Van't Hoff's equation for a chemical reaction under equilibrium is given by standard reaction enthalpy at temperature T and K is the equilibrium constant . Predict how K will vary with temperature for an exothermic

Options

A. K decreases as the temperature rises
B. K is remain unchanged
C. K increases as the temperature rise
D. Information is not sufficient

Answer

To determine how the equilibrium constant \( K \) varies with temperature for an exothermic reaction, we can use Van't Hoff's equation, which relates the change in the equilibrium constant with temperature to the standard reaction enthalpy and entropy.

### Step-by-Step Solution:

1. **Understand Van't Hoff's Equation**:
The Van't Hoff equation is given by:
\[ \ln K = -\frac{\Delta H^\circ}{R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{R} \]
where:
- \( K \) is the equilibrium constant,
- \( \Delta H^\circ \) is the standard reaction enthalpy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta S^\circ \) is the standard reaction entropy.

2. **Identify the Nature of the Reaction**:
For an exothermic reaction, the standard reaction enthalpy \( \Delta H^\circ \) is negative (i.e., \( \Delta H^\circ < 0 \)).

3. **Analyze the Equation**:
Since \( \Delta H^\circ \) is negative, the term \(-\frac{\Delta H^\circ}{R}\) becomes positive. This means that the slope of the line when plotting \( \ln K \) against \( \frac{1}{T} \) is positive.

4. **Graphical Interpretation**:
- On the graph, as \( T \) increases, \( \frac{1}{T} \) decreases.
- Since \( \ln K \) is positively correlated with \( \frac{1}{T} \), an increase in \( T \) (which decreases \( \frac{1}{T} \)) will lead to a decrease in \( \ln K \).

5. **Conclude the Effect on \( K \)**:
If \( \ln K \) decreases, then \( K \) itself must also decrease. Therefore, for an exothermic reaction, as the temperature increases, the equilibrium constant \( K \) decreases.

### Final Answer:
For an exothermic reaction, the equilibrium constant \( K \) decreases as the temperature rises.