Pre-exponential factors of two different reactions of same order...

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 \mathrm{~kJ} \mathrm{~mol}^{-1}

. If

\mathrm{k}_{1}

and

\mathrm{k}_{2}

are the rate constants of first and second reaction respectively at 300 K, then

\ln \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}

will be

\_\_\_\_

. (nearest integer)

\left[\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]

Correct answer

Step-by-step solution

Given:

A_1 = A_2

E_{a1} = E_{a2} + 20

kJ/mol, T = 300 K, R = 8.3 J K⁻¹ mol⁻¹
Using Arrhenius equation:

k = Ae^{-E_a/RT}

\frac{k_2}{k_1} = \frac{e^{-E_{a2}/RT}}{e^{-E_{a1}/RT}} = e^{(E_{a1} - E_{a2})/RT}

E_{a1} - E_{a2} = 20

kJ/mol = 20000 J/mol

\frac{k_2}{k_1} = e^{\frac{20000}{8.3 \times 300}} = e^{\frac{20000}{2490}}

\ln\frac{k_2}{k_1} = \frac{20000}{2490} = 8.03 \approx 8

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