Question

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)}

• A. $-|\Delta H^\circ|$
• B. $|\Delta H^\circ|$
• C. $\Delta S^\circ$
• D. $-\Delta S^\circ$
• E. $\Delta G^\circ$

Hint:

Slope of $\ln K$ vs $1/T$ is $-\Delta H^\circ/R$.

Answer 1

The Correct Option is A

Concept:
Van’t Hoff equation: \[ \ln K = -\frac{\Delta H^\circ}{R}\cdot \frac{1}{T} + \frac{\Delta S^\circ}{R} \]

Step 1: Slope identification.
\[ \text{slope} = -\frac{\Delta H^\circ}{R} \]

Step 2: Given condition.
$\Delta H^\circ < 0$ so slope is positive. \[ \text{slope} \propto -\Delta H^\circ = |\Delta H^\circ| \]