A schematic plot of versus for a specific reversible gaseous reaction yields a straight line with a — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

The Van't Hoff equation in integrated form is $\ln K_{eq} = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T}\right) + \frac{\Delta S^\circ}{R}$. The slope of the plot of $\ln K_{eq}$ against $1/T$ is exactly $-\frac{\Delta H^\circ}{R}$. For the slope to be positive, $\Delta H^\circ$ must be a negative value ($\Delta H^\circ < 0$). A negative enthalpy change signifies an exothermic reaction.