A schematic plot of ln K_{eq} versus inverse | vedclass

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

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endothermic

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

$A$ schematic plot of $\ln K_{eq}$ versus inverse of temperature $(1/T)$ for a reaction is shown below. The reaction must be

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