = 4\bigl(k\,[CO]_1^{2}\bigr) \quad\ldots(3)$$

Comparing (3) with (2),

$$r_2 = 4\,r_1$$.

Thus, doubling the concentration of carbon monoxide increases the rate of reaction by a factor of 4.

Option C which is: increase by a factor of 4

The Arrhenius equation is written as $$k = A e^{-E/RT}$$ where:

• $$k$$ is the temperature-dependent rate constant.
• $$A$$ is the pre-exponential (frequency) factor that contains steric and collision-frequency terms.
• $$E$$ is the activation energy of the reaction (in $$\text{J mol}^{-1}$$).
• $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

By definition, the activation energy $$E$$ is the minimum energy that reacting molecules must possess for a successful (effective) collision leading to product formation. Hence:

• Molecules having energy $$\lt E$$ do not possess sufficient energy to cross the energy barrier, so they do not react.
• Molecules having energy $$\ge E$$ are able to overcome the barrier and can react if correctly oriented.

Therefore, $$E$$ represents the energy below which the colliding molecules will not react.

Checking each option:

Option A: “energy above which all the colliding molecules will react” — incorrect, because even with energy $$\gt E$$ molecules still need proper orientation; “all” is too strong.

Option B: “energy below which colliding molecules will not react” — correct; this matches the definition of activation energy.

Option C: “total energy of the reacting molecules at temperature $$T$$” — incorrect; total molecular energy is unrelated to $$E$$.

Option D: “fraction of molecules with energy greater than the activation energy” — incorrect; that fraction is given by $$e^{-E/RT}$$, not by $$E$$ itself.

Hence, the correct choice is Option B which is: the energy below which colliding molecules will not react.

The mechanism contains two elementary steps:

$$NO + Br_2 \;\rightleftharpoons\; NOBr_2 \quad\text{(fast equilibrium)}$$
$$NOBr_2 + NO \;\longrightarrow\; 2\,NOBr \quad\text{(slow - rate-determining)}$$

For an elementary step, the rate law is written directly from its molecularity. Because the second step is slow, the overall rate is controlled by this step:

$$\text{rate} = k_2\,[NOBr_2]\,[NO]$$

The concentration $$[NOBr_2]$$ is not an experimental variable, so we eliminate it using the fast equilibrium of step 1. For the reversible step 1, write the equilibrium constant:

$$K = \frac{[NOBr_2]}{[NO]\,[Br_2]} \;\;\Longrightarrow\;\; [NOBr_2] = K\,[NO]\,[Br_2]$$

Substitute this expression into the rate law obtained from the slow step:

$$\text{rate} = k_2\,[NO]\,(K\,[NO]\,[Br_2]) = (k_2K)\,[NO]^2\,[Br_2]$$

The exponent of $$[NO]$$ in the final rate law is $$2$$, so the reaction is second-order with respect to nitric oxide.

Option D which is: 2

For any reaction that follows the Arrhenius equation
$$\text{}$$
$$k = A\,e^{-\dfrac{E_a}{RT}}$$
$$\text{}$$
taking natural logarithms for two different temperatures $$T_1$$ and $$T_2$$ gives

$$\ln\dfrac{k_2}{k_1}= -\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$$ $$-(1)$$

Given data
$$\text{}$$
$$k_1 = 1.5\times10^{3}\,\text{s}^{-1},\; T_1 = 27^{\circ}\text{C}=300\,\text{K}$$
$$k_2 = 4.5\times10^{3}\,\text{s}^{-1},\; E_a = 60\,\text{kJ mol}^{-1}= 60000\,\text{J mol}^{-1}$$
$$R = 8.3\,\text{J K}^{-1}\text{ mol}^{-1}$$

First evaluate $$\ln\dfrac{k_2}{k_1}$$:
$$\text{}$$
$$\dfrac{k_2}{k_1}=\dfrac{4.5}{1.5}=3$$
$$\text{}$$
$$\ln 3 = 2.303\log 3 = 2.303 \times 0.48 \approx 1.105$$ $$-(2)$$

Substituting $$E_a$$, $$R$$ and equation $$(2)$$ into equation $$(1)$$:

$$1/T_2 - 1/T_1 = -\dfrac{R}{E_a}\,\ln\dfrac{k_2}{k_1}$$
$$\text{}$$
$$\text{}$$
$$1/T_2 = \dfrac{1}{300}\;-\;\dfrac{8.3}{60000}\times 1.105$$

Compute each term
$$\text{}$$
$$\dfrac{1}{300}=3.333\times10^{-3}\,\text{K}^{-1}$$
$$\text{}$$
$$\dfrac{8.3}{60000}=1.383\times10^{-4}\,\text{K}^{-1}$$
$$\text{}$$
Product with $$1.105$$:
$$\text{}$$
$$1.383\times10^{-4}\times1.105 \approx 1.522\times10^{-4}\,\text{K}^{-1}$$

Therefore
$$1/T_2 = 3.333\times10^{-3} - 1.522\times10^{-4}$$
$$\text{}$$
$$1/T_2 = 3.181\times10^{-3}\,\text{K}^{-1}$$

Invert to find $$T_2$$:
$$\text{}$$
$$T_2 = \dfrac{1}{3.181\times10^{-3}} \approx 3.14\times10^{2}\,\text{K} = 314\,\text{K}$$

Convert to Celsius:
$$\text{}$$
$$T_2(^{\circ}\text{C}) = 314 - 273 = 41^{\circ}\text{C}$$

Hence, the required temperature is 41 °C (nearest integer).

