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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 temperature range between $T_1$ and $T_2$, it is readily observed that:
A
$K_p > K_p'$
B
$K_p < K_p'$
C
$K_p = K_p'$
D
$K_p = \frac{1}{K_p'}$
Detailed Solution
For an exothermic reaction, the value of $K_p$ decreases on increasing temperature.
So $K_p > K_p'$ (with $T_2 > T_1$).
So $K_p > K_p'$ (with $T_2 > T_1$).