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Question

Chemistry Question on Thermodynamics

A plot of ln K against 1T\frac{1}{T} (abscissa) is expected to be a straight line with intercept on ordinate axis equal

A

\DelatS∘2.303R\frac{\Delat S^{\circ}}{2.303 R}

B

\DelatS∘R\frac{\Delat S^{\circ}}{R}

C

−\DelatS∘R - \frac{\Delat S^{\circ}}{R}

D

R×ΔS∘R \times \Delta S^{\circ}

Answer

\DelatS∘R\frac{\Delat S^{\circ}}{R}

Explanation

Solution

RT In K=−ΔG∘=TΔS∘−ΔH∘RT \, In \, K = - \Delta G^{\circ} = T \Delta S^{\circ} - \Delta H^{\circ} In K=ΔS∘R−ΔH∘RT In \, K = \frac{\Delta S^{\circ}}{R} - \frac{\Delta H^{\circ}}{RT} Thus, a plot of ln K versus 1 T (abscissa) will be straight line with slope equal to −ΔH∘R \frac{-\Delta H^{\circ}}{R} and intercept ΔS∘R \frac{\Delta S^{\circ}}{R}