For a reaction,if the graph of ln K_{eq} vers | vedclass

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(B) The van't Hoff equation is given by: $\ln K_{eq} = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{T}\right) + \frac{\Delta S^{\circ}}{R}$.Comparing th

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Exothermic

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(B) The van't Hoff equation is given by: $\ln K_{eq} = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{T}\right) + \frac{\Delta S^{\circ}}{R}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln K_{eq}$ and $x = 1/T$,the slope $m = -\frac{\Delta H^{\circ}}{R}$.From the given graph,the slope is positive (as $\ln K_{eq}$ increases with $1/T$).Since the slope is positive,$-\frac{\Delta H^{\circ}}{R} > 0$,which implies $\Delta H^{\circ} $A$ negative enthalpy change $(\Delta H^{\circ} < 0)$ indicates an exothermic reaction.

For a reaction,if the graph of $\ln K_{eq}$ versus $1/T$ is plotted as shown,the reaction must be:

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