For a given exothermic reaction, $K_p$ and $K'_p$ are the equilibrium constants at temperatures $T_1$ and $T_2

Question

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 a temperature range between $T_1$ and $T_2$, it is readily observed that (assuming $T_2 > T_1$):

Accepted Answer

According to the van't Hoff equation: $\log \frac{K'_p}{K_p} = \frac{\Delta H}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$ For an exothermic reaction, $\Delta H$ is negative ( $\Delta H < 0$ ). Assuming $T_2 > T_1$ , the term $\left( \frac{1}{T_1} - \frac{1}{T_2} \right)$ is positive. Therefore, the right side of the equation becomes negative, meaning $\log \frac{K'_p}{K_p} < 0$ . This implies $\frac{K'_p}{K_p} < 1$ , or $K_p > K'_p$ .

Alternatively, according to Le Chatelier's principle, an increase in temperature for an exothermic reaction shifts the equilibrium in the backward direction to counteract the added heat. This leads to a decrease in the value of the equilibrium constant. Hence, the equilibrium constant $K'_p$ at the higher temperature $T_2$ will be less than $K_p$ at the lower temperature $T_1$ .

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