Question

For a reversible endothermic chemical reaction with constant heat of reaction over the operating temperature range, \(K\) is the thermodynamic equilibrium constant. Which one of the following figures shows the CORRECT dependence of \(K\) on temperature \(T\)?

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

- Use the van’t Hoff equation for equilibrium constant dependence on temperature.
- Endothermic (\(\Delta H^\circ>0\)) → negative slope in \(\ln K\) vs \(1/T\).
- Exothermic (\(\Delta H^\circ<0\)) → positive slope in \(\ln K\) vs \(1/T\).

Answer 1

The Correct Option is A

Step 1: Recall the van’t Hoff equation: \[ \ln K = -\frac{\Delta H^\circ}{R}\cdot \frac{1}{T} + \frac{\Delta S^\circ}{R}. \] Step 2: For an endothermic reaction, \(\Delta H^\circ>0\). Thus the slope of the line in the plot of \(\ln K\) vs \(1/T\) is: \[ \text{slope} = -\frac{\Delta H^\circ}{R}<0. \] Step 3: Therefore, \(\ln K\) decreases linearly as \(1/T\) increases (negative slope straight line). Step 4: This corresponds to Option (A). \[ \boxed{(A) \; \ln K \; \text{decreases linearly with } 1/T} \]