For two reactions,the values of the pre-expon | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.For two reactions with the same pre-exponential factor $A$,the ratio of rate constants i

Vedclass · Accepted Answer

$2.16 	imes 10^4$

Vedclass · Answer

(D) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.For two reactions with the same pre-exponential factor $A$,the ratio of rate constants is $\frac{k_1}{k_2} = \frac{A e^{-E_{a1} / RT}}{A e^{-E_{a2} / RT}} = e^{(E_{a2} - E_{a1}) / RT}$.Given $\Delta E_a = E_{a2} - E_{a1} = 24.9 \, kJ \, mol^{-1} = 24900 \, J \, mol^{-1}$.$T = 300 \, K$ and $R = 8.314 \, J \, K^{-1} \, mol^{-1}$.$\frac{k_1}{k_2} = e^{24900 / (8.314 	imes 300)} = e^{24900 / 2494.2} \approx e^{9.983} \approx 21655 \approx 2.16 	imes 10^4$.

For two reactions,the values of the pre-exponential factor are the same. However,the difference between their activation energy values is $24.9 \, kJ \, mol^{-1}$. The ratio of their rate constants at $300 \, K$ is ....

Similar Questions

Which of the following graphs of $\log \, K \rightarrow 1/T$ can be used to calculate the activation energy?

The activation energy of a reaction is $94.14 \, kJ \, mol^{-1}$ and the rate constant at $310 \, K$ is $10^{-2} \, s^{-1}$. What is the value of the frequency factor $(A)$?

Who discovered collision theory? On which the collision theory is based?

An endothermic reaction with high activation energy for the forward reaction is given by the diagram:

At $27^{\circ}C$ in the presence of a catalyst,the activation energy of a reaction is lowered by $10 \ kJ \ mol^{-1}$. The logarithm ratio of $\frac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is .... (Consider that the frequency factor for both the reactions is the same)

Vedclass Test Series

Exam Paper Generator

Online Exam Module