A plot of ln K against frac{1}{T} (x-axis) is | vedclass

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(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by $\Delta G^{\circ} = -RT \ln K$.Since $\Delta G

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$\frac{\Delta S^{\circ}}{R}$

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(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by $\Delta G^{\circ} = -RT \ln K$.Since $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$,we can write $-RT \ln K = \Delta H^{\circ} - T \Delta S^{\circ}$.Dividing by $-RT$,we get $\ln K = \frac{\Delta S^{\circ}}{R} - \frac{\Delta H^{\circ}}{RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln K$ and $x = \frac{1}{T}$,the slope $m = -\frac{\Delta H^{\circ}}{R}$ and the intercept $c = \frac{\Delta S^{\circ}}{R}$.

$A$ plot of $\ln K$ against $\frac{1}{T}$ ($x$-axis) is expected to be a straight line,with intercept on $Y$-axis equal to

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