(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

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(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

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

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

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