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.

In first-order kinetics, the concentration at time $$t$$ is given by $$ C = C_0 e^{-kt} $$.

Here the initial concentration is $$C_0 = 16$$ mg/mL and after $$t = 12$$ months the concentration is $$C = 4$$ mg/mL.

Substituting these values gives $$4 = 16 e^{-12k}$$, which implies $$e^{-12k} = \frac{1}{4}$$ and so $$k = \frac{\ln 4}{12} = \frac{2\ln 2}{12} = \frac{\ln 2}{6}$$.

The drug becomes ineffective after 50% decomposition, that is when $$C = \frac{C_0}{2} = 8$$ mg/mL.

The half-life for first-order kinetics is $$t_{1/2} = \frac{\ln 2}{k} = \frac{\ln 2}{\ln 2/6} = 6 \text{ months}$$, so the expiry time is 6 months.

The correct answer is Option 2: 6.

For the gas-phase decomposition $$A(g) \rightarrow B(g)+C(g)$$ at constant temperature and volume, the total pressure is directly proportional to the total number of moles present at any instant.

Let the reaction start with $$a$$ moles of $$A$$ and no $$B$$ or $$C$$.
At time $$t$$, let $$x$$ moles of $$A$$ decompose.

Moles present at time $$t$$:
$$A : a-x$$
$$B : x$$
$$C : x$$

Total moles at time $$t$$ are $$n_t = a-x + x + x = a + x$$.

If $$f = \dfrac{RT}{V}$$ (a common proportionality factor), then

initial pressure, $$P_0 = f a$$,
pressure at time $$t$$, $$P_t = f (a + x) = P_0 + f x$$ $$-(1)$$

From $$(1)$$, $$x = \dfrac{P_t - P_0}{f}$$ $$-(2)$$

When the reaction goes to completion ($$t \rightarrow \infty$$), $$x = a$$.
Total moles then are $$a + a = 2a$$, so the final pressure is

$$P_{\infty} = f(2a) = 2 P_0$$ $$\Longrightarrow P_0 = \dfrac{P_{\infty}}{2}$$ $$-(3)$$

The first-order rate law is $$k = \dfrac{1}{t} \ln \dfrac{\text{initial concentration of }A}{\text{concentration of }A\text{ at time }t}$$, i.e.

$$k = \dfrac{1}{t} \ln \dfrac{a}{a - x}$$ $$-(4)$$

Using $$(2)$$, $$a - x = \dfrac{P_0}{f} - \dfrac{P_t - P_0}{f} = \dfrac{2P_0 - P_t}{f}$$.

Thus,
$$\dfrac{a}{a - x} = \dfrac{\dfrac{P_0}{f}}{\dfrac{2P_0 - P_t}{f}} = \dfrac{P_0}{2P_0 - P_t}$$ $$-(5)$$

Substitute $$P_0 = \dfrac{P_{\infty}}{2}$$ from $$(3)$$ into $$(5)$$:

$$\dfrac{a}{a - x} = \dfrac{\dfrac{P_{\infty}}{2}}{P_{\infty} - P_t} = \dfrac{P_{\infty}}{2(P_{\infty} - P_t)}$$ $$-(6)$$

Insert $$(6)$$ into the rate law $$(4)$$:

$$k = \dfrac{1}{t} \ln \left[ \dfrac{P_{\infty}}{2(P_{\infty} - P_t)} \right]$$

This matches Option C.

Hence, the correct expression for the first-order rate constant is
Case C: $$\;k = \dfrac{1}{t} \ln \dfrac{p_{\infty}}{2(p_{\infty} - P_t)}$$.

For a first-order reaction the half-life is related to the rate constant by
$$t_{1/2} = \frac{0.693}{k}$$

Rate constants
For reactant $$A$$:
$$k_A = \frac{0.693}{t_{1/2\,(A)}} = \frac{0.693}{10\,\text{min}} = 0.0693\,\text{min}^{-1}$$

For reactant $$B$$:
$$k_B = \frac{0.693}{t_{1/2\,(B)}} = \frac{0.693}{40\,\text{min}} = 0.017325\,\text{min}^{-1}$$

First-order concentration expressions
$$[A] = [A]_0\,e^{-k_A t}$$
$$[B] = [B]_0\,e^{-k_B t}$$

The initial concentrations are related by
$$[A]_0 = 8[B]_0$$ $$-(1)$$

We need the time $$t$$ at which the instantaneous concentrations become equal:
$$[A] = [B]$$ $$-(2)$$

Substitute $$(1)$$ and the exponential expressions into $$(2)$$:
$$[A]_0\,e^{-k_A t} = [B]_0\,e^{-k_B t}$$
$$8[B]_0\,e^{-k_A t} = [B]_0\,e^{-k_B t}$$

Cancel $$[B]_0$$ to obtain
$$8\,e^{-k_A t} = e^{-k_B t}$$

Take natural logarithm on both sides:
$$\ln 8 - k_A t = -k_B t$$

Rearrange for $$t$$:
$$\ln 8 = (k_A - k_B)\,t$$

Insert numerical values:
$$\ln 8 = \ln 2^3 = 3\ln 2 = 3(0.693) = 2.079$$
$$k_A - k_B = 0.0693 - 0.017325 = 0.051975\,\text{min}^{-1}$$

Therefore
$$t = \frac{2.079}{0.051975} \approx 40\,\text{min}$$

Hence, the concentrations of $$A$$ and $$B$$ become equal after $$40$$ minutes.
Option D.

The energy profile of an elementary reaction relates the activation energies of the forward ($$E_a^{\,f}$$) and backward ($$E_a^{\,b}$$) steps to the enthalpy change $$\Delta H$$ of the reaction through the relation
$$\Delta H = E_a^{\,f} - E_a^{\,b}\; -(1)$$

For the uncatalysed reaction we have
$$E_a^{\,f} = 180\ \text{kJ mol}^{-1},\qquad E_a^{\,b} = 200\ \text{kJ mol}^{-1}$$

Substituting these values in $$(1)$$:
$$\Delta H = 180 - 200 = -20\ \text{kJ mol}^{-1}$$
(The negative sign shows the reaction is exothermic.)

The catalyst lowers the activation energy of both directions by the same amount, 100 kJ mol$$^{-1}$$:
$$E_{a,\ \text{cat}}^{\,f} = 180 - 100 = 80\ \text{kJ mol}^{-1}$$
$$E_{a,\ \text{cat}}^{\,b} = 200 - 100 = 100\ \text{kJ mol}^{-1}$$

Using $$(1)$$ again after catalysis:
$$\Delta H_{\text{cat}} = 80 - 100 = -20\ \text{kJ mol}^{-1}$$
Thus the enthalpy change remains exactly the same.

Now evaluate each statement:

Option A: A catalyst does not change the thermodynamic parameters such as Gibbs free energy change $$\Delta G$$. This is correct.

Option B: A catalyst cannot alter $$\Delta G$$, so it cannot convert a non-spontaneous ($$\Delta G \gt 0$$) reaction into a spontaneous one. This statement is incorrect.

Option C: We found $$\Delta H = -20\ \text{kJ mol}^{-1}$$, not $$+20\ \text{kJ mol}^{-1}$$. This statement is incorrect.

Option D: The enthalpy change is the same before and after catalysis. Therefore this statement is incorrect.

Hence, only Option A is correct.

$$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$$. Find final pressure when 50% reacts.

Initial: P₀ (N₂O₅ only)

At 50% decomposition: N₂O₅ = P₀/2, NO₂ = 2(P₀/2) = P₀, O₂ = ½(P₀/2) = P₀/4

Total = P₀/2 + P₀ + P₀/4 = 7P₀/4

The correct answer is Option 4: $$\frac{7}{4}$$ times initial pressure.

The given rate law is $$r = k[A]^n[B]^m$$, where $$r$$ is the rate, $$k$$ is the rate constant, and $$n,m$$ are the orders with respect to A and B respectively.

Initial rate $$r_1$$ is therefore
$$r_1 = k[A]^n[B]^m \quad -(1)$$

For the new experiment, the concentration changes are:
A is doubled  →  $$[A]_{\text{new}} = 2[A]$$
B is halved  →  $$[B]_{\text{new}} = \frac{1}{2}[B]$$

Substituting these new concentrations into the rate law gives the new rate $$r_2$$:
$$\begin{aligned} r_2 &= k\,(2[A])^n\!\left(\frac{1}{2}[B]\right)^m \\ &= k\,2^n[A]^n \; 2^{-m}[B]^m \\ &= k\,2^{\,n-m}[A]^n[B]^m \quad -(2) \end{aligned}$$

Divide equation $$(2)$$ by equation $$(1)$$ to obtain the required ratio:
$$\frac{r_2}{r_1} = \frac{k\,2^{\,n-m}[A]^n[B]^m}{k[A]^n[B]^m} = 2^{\,n-m}$$

Thus, $$\dfrac{r_2}{r_1} = 2^{\,n-m}$$.

The correct choice is Option A.

Volume reduced to 1/3: concentrations triple. Rate = k[A][B] → new rate = k(3[A])(3[B]) = 9k[A][B] = 9× original rate. x = 9.

The correct answer is Option 3: 9.

We need to identify which statement is NOT true for radioactive decay.

Option 1: Decay constant increases with increase in temperature.

Radioactive decay is a nuclear process and is independent of temperature. The decay constant does NOT change with temperature. This statement is NOT TRUE.

Option 2: Amount of radioactive substance remained after three half lives is 1/8th of original amount.

After n half-lives, the remaining amount is $$(1/2)^n$$ of the original. After 3 half-lives: $$(1/2)^3 = 1/8$$. This is TRUE.

Option 3: Decay constant does not depend upon temperature.

This is TRUE as radioactive decay is a nuclear process independent of external conditions.

Option 4: Half life is $$\ln 2$$ times of mean life.

$$t_{1/2} = \frac{\ln 2}{\lambda} = \ln 2 \times \tau$$ where $$\tau = 1/\lambda$$ is the mean life. This is TRUE.

The correct answer is Option 1: Decay constant increases with increase in temperature (this is false).

The reaction $$A_2 + B_2 \rightarrow 2AB$$ follows the given mechanism and we need to find the overall order.

We note that the slow step is $$A + B_2 \xrightarrow{k_2} AB + B$$, and the rate law is Rate = $$k_2[A][B_2]$$.

Next, the first step is a fast equilibrium: $$A_2 \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} A + A$$. At equilibrium, $$K = \frac{k_1}{k_{-1}} = \frac{[A]^2}{[A_2]}$$, so $$[A] = \sqrt{K[A_2]} = \sqrt{\frac{k_1}{k_{-1}}} \cdot [A_2]^{1/2}$$.

Substituting this expression into the rate law gives Rate = $$k_2 [A][B_2] = k_2 \sqrt{\frac{k_1}{k_{-1}}} [A_2]^{1/2} [B_2]$$ and $$= k'[A_2]^{1/2}[B_2]^1$$.

Therefore the overall order is $$\frac{1}{2} + 1 = 1.5$$.

