Which In K vs 1/T plot is correct for an equilibrium that shits toward reactants at higher temperatures?

Options

A.
B.
C.
D.

Answer

To solve the question regarding the plot of ln K versus 1/T for an equilibrium that shifts toward reactants at higher temperatures, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction Shift**:
- When the equilibrium shifts toward the reactants at higher temperatures, it indicates that the reaction is exothermic. According to Le Chatelier's principle, increasing the temperature will favor the endothermic direction, which in this case is the reverse reaction (toward reactants).

2. **Relating Equilibrium Constant to Temperature**:
- The relationship between the equilibrium constant (K) and temperature (T) can be described by the van 't Hoff equation:
\[ \ln K = -\frac{\Delta H^\circ}{R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{R} \]
- Here, \( \Delta H^\circ \) is the change in enthalpy, \( R \) is the gas constant, and \( \Delta S^\circ \) is the change in entropy.

3. **Determining the Sign of ΔH**:
- Since the equilibrium shifts toward the reactants at higher temperatures, we conclude that the reaction is exothermic, which means \( \Delta H^\circ < 0 \).

4. **Analyzing the Slope**:
- In the equation \( \ln K = -\frac{\Delta H^\circ}{R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{R} \), the slope (m) is given by:
\[ m = -\frac{\Delta H^\circ}{R} \]
- Since \( \Delta H^\circ \) is negative, the slope \( m \) will be positive.

5. **Graph Interpretation**:
- The plot of \( \ln K \) versus \( \frac{1}{T} \) will yield a straight line with a positive slope. This means that as \( \frac{1}{T} \) increases (which corresponds to decreasing temperature), \( \ln K \) also increases, indicating that K decreases as temperature increases, consistent with the shift toward reactants.

6. **Conclusion**:
- Therefore, the plot of \( \ln K \) versus \( \frac{1}{T} \) for an equilibrium that shifts toward reactants at higher temperatures will have a positive slope.