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Q.

For a given exothermic reaction, $K_{p}$ and $K_{p}^{'}$ are the equilibrium constants at temperatures $T_{1} and T_{2},$ respectively. Assuming that heat of reaction is constant in temperature range between $T_{1} a n d T_{2}$ , it is readily observed that

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a

$K_{p} > K_{p}^{'}$

b

$K_{p} < K_{p}^{'}$

c

$K_{p} = K_{p}^{'}$

d

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

answer is A.

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Detailed Solution

$\log \frac{K_{p}^{'}}{K_{p}} = - \frac{Δ H}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

$For exothermic reaction, Δ H = - ve i.e. heat is evolved. The$ $temperature T_{2} is higher than T_{1} .$

$Thus, (\frac{1}{T_{2}} - \frac{1}{T_{1}}) is negative.$

$so, \log K_{p}^{'} - \log K_{p} = - ve or \log K_{p} > \log K_{p}^{'}$

$or K_{p} > K_{p}^{'}$

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For a given exothermic reaction, Kp and Kp' are the equilibrium constants at temperatures T1 and T2,respectively. Assuming that heat of reaction is constant in temperature range between T1 and T2, it is readily observed that