The correct answer is Option D) 1.5.

The empirical Arrhenius equation for the temperature dependence of a rate constant is
$$k = A \; e^{-\;E_a/RT}$$
where $$k$$ is the rate constant, $$A$$ is the pre-exponential (frequency) factor, $$E_a$$ is the activation energy, $$R$$ is the gas constant and $$T$$ is the absolute temperature.

Statement A: “The Arrhenius equation holds true only for an elementary homogeneous reaction.”
The equation is observed to describe a very wide range of reactions—elementary as well as complex, homogeneous as well as heterogeneous. Hence Statement A is incorrect.

Statement B: “The unit of $$A$$ is same as that of $$k$$ in the Arrhenius equation.”
In $$k = A\,e^{-E_a/RT}$$ the exponential factor is dimensionless. Therefore $$A$$ must possess exactly the same units as $$k$$ so that the product on the right retains the units of a rate constant. Statement B is correct.

Statement C: “At a given temperature, a low activation energy means a fast reaction.”
For fixed $$T$$, the exponential term is $$e^{-E_a/RT}$$. A smaller $$E_a$$ gives a larger value of the exponential term, hence a larger $$k$$ and therefore a faster reaction. Statement C is correct.

Statement D: “$$A$$ and $$E_a$$ as used in the Arrhenius equation depend on temperature.”
In the usual Arrhenius treatment, both $$A$$ and $$E_a$$ are taken to be temperature-independent over a moderate temperature range. They may vary slightly in reality, but the statement that they depend on temperature is not accepted in the standard model. Statement D is incorrect.

Statement E: “When $$E_a \gg RT$$, $$A$$ and $$E_a$$ become interdependent.”
Even if $$E_a$$ is much larger than $$RT$$, the Arrhenius parameters remain independent constants obtained from the intercept and slope of an Arrhenius plot. Statement E is incorrect.

Therefore, only Statements B and C are correct.

Correct answer: Option C (B and C Only).

For a zero-order reaction $$A \rightarrow \text{products}$$, the rate is independent of concentration:

Rate law : $$\frac{d[A]}{dt} = -k$$

On integrating from $$t = 0$$ (concentration $$[A]_0$$) to any time $$t$$ (concentration $$[A]_t$$) we obtain the integrated equation

$$[A]_t = [A]_0 - k\,t \quad -(1)$$

Half-life $$t_{1/2}$$ is the time at which $$[A]_t = \frac{[A]_0}{2}$$. Substituting into equation $$(1)$$ gives the zero-order half-life formula:

$$t_{1/2} = \frac{[A]_0}{2k} \quad -(2)$$

The question states that for $$[A]_0 = 2.0\;\text{mol L}^{-1}$$, $$t_{1/2} = 1\;\text{h}$$. Using $$-(2)$$ to find the rate constant $$k$$:

$$k = \frac{[A]_0}{2\,t_{1/2}} = \frac{2.0}{2 \times 1\;\text{h}} = 1.0\;\text{mol L}^{-1}\text{h}^{-1}$$

Now we need the time required for the concentration to drop from $$0.50$$ to $$0.25\;\text{mol L}^{-1}$$. Let this time be $$t$$ and take $$[A]_0 = 0.50\;\text{mol L}^{-1}$$ in equation $$(1)$$:

$$[A]_t = 0.50 - k\,t$$

Set $$[A]_t = 0.25\;\text{mol L}^{-1}$$ and $$k = 1.0\;\text{mol L}^{-1}\text{h}^{-1}$$:

$$0.25 = 0.50 - 1.0 \times t$$

$$t = 0.50 - 0.25 = 0.25\;\text{h}$$

Convert $$0.25$$ hour to minutes:

$$0.25\;\text{h} \times 60\;\frac{\text{min}}{\text{h}} = 15\;\text{min}$$

Hence, the time required is 15 minutes.

The correct choice is Option C (15 min).

We need to find the time for 50% conversion using first order kinetics.

At T = 1000 K, the activation energy is 191.48 kJ/mol = 191480 J/mol, the frequency factor is A = 10^{20}, and R = 8.314 J K⁻¹ mol⁻¹.

We start by calculating the rate constant using the Arrhenius equation:

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

Next, we compute the exponent:

$$\frac{E_a}{RT} = \frac{191480}{8.314 \times 1000} = \frac{191480}{8314} = 23.03$$

This gives:

$$k = 10^{20} \times e^{-23.03}$$

Since ln(10) = 2.303, we have:

$$e^{-23.03} = e^{-10 \times 2.303} = 10^{-10}$$

Substituting back yields:

$$k = 10^{20} \times 10^{-10} = 10^{10} \text{ s}^{-1}$$

For 50% conversion, the half-life in first order kinetics is given by:

$$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \text{ s}$$

Converting this to picoseconds using 1 ps = 10^{-12} s:

$$t_{1/2} = \frac{6.93 \times 10^{-11}}{10^{-12}} = 69.3 \text{ ps}$$

Rounding to the nearest integer gives a half-life of approximately 69 picoseconds.

We need to find the overall activation energy given that the overall rate constant $$k = \sqrt{\frac{k_1 k_3}{k_2}}$$ and the activation energies of the three steps are 60, 30, and 10 kJ/mol respectively.

The Arrhenius equation states that each rate constant depends on its activation energy according to $$k_i = A_i \, e^{-E_{a_i}/(RT)}$$, and the overall rate constant also takes the form $$k = A \, e^{-E_a/(RT)}$$, where $$E_a$$ is the overall activation energy.

Since $$k = \sqrt{\frac{k_1 k_3}{k_2}} = \left(\frac{k_1 k_3}{k_2}\right)^{1/2}$$, taking the natural logarithm of both sides gives $$\ln k = \frac{1}{2}\bigl(\ln k_1 + \ln k_3 - \ln k_2\bigr).$$

Substituting $$\ln k_i = \ln A_i - \frac{E_{a_i}}{RT}$$ into this expression yields

$$\ln k = \frac{1}{2}\Bigl[\bigl(\ln A_1 - \frac{E_{a_1}}{RT}\bigr) + \bigl(\ln A_3 - \frac{E_{a_3}}{RT}\bigr) - \bigl(\ln A_2 - \frac{E_{a_2}}{RT}\bigr)\Bigr].$$

Rearranging then leads to $$\ln k = \frac{1}{2}\left(\ln A_1 + \ln A_3 - \ln A_2\right) - \frac{1}{2RT}\left(E_{a_1} + E_{a_3} - E_{a_2}\right).$$

By comparing this with $$\ln k = \ln A - \frac{E_a}{RT}$$, it follows that $$E_a = \frac{1}{2}\left(E_{a_1} + E_{a_3} - E_{a_2}\right).$$

Substituting the given values $$E_{a_1} = 60$$, $$E_{a_2} = 30$$, and $$E_{a_3} = 10$$ kJ/mol into this formula gives $$E_a = \frac{1}{2}(60 + 10 - 30) = \frac{1}{2}(40) = 20 \, \text{kJ mol}^{-1}.$$

The answer is 20 kJ mol$$^{-1}$$.

Evaluate two statements about $$N(CH_3)_3$$ and $$P(CH_3)_3$$ as ligands.

Statement I: "$$N(CH_3)_3$$ and $$P(CH_3)_3$$ can act as ligands to form transition metal complexes."

CORRECT. Both have a lone pair on the central atom (N or P) that can be donated to a transition metal, forming coordinate bonds. They are well-known ligands in coordination chemistry.

Statement II: "The nature of bonding of $$N(CH_3)_3$$ and $$P(CH_3)_3$$ is always same with transition metals."

INCORRECT. $$N(CH_3)_3$$ acts primarily as a $$\sigma$$-donor (donates its lone pair). $$P(CH_3)_3$$, however, can act as both a $$\sigma$$-donor and a $$\pi$$-acceptor — phosphorus has available empty d-orbitals that can accept electron density from filled metal d-orbitals ($$\pi$$-back bonding). This difference in bonding nature is significant.

The correct answer is Option (1): Statement I is correct but Statement II is incorrect.

We need to find the correct IUPAC name of $$[PtBr_2(PMe_3)_2]$$.

To determine this, we recall that ligands are named in alphabetical order ignoring multiplicative prefixes, with simple ligands taking prefixes like di- and tri-, while more complex ligands use bis- or tris- within parentheses; anionic ligands end in “-o” (for example, bromo for $$Br^-$$); and the metal name is followed by its oxidation state in Roman numerals.

First, the bromide anion $$Br^-$$ serves as an anionic ligand named “bromo,” and since there are two bromides, we use “dibromo.” The neutral ligand $$PMe_3$$ (trimethylphosphine) is already composite (it contains the prefix tri-), so we apply “bis” and enclose the ligand name in parentheses to obtain “bis(trimethylphosphine).” Balancing the charges via $$x + 2(-1) + 2(0) = 0$$ shows that $$x = +2$$, corresponding to platinum(II).

Next, arranging the ligand names alphabetically places “bromo” before “trimethylphosphine,” giving the name $$\text{dibromobis(trimethylphosphine)platinum(II)}$$.

The correct answer is Option (3): dibromobis(trimethylphosphine)platinum(II).

The overall rate of a multi-step reaction is governed by the slow (rate-determining) step. In the given mechanism, the slow step is

$$N_2O_2(g)+H_2(g)\;\xrightarrow{k_2}\;N_2O(g)+H_2O(g)$$

Hence the instantaneous rate is

$$\text{rate}=k_2\,[N_2O_2]\,[H_2]\;-(1)$$

The species $$N_2O_2$$ is an intermediate; its concentration must be expressed in terms of the reactants that appear in the overall balanced equation. This is done using the preceding fast equilibrium

$$2NO(g)\;\rightleftharpoons\;N_2O_2(g)$$

For this equilibrium, the equilibrium constant $$K$$ is

$$K=\frac{[N_2O_2]}{[NO]^2}\;-(2)$$

Rearranging $$-(2)$$ gives

$$[N_2O_2]=K\,[NO]^2\;-(3)$$

Substituting $$-(3)$$ into the rate expression $$-(1)$$:

$$\text{rate}=k_2\,K\,[NO]^2\,[H_2]$$

Since $$k_2K$$ is a constant, we can write

$$\text{rate}=k_{\text{obs}}\,[NO]^2\,[H_2]$$

Thus, the reaction is

• second order with respect to $$NO$$
• first order with respect to $$H_2$$

Total (overall) order = $$2+1=3$$.

Therefore, the order of the reaction is 3.

For the first-order gas phase reaction $$A(g) \rightarrow B(g) + C(g)$$, we need to find the rate constant expression in terms of pressures.

Set up pressure relationships.

At $$t = 0$$: only A is present with pressure $$P_i$$.

At time $$t$$: let $$x$$ be the decrease in pressure of A. Then:

- Pressure of A: $$P_A = P_i - x$$

- Pressure of B: $$P_B = x$$

- Pressure of C: $$P_C = x$$

- Total pressure: $$P_t = (P_i - x) + x + x = P_i + x$$

Express $$P_A$$ in terms of measurable quantities.

