The equilibrium constants for `A_2 (g) hArr 2A(g)` at 400 k and 600 k are `1xx 10^(-8) and 1 xx 10^(-2)` respectively . The reaction is

Options

A. Exothermic
B. Endothermic
C. May be exodhermic or endothermic
D. No heat is evolved or abosorbed

Answer

To determine whether the reaction \( A_2 (g) \rightleftharpoons 2A(g) \) is exothermic or endothermic based on the given equilibrium constants at two different temperatures, we can follow these steps:

### Step 1: Identify the given data
We have the equilibrium constants at two temperatures:
- At \( T_1 = 400 \, K \), \( K_1 = 1 \times 10^{-8} \)
- At \( T_2 = 600 \, K \), \( K_2 = 1 \times 10^{-2} \)

### Step 2: Analyze the change in equilibrium constant with temperature
We observe that as the temperature increases from \( 400 \, K \) to \( 600 \, K \), the equilibrium constant \( K \) increases from \( 1 \times 10^{-8} \) to \( 1 \times 10^{-2} \). This indicates that the reaction favors the formation of products \( 2A \) at higher temperatures.

### Step 3: Apply the Van't Hoff equation
The Van't Hoff equation relates the change in the equilibrium constant with temperature to the enthalpy change (\( \Delta H \)) of the reaction:
\[ \ln \left( \frac{K_2}{K_1} \right) = \frac{\Delta H}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where \( R \) is the universal gas constant.

### Step 4: Calculate \( \ln K_2 \) and \( \ln K_1 \)
- Calculate \( \ln K_1 \):
\[ \ln K_1 = \ln(1 \times 10^{-8}) = -8 \ln(10) \approx -18.42 \]
- Calculate \( \ln K_2 \):
\[ \ln K_2 = \ln(1 \times 10^{-2}) = -2 \ln(10) \approx -4.61 \]

### Step 5: Substitute into the Van't Hoff equation
Now we can substitute these values into the equation:
\[ \ln K_2 - \ln K_1 = -4.61 - (-18.42) = 13.81 \]
Now, we need to calculate \( \frac{1}{T_1} - \frac{1}{T_2} \):
\[ \frac{1}{T_1} = \frac{1}{400} = 0.0025 \, K^{-1} \]
\[ \frac{1}{T_2} = \frac{1}{600} \approx 0.00167 \, K^{-1} \]
\[ \frac{1}{T_1} - \frac{1}{T_2} = 0.0025 - 0.00167 = 0.00083 \, K^{-1} \]

### Step 6: Rearranging for \( \Delta H \)
Now we can rearrange the Van't Hoff equation to solve for \( \Delta H \):
\[ \Delta H = R \cdot \left( \ln K_2 - \ln K_1 \right) \cdot \left( \frac{1}{T_1} - \frac{1}{T_2} \right)^{-1} \]
Substituting \( R \approx 8.314 \, J \cdot K^{-1} \cdot mol^{-1} \):
\[ \Delta H = 8.314 \cdot 13.81 \cdot \frac{1}{0.00083} \approx 8.314 \cdot 13.81 \cdot 1204.82 \approx 1.1 \times 10^5 \, J/mol \]

### Step 7: Conclusion
Since \( \Delta H \) is positive, this indicates that the reaction absorbs heat, meaning the reaction is endothermic.

### Final Answer
The reaction \( A_2 (g) \rightleftharpoons 2A(g) \) is endothermic.