For a first-order reaction, the half-life remains constant.

The number of half-lives elapsed in $$6\ \text{hours}$$ is

$$n=\frac{t}{t_{1/2}}=\frac{6}{3}=2$$

After the first half-life:

$$100%\rightarrow 50%$$

After the second half-life:

$$50%\rightarrow 25%$$

Alternatively,

$$\text{Amount remaining}=\frac{\text{Initial amount}}{2^n}=\frac{100}{2^2}=25%$$

Therefore, the percentage of sucrose remaining after $$6\ \text{hours}$$ is

$$25%$$

Hence, the correct answer is $$25$$.

We consider two first-order reactions, $$A \rightarrow B$$ and $$C \rightarrow D$$. The rate constant for the first reaction at 500 K is double its rate constant at 300 K, so $$k_{1,500} = 2 k_{1,300}$$. Since 50% of the first reaction completes in 2 hours at 500 K and the half-life for a first-order reaction is given by $$t_{1/2} = \frac{\ln 2}{k}$$, it follows that $$2 = \frac{\ln 2}{k_{1,500}}$$ and hence $$k_{1,500} = \frac{\ln 2}{2}\ \text{hour}^{-1}$$. The activation energy of the second reaction is half that of the first reaction, so $$E_{a2} = \frac{1}{2} E_{a1}$$, and at 500 K the rate constant of the second reaction is twice that of the first reaction, giving $$k_{2,500} = 2k_{1,500}\,. $$

We need to find the rate constant of the second reaction at 300 K, denoted $$k_{2,300}$$, and express it as an integer multiplied by $$10^{-1}\ \text{hour}^{-1}$$. We begin by using the Arrhenius equation, $$k = A e^{-E_a / (R T)}\,, $$ where $$k$$ is the rate constant, $$A$$ is the pre-exponential factor, $$E_a$$ is the activation energy, $$R$$ is the gas constant, and $$T$$ is the absolute temperature.

Since for the first reaction at 300 K and 500 K we have $$k_{1,300} = A_1 e^{-E_{a1}/(R\cdot 300)}$$ and $$k_{1,500} = A_1 e^{-E_{a1}/(R\cdot 500)}$$ and it is given that $$k_{1,500} = 2 k_{1,300}$$, we set $$A_1 e^{-E_{a1}/(500R)} = 2 A_1 e^{-E_{a1}/(300R)}$$. Dividing by $$A_1$$ and taking natural logarithms gives $$-\frac{E_{a1}}{500R} = \ln 2 \;-\;\frac{E_{a1}}{300R}\,. $$ Rearranging yields $$\frac{E_{a1}}{R}\Bigl(-\tfrac{1}{500}+\tfrac{1}{300}\Bigr)=\ln 2,\quad -\tfrac{1}{500}+\tfrac{1}{300}=\tfrac{1}{750},$$ so $$\frac{E_{a1}}{R}=750\ln 2\,. $$

Substituting the half-life condition at 500 K, $$k_{1,500}=\frac{\ln 2}{2}$$, into the Arrhenius form $$k_{1,500}=A_1 e^{-E_{a1}/(500R)}$$ and using $$\frac{E_{a1}}{R}=750\ln 2$$ gives $$\frac{\ln 2}{2}=A_1 e^{-(750\ln 2)/(500)}=A_1 e^{-(3/2)\ln 2}=A_1\cdot 2^{-3/2}=\frac{A_1}{2\sqrt2}\,. $$ Solving for $$A_1$$ yields $$A_1=\ln 2\cdot\sqrt2\,. $$

Since $$k_{2,500}=2k_{1,500}$$ and $$k_{1,500}=\frac{\ln 2}{2}$$, it follows that $$k_{2,500}=2\cdot\frac{\ln 2}{2}=\ln 2\ \text{hour}^{-1}\,. $$

For the second reaction the Arrhenius equation at 300 K and 500 K gives $$k_{2,300}=A_2 e^{-E_{a2}/(300R)},\quad k_{2,500}=A_2 e^{-E_{a2}/(500R)},$$ so $$\frac{k_{2,500}}{k_{2,300}}=e^{E_{a2}/R\bigl(-\frac{1}{500}+\frac{1}{300}\bigr)}=e^{(E_{a2}/R)(1/750)}\,. $$ Since $$E_{a2}=\tfrac12E_{a1}$$ and $$E_{a1}/R=750\ln 2$$, we have $$E_{a2}/R=375\ln 2$$ and therefore $$k_{2,300}=k_{2,500}\,e^{-(375\ln 2)/(750)}=\ln 2\;e^{-(1/2)\ln 2}=\ln 2\cdot2^{-1/2}=\frac{\ln 2}{\sqrt2}\,. $$

Using $$\ln 2\approx0.693147$$ and $$\sqrt2\approx1.414214$$ gives $$k_{2,300}=\frac{0.693147}{1.414214}\approx0.4900\ \text{hour}^{-1}=4.900\times10^{-1}\ \text{hour}^{-1},$$ and the nearest integer to 4.900 is 5.

Therefore, the rate constant for the second reaction at 300 K is $$5\times10^{-1}\ \text{hour}^{-1}\,.$$