From $$P_t = P_i + x$$: $$x = P_t - P_i$$

$$ P_A = P_i - x = P_i - (P_t - P_i) = 2P_i - P_t $$

Write the first-order rate constant expression.

For a first-order reaction: $$k = \frac{2.303}{t}\log\frac{[A]_0}{[A]_t}$$

In terms of pressures (since pressure is proportional to concentration for ideal gases):

$$ k = \frac{2.303}{t}\log\frac{P_i}{P_A} = \frac{2.303}{t}\log\frac{P_i}{2P_i - P_t} $$

The correct answer is Option (1): $$k = \frac{2.303}{t}\log\frac{P_i}{2P_i - P_t}$$.

Count complexes with an even number of electrons in $$t_{2g}$$ orbitals (weak field / high spin octahedral, since all use $$H_2O$$).

Analysis (all are weak-field/high-spin octahedral):

- $$[Fe(H_2O)_6]^{2+}$$: Fe²⁺ = d⁶. High spin: $$t_{2g}^4 e_g^2$$. $$t_{2g}$$ electrons = 4 (even).

- $$[Co(H_2O)_6]^{2+}$$: Co²⁺ = d⁷. High spin: $$t_{2g}^5 e_g^2$$. $$t_{2g}$$ electrons = 5 (odd).

- $$[Co(H_2O)_6]^{3+}$$: Co³⁺ = d⁶. Low spin (Co³⁺ is strong enough): $$t_{2g}^6 e_g^0$$. $$t_{2g}$$ electrons = 6 (even).

- $$[Cu(H_2O)_6]^{2+}$$: Cu²⁺ = d⁹. $$t_{2g}^6 e_g^3$$. $$t_{2g}$$ electrons = 6 (even).

- $$[Cr(H_2O)_6]^{2+}$$: Cr²⁺ = d⁴. High spin: $$t_{2g}^3 e_g^1$$. $$t_{2g}$$ electrons = 3 (odd).

Even $$t_{2g}$$ count: Fe²⁺ (4), Co³⁺ (6), Cu²⁺ (6) → 3 complexes.

The correct answer is Option (2): 3.

$$CoCl_3 \cdot nNH_3$$ gives 2 mol AgCl with excess AgNO₃. Find $$x + n$$ where $$x$$ is the oxidation state of Co.

Since 2 mol AgCl is produced, 2 Cl⁻ ions are outside the coordination sphere and 1 Cl is inside as a ligand, giving the formula $$[Co(NH_3)_nCl]Cl_2$$.

The charge on the complex ion is +2, where the inner Cl carries a -1 charge and NH₃ is neutral, so $$x + 0 \times n + (-1) = +2 \implies x = +3$$.

Assuming octahedral geometry with coordination number 6 and ligands consisting of n NH₃ molecules plus 1 Cl, we have $$n + 1 = 6$$, hence $$n = 5$$.

Therefore, $$x + n = 3 + 5 = 8$$.

The correct answer is Option (2): 8.

A) Since Cr is in +3 oxidation state, the electronic configuration is 4s2 3d3. That means 3 unpaired electrons are present. Hence BM =sqrt(n*(n+2)) = 3.87

B) Since Ni is in +2 oxidation state, the electronic configuration is 4s0 3d8. That means 2 unpaired electrons are present. Hence BM =sqrt(n*(n+2)) = 2.83

C) Since Co is in +3 oxidation state, the electronic configuration is 4s0 3d4. That means 3 unpaired electrons are present. Hence BM =sqrt(n*(n+2)) = 4.90

D) Since Ni is in +2 oxidation state but CN is a strong field ligand, the electronic configuration is 4s0 3d8 with all electrons paired. That means 0 unpaired electrons are present . Hence BM =sqrt(n*(n+2)) = 0

The colour shown by a transition-metal complex depends on the energy gap $$\Delta_{oct}$$ between its two d-electron sets. The light absorbed has frequency $$\nu$$ (or wavenumber $$\tilde{\nu}=1/\lambda$$) given by $$\Delta_{oct}=h\nu=hc\tilde{\nu}$$. Therefore:

larger $$\Delta_{oct}\;\Longrightarrow\;$$ higher energy $$\Longrightarrow\;$$ higher frequency $$\Longrightarrow\;$$ higher wavenumber of light absorbed.

Two main factors decide the magnitude of $$\Delta_{oct}$$:

1. Nature of ligands (spectrochemical series)   Weak-field ligands give small $$\Delta_{oct}$$, strong-field ligands give large $$\Delta_{oct}$$.
    $$Cl^- \lt H_2O \lt NH_3 \lt CN^-$$ in ligand strength.

2. Oxidation state of the metal ion   For the same ligands, a higher positive charge on the metal pulls the ligands closer, increasing $$\Delta_{oct}$$.

Let us analyse each complex.

Co is in $$+3$$ oxidation state (because $$x + (-1) = +2 \Rightarrow x = +3$$). It has five moderate ligands $$NH_3$$ and one weak ligand $$Cl^-$$, so the overall ligand field is weaker than that produced by six $$NH_3$$ molecules.

Co is again $$+3$$. The ligand $$CN^-$$ is a very strong-field ligand. Hence $$\Delta_{oct}$$ is the largest among the four complexes.

Co is $$+3$$. There are five $$NH_3$$ (stronger than $$H_2O$$) and one $$H_2O$$. Because $$H_2O$$ is stronger than $$Cl^-$$, the average ligand field here is stronger than in Case A but weaker than in Case B.

Cu is only in the $$+2$$ state, and the ligands are the medium-weak $$H_2O$$. The smaller oxidation state and weaker ligand field give the smallest $$\Delta_{oct}$$. (The geometry is tetragonally distorted octahedral or square-planar, but either way the splitting is smaller than that for the Co$$^{3+}$$ complexes.)

Combining the above comparisons:

$$\Delta_{oct}(D) \lt \Delta_{oct}(A) \lt \Delta_{oct}(C) \lt \Delta_{oct}(B)$$

Since wavenumber of absorbed light is directly proportional to $$\Delta_{oct}$$, the same order applies to $$\tilde{\nu}\,(\text{absorbed})$$:

$$[Cu(H_2O)_4]^{2+} \; \lt \; [CoCl(NH_3)_5]^{2+} \; \lt \; [Co(NH_3)_5(H_2O)]^{3+} \; \lt \; [Co(CN)_6]^{3-}$$

Hence the correct order is D < A < C < B, which corresponds to Option D.

We need the complex with the least paramagnetic behaviour, i.e., the fewest unpaired electrons. All given complexes are octahedral with $$H_2O$$ (a weak field ligand), so we use high-spin configurations.

$$[Cr(H_2O)_6]^{2+}$$: $$Cr^{2+}$$ is $$d^4$$. High spin: $$t_{2g}^3 e_g^1$$. Unpaired electrons = 4.

$$[Fe(H_2O)_6]^{2+}$$: $$Fe^{2+}$$ is $$d^6$$. High spin: $$t_{2g}^4 e_g^2$$. Unpaired electrons = 4.

$$[Co(H_2O)_6]^{2+}$$: $$Co^{2+}$$ is $$d^7$$. High spin: $$t_{2g}^5 e_g^2$$. Unpaired electrons = 3.

$$[Mn(H_2O)_6]^{2+}$$: $$Mn^{2+}$$ is $$d^5$$. High spin: $$t_{2g}^3 e_g^2$$. Unpaired electrons = 5.

$$[Co(H_2O)_6]^{2+}$$ has the fewest unpaired electrons (3), so it exhibits the least paramagnetic behaviour.

The correct answer is Option (3): $$[Co(H_2O)_6]^{2+}$$.

We need to evaluate the Assertion (A) and Reason (R) about the complex $$[Co(en)_2Cl_2]^+$$.

Analysis of Assertion (A):

"The total number of geometrical isomers shown by $$[Co(en)_2Cl_2]^+$$ is three."

To determine geometrical isomers, we need to understand the structure:

- Co$$^{3+}$$ is the central metal ion (cobalt in +3 oxidation state: $$d^6$$ configuration)

- $$en$$ (ethylenediamine) is a bidentate ligand that occupies 2 coordination positions

- Two en ligands occupy $$2 \times 2 = 4$$ positions, and two Cl$$^-$$ ligands occupy 2 positions, totaling 6 positions (octahedral)

For an octahedral complex of the type $$[M(AA)_2X_2]$$ (where AA is a bidentate ligand and X is a monodentate ligand), the possible geometrical isomers are:

1. cis isomer: The two Cl ligands are adjacent to each other (at 90 degrees)

2. trans isomer: The two Cl ligands are opposite to each other (at 180 degrees)

There are only 2 geometrical isomers (cis and trans), NOT 3. The assertion that there are three geometrical isomers is incorrect.

Analysis of Reason (R):

"$$[Co(en)_2Cl_2]^+$$ has an octahedral geometry."

Co$$^{3+}$$ has a coordination number of 6 in this complex (2 bidentate en ligands providing 4 donor atoms + 2 Cl atoms = 6 donor atoms). With 6 donor atoms, the geometry is indeed octahedral. This is consistent with the $$d^2sp^3$$ hybridization of Co$$^{3+}$$ (or inner orbital complex with strong field en ligands).

The reason is correct.

Conclusion: Assertion (A) is incorrect (there are only 2 geometrical isomers, not 3), but Reason (R) is correct (the complex does have octahedral geometry).

The correct answer is Option (2): (A) is not correct but (R) is correct.

We need to find the correct order of ligands arranged in increasing field strength according to the spectrochemical series.

The spectrochemical series (increasing field strength) is:

$$I^- < Br^- < S^{2-} < Cl^- < N_3^- < F^- < OH^- < ox^{2-} < H_2O < NCS^- < CH_3CN < py < NH_3 < en < bipy < phen < NO_2^- < PPh_3 < CN^- < CO < NO^+$$

Check each option:

Option A: $$F^- < Br^- < I^- < NH_3$$ - Incorrect. The order among halides is $$I^- < Br^- < Cl^- < F^-$$, not $$F^- < Br^- < I^-$$.

Option B: $$Br^- < F^- < H_2O < NH_3$$ - This follows the spectrochemical series correctly: $$Br^-$$ is weaker than $$F^-$$, $$F^-$$ is weaker than $$H_2O$$, and $$H_2O$$ is weaker than $$NH_3$$. This is correct.

Option C: $$H_2O < OH^- < CN^- < NH_3$$ - Incorrect. $$CN^-$$ is a stronger field ligand than $$NH_3$$, so $$NH_3 < CN^-$$, not $$CN^- < NH_3$$.

Option D: $$Cl^- < OH^- < Br^- < CN^-$$ - Incorrect. $$Br^-$$ is a weaker field ligand than $$Cl^-$$, so $$Br^-$$ should come before $$Cl^-$$.

The correct answer is Option B.

