JEE Chemical Kinetics Questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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: $$\;k = \dfrac{1}{t} \ln \dfrac{p_{\infty}}{2(p_{\infty} - P_t)}$$.

We need to determine the correct statements regarding the effect of temperature on the spontaneity of chemical reactions based on the Gibbs free energy relationship:

$$\Delta G = \Delta H - T\Delta S$$

For a reaction to be spontaneous, the change in Gibbs free energy must be negative ($$\Delta G < 0$$). If $$\Delta G > 0$$, the reaction is non-spontaneous.

Analysis of the Statements:

• Statement (A): $$\Delta H = (+)$$, $$\Delta S = (-)$$, at any $$T$$  $$\rightarrow$$ Non-spontaneous

  Substituting the signs into the Gibbs equation: $$\Delta G = (+) - T(-)\ = (+) + (+)\ = (+)$$. Since $$\Delta G$$ is always positive regardless of temperature, the reaction is permanently non-spontaneous. (This statement is TRUE)

• Statement (B): $$\Delta H = (+)$$, $$\Delta S = (+)$$, at low $$T$$  $$\rightarrow$$ Spontaneous

  When both are positive: $$\Delta G = (+) - T(+)$$. At low temperatures, the magnitude of $$T\Delta S$$ is small, so the positive $$\Delta H$$ dominates, making $$\Delta G > 0$$ (non-spontaneous). It only becomes spontaneous at high temperatures. (This statement is FALSE)

• Statement (C): $$\Delta H = (-)$$, $$\Delta S = (-)$$, at low $$T$$  $$\rightarrow$$ Non-spontaneous

  When both are negative: $$\Delta G = (-) - T(-)\ = (-) + T(+)$$. At low temperatures, the magnitude of $$T\Delta S$$ is small, so the negative $$\Delta H$$ term dominates, making $$\Delta G < 0$$ (spontaneous). (This statement is FALSE)

• Statement (D): $$\Delta H = (-)$$, $$\Delta S = (+)$$, at any $$T$$  $$\rightarrow$$ Spontaneous

  Substituting the signs into the equation: $$\Delta G = (-) - T(+)\ = (-) + (-)\ = (-)$$. Since $$\Delta G$$ is always negative regardless of temperature, the reaction is permanently spontaneous. (This statement is TRUE)

Conclusion:

Statements (A) and (D) correctly depict the thermodynamic criteria for reaction spontaneity.

Answer: Option C — (A) and (D) only

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 energy profile of a multi-step reaction is determined by both the activation energy of each step and the enthalpy change associated with it.

The speed of an elementary step depends on its activation energy $$(E_a)$$. A slow step has a large activation energy and therefore corresponds to the highest peak on the energy profile. A fast step has a comparatively smaller activation energy and is represented by a lower peak.

The sign of $$\Delta H$$ determines the relative energy levels of the reactants and products of each step:

• If $$\Delta H > 0$$ (endothermic), the product of that step lies at a higher energy than the reactant.
• If $$\Delta H < 0$$ (exothermic), the product of that step lies at a lower energy than the reactant.

For the given reaction:

• $$A \rightarrow B$$ is slow with $$\Delta H > 0$$.
  • The first transition state must therefore be the highest peak on the diagram.
  • Intermediate $$B$$ must lie above reactant $$A$$ in energy.
• $$B \rightarrow C$$ is fast with $$\Delta H < 0$$.
  • The second peak should be lower than the first.
  • Intermediate $$C$$ must lie below $$B$$.
• $$C \rightarrow D$$ is fast with $$\Delta H < 0$$.
  • The third peak should also be relatively small.
  • Product $$D$$ must lie below $$C$$.

The problem also states that there is a net evolution of heat, meaning the overall reaction is exothermic. Hence, the final product $$D$$ must be at a lower energy level than the initial reactant $$A$$.

The first graph satisfies all these conditions:

• The first peak is the highest, representing the slow rate-determining step.
• Intermediate $$B$$ is at a higher energy than $$A$$, consistent with $$\Delta H > 0$$.
• The second and third peaks are lower, corresponding to the two fast steps.
• The energy decreases successively from $$B$$ to $$C$$ and from $$C$$ to $$D$$, matching the negative values of $$\Delta H$$ for the last two steps.
• The final product $$D$$ lies below $$A$$, indicating an overall exothermic reaction.

