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Let the activation energies of two reactions be $E_1$ and $E_2$ where $E_1>E_2$. If temperature of the system is increased from $T_1$ to $T_2$, the rate constant of the reactions changes from $k_1$ to $k_1'$ in the first reaction and from $k_2$ to $k_2'$ in the second reaction. Which of the following expression is correct?

(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 $\frac{k_1'}{k_1} > \frac{k_2'}{k_2}$
$\Longrightarrow \frac{k_1'}{k_1} > \frac{k_2'}{k_2}$
answered Feb 27, 2014 by mosymeow_1
edited Mar 15, 2014 by balaji.thirumalai

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