Step 1: Identify the product “A” in the first reaction. In the reaction $$NiS + HNO_3 + HCl \;\longrightarrow\; A + NO + S + H_2O$$ nitrate ion acts as an oxidising agent, converting sulfide to elemental sulfur (S) and itself reduced to NO. The nickel remains in the +2 oxidation state and combines with two chloride ions. Hence $$A \;=\; NiCl_2.$$

Step 2: Write the second reaction with the structure of the diacetyl dioxime ligand. Diacetyl dioxime has the formula $$CH_3C(=N-OH)C(=N-OH)CH_3,$$ and reacts with $$NiCl_2$$ in the presence of ammonium hydroxide as follows: $$NiCl_2 + 2\;NH_4OH + CH_3C(=N-OH)C(=N-OH)CH_3 \;\longrightarrow\; B + NH_4Cl + H_2O.$$ In fact, one molecule of diacetyl dioxime provides two oxime groups that chelate to Ni$$^{2+}$$, giving the neutral complex $$B = Ni[CH_3C(=NOH)C(=NOH)CH_3]_2.$$

Step 3: Draw or visualise the structure of product $$B$$. It is a square-planar complex in which each diacetyl dioxime ligand is bidentate, coordinating through the two nitrogen atoms. Each ligand retains two -OH groups, which form intramolecular hydrogen bonds to the adjacent C=N nitrogens: Intramolecular H-bonds: each $$-OH$$ proton is hydrogen-bonded to the adjacent $$=N$$ atom.

Step 4: Count all the protons in $$B$$. Each ligand $$CH_3C(=NOH)C(=NOH)CH_3$$ has: • Two methyl groups, each with 3 protons → total 6 methyl protons per ligand. • Two oxime -OH protons. Since there are two ligands, total protons in $$B$$ = $$2\times(6\;{\rm methyl}) + 2\times(2\;{\rm OH}) = 12\;{\rm CH_3} + 4\;{\rm OH} = 16\;{\rm H}.$$

Step 5: Determine which protons are involved in hydrogen bonding. The four $$-OH$$ protons are tied up in intramolecular hydrogen bonds to the adjacent $$=N$$ groups. Therefore these 4 protons are involved in hydrogen bonding.

Step 6: Calculate the number of protons not involved in hydrogen bonding. Subtracting the 4 hydrogen-bonded protons from the total 16 gives: $$16 - 4 \;=\; 12.$$

Final Answer: The number of protons in product $$B$$ that do not participate in hydrogen bonding is 12.

We seek the time required for [A] to drop from 1 M to 0.1 M under the rate law $$k[A]^{1/2}[B]^{1/2}$$, with $$[A]_0 = [B]_0 = 1$$ M and $$k = 4.6 \times 10^{-2}$$ s$$^{-1}$$.

Since $$[A]_0 = [B]_0$$ and they appear symmetrically in the rate expression, [A] remains equal to [B] at all times, which gives $$rate = k[A]^{1/2}[A]^{1/2} = k[A].$$ This effective rate law is thus first order in [A].

Using the first-order integrated rate law, we write $$\ln\frac{[A]_0}{[A]} = kt.$$ Substituting the values yields $$\ln\frac{1}{0.1} = 4.6 \times 10^{-2} \times t,$$ so $$\ln 10 = 4.6 \times 10^{-2} \times t.$$

Since $$\ln 10 = 2.303,$$ it follows that $$2.303 = 4.6 \times 10^{-2} \times t,$$ and hence $$t = \frac{2.303}{0.046} = 50.07 \approx 50 \text{ sec}.$$

The correct answer is 50.

Since the reaction $$2A_{(g)} + B_{(g)} \rightarrow C_{(g)}$$ is elementary, its rate law is $$ r = k \cdot P_A^2 \cdot P_B $$.

At $$P_A = 1.5$$ atm and $$P_B = 0.7$$ atm one finds $$ r_1 = k \times (1.5)^2 \times 0.7 = k \times 2.25 \times 0.7 = 1.575k $$.

When $$P_C$$ has increased by 0.5 atm, stoichiometry shows that $$P_A$$ decreases by $$2 \times 0.5 = 1.0$$ atm and $$P_B$$ decreases by $$1 \times 0.5 = 0.5$$ atm, giving $$P_A = 0.5$$ atm and $$P_B = 0.2$$ atm.

Under these conditions, $$ r_2 = k \times (0.5)^2 \times 0.2 = k \times 0.25 \times 0.2 = 0.05k $$.

Dividing these rates gives $$ \frac{r_1}{r_2} = \frac{1.575k}{0.05k} = \frac{1.575}{0.05} = 31.5 $$. Therefore, $$r_1 : r_2 = 31.5 = 315 \times 10^{-1}$$, so the answer is 315.

$$2N_2O_5 \rightarrow 4NO_2 + O_2$$

Rate of disappearance of N$$_2$$O$$_5$$: $$-\frac{\Delta[N_2O_5]}{\Delta t} = \frac{3 - 2.75}{30} = \frac{0.25}{30}$$ mol L$$^{-1}$$ min$$^{-1}$$.

Rate of formation of NO$$_2$$ = $$\frac{4}{2} \times$$ rate of disappearance of N$$_2$$O$$_5$$:

$$= 2 \times \frac{0.25}{30} = \frac{0.5}{30} = \frac{1}{60} \approx 0.01667$$ mol L$$^{-1}$$ min$$^{-1}$$.

$$= 16.67 \times 10^{-3} \approx 17 \times 10^{-3}$$ mol L$$^{-1}$$ min$$^{-1}$$.

Therefore, $$x = \boxed{17}$$.

The rate expression $$r = kA$$ shows that the rate is directly proportional to the concentration of $$A$$, so the reaction follows first-order kinetics.

For a first-order reaction the integrated rate law is
$$\ln\!\left(\frac{[A]_0}{[A]}\right)=kt$$ $$-(1)$$
and the half-life is related to the rate constant by
$$t_{1/2} = \frac{0.693}{k}$$ $$-(2)$$

50 % decomposition means $$[A] = \tfrac{1}{2}[A]_0$$. By definition this is the half-life, so
$$t_{1/2}=120\ \text{min}$$

Using $$(2)$$,
$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{120\ \text{min}} = 0.005775\ \text{min}^{-1}$$

90 % decomposition leaves $$10\%$$ of the initial concentration, so $$[A]=0.1[A]_0$$.
Substitute into $$(1)$$:

$$\ln\!\left(\frac{[A]_0}{0.1[A]_0}\right)=kt$$
$$\ln 10 = k\,t$$
$$2.303 = k\,t$$

Solve for $$t$$:
$$t = \frac{2.303}{k}= \frac{2.303}{0.005775\ \text{min}^{-1}} = 3.992\times10^{2}\ \text{min} \approx 399\ \text{min}$$

Therefore, the time required for 90 % decomposition of $$A$$ is about 399 minutes.

$$A(g) \to 2B(g) + C(g)$$. Initial: P₀, 0, 0. At time t: P₀-x, 2x, x. Total = P₀+2x.

Complete decomposition: P₀+2P₀ = 3P₀ = 300 ⟹ P₀ = 100 Torr.

At t=23s: P₀+2x = 200 ⟹ 2x = 100 ⟹ x = 50. So P_A = 100-50 = 50 Torr.

$$k = \frac{1}{t}\ln\frac{P_0}{P_A} = \frac{1}{23}\ln\frac{100}{50} = \frac{\ln 2}{23} = \frac{2.303\log 2}{23} = \frac{2.303\times0.301}{23} = \frac{0.6932}{23} = 0.03014$$ s⁻¹.

$$\approx 3 \times 10^{-2}$$ s⁻¹.

The answer is $$\boxed{3}$$.

Using the total pressure at $$t = 115\text{ s}$$:

$$0.1 + 2x = 0.28$$ 

$$2x = 0.18 \implies x = 0.09 \text{ atm}$$

Now, find the remaining pressure of reactant $$\text{A}$$ at $$t = 115\text{ s}$$:

$$P_\text{A} = P_0 - x = 0.1 - 0.09 = 0.01 \text{ atm}$$

3. Calculate the Rate Constant ($$k$$)

Using the integrated first-order rate equation:

$$k = \frac{2.303}{t} \log_{10}\left(\frac{P_0}{P_\text{A}}\right)$$

Substitute the values ($$t = 115$$, $$P_0 = 0.1$$, $$P_\text{A} = 0.01$$):

$$k = \frac{2.303}{115} \log_{10}\left(\frac{0.1}{0.01}\right)$$

$$k = \frac{2.303}{115} \log_{10}(10) = \frac{2.303}{115} \times 1$$

$$k \approx 0.0200 \text{ s}^{-1}$$

$$0.0200 \text{ s}^{-1} = 2 \times 10^{-2}\text{ s}^{-1}$$

The amount of a radioactive isotope remaining after time $$t$$ follows the decay formula $$N = N_0\left(\frac{1}{2}\right)^{t/T_{1/2}},$$ where $$N_0$$ is the initial amount, $$N$$ is the current amount, and $$T_{1/2}$$ is the half-life.

In this wood sample the ratio of $$\,^{14}C/\,^{12}C$$ has decreased to $$\tfrac{1}{8}$$ of the atmospheric value, so we set $$\frac{N}{N_0} = \frac{1}{8} = \left(\frac{1}{2}\right)^{t/T_{1/2}}.$$

Noting that $$\frac{1}{8} = \left(\frac{1}{2}\right)^3$$ shows that $$\tfrac{t}{T_{1/2}} = 3,$$ which means three half-lives have elapsed.

Since the half-life $$T_{1/2}$$ is 5730 years, the age of the sample is $$t = 3 \times 5730 = 17190 \text{ years}.$$

The answer is 17190 years.

We need to find the total number of unpaired electrons in $$[Co(NH_3)_6]^{3+}$$ and $$[NiCl_4]^{2-}$$.

For $$[Co(NH_3)_6]^{3+}$$:

Cobalt is in +3 oxidation state: $$Co^{3+}$$ has configuration $$[Ar]3d^6$$.

$$NH_3$$ is a strong field ligand, so this is a low-spin octahedral complex.

In a strong octahedral field, the 6 d-electrons fill the $$t_{2g}$$ orbitals completely: $$t_{2g}^6 e_g^0$$.

Number of unpaired electrons = 0.

For $$[NiCl_4]^{2-}$$:

Nickel is in +2 oxidation state: $$Ni^{2+}$$ has configuration $$[Ar]3d^8$$.

$$Cl^-$$ is a weak field ligand, so this is a tetrahedral complex (common for $$Ni^{2+}$$ with weak field ligands).

In a tetrahedral field with $$d^8$$ configuration, the splitting is: $$e^4 t_2^4$$.

Number of unpaired electrons = 2.

Total unpaired electrons = $$0 + 2 = 2$$.

The answer is $$\boxed{2}$$.

Given data for $$2HI_{(g)} \to H_{2(g)} + I_{2(g)}$$:

Let rate = $$k[HI]^n$$.