The second graph is incorrect because its third peak is higher than the first peak, implying that the $$C \rightarrow D$$ step has the largest activation energy and is the slowest step, which contradicts the given information.

Hence, the correct energy profile is Option (A).

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.

To determine the correct order of activation energies, we use the relationship between the rate constant $$k$$, temperature $$T$$, and activation energy $$E_a$$ given by the Arrhenius equation.

The Arrhenius equation is

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

where

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

To relate this equation to the given graph of $$\log k$$ versus $$1/T$$, we take the logarithm of both sides.

This gives

$$\log k=\log A-\frac{E_a}{2.303R}\left(\frac{1}{T}\right).$$

This equation is of the form

$$y=mx+c,$$

where

• $$y=\log k,$$
• $$x=\dfrac{1}{T},$$
• the intercept is $$\log A,$$
• the slope is $$-\dfrac{E_a}{2.303R}.$$

The magnitude of the slope is directly proportional to the activation energy.

$$\left|\text{slope}\right|=\frac{E_a}{2.303R}.$$

Hence, a steeper line corresponds to a larger activation energy, while a flatter line corresponds to a smaller activation energy.

From the graph,

• Line 2 is the steepest, so it has the highest activation energy.
• Line 1 has intermediate steepness, so it has an intermediate activation energy.
• Line 3 is the least steep, so it has the lowest activation energy.

Therefore,

$$|slope_2|>|slope_1|>|slope_3|.$$

Hence, the correct order of activation energies is

$$E_{a2}>E_{a1}>E_{a3}.$$

Therefore, the correct answer is Option (A).

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).

By analyzing the given concentration versus time graph, we observe that:

• At $$t = 0,\text{s}$$, the initial concentration is $$[A]_0 = 50,\text{g L}^{-1}.$$
• At $$t = 10,\text{s}$$, the concentration becomes $$[A] = 25,\text{g L}^{-1}.$$

Since the concentration decreases from $$50,\text{g L}^{-1}$$ to $$25,\text{g L}^{-1}$$ in $$10,\text{s}$$, the half-life of the reaction is

$$t_{1/2} = 10,\text{s}.$$

A constant half-life is characteristic of a first-order reaction.

For a first-order reaction,

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

The relationship between the rate constant and half-life is

$$k=\frac{2.303\log 2}{t_{1/2}}.$$

Substituting this expression for $$k$$ into the integrated rate equation,

$$t=\frac{t_{1/2}}{\log 2}\log\left(\frac{[A]_0}{[A]}\right).$$

Using the given data,

• $$t_{1/2}=10,\text{s}$$
• $$[A]_0=50,\text{g L}^{-1}$$
• $$[A]=2.5,\text{g L}^{-1}$$
• $$\log 2=0.3010$$

we get

$$t=\frac{10}{0.3010}\log\left(\frac{50}{2.5}\right).$$

Since

$$\frac{50}{2.5}=20,$$

the equation becomes

$$t=\frac{10}{0.3010}\log(20).$$

Using the logarithmic identity,

$$\log(20)=\log(2\times10)=\log2+\log10=0.3010+1=1.3010,$$

we obtain

$$t=\frac{10}{0.3010}\times1.3010,$$

$$t=\frac{13.010}{0.3010}\approx43.22,\text{s}.$$

Rounding to the nearest integer,

$$\boxed{43}$$

Hence, the required time is

$$\boxed{43\ \text{s}}.$$

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$$^{-1}$$.

The graph of $$t_{1/2}$$ versus $$[A]_0$$ is a straight line passing through the origin.

$$t_{1/2}\propto [A]_0$$

This is characteristic of a zero-order reaction.

For a zero-order reaction,

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

Comparing with the equation of a straight line,

$$\text{Slope}=\frac{1}{2k}$$

Given,

$$\text{Slope}=76.92$$

Therefore,

$$\frac{1}{2k}=76.92$$

$$2k=\frac{1}{76.92}$$

$$k=\frac{1}{2\times 76.92}=\frac{1}{153.84}=0.00650\ \text{mol L}^{-1}\text{min}^{-1}$$

For a zero-order reaction,

$$[A]_t=[A]_0-kt$$

Given,

$$[A]_0=2.5\ \text{mol L}^{-1}$$

$$t=10\ \text{min}$$

Substituting,

$$[A]_{10}=2.5-(0.00650\times 10)$$

$$[A]_{10}=2.5-0.065$$

$$[A]_{10}=2.435\ \text{mol L}^{-1}$$

Expressing the concentration in the form $$x\times 10^{-3}\ \text{mol L}^{-1}$$,

$$2.435\ \text{mol L}^{-1}=2435\times 10^{-3}\ \text{mol L}^{-1}$$

Hence, the required value is

$$\boxed{2435}$$

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.

