Question

Which of the following graphs correctly represents the temperature dependence of equilibrium constant K, for an
endothermic reaction?

• A. None of these
• B.
• C.
• D.

Answer 1

The correct option is D

we know, $△G^o=-RT\ln K$
(K= equilibrium constant) and
$△G^o=△H^o-T△S^o$, so
$-RT\ln K=△H^o-T△S^o$
or
$\ln K=-\frac{△H^o}{RT}+\frac{△S^o}{R}$ -(eq.1)
The above equation relates equilibrium constant with temperature, on plotting $\left(\ln K\right)$ versus $\left(\frac{1}{T}\right)$ slope m is $m=-\frac{△H^o}{RT}$
for endothermic reaction $△H^o=+ve$ $\therefore$ slope is negative, hence option (b) is correct

Note: When we differentiate the (eq.1) equation with respect to T, we get van't Hoff equation:
$\frac{\ln K}{dT}=\frac{△H^o}{R{T}^2}$