Using experiments 1 and 2:

$$\frac{3.0 \times 10^{-3}}{7.5 \times 10^{-4}} = \left(\frac{0.01}{0.005}\right)^n$$

$$4 = 2^n$$

$$n = 2$$

Verification with experiments 2 and 3:

$$\frac{1.2 \times 10^{-2}}{3.0 \times 10^{-3}} = \left(\frac{0.02}{0.01}\right)^n$$

$$4 = 2^n$$

$$n = 2$$ ✓

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

We need to find the overall activation energy when the overall rate constant $$K = \frac{K_1 K_2}{K_3}$$.

Recall the Arrhenius equation.

The Arrhenius equation relates the rate constant to activation energy:

$$K_i = A_i \, e^{-Ea_i/RT}$$

Express the overall rate constant.

$$K = \frac{K_1 K_2}{K_3} = \frac{A_1 e^{-Ea_1/RT} \cdot A_2 e^{-Ea_2/RT}}{A_3 e^{-Ea_3/RT}}$$

$$= \frac{A_1 A_2}{A_3} \cdot e^{-(Ea_1 + Ea_2 - Ea_3)/RT}$$

Identify the overall activation energy.

Comparing with $$K = A \, e^{-Ea/RT}$$, the overall activation energy is:

$$Ea = Ea_1 + Ea_2 - Ea_3$$

Substitute the values.

$$Ea = 40 + 50 - 60 = 30 \text{ kJ/mol}$$

The answer is 30 kJ/mol.

Ambidentate ligands are those that can coordinate through two different donor atoms.

$$NO_2^-$$: Can bind through N (nitro) or O (nitrito) — ambidentate ✓

$$SCN^-$$: Can bind through S (thiocyanato) or N (isothiocyanato) — ambidentate ✓

$$C_2O_4^{2-}$$: Bidentate ligand (binds through two O atoms of the same type) — not ambidentate ✗

$$NH_3$$: Monodentate, binds only through N — not ambidentate ✗

$$CN^-$$: Can bind through C (cyano) or N (isocyano) — ambidentate ✓

$$SO_4^{2-}$$: Not an ambidentate ligand ✗

$$H_2O$$: Monodentate, binds through O — not ambidentate ✗

Number of ambidentate ligands = 3 ($$NO_2^-, SCN^-, CN^-$$).

The answer is $$\boxed{3}$$.

We need to find the fraction of radioisotopic bromine-82 remaining after one day (24 hours), given its half-life is 36 hours.

The formula for radioactive decay is $$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/t_{1/2}},$$ where $$t_{1/2} = 36\text{ hours}$$ and $$t = 24\text{ hours}.$$ Hence, $$\frac{t}{t_{1/2}} = \frac{24}{36} = \frac{2}{3},$$ so $$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{2/3}.$$

Taking logarithm (base 10) of this expression gives

$$\log\left(\frac{N}{N_0}\right) = \frac{2}{3}\times\log\left(\frac{1}{2}\right) = \frac{2}{3}\times(-0.3010) = -0.2007.$$

Thus,

$$\frac{N}{N_0} = \text{antilog}(-0.2007) = \frac{1}{\text{antilog}(0.2007)}$$

and since antilog(0.2006) = 1.587,

$$\frac{N}{N_0} = \frac{1}{1.587} = 0.6301 \approx 63\times10^{-2}.$$

Therefore, the fraction remaining is 63 × 10−2.

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

$$ r = k[A] $$

Since the rate is proportional to concentration, the ratio of rates at two times gives the ratio of concentrations:

$$ \frac{r_1}{r_2} = \frac{[A]_1}{[A]_2} $$

Given: $$r_1 = 0.04$$ at $$t_1 = 10$$ min, $$r_2 = 0.03$$ at $$t_2 = 20$$ min.

$$ \frac{[A]_{10}}{[A]_{20}} = \frac{0.04}{0.03} = \frac{4}{3} $$

For a first order reaction, the rate constant is:

$$ k = \frac{2.303}{t_2 - t_1} \log\frac{[A]_1}{[A]_2} = \frac{2.303}{10} \log\frac{4}{3} $$

$$ k = \frac{2.303}{10}(\log 4 - \log 3) = \frac{2.303}{10}(2\log 2 - \log 3) $$

$$ k = \frac{2.303}{10}(2 \times 0.3010 - 0.4771) = \frac{2.303}{10}(0.6020 - 0.4771) $$

$$ k = \frac{2.303 \times 0.1249}{10} = \frac{0.2877}{10} = 0.02877 \text{ min}^{-1} $$

The half-life for a first order reaction is:

$$ t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02877} \approx 24.09 \text{ min} $$

Rounding to the nearest integer: $$t_{1/2} \approx 24$$ minutes.

Therefore, the answer is $$\boxed{24}$$.

For a first-order reaction: $$t = \frac{2.303}{k}\log\frac{1}{1-x}$$

For 99.9% completion ($$x = 0.999$$):

$$t_{99.9\%} = \frac{2.303}{k}\log\frac{1}{0.001} = \frac{2.303}{k} \times 3 = \frac{6.909}{k}$$

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

$$\frac{t_{99.9\%}}{t_{1/2}} = \frac{6.909}{0.693} \approx 9.97 \approx 10$$

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

This question is about the iodine clock reaction, where iodide ions react with H$$_2$$O$$_2$$.

The reaction:

$$\text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \to \text{I}_2 + 2\text{H}_2\text{O}$$

Identifying the indicator X:

The indicator forms a blue colored complex with a compound present in the solution. The well-known indicator that forms a deep blue complex is starch.

Identifying compound A:

Starch forms a blue complex specifically with iodine (I$$_2$$), which is the product of the reaction between iodide and hydrogen peroxide.

The appearance of the blue color indicates that free iodine has been produced, which is used to study the rate of the reaction.

Therefore, indicator X = Starch and compound A = Iodine.

The correct answer is Option 1: Starch and iodine.

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).

Hydrogen peroxide (H$$_2$$O$$_2$$) is unstable and decomposes when exposed to light, heat, or catalysts:

$$2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$$

Light (particularly UV light) provides energy that breaks the weak O-O bond in H$$_2$$O$$_2$$ (bond energy approximately 146 kJ/mol), initiating decomposition. We need to identify which option can reduce this photodecomposition.

Urea (NH$$_2$$CONH$$_2$$) can stabilize H$$_2$$O$$_2$$ by forming hydrogen-bonded complexes, but it primarily works against thermal and catalytic decomposition rather than specifically blocking light.

Alkaline conditions actually accelerate the decomposition of H$$_2$$O$$_2$$ because the hydroperoxide ion (HO$$_2^-$$) formed in basic solution is more reactive.

Now, H$$_2$$O$$_2$$ is stored in dark-coloured (brown or amber) glass containers. The coloured glass absorbs UV and visible light, preventing it from reaching the H$$_2$$O$$_2$$, thereby directly reducing photodecomposition.

Dust particles can contain metal ions (such as Fe$$^{2+}$$, Mn$$^{2+}$$) that act as catalysts for H$$_2$$O$$_2$$ decomposition, so dust would increase decomposition rather than reduce it.

Hence, the correct answer is Option 3: Glass containers.

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.

No.of intermediates can be found by no. of stable compound which is shown by the low energy level. No. of activated complex would be find from the high level energy state which is 3 in the case of shown graph. Rate determining step would be slowest step in the reaction which is the 3rd step in case of given graph

We are given the Freundlich adsorption isotherm plotted as a straight line: $$y = 3x + 2.505$$.

The Freundlich equation is: $$\frac{x}{m} = K \cdot p^{1/n}$$

Taking logarithm:

$$\log\frac{x}{m} = \log K + \frac{1}{n}\log p$$

Comparing with $$y = 3x + 2.505$$:

Here $$y = \log(x/m)$$ and $$x$$-axis = $$\log p$$.

Slope = $$\frac{1}{n} = 3$$

Intercept = $$\log K = 2.505$$

$$\frac{1}{n} = 3$$ and $$\log K = 2.505$$.

The correct answer is Option C: $$3$$ and $$2.505$$.

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$$^{-1}$$.

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.

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.

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.

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.

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.

• A. Incorrect: The reactants diffuse over the surface of the catalyst, not the other way around.
• B. Correct: Reactant molecules approach and get adsorbed onto the active sites on the catalyst's surface.
• C. Correct: Adsorbed reactants react chemically on the surface to form an activated intermediate complex.
• D. Correct: Modern adsorption theory is explicitly a combination of the old adsorption theory (surface concentration) and the intermediate compound formation theory.
• 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).

This question is about the iodine clock reaction, which involves the reaction of iodide ions with hydrogen peroxide ($$H_2O_2$$). Let us analyze each statement:

(A) Always use freshly prepared starch solution.

This is correct. Starch solution degrades over time due to bacterial decomposition and may not give a sharp blue-black colour with iodine. Freshly prepared starch is essential for reliable endpoint detection.

(B) Always keep the concentration of sodium thiosulphate solution less than that of KI solution.

This is correct. In the iodine clock reaction, sodium thiosulphate ($$Na_2S_2O_3$$) reacts with the iodine produced and keeps the solution colourless until all the thiosulphate is consumed. The thiosulphate concentration must be less than KI concentration so that iodine eventually accumulates and turns the starch indicator blue.

(C) Record the time immediately after the appearance of blue colour.

This is correct. The appearance of blue colour marks the endpoint — all thiosulphate has been consumed and free iodine reacts with starch. The time should be recorded at this point.

(D) Record the time immediately before the appearance of blue colour.

This is incorrect. You cannot predict the exact moment before the colour appears. The time is recorded when the blue colour actually appears.

(E) Always keep the concentration of sodium thiosulphate solution more than that of KI solution.

This is incorrect. If thiosulphate concentration exceeds KI concentration, the blue colour may never appear during the observation period, making the experiment impractical.

The correct statements are A, B, and C.

Therefore, the correct answer is Option A: A, B, C only.

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.

We are given the critical temperatures of four gases — He (5.2 K), $$CH_4$$ (190 K), $$CO_2$$ (304.2 K), and $$NH_3$$ (405.5 K) — and asked which gas shows the least adsorption on charcoal.

Adsorption of a gas on a solid surface depends on how easily the gas can be liquefied. A gas that is more easily liquefied (i.e., has stronger intermolecular forces) will be adsorbed more readily on the surface of charcoal. The critical temperature of a gas is a direct measure of the strength of its intermolecular forces: the higher the critical temperature, the stronger the intermolecular attractions, and the more easily the gas can be liquefied.

Now, among the given gases, $$NH_3$$ has the highest critical temperature (405.5 K), meaning it has the strongest intermolecular forces (due to hydrogen bonding) and is most easily liquefied — so it will be adsorbed the most. $$CO_2$$ (304.2 K) and $$CH_4$$ (190 K) follow in decreasing order of adsorption.

Helium has the lowest critical temperature of just 5.2 K, indicating extremely weak van der Waals forces. It is the hardest gas to liquefy among the four, and therefore it will show the least adsorption on charcoal.

Hence, the correct answer is Option A.

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.

We need to match each industrial process/reaction with its correct catalyst.

