Chemistry - Chemical Equilibrium Question with Solution | TestHub

The reaction shifts towards the reactant side when temperature increases, so the forward reaction is exothermic ($\Delta H<0$).

According to the Van't Hoff equation:

$$\ln K=-\frac{\Delta H}{R}\left(\frac{1}{T}\right)+C$$

For an exothermic reaction, $\Delta H$ is negative, hence the slope $-\frac{\Delta H}{R}$ is positive.

Therefore, a plot of $\ln K$ versus $1/T$ must be a straight line with a positive slope.

Among the given graphs, Option (C) shows a positive slope.