(a) $\frac{k_1'}{k_1} = \frac{k_2'}{k_2}$

(b) $\frac{k_1'}{k_1} > \frac{k_2'}{k_2}$

(c) $\frac{k_1'}{k_1} < \frac{k_2'}{k_2}$

(d) $\frac{k_1'}{k_1} = \frac{k_2'}{k_2}$ = 0

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

(a) $\frac{k_1'}{k_1} = \frac{k_2'}{k_2}$

(b) $\frac{k_1'}{k_1} > \frac{k_2'}{k_2}$

(c) $\frac{k_1'}{k_1} < \frac{k_2'}{k_2}$

(d) $\frac{k_1'}{k_1} = \frac{k_2'}{k_2}$ = 0

0 votes

Answer: $\frac{k_1'}{k_1} > \frac{k_2'}{k_2}$

We know, log $\frac{k_2}{k_1} = \frac{E_a}{2.303 R}[\frac{T_2 - T_1}{T_1T_2}]$

i.e., if $E_a$ is more, then log $\frac{k_2}{k_1}$ is more

If $E_1> E_2$,

$\Longrightarrow$ log $\frac{k_1'}{k_1} > \frac{k_2'}{k_2}$

$\Longrightarrow \frac{k_1'}{k_1} > \frac{k_2'}{k_2}$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...