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 $T_1$.