Activation energies of two reactions are $E_{a}$ and $E_{a}^{'}$ with $E_{a} > E_{a}^{'}$ . If the temperature of the reacting systems is increased from $T_{1}$ to $T_{2}$ ( $k^{'}$ is rate constant at higher temperature).

\frac{k^{'}}{k_{1}} = \frac{k_{2}^{'}}{k_{2}}

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\frac{k^{'}}{k_{1}} < \frac{k_{2}^{'}}{k_{2}}

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\frac{k^{'}}{k_{1}} > \frac{k_{2}^{'}}{k_{2}}

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\frac{k^{'}}{k_{1}} > \frac{{2 k}_{2}^{'}}{k_{2}}

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Solution

The correct option is C $\frac{k^{'}}{k_{1}} > \frac{k_{2}^{'}}{k_{2}}$
For first reaction:
log $k_{1} = l o g A - E_{a} / R T_{1}$
log $k_{2} = l o g A - E_{a} / R T_{2}$
$∴ l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
Similarly, for second reaction,
log $\frac{k_{2}^{'}}{k_{1}} = \frac{E_{a}^{'}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
From equation (i) and (ii),
$k_{2} / k_{1} \propto E_{a}$ and $k_{2}^{'} / k_{1}^{'} \propto E_{a}^{'}$
Since $E_{a} > E_{a}^{'}$
$∴ \frac{k_{2}}{k_{1}} > \frac{k_{2}^{'}}{k_{1}} \Rightarrow \frac{k_{1}^{'}}{k_{1}} > \frac{k_{1}^{'}}{k_{2}^{'}}$

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