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 $K_p > K_p'$.