Each question contains STATEMENT-1 (Assertion) and STATEMENT-2( Reason).
Examine the statements carefully and mark the correct answer according to the instruction given below:
STATEMENT-1: The equilibrium constant of the exothermic reaction at high temperature decreases.
STATEMENT-2: Since In `(K_(2))/(K_(1))=(DeltaH^(@))/(R)[(1)/(T_(1))-(1)/(T_(2))]` and for exothermic reaction ,
`DeltaH^(@)=`-ve and thereby, `(K_(2))/(K_(1))lt1`

Options

A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1
B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1
C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE
D. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Answer

To solve the given problem, we need to analyze both statements and determine their validity and relationship.

### Step-by-Step Solution:

1. **Understanding the Statements:**
- **Statement 1:** The equilibrium constant of the exothermic reaction at high temperature decreases.
- **Statement 2:** Since \( \ln \left( \frac{K_2}{K_1} \right) = \frac{\Delta H^\circ}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \) and for an exothermic reaction, \( \Delta H^\circ \) is negative, thereby \( \frac{K_2}{K_1} < 1 \).

2. **Analyzing Statement 1:**
- In an exothermic reaction, heat is released, which means the enthalpy change (\( \Delta H^\circ \)) is negative.
- According to Le Chatelier's principle, increasing the temperature of an exothermic reaction shifts the equilibrium to the left (toward the reactants), resulting in a decrease in the equilibrium constant (\( K \)).

3. **Analyzing Statement 2:**
- The Van't Hoff equation relates the change in the equilibrium constant with temperature:
\[ \ln \left( \frac{K_2}{K_1} \right) = \frac{\Delta H^\circ}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
- Here, \( K_1 \) is the equilibrium constant at temperature \( T_1 \) and \( K_2 \) is at temperature \( T_2 \). If \( T_2 > T_1 \) (high temperature), then:
- Since \( \Delta H^\circ < 0 \) (for exothermic reactions), the right side of the equation becomes negative, implying that \( \ln \left( \frac{K_2}{K_1} \right) < 0 \).
- This indicates \( K_2 < K_1 \), confirming that the equilibrium constant decreases with an increase in temperature for exothermic reactions.

4. **Conclusion:**
- Both statements are true.
- Statement 2 correctly explains Statement 1 using the Van't Hoff equation.

### Final Answer:
Both statements are true, and Statement 2 is the correct explanation for Statement 1.