Case A: $$[CoCl(NH_3)_5]^{2+}$$

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.

Case B: $$[Co(CN)_6]^{3-}$$

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

Case C: $$[Co(NH_3)_5(H_2O)]^{3+}$$

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.

Case D: $$[Cu(H_2O)_4]^{2+}$$

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.
$$\text{}$$
No. of activated complex would be find from the high level energy state which is 3 in the case of shown graph.
$$\text{}$$
Rate determining step would be slowest step in the reaction which is the 3rd step in case of given graph.
$$\text{}$$
Hence, Option B is correct.

The substrate is triphenylmethyl chloride, $$\left(C_6H_5\right)_3CCl$$.

Due to the presence of three bulky phenyl groups around the carbon atom, backside attack by the nucleophile is highly hindered. Therefore, an $$S_N2$$ mechanism is not feasible.

When the carbon-chlorine bond breaks, the triphenylmethyl carbocation,

$$\left(C_6H_5\right)_3C^+$$

is formed.

This carbocation is highly stable because the positive charge is extensively delocalized over the three phenyl rings through resonance.

Hence, the reaction proceeds through an $$S_N1$$ mechanism.

In an $$S_N1$$ reaction, the slow and rate-determining step is the formation of the carbocation.

Therefore, the rate depends only on the concentration of the substrate and is independent of the concentration of the nucleophile.

The rate law is

$$\text{Rate}=k\left[\left(C_6H_5\right)_3CCl\right]$$

Thus, the reaction is first order with respect to the substrate and zero order with respect to $$OH^-$$.

A plot of rate versus substrate concentration must therefore be a straight line passing through the origin.

A plot of rate versus $$OH^-$$ concentration must be a horizontal line because the rate is independent of the nucleophile concentration.

If Option A shows a horizontal line against

$$\left[\left(C_6H_5\right)_3CCl\right]$$

then it cannot be correct, since that would imply zero-order dependence on the substrate.

The graph that correctly represents the rate law

$$\text{Rate}=k\left[\left(C_6H_5\right)_3CCl\right]$$

is the straight-line graph of rate versus substrate concentration.

Hence, the correct graph is Option C.

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 $$2.505$$.

Explanation

Given,

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

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

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

Converting temperatures into Kelvin,

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

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

Using the Arrhenius equation,

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

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

Therefore,

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

Cancelling $$A$$ and taking logarithms,

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

Substituting the given values,

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

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

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

For a reaction,

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

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

Using the catalysed pathway,

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

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

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

Hence, the activation energy of the catalysed backward reaction is

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

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

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

or equivalently,

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

Using two data points from the table,

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

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

Taking the ratio,

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

Substituting the values,

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

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

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

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

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

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

Equating the exponents,

$$1=n-1$$

$$n=2$$

Thus, the reaction is second order.

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

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

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

Hence, the correct answer is

$$n=2$$

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

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

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

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

Recall the zero-order kinetics formula.

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

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

Find the rate constant.

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

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

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

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

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

The correct answer is 75.

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

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

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

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

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

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

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

Comparing exponents:

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

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

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

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

Using the Arrhenius equation:

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

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

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

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

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

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

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

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

Relation between units and order.

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

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

Matching with given units:

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

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

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

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

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

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

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

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

Converting to log base 10:

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

Hence, the answer is $$2$$.

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

First order kinetics formula:

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

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

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

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

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

The balanced reaction is:

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

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

Change in concentration = 10 - 8.8 = 1.2 M

Time = 30 min = 1800 s

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

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

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

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

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

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

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

Case A:

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

Case B:

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

Case C:

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

Case D:

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

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

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

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

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

$$\text{}$$

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

Conclusion

The correct statements are B, C, and D.

The total number of correct statements is 3.

Correct Answer: 3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Number of incorrect statements = 1 (only C).

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.

