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 the heat of reaction is constant in the temperature range between $T_1$ and $T_2$, it is readily observed that: (Assume $T_2 > T_1$)

Accepted Answer

According to Le Chatelier's principle, for an exothermic reaction, an increase in temperature shifts the equilibrium in the backward direction, which decreases the value of the equilibrium constant. Since $T_2 > T_1$, the equilibrium constant at the higher temperature ($K_p'$) will be less than the equilibrium constant at the lower temperature ($K_p$). Therefore, $K_p > K_p'$. Alternatively, using the van't Hoff equation: $\log \frac{K_p'}{K_p} = \frac{\Delta H}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$ Since $T_2 > T_1$, the term $(T_2 - T_1)$ is positive. For an exothermic reaction, $\Delta H$ is negative. Thus, the entire RHS is negative, making $\log \frac{K_p'}{K_p} < 0$. This implies $\frac{K_p'}{K_p} < 1$, or $K_p > K_p'$.

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