Question

For a given exothermic reaction $K_p$ and ${K^{{}^{\prime }}}_p$ are the equilibrium constants at temperature $T_1$ and $T_2$ respectively. Assuming that heat of reaction is constant in temperature range between 183. $T_1$ and $T_2$ it is readily observed that:

• A. $K_p={K^{{}^{\prime }}}_p$
• B. $K_p=1/{K^{{}^{\prime }}}_p$
• C. $K_p>{K^{{}^{\prime }}}_p$
• D. $K_p<{K^{{}^{\prime }}}_p$

Answer 1

Answer: (C)

$K_p>{K^{{}^{\prime }}}_p$

The equilibrium constant at two different temperatures for thermodynamic process is given by

$\log \frac{K_2}{K_1}=\frac{\Delta H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

Here $K_1$ and $K_2$ are replaced by $K_p$ and $K_p^{\prime }$

Therefore,

$\log \frac{K_p^{\prime }}{K_p}=\frac{\Delta H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

For exothermic reaction, $T_2>T_1$ and $\Delta H=-ve$

$K_p>K_p^{\prime }$

Option C is correct.