The reaction

$$2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$$

is the Contact Process for manufacturing sulphuric acid and uses $$V_2O_5$$ (vanadium pentoxide) as the catalyst, so A matches with (III).

The reaction

$$4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)$$

is the Ostwald Process for manufacturing nitric acid and uses Pt(s)-Rh(s) (platinum-rhodium gauze) as the catalyst, so B matches with (II).

The reaction

$$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$$

is the Haber Process for manufacturing ammonia and uses Fe(s) (finely divided iron with promoters like $$Al_2O_3$$ and $$K_2O$$) as the catalyst, so C matches with (I).

The reaction Vegetable oil(l) + $$H_2$$ $$\rightarrow$$ Vegetable ghee(s) is the hydrogenation of vegetable oils (hardening of oils) and uses Ni(s) (finely divided nickel), known as the Sabatier-Senderens reaction, so D matches with (IV).

Hence, the correct answer is Option B: A-III, B-II, C-I, D-IV.

We need to match each reaction with its catalyst. List-II contains catalysts used in important industrial and laboratory reactions.

A. $$4NH_3(g) + 5O_2(g) \to 4NO(g) + 6H_2O(g)$$

This is the Ostwald process for manufacturing nitric acid. The catalyst used is Platinum (Pt).

$$\Rightarrow$$ A matches with III (Pt(s))

B. $$N_2(g) + 3H_2(g) \to 2NH_3(g)$$

This is the Haber process for synthesis of ammonia. The catalyst used is Iron (Fe).

$$\Rightarrow$$ B matches with IV (Fe(s))

C. $$C_{12}H_{22}O_{11}(aq) + H_2O(l) \to C_6H_{12}O_6 \text{ (Glucose)} + C_6H_{12}O_6 \text{ (Fructose)}$$

This is the hydrolysis (inversion) of sucrose. The catalyst used is dilute sulphuric acid ($$H_2SO_4$$).

$$\Rightarrow$$ C matches with II ($$H_2SO_4(l)$$)

D. $$2SO_2(g) + O_2(g) \to 2SO_3(g)$$

This is the Contact process for manufacturing sulphuric acid. While the primary catalyst is $$V_2O_5$$, the older Lead Chamber process uses NO(g) as a catalyst for the oxidation of $$SO_2$$.

$$\Rightarrow$$ D matches with I (NO(g))

The correct matching is: A-III, B-IV, C-II, D-I

Therefore, the correct answer is Option C.

The four rate expressions given in LIST-I resemble the Michaelis-Menten form that is often written for enzyme catalysis or surface-mediated (Langmuir-Hinshelwood) kinetics. In general, the rate of decomposition of $$X$$ is expressed as a function of its concentration $$[X]$$ and a saturation constant $$X_s$$.

$$\textbf{General form :}\qquad \text{rate}=\frac{k[X]^m}{X_s+[X]}$$
Here $$m=1$$ or $$2$$ depending on the mechanism.

To match each rate law with its graphical profile we analyse its limiting behaviour in the two extreme regimes:
(1) $$[X]\ll X_s$$ (very dilute substrate)
(2) $$[X]\gg X_s$$ (substrate present in large excess)

• When $$[X]\ll X_s$$, $$\;X_s+[X]\approx X_s\Rightarrow\text{rate}\approx\dfrac{k}{X_s}[X]$$ (first-order).
• When $$[X]\gg X_s$$, $$\;X_s+[X]\approx[X]\Rightarrow\text{rate}\approx k$$ (zero-order plateau).
Hence the plot of rate vs. $$[X]$$ rises linearly at first and then levels off to a constant value - the typical saturation (rectangular-hyperbola) curve. That overall curve is denoted in LIST-II by profile P.

Because the initial concentration is far smaller than $$X_s$$ we may replace $$X_s+[X]$$ by $$X_s$$ right from the start:
$$\text{rate}\approx\dfrac{k}{X_s}[X]$$
Thus the graph is a straight line through the origin whose slope is $$k/X_s$$ - a purely first-order dependence. In LIST-II that simple linear plot is labelled Q.

Now $$X_s+[X]\approx[X]$$, giving
$$\text{rate}\approx k\;(\text{independent of }[X])$$
Hence the rate is essentially constant; the profile is a horizontal line (zero-order region) tagged as S in LIST-II.

For very large $$[X]$$, $$X_s+[X]\approx[X]$$, so
$$\text{rate}\approx\frac{k[X]^2}{[X]}=k[X]$$
The rate once again varies linearly with concentration, but its slope is simply $$k$$ (different numerical value from Case II). The corresponding straight-line profile is marked T in LIST-II.

Collecting the matches:

I → P (general saturation curve)
II → Q (first-order straight line at low $$[X]$$)
III → S (zero-order plateau at high $$[X]$$)
IV → T (first-order straight line obtained from the $$[X]^2$$ law at high $$[X]$$)

Therefore, the correct option is:
Option A which is: I → P; II → Q; III → S; IV → T

We need to find the percentage of original activity remaining after 83 days for a radioactive element with a half-life of 200 days. Half-life (t$$_{1/2}$$) = 200 days, time elapsed (t) = 83 days, antilog 0.125 = 1.333, antilog 0.693 = 4.93.

The decay constant is given by $$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{200}$$. Using the radioactive decay formula $$\frac{A}{A_0} = e^{-\lambda t}$$ and taking log$$_{10}$$ of the ratio, we get $$\log_{10}\left(\frac{A_0}{A}\right) = \frac{\lambda t}{2.303} = \frac{0.693 \times 83}{200 \times 2.303}$$ which yields $$\log_{10}\left(\frac{A_0}{A}\right) = \frac{57.519}{460.6} = 0.1249 \approx 0.125$$.

Using the given antilog value, $$\frac{A_0}{A} = \text{antilog}(0.125) = 1.333$$ so $$\frac{A}{A_0} = \frac{1}{1.333} = 0.750$$ and hence $$\% \text{ activity remaining} = 0.750 \times 100 = 75\%$$. Hence, the answer is 75.

The given reaction is:

$$2NO + 2H_2 \rightarrow N_2 + 2H_2O$$

This reaction has been studied at 800°C, and the mechanism is well-established. In the first, slow, rate-determining step the following occurs: $$2NO + H_2 \rightarrow N_2O + H_2O$$. A fast second step then follows: $$N_2O + H_2 \rightarrow N_2 + H_2O$$.

Since the first step is rate-determining, the rate law is determined by this step, giving: $$\text{Rate} = k[NO]^2[H_2]$$.

From the rate law, the order with respect to NO is the exponent of [NO], which is 2. Therefore, the order of the reaction with respect to NO is $$\textbf{2}$$.

We need to find the rate constant for a first-order reaction where the concentration of A becomes half in 70 minutes. The concentration of A becomes half of its initial concentration in 70 minutes, which means the half-life is given by $$t_{1/2} = 70 \text{ minutes} = 70 \times 60 = 4200 \text{ seconds}$$.

For a first-order reaction the rate constant is calculated using $$k = \frac{0.693}{t_{1/2}}$$. Substituting the value of the half-life gives $$k = \frac{0.693}{4200}$$, which evaluates to $$k = 1.65 \times 10^{-4} \text{ s}^{-1}$$. This can also be expressed as $$k = 165 \times 10^{-6} \text{ s}^{-1}$$. Therefore, $$x = 165$$, and the answer is 165.

The reaction considered is a first order reaction: $$A \to B$$, with half-life $$t_{1/2} = 0.3010$$ min and time $$t = 2.0$$ min.

For a first order reaction, the half-life is given by $$t_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}$$, so that $$k = \frac{0.693}{0.3010} = \frac{\ln 2}{0.3010}$$.

Since $$\ln 2 = 2.303 \times \log 2 = 2.303 \times 0.3010 = 0.6932$$, this gives $$k = \frac{2.303 \times 0.3010}{0.3010} = 2.303 \text{ min}^{-1}$$.

In order to find the concentration ratio, we use the first order rate expression $$\ln\frac{[A_0]}{[A]} = kt$$, which leads to $$\ln\frac{[A_0]}{[A]} = 2.303 \times 2.0 = 4.606$$.

Converting to a common logarithm, we write $$2.303 \times \log\frac{[A_0]}{[A]} = 4.606$$, hence $$\log\frac{[A_0]}{[A]} = \frac{4.606}{2.303} = 2$$.

Therefore, $$\frac{[A_0]}{[A]} = 10^2 = 100$$.

This shows that the ratio of the initial concentration to the concentration at time 2.0 min is 100.

We are given the Arrhenius equation: $$k = (6.5 \times 10^{12} \text{ s}^{-1})e^{-26000\text{ K}/T}$$. The Arrhenius equation is: $$k = Ae^{-E_a/(RT)}$$, so comparing the exponents gives $$\frac{E_a}{RT} = \frac{26000}{T}$$.

From this, $$\frac{E_a}{R} = 26000 \text{ K}$$ and $$E_a = 26000 \times R = 26000 \times 8.314 \text{ J mol}^{-1}$$.

Therefore, $$E_a = 26000 \times 8.314 = 216164 \text{ J mol}^{-1}$$ and $$E_a = 216.164 \text{ kJ mol}^{-1}$$. Rounding to the nearest integer: $$E_a = 216$$ kJ mol$$^{-1}$$.

We need to determine the order of the decomposition reaction of compound A using the relationship between half-life and initial pressure.

For a reaction of order $$ n $$, the half-life is related to the initial concentration (or pressure) by:

$$t_{1/2} \propto P_0^{1-n}$$

This gives us:

$$\frac{t_{1/2,1}}{t_{1/2,2}} = \left(\frac{P_{0,1}}{P_{0,2}}\right)^{1-n}$$

First half-life: $$ t_{1/2,1} = 240 \text{ s} = 4 \text{ min} $$ at $$ P_{0,1} = 500 \text{ Torr} $$

Second half-life: $$ t_{1/2,2} = 4.0 \text{ min} = 240 \text{ s} $$ at $$ P_{0,2} = 250 \text{ Torr} $$

$$\frac{240}{240} = \left(\frac{500}{250}\right)^{1-n}$$

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

Since $$ 2^{1-n} = 1 = 2^0 $$, we get:

$$1 - n = 0$$

$$n = 1$$

The order of the reaction is 1.

To find the half-life of the first-order decomposition of azomethane, we start by determining the rate constant (k) from the provided graph. For a first-order reaction, the integrated rate equation relating partial pressure to time is ln(p/p₀) = -kt.

When you plot ln(p/p₀) on the y-axis against time (t) on the x-axis, the relationship mirrors the equation of a straight line (y = mx) where the slope (m) is equal to -k.

The given graph indicates that the slope is -3.465 × 10⁴. By setting this equal to -k, we determine that the rate constant is k = 3.465 × 10⁴ s⁻¹.

Next, we apply the standard half-life formula for a first-order reaction, which is t₁/₂ = 0.693 / k.

