An endothermic reaction is represented by the graph :

Options

A.
B.
C.
D.

Answer

To solve the question regarding the representation of an endothermic reaction through a graph, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Thermodynamic Equation**:
We start with the fundamental thermodynamic equation:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
where:
- \(\Delta G^\circ\) = standard free energy change
- \(\Delta H^\circ\) = standard enthalpy change
- \(T\) = temperature
- \(\Delta S^\circ\) = standard entropy change

2. **Relate \(\Delta G^\circ\) to the Equilibrium Constant**:
The standard free energy change is related to the equilibrium constant \(K_p\) as follows:
\[ \Delta G^\circ = -RT \ln K_p \]
where \(R\) is the universal gas constant.

3. **Substitute into the Thermodynamic Equation**:
By substituting the expression for \(\Delta G^\circ\) into the first equation, we get:
\[ -RT \ln K_p = \Delta H^\circ - T \Delta S^\circ \]

4. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \ln K_p = -\frac{\Delta H^\circ}{R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{R} \]
This can be recognized as the equation of a straight line \(y = mx + c\), where:
- \(y = \ln K_p\)
- \(x = \frac{1}{T}\)
- \(m = -\frac{\Delta H^\circ}{R}\) (slope)
- \(c = \frac{\Delta S^\circ}{R}\) (y-intercept)

5. **Determine the Slope for Endothermic Reactions**:
For an endothermic reaction, \(\Delta H^\circ\) is positive. Therefore, the slope \(m\) will be negative:
\[ m = -\frac{\Delta H^\circ}{R} < 0 \]

6. **Analyze the Graph Options**:
Since we are looking for a graph where \(\ln K_p\) is plotted against \(\frac{1}{T}\) with a negative slope, we need to identify which graph shows a downward trend as \(x\) (or \(\frac{1}{T}\)) increases.

7. **Select the Correct Graph**:
After analyzing the options, the graph that shows a negative slope is the correct representation of an endothermic reaction.

### Conclusion:
The graph that represents an endothermic reaction is the one with a negative slope, indicating that as the temperature increases (or \(\frac{1}{T}\) decreases), the value of \(\ln K_p\) also decreases.