Chemical Kinetics and Nuclear ChemistryMedium
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
$\frac{K_{\text{catalyst~}}}{K_{\text{uncatalyst~}}} = e^{\frac{\Delta E_{a}}{RT}}$
$${ln\frac{K_{\text{catalyst~}}}{K_{\text{uncatalyst~}}} = \frac{\Delta E_{a}}{RT} }{log\frac{K_{\text{catalyst~}}}{K_{\text{uncatalyst~}}} = \frac{\Delta E_{a}}{2.303RT} }{= \frac{10 \times 1000}{2.303 \times 8.314 \times 300} }{log\frac{K_{\text{catalyst~}}}{K_{\text{uncatalyst~}}} = 1.741}$$
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