Substituting our calculated value for k into the formula gives: t₁/₂ = 0.693 / (3.465 × 10⁴)

Dividing 0.693 by 3.465 gives 0.2. Therefore, the equation simplifies to: t₁/₂ = 0.2 × 10⁻⁴ s

To match the format requested in the question, we rewrite this in standard scientific notation: t₁/₂ = 2 × 10⁻⁵ s

The right answer is 2.

We are given the rate constant equation for a first order reaction:

$$\ln k = 33.24 - \frac{2.0 \times 10^4 \text{ K}}{T}$$

We need to compare this with the Arrhenius equation in its logarithmic form.

The Arrhenius equation is:

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

Taking the natural logarithm:

$$\ln k = \ln A - \frac{E_a}{RT}$$

$$\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$$

Comparing with the given equation $$\ln k = 33.24 - \frac{2.0 \times 10^4}{T}$$, we identify:

$$\frac{E_a}{R} = 2.0 \times 10^4 \text{ K}$$

Therefore:

$$E_a = R \times 2.0 \times 10^4 \text{ K}$$

$$E_a = 8.3 \text{ J K}^{-1} \text{ mol}^{-1} \times 2.0 \times 10^4 \text{ K}$$

$$E_a = 16.6 \times 10^4 \text{ J mol}^{-1}$$

$$E_a = 166 \times 10^3 \text{ J mol}^{-1}$$

$$E_a = 166 \text{ kJ mol}^{-1}$$

Therefore, the activation energy is 166 kJ mol$$^{-1}$$.

We are given that the reaction is first order in X and zero order in Y. So the rate law is:

$$\text{Rate} = k[X]^1[Y]^0 = k[X]$$

From Experiment I, we have $$[X] = 0.1$$ and Rate $$= 2 \times 10^{-3}$$. Substituting into the rate law:

$$2 \times 10^{-3} = k \times 0.1$$

$$k = \frac{2 \times 10^{-3}}{0.1} = 2 \times 10^{-2} \text{ min}^{-1}$$

We can verify this using Experiment IV, where $$[X] = 0.1$$ and $$[Y] = 0.2$$. The predicted rate is $$k \times 0.1 = 2 \times 10^{-2} \times 0.1 = 2 \times 10^{-3}$$ mol L$$^{-1}$$ min$$^{-1}$$, which matches the given rate. This confirms the rate is independent of [Y].

Now, from Experiment II, the rate is $$4 \times 10^{-3}$$ and $$[X] = L$$:

$$4 \times 10^{-3} = 2 \times 10^{-2} \times L$$

$$L = \frac{4 \times 10^{-3}}{2 \times 10^{-2}} = 0.2$$

From Experiment III, $$[X] = 0.4$$ and Rate $$= M \times 10^{-3}$$:

$$M \times 10^{-3} = 2 \times 10^{-2} \times 0.4 = 8 \times 10^{-3}$$

$$M = 8$$

The required ratio is:

$$\frac{M}{L} = \frac{8}{0.2} = 40$$

Hence, the correct answer is 40.

We need to find the ratio of rate constants when a catalyst reduces the activation energy by 10 kJ mol⁻¹.

According to the Arrhenius equation, the rate constant with catalyst is given by $$k_{cat} = A \cdot e^{-(E_a - 10000)/(RT)}$$ and the rate constant without catalyst is $$k_{uncat} = A \cdot e^{-E_a/(RT)}$$. The pre-exponential factor $$A$$ is the same in both cases.

Taking the ratio of these expressions yields $$\frac{k_{cat}}{k_{uncat}} = \frac{e^{-(E_a - 10000)/(RT)}}{e^{-E_a/(RT)}} = e^{10000/(RT)}$$.

Substituting the gas constant and temperature gives $$x = \frac{10000}{RT} = \frac{10000}{8.31 \times 300} = \frac{10000}{2493} = 4.011$$.

Rounding to the nearest integer, we obtain $$x \approx 4$$, so the value of $$x$$ is 4.

We need to find the value of $$x$$ such that the time for 67% completion of a first-order reaction equals $$x \times 10^{-1}$$ times the half-life.

For a first-order reaction, the time for a given fraction of completion is given by $$t = \frac{2.303}{k} \log\left(\frac{1}{1 - \text{fraction}}\right).$$ When 67% of the reaction is complete, the fraction remaining is $$1 - 0.67 = 0.33 = \frac{1}{3},$$ so $$t_{67\%} = \frac{2.303}{k} \log\left(\frac{1}{1/3}\right) = \frac{2.303}{k} \log 3.$$

The half-life for a first-order reaction is $$t_{1/2} = \frac{0.693}{k}.$$ Hence, $$\frac{t_{67\%}}{t_{1/2}} = \frac{2.303 \times \log 3}{0.693}.$$ Substituting $$\log 3 = 0.4771$$ gives $$\frac{t_{67\%}}{t_{1/2}} = \frac{2.303 \times 0.4771}{0.693}.$$

Since $$2.303 \times 0.4771 = 1.0988,$$ we get $$\frac{1.0988}{0.693} = 1.5855 \approx 1.6.$$ Writing $$t_{67\%} = x \times 10^{-1} \times t_{1/2}$$ then implies $$x \times 10^{-1} = 1.6,$$ so $$x = 16.$$

Therefore, the answer is 16.

For the reaction $$\gamma_1 A + \gamma_2 B \to \gamma_3 C + \gamma_4 D$$, we need to find the rate of reaction.

The concentration of C changes from 10 mmol dm$$^{-3}$$ to 20 mmol dm$$^{-3}$$ in 10 s, so $$\frac{d[C]}{dt} = \frac{20 - 10}{10} = 1 \text{ mmol dm}^{-3} \text{s}^{-1}$$.

The rate of appearance of D is given as 9 mmol dm$$^{-3}$$ s$$^{-1}$$:
$$\frac{d[D]}{dt} = 9 \text{ mmol dm}^{-3} \text{s}^{-1}$$ Since the rate of appearance of D equals 1.5 times the rate of disappearance of B, it follows that
$$\frac{d[D]}{dt} = 1.5 \times \left(-\frac{d[B]}{dt}\right) \implies -\frac{d[B]}{dt} = \frac{9}{1.5} = 6 \text{ mmol dm}^{-3} \text{s}^{-1}$$ Furthermore, as the rate of disappearance of B is twice the rate of disappearance of A, we have
$$-\frac{d[B]}{dt} = 2 \times \left(-\frac{d[A]}{dt}\right) \implies -\frac{d[A]}{dt} = \frac{6}{2} = 3 \text{ mmol dm}^{-3} \text{s}^{-1}$$.

For the general reaction $$\gamma_1 A + \gamma_2 B \to \gamma_3 C + \gamma_4 D$$, the rate of reaction $$r$$ is defined as:
$$r = \frac{1}{\gamma_1}\left(-\frac{d[A]}{dt}\right) = \frac{1}{\gamma_2}\left(-\frac{d[B]}{dt}\right) = \frac{1}{\gamma_3}\frac{d[C]}{dt} = \frac{1}{\gamma_4}\frac{d[D]}{dt}$$ Since all these expressions must be equal, the coefficients are proportional to the individual rates:
$$\gamma_1 : \gamma_2 : \gamma_3 : \gamma_4 = 3 : 6 : 1 : 9$$

Using any of the equivalent expressions yields:
$$r = \frac{1}{\gamma_3}\frac{d[C]}{dt} = \frac{1}{1} \times 1 = 1 \text{ mmol dm}^{-3} \text{s}^{-1}$$ Verification with other species confirms consistency:
$$r = \frac{1}{\gamma_1}\left(-\frac{d[A]}{dt}\right) = \frac{1}{3} \times 3 = 1$$ ✓
$$r = \frac{1}{\gamma_2}\left(-\frac{d[B]}{dt}\right) = \frac{1}{6} \times 6 = 1$$ ✓
$$r = \frac{1}{\gamma_4}\frac{d[D]}{dt} = \frac{1}{9} \times 9 = 1$$ ✓

The correct answer is $$\mathbf{1}$$ mmol dm$$^{-3}$$ s$$^{-1}$$.

We are given that for the reaction $$A \rightarrow 2B + C$$, the half-lives are 100 s and 50 s when the concentrations of A are 0.5 and 1.0 mol L$$^{-1}$$ respectively.

For an nth-order reaction, the half-life is related to initial concentration by:

$$t_{1/2} \propto [A_0]^{1-n}$$

The ratio of half-lives is:

$$\frac{t_{1/2}^{(1)}}{t_{1/2}^{(2)}} = \left(\frac{[A_0]_1}{[A_0]_2}\right)^{1-n}$$

$$\frac{100}{50} = \left(\frac{0.5}{1.0}\right)^{1-n}$$

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

Rewriting,

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

It follows that

$$1 = n - 1$$

$$n = 2$$

Therefore, the correct answer is 2.

We are given that for a rise of 9 K in temperature, the rate constant doubles. We need to find the activation energy at T = 300 K.

Using the Arrhenius equation in logarithmic form:

$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

Substituting the given values $$T_1 = 300$$ K, $$T_2 = 309$$ K, and $$\frac{k_2}{k_1} = 2$$ yields

$$\log 2 = \frac{E_a}{2.303 \times 8.3}\left(\frac{1}{300} - \frac{1}{309}\right)$$

Noting that

$$\frac{1}{300} - \frac{1}{309} = \frac{309 - 300}{300 \times 309} = \frac{9}{92700}$$

This leads to

$$0.30 = \frac{E_a}{2.303 \times 8.3} \times \frac{9}{92700}$$

$$0.30 = \frac{E_a \times 9}{19.1149 \times 92700}$$

Next,

$$19.1149 \times 92700 = 1772,951.23$$

Approximating: $$2.303 \times 8.3 = 19.1149$$

$$19.1149 \times 92700 \approx 1,772,951$$

Solving for $$E_a$$:

$$E_a = \frac{0.30 \times 1,772,951}{9}$$

$$E_a = \frac{531,885.3}{9}$$

$$E_a = 59,098.4 \text{ J mol}^{-1}$$

Converting to kJ,

$$E_a \approx 59.1 \text{ kJ mol}^{-1}$$

Therefore, the activation energy is approximately 59 kJ mol$$^{-1}$$.

The relationship is defined as

$$k = A e^{-E_a / RT}$$, where $$A$$ represents the pre-exponential factor (frequency of collisions), $$E_a$$ is the activation energy, $$R$$

is the universal gas constant ($$8.314 \text{ J K}^{-1} \text{mol}^{-1}$$), and $$T$$ is the temperature in Kelvin ($$K$$). So the graph will be increasing exponentially. Thus, the right option is D.

Taken from NCERT:

The slope of the straight line gives the value of .$$\frac{\ 1}{n}$$ The intercept on the y-axis gives the value of log k.

Freundlich isotherm explains the behaviour of adsorption in an approximate manner. The

factor $$\frac{\ 1}{n}$$ can have values between 0 and 1, (probable range 0.1 to 0.5)

First, we recall the definition of the rate law for a reaction of overall order $$n$$. For a simple case where the rate depends on a single concentration $$[A]$$, we write

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

Here, $$k$$ is the rate constant whose units we are asked to find, $$[A]$$ is the molar concentration of the reactant, and $$n$$ is the order of the reaction.

