Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceed 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 \(\ln \frac{k_{2}}{k_{1}}\) will be___. (nearest integer) [\(R=8.3 \text{ J K}^{-1} \text{ mol}^{-1}\)]
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Correct Answer: 8
Solution and Explanation
\[ k_1 = A e^{-E_{a1}/RT} \] \[ k_2 = A e^{-E_{a2}/RT} \] Step 2: Calculate Ratio
\[ \frac{k_2}{k_1} = \frac{A e^{-E_{a2}/RT}}{A e^{-E_{a1}/RT}} = e^{-(E_{a2} - E_{a1})/RT} = e^{(E_{a1} - E_{a2})/RT} \] Taking natural log: \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_{a1} - E_{a2}}{RT} \] Step 3: Substitute Values
Given \(E_{a1} - E_{a2} = 20 \text{ kJ/mol} = 20000 \text{ J/mol}\). \(T = 300\) K, \(R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}\). \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{20000}{8.3 \times 300} \] \[ = \frac{20000}{2490} \approx 8.032 \] Final Answer: Nearest integer is 8.
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