The activation energies of two reactions are $E_{a 1}$ and $E_{a 2}$ with $E_{a 1} > E_{a 2}$ . If the temperature of the reacting system is increased from T to T', which of the following is correct?

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$\frac{K_{1}^{'}}{K_{1}} < \frac{K_{2}^{'}}{K_{2}}$

$\frac{K_{1}^{'}}{K_{1}} < 2 \frac{K_{2}^{'}}{K_{2}}$

$\frac{K_{1}^{'}}{K_{1}} > \frac{K_{2}^{'}}{K_{2}}$

$\frac{K_{1}^{'}}{K_{1}} = \frac{K_{2}^{'}}{K_{2}}$

answer is C.

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Detailed Solution

$\ln \frac{K'}{K} = \frac{E_{a}}{R} (\frac{1}{T} - \frac{1}{T'}) = \frac{E_{a}}{R} \times \frac{ΔT}{TT'}$

$ΔT$ is positive, therefore greater the $E_{a}$ value, greater will be value of $\frac{K'}{K}$

$∴ \frac{K_{1}^{'}}{K_{1}} > \frac{K_{2}^{'}}{K_{2}}$

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