Question:medium
In the $\ln K$ vs $\frac{1}{T}$ plot of a chemical process having $\Delta S^\circ > 0$ and $\Delta H^\circ < 0$, the slope is proportional to (where $K$ is equilibrium constant)}
In the $\ln K$ vs $\frac{1}{T}$ plot of a chemical process having $\Delta S^\circ > 0$ and $\Delta H^\circ < 0$, the slope is proportional to (where $K$ is equilibrium constant)}
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Slope of $\ln K$ vs $1/T$ is $-\Delta H^\circ/R$.
Updated On: May 1, 2026
- $-|\Delta H^\circ|$
- $|\Delta H^\circ|$
- $\Delta S^\circ$
- $-\Delta S^\circ$
- $\Delta G^\circ$
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The Correct Option is A
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