The schematic plot of In $K_{\mathrm{eq}}$ versus inverse of temperature for a reaction is shown in the figure. The reaction must be

 

The variation of standard equilibrium constant with temperature is given by

$\frac{\mathrm{dln}{K}_{\mathrm{eq}}°}{\mathrm{dT}}=\frac{\Delta H°}{R{T}^2}$ which on integeration gives $\ln K_{\mathrm{eq}}°=-\frac{\Delta H°}{RT}+$ constant

The slope of the graph between $\ln K_{\mathrm{eq}}°$ and $1/T$ shown in the figure is positive. The slope as given by the above equation is $-\Delta H/R$ This means $∆\mathrm{H}$ is negative, hence, the reaction is exothermic.