Now we analyse the units of each quantity one by one.

We know that the numerical value of rate is change in concentration divided by time. Therefore, in SI-compatible chemical units,

$$[\text{Rate}]=\dfrac{\text{mol L}^{-1}}{\text{s}}=\text{mol L}^{-1}\,\text{s}^{-1}$$

Next, the concentration $$[A]$$ itself has the unit $$\text{mol L}^{-1}$$. Raising this to the power $$n$$ gives

$$[A]^n=\bigl(\text{mol L}^{-1}\bigr)^n=\text{mol}^n\,\text{L}^{-n}$$

We now substitute these unit expressions into the rate law. Isolating the unit of $$k$$ from

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

we have

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

Substituting, we obtain

$$[k]=\dfrac{\text{mol L}^{-1}\,\text{s}^{-1}}{\text{mol}^n\,\text{L}^{-n}}$$

To simplify, we divide the powers of identical units algebraically:

The exponent of $$\text{mol}$$ becomes $$1-n$$ because $$1-n=1-(n)$$.

The exponent of $$\text{L}$$ becomes $$-1-(-n)=n-1$$.

The exponent of $$\text{s}$$ remains $$-1$$ because no time unit appears in the denominator.

Hence, after simplification,

$$[k]=\text{mol}^{\,1-n}\,\text{L}^{\,n-1}\,\text{s}^{-1}$$

Comparing this with the options given, we find that this expression matches the unit written in Option C.

Hence, the correct answer is Option C.

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 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 are dealing with the reversible chemical change $$A \rightleftharpoons B$$. The two simultaneous reactions involved are

$$A \xrightarrow{k_f} B \quad\text{(forward)}$$

$$B \xrightarrow{k_r} A \quad\text{(reverse)}$$

At any instant the rate of the forward reaction is given, by definition, as

$$r_f = k_f\,[A]$$

and the rate of the reverse reaction is

$$r_r = k_r\,[B]$$

Initially, that is at time $$t = 0$$, only reactant $$A$$ is present while product $$B$$ has not yet been formed. Hence we have

$$[A] = [A]_0 \quad\text{and}\quad [B] = 0$$

Substituting these concentrations in the two rate expressions gives

$$r_f(t=0) = k_f\,[A]_0 \;>\; 0$$

$$r_r(t=0) = k_r\,[B]_{t=0} = k_r\times 0 = 0$$

So the forward rate starts from a finite positive value, while the reverse rate starts from zero.

As time passes, $$A$$ is consumed and its concentration $$[A]$$ continuously falls. According to the equation $$r_f = k_f\,[A]$$, the forward rate therefore keeps decreasing smoothly from its initial value. Simultaneously, $$B$$ is being produced, so $$[B]$$ keeps rising from zero. From $$r_r = k_r\,[B]$$ we see that the reverse rate increases smoothly from zero.

This opposite monotonic behaviour continues until a particular moment when

$$r_f = r_r$$

This equality of the two opposing rates is the very definition of chemical equilibrium. Once equilibrium is reached, the concentrations $$[A]$$ and $$[B]$$ no longer change with time, so both $$r_f$$ and $$r_r$$ remain constant and equal thereafter. Consequently each rate curve attains a horizontal plateau, the two plateaux being super-imposed because the numerical values of the rates are identical at equilibrium.

Putting all these deductions together, the required plot of rate versus time must possess the following characteristic features:

1. Forward curve (labelled a): starts at a non-zero value on the ordinate and falls asymptotically to a constant value.
2. Reverse curve (labelled b): starts exactly from the origin on the ordinate, rises smoothly and approaches the same constant value.
3. The two curves intersect exactly once, at the equilibrium point, and then run parallel (actually coincident) afterwards because both rates have become equal and time-independent.

Among the four sketches provided in the question, graph (2) alone shows a curve a that descends from a positive initial value, a curve b that ascends from zero, the two meeting once and then remaining flat and equal. Hence graph (2) uniquely satisfies every theoretical requirement we have just derived.

Hence, the correct answer is Option 2.

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.

For chemical reactions we use 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.

We are told that the reaction proceeds with the same rate (hence the same rate constant) under two different conditions:

$$A \xrightarrow{700\,\text{K}} \text{Product} \quad\text{(uncatalysed)}$$

$$A \xrightarrow{500\,\text{K, catalyst}} \text{Product} \quad\text{(catalysed)}$$

Let the activation energies be $$E_1$$ (uncatalysed) and $$E_2$$ (catalysed). Because the rates are equal and the problem states that the pre-exponential factor $$A$$ remains unchanged, we write

$$A\,e^{-E_1/(R\;700)} \;=\; A\,e^{-E_2/(R\;500)}.$$

Cancelling the common factor $$A$$ and taking natural logarithms on both sides gives

$$-\dfrac{E_1}{R\;700} \;=\; -\dfrac{E_2}{R\;500}.$$

Multiplying through by $$-R$$ removes the negatives and the gas constant:

$$\dfrac{E_1}{700} \;=\; \dfrac{E_2}{500}.$$

So we have a simple proportion:

$$E_2 \;=\; \dfrac{500}{700}\,E_1 \;=\; \dfrac{5}{7}\,E_1.$$

The question also states that the catalyst lowers the activation energy by $$30\;\text{kJ mol}^{-1}$$, so

$$E_2 \;=\; E_1 \;-\; 30.$$

Substituting the proportional relation $$E_2 = \dfrac{5}{7}E_1$$ into the reduction statement gives

$$\dfrac{5}{7}E_1 = E_1 - 30.$$

Bringing like terms together:

$$E_1 - \dfrac{5}{7}E_1 = 30.$$

The left side simplifies because $$1 - \dfrac{5}{7} = \dfrac{2}{7}$$, so

$$\dfrac{2}{7}E_1 = 30.$$

Solving for $$E_1$$:

$$E_1 = 30 \times \dfrac{7}{2} = 105\;\text{kJ mol}^{-1}.$$

Finally, the catalysed activation energy is

$$E_2 = E_1 - 30 = 105 - 30 = 75\;\text{kJ mol}^{-1}.$$

Hence, the correct answer is Option A.

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.

For any chemical reaction we start with 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.

Taking natural logarithm on both sides, we obtain

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

This expression has the general form $$y = c + mx$$ with

$$y = \ln k,\qquad x = \frac{1}{T},\qquad m = -\frac{E_a}{R},\qquad c = \ln A.$$

Hence, a plot of $$\ln k$$ (vertical axis) against $$1/T$$ (horizontal axis) must be a straight line whose slope $$m$$ is

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

Looking at the graph provided in the question, two clearly readable points are chosen:

$$\bigl(x_1,\,y_1\bigr) = \bigl(0.5,\;3.0\bigr),\qquad \bigl(x_2,\,y_2\bigr) = \bigl(1.5,\;1.0\bigr)$$

(All numbers are taken directly from the straight line in the figure.) The slope is therefore

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1.0 \;-\; 3.0}{1.5 \;-\; 0.5} = \frac{-2.0}{1.0} = -2.$$

The numerical value of the slope is thus $$m = -2$$. Substituting this value into the relation $$m = -E_a/R$$, we have

$$-\frac{E_a}{R} = -2 \;\Longrightarrow\; E_a = 2R.$$

If desired, one may insert the value $$R = 8.314\times10^{-3}\,\text{kJ mol}^{-1}\text{K}^{-1}$$ to find

$$E_a = 2 \times 8.314 \times 10^{-3}\,\text{kJ mol}^{-1} = 1.66 \times 10^{-2}\,\text{kJ mol}^{-1},$$

but the option list already expresses the answer directly in terms of $$R$$, so we simply choose the multiple that we have obtained.

Hence, the correct answer is Option D.

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.

First let us recall the definition of activation enthalpy. For a given elementary step, the activation enthalpy is the difference in enthalpy between the reactants and the top of the energy barrier (the transition state). Symbolically we write

$$E_a = H_{\text{TS}} - H_{\text{reactants}}$$

where $$H_{\text{TS}}$$ is the enthalpy of the transition state and $$H_{\text{reactants}}$$ is the enthalpy of the reacting species at that point in the mechanism.

According to the diagram supplied in the question (enthalpy on the vertical axis and reaction progress on the horizontal axis) there are two possible pathways that start from the common reactant ensemble $$A+B$$ and end in the two alternative products $$C$$ and $$D$$. The salient numerical features of the plot can be stated in words:

• Product $$C$$ lies lower on the enthalpy scale than product $$D$$. Hence we have

$$H_C < H_D$$

so that the formation of $$C$$ is thermodynamically more favourable.

• The peak (transition state) on the pathway leading to $$D$$ is lower than the peak on the pathway leading to $$C$$. Writing the two activation enthalpies as $$E_{a,D}$$ and $$E_{a,C}$$, the diagram clearly shows

$$E_{a,D} < E_{a,C}$$

In fact, the numerical vertical difference between the two peaks is labelled as $$5\ \text{kJ mol}^{-1}$$, but the higher peak belongs to the route that forms $$C$$, not the one that forms $$D$$. Quantitatively:

$$E_{a,C}=E_{a,D}+5\ \text{kJ mol}^{-1}$$

Combining the last two statements gives us the strict inequality

$$E_{a,C} > E_{a,D}$$

Now we analyse each option with these relations in mind.

Option A. “Formation of A and B from C has highest enthalpy of activation.” To go backward from $$C$$ to $$A+B$$, the system must climb from the low level $$H_C$$ up to the higher transition state situated on the $$A+B \to C$$ curve. Since $$H_C$$ is the lowest point on the entire diagram, this backward climb is indeed the largest uphill journey shown. Hence Option A is a correct statement.

Option B. “D is kinetically stable product.” Kinetic stability (greater rate of formation) is governed by the magnitude of $$E_a$$; the smaller the activation enthalpy, the faster the product forms. Because $$E_{a,D} < E_{a,C}$$, product $$D$$ forms faster and is therefore the kinetically favoured product. Option B is therefore correct.

Option C. “C is the thermodynamically stable product.” Thermodynamic stability depends on the final enthalpy level. Since $$H_C < H_D$$, product $$C$$ is indeed more stable in a thermodynamic sense. So Option C is correct.

Option D. “Activation enthalpy to form C is 5 kJ mol$$^{-1}$$ less than that to form D.” We have already established the quantitative relation $$E_{a,C}=E_{a,D}+5\ \text{kJ mol}^{-1}$$, which is algebraically equivalent to

$$E_{a,C}-E_{a,D}=+5\ \text{kJ mol}^{-1}$$

or, rearranged,

$$E_{a,C} > E_{a,D}$$

This means the activation enthalpy for $$C$$ is larger, not smaller, by exactly $$5\ \text{kJ mol}^{-1}$$. Therefore the wording in Option D directly contradicts the numerical data of the diagram, making Option D an incorrect statement.

Since the problem asks us to identify the incorrect statement, we must select Option D.

Hence, the correct answer is Option 4.