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 temperature range between $T_{1}$ and $T_{2}$ , it is readily observed that
$\left(T_{2} > T_{1}\right)$

• A. $K_{P}=\frac{1}{K_{P}^{'}}$
• B. $K_{P} < K_{P}^{'}$
• C. $K_{P}=K_{P}^{'}$
• D. $K_{P}>K_{P}^{'}$

Answer 1

Answer: (D)

$log \frac{k_{p}}{k p^{'}} = \frac{\Delta H}{2.303 R} \left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right)$
for exothermic reaction
$\Delta H=-ve$
$T_{2}>T_{1}$
In exothermic reactions on increasing temperature value of equilibrium decreases.
So, $K_{p}>K_{p}^{'}$