Which of the plots of ln K vs `(1/T)` is/are correct?

Options

A.
B.
C.
D.

Explanation

To solve the question regarding the plot of ln K versus (1/T), we will follow a systematic approach based on the thermodynamic equations provided.

### Step-by-Step Solution:

1. **Understand the Relationship**: We start with two important thermodynamic equations:
- \( \Delta G^0 = -RT \ln K \) (1)
- \( \Delta G^0 = \Delta H^0 - T \Delta S^0 \) (2)

2. **Equate the Two Equations**: Since both equations are equal to \( \Delta G^0 \), we can set them equal to each other:
\[ -RT \ln K = \Delta H^0 - T \Delta S^0 \]

3. **Rearranging the Equation**: We can rearrange this equation to isolate \( \ln K \):
\[ -RT \ln K = \Delta H^0 - T \Delta S^0 \]
Dividing through by \(-RT\):
\[ \ln K = -\frac{\Delta H^0}{RT} + \frac{\Delta S^0}{R} \]

4. **Identify the Form of the Equation**: The equation can be rewritten in the form of a straight line \( y = mx + c \):
- Here, \( y = \ln K \)
- \( m = -\frac{\Delta H^0}{R} \)
- \( x = \frac{1}{T} \)
- \( c = \frac{\Delta S^0}{R} \)

5. **Determine the Slope and Intercept**:
- The slope \( m \) is negative, indicating that as \( \frac{1}{T} \) increases, \( \ln K \) decreases.
- The intercept on the y-axis is positive if \( \Delta S^0 \) is positive.

6. **Plotting the Graph**:
- On the x-axis, we plot \( \frac{1}{T} \).
- On the y-axis, we plot \( \ln K \).
- The line will slope downwards (negative slope) and intersect the y-axis at \( \frac{\Delta S^0}{R} \).

7. **Conclusion**: Based on the analysis, the correct plot of \( \ln K \) versus \( \frac{1}{T} \) will show a downward slope, confirming that as temperature increases (or \( \frac{1}{T} \) decreases), the equilibrium constant \( K \) decreases.