Chemical Kinetics and Nuclear ChemistryHard
Question
At $27^{\circ}C$ in presence of a catalyst, activation energy of a reaction is lowered by $10\text{ }kJ{\text{ }mol}^{- 1}$. The logarithm ratio of $\frac{k\text{~(catalysed)~}}{k\text{~(uncatalysed)~}}$ is ....
(Consider that the frequency factor for both the reactions is same)
Options
A.17.41
B.1.741
C.3.482
D.0.1741
Solution
$\frac{K_{\text{catalyst~}}}{K_{\text{uncatalyst~}}} = e^{\frac{\Delta E_{a}}{RT}}$
$${ln\frac{K_{\text{catalyst~}}}{K_{\text{uncatalyst~}}} = \frac{\Delta E_{a}}{RT} }{log\frac{K_{\text{catalyst~}}}{K_{\text{uncatalyst~}}} = \frac{\Delta E_{a}}{2.303RT} }{= \frac{10 \times 1000}{2.303 \times 8.314 \times 300} }{log\frac{K_{\text{catalyst~}}}{K_{\text{uncatalyst~}}} = 1.741}$$
Create a free account to view solution
View Solution FreeTopic: Chemical Kinetics and Nuclear Chemistry·Practice all Chemical Kinetics and Nuclear Chemistry questions
More Chemical Kinetics and Nuclear Chemistry Questions
In a certain reaction, 10% of the reactant decomposes in one hour, 20% in two hours, 30% in three hours, and so on. The ...What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20oC to 35oC ? (R =...A → products (First order reaction).Three sets of experiment were performed for a reaction under similar experimental co...A reaction takes place in three steps having rate constants K1, K2, K3 respectively. The overall rate constant K = . If ...Match the graphical study with the order of the reactions : I : Rate II : Half life III : (a - x)-1...