First slide

Factors influencing equilibrium constant

Question

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

Moderate

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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