Question

If for a particular reversible reaction at:
$K_C=57$ at ${355}^{\circ }C$ and $K_C=69$ at ${450}^{\circ }C$ then:

• A. $\Delta H<0$
• B. $\Delta H>0$
• C. $\Delta H=0$
• D. $\Delta H$ whose sign can't be determined

Answer 1

The correct option is B $\Delta H>0$

With increase in temperature, $K_C$ increases, this means the forward reaction increases with increase in temperature. Hence, forward reaction is endothermic.

Using equation
$\log \frac{K_2}{K_1}=\frac{\Delta H^{\circ }}{2.303R}(\frac{1}{T_1}-\frac{1}{T_2})\log \frac{69}{57}=\frac{\Delta H^{\circ }}{2.303R}(\frac{1}{628}-\frac{1}{723})$
From here, you will get
$\Delta H^{\circ }=+ve$