For a pseudo-first-order reaction A H + B , the reaction was conducted in two different solvents, S1 and S2. The rate constants k 1 and k 2 were found to follow: ln k = ln A − RT Ea . If E a 1

Question

For a pseudo-first-order reaction A H + ​ B , the reaction was conducted in two different solvents, S1 and S2. The rate constants k 1 and k 2 were found to follow: ln k = ln A − RT Ea ​ . If E a 1 ​ > E a 2 ​ and at T 0 ​ , k 1 ​ = k 2 ​ , then:

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Concept:

Arrhenius Equation comparison, Temperature sensitivity of reactions with different activation energies.

Solution:

The Arrhenius equation is given by $k = A e^{- E_{a} / RT}$ .

Comparing two reactions, we have:

$ln (\frac{k _{1}}{k _{2}}) = ln (\frac{A _{1}}{A _{2}}) - \frac{E _{a 1} - E _{a 2}}{RT}$

Since $k_{1} = k_{2}$ at $T_{0}$ , we have:

$0 = ln (\frac{A _{1}}{A _{2}}) - \frac{Δ E _{a}}{R T _{0}}$

This implies:

$ln (\frac{A _{1}}{A _{2}}) = \frac{Δ E _{a}}{R T _{0}}$

Substituting this back into the first equation:

$ln (\frac{k _{1}}{k _{2}}) = \frac{Δ E _{a}}{R T _{0}} - \frac{Δ E _{a}}{RT} = \frac{Δ E _{a}}{R} (\frac{1}{T _{0}} - \frac{1}{T})$

Given $Δ E_{a} = E_{a 1} - E_{a 2} > 0$ :

If $T > T_{0}$ , then $(\frac{1}{T _{0}} - \frac{1}{T}) > 0 ⟹ ln (\frac{k _{1}}{k _{2}}) > 0 ⟹ k_{1} > k_{2}$ . Therefore, (A) is correct.

If $T < T_{0}$ , then $(\frac{1}{T _{0}} - \frac{1}{T}) < 0 ⟹ ln (\frac{k _{1}}{k _{2}}) < 0 ⟹ k_{2} > k_{1}$ . Therefore, (B) is incorrect.

$E_{a}$ is a measure of temperature sensitivity (the slope of the Arrhenius plot). A larger $E_{a}$ means higher sensitivity. Therefore, (C) is correct.

The slope of $ln (\frac{k _{1}}{k _{2}})$ versus $\frac{1}{T}$ is $- \frac{Δ E _{a}}{R}$ . Since $Δ E_{a} > 0$ , the slope is negative. Therefore, (D) is incorrect.

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Doubts & Discussion

JEE Advanced 2027 (Class 12)