Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second rea

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Solution

Q.
Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 kJ ${mol}^{- 1}$ . If $k ₁ and k ₂$ are the rate constants of first and second reaction respectively at 300 K, then ln $(k ₂ / k ₁)$ will be ________.

(nearest integer) $[R = 8.3 J K⁻¹ mol⁻¹]$ [2026]

Ans.
(8)

$A \overset{R^{x_{n} (1)}}{\to} product E_{1}$

$B \overset{R^{x_{n} (2)}}{\to} product E_{2}$

$Assuming 'A' same for both reaction.$

$\ln k_{1} = \ln A - \frac{E_{1}}{300 R}$

$\ln k_{2} = \ln A - \frac{E_{2}}{300 R}$

$\ln (\frac{k_{2}}{k_{1}}) = \frac{E_{1} - E_{2}}{300 R} = \frac{20 \times 1000}{300 R}$

$= 8.032$

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