For an $$\mathrm{n^{th}}$$ order reaction, the half-life relation is:

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

Thus:

$$\mathrm{\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left(\frac{P_1}{P_2}\right)^{1-n}}$$

Substituting the given values:

$$\mathrm{\frac{340}{170} = \left(\frac{55.5}{27.8}\right)^{1-n}}$$

Simplifying:

$$\mathrm{2 \approx 2^{\,1-n}}$$

For this equality to hold:

$$\mathrm{1-n = 1}$$

$$\mathrm{n = 0}$$

Thus, the reaction is:

$$\mathrm{Zero\ Order}$$

Correct Answer: $$\mathrm{0}$$

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$$^{-1}$$.

The Arrhenius equation is:

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

The graph is plotted between $$\mathrm{\ln k}$$ and $$\mathrm{\frac{10^3}{T}}$$.

Rewriting the Arrhenius equation according to the given axes:

$$\mathrm{\ln k = -\left(\frac{E_a}{10^3R}\right)\left(\frac{10^3}{T}\right) + \ln A}$$

Comparing with:

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

Slope of the graph is:

$$\mathrm{-\frac{E_a}{10^3R}}$$

Given slope:

$$\mathrm{-18.5}$$

Thus:

$$\mathrm{-18.5 = -\frac{E_a}{10^3 \times R}}$$

Substituting:

$$\mathrm{R = 8.31\ J\,K^{-1}\,mol^{-1}}$$

$$\mathrm{18.5 = \frac{E_a}{1000 \times 8.31}}$$

$$\mathrm{E_a = 18.5 \times 1000 \times 8.31}$$

$$\mathrm{E_a = 153735\ J\,mol^{-1}}$$

Converting into $$\mathrm{kJ\,mol^{-1}}$$:

$$\mathrm{E_a = 153.735\ kJ\,mol^{-1}}$$

Rounding to nearest integer:

$$\mathrm{154}$$

Correct Answer: $$\mathrm{154}$$

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.

The slope of the graph is given by:

$$\mathrm{\frac{-E_a}{R}}$$

From the graph:

$$\mathrm{Slope = \frac{0-20}{5}}$$

$$\mathrm{Slope = -4}$$

Thus:

$$\mathrm{\frac{-E_a}{R} = -4}$$

Given:

$$\mathrm{R = 2}$$

Therefore:

$$\mathrm{E_a = 2 \times 4}$$

$$\mathrm{E_a = 8}$$

Hence, the required integer is:

$$\mathrm{8}$$

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 Freundlich adsorption isotherm is given by:

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

where:

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

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

$$\mathrm{P = pressure}$$

Taking logarithm on both sides:

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

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

Comparing with:

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

gives:

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

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

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

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

Hence, the graph must be a straight line between:

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

with positive slope and positive intercept.

Graph (a) plots:

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

but gives a curve instead of a straight line.

Hence, it is incorrect.

Graph (b) plots:

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

which is not the required relation.

Hence, incorrect.

Graph (c) also plots:

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

Hence, incorrect.

Graph (d) correctly shows a straight line between:

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

with positive intercept.

Thus, graphs not obeying Freundlich adsorption isotherm are:

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

Total number of incorrect graphs:

$$\mathrm{3}$$

For a zero-order reaction:

$$\mathrm{Rate = k[A]^0 = k}$$

the rate is independent of concentration.

Hence, in graph (a), rate remains constant with time, giving a horizontal line.

For zero-order reactions, half-life is:

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

Thus:

$$\mathrm{t_{1/2} \propto [A]_0}$$

Therefore, graph (b) gives a straight line passing through the origin and represents zero-order kinetics.

For a first-order reaction:

$$\mathrm{Rate = k[A]}$$

The integrated rate equation is:

$$\mathrm{[A] = [A]_0 e^{-kt}}$$

Thus, concentration decreases exponentially with time.

Hence, graph (c) represents first-order kinetics.

For first-order reactions:

$$\mathrm{Rate \propto [A]}$$

Therefore, graph (e), which is a straight line between rate and concentration passing through the origin, also represents first-order kinetics.

Graph (d) shows constant concentration with time.

This means concentration does not decrease and effectively no reaction occurs.

Therefore:

$$\mathrm{(a)\ and\ (b)\ \rightarrow\ Zero\ Order}$$

$$\mathrm{(c)\ and\ (e)\ \rightarrow\ First\ Order}$$

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