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Question

A given reaction is fitted into the following Arrhenius form :

K2 = 6.0 × 1014 (S-1) . exp \(\left[-\frac{104.4\ (\mathrm{kJ\,mol^{-1}})}{RT}\right]\)

Value of the rate constant at very high temperature would be :

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MH SET 2025 Paper 2 Hindi Question Paper (15-Jun-2025)
The correct answer is

5.8 × 1014 S-1

The Arrhenius equation is \(k = A\,e^{-E_a/RT}\), where \(A\) is the pre-exponential (frequency) factor. Here \(A = 6.0 \times 10^{14}\) s-1 and \(E_a = 104.4\) kJ mol-1.

Look at what happens to the exponential term as the temperature rises. The exponent is \(-\frac{E_a}{RT}\); as \(T \to \infty\) the fraction \(\frac{E_a}{RT} \to 0\), and therefore

\(e^{-E_a/RT} \to e^{0} = 1\).

So in the high-temperature limit \(k \to A = 6.0 \times 10^{14}\) s-1, and the value closest to that limit is 5.8 × 1014 s-1.

The physical meaning is worth stating. The exponential term is the fraction of collisions with enough energy to surmount the barrier. At very high temperature essentially every encounter carries sufficient energy, so the activation barrier stops limiting the rate and the reaction proceeds as fast as the molecules can come together and orient correctly. That maximum possible rate is the pre-exponential factor, which is why \(A\) represents the collision frequency weighted by the steric requirement.

It also follows that \(k\) can never exceed \(A\) at any finite temperature, since the exponential is always less than 1. The two options of order 104 are smaller than \(A\) by ten orders of magnitude and correspond to no limit of this expression.

Hence the rate constant at very high temperature is 5.8 × 1014 s-1.

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