Variation of equilibrium constan K with temperature is given by van't Hoff equation
`InK=(Delta_(r)S^(@))/R-(Delta_(r)H^(@))/(RT)`
for this equation, `(Delta_(r)H^(@))` can be evaluated if equilibrium constants `K_(1)` and `K_(2)` at two temperature `T_(1)` and `T_(2)` are known.
`log(K_(2)/K_(1))=(Delta_(r)H^(@))/(2.303R)[1/T_(1)-1/T_(2)]
`The equilibrium constant `Kp` for the following reaction is `1` at `27^(@)C` and `4` at `47^(@)C.`
`A(g)hArrB(g)+C(g)`
For the reaction calculate enthalpy change for the
`B(g)+C(g)hArrA(g)`
`(Given : R=2cal//mol-K)`

Options

A. `-13.31Kcal//mol`
B. `13.31Kcal//mol`
C. `-19.2Kcal//mol`
D. `-55.63Kcal//mol`

Answer

To solve the problem regarding the enthalpy change for the reaction \( B(g) + C(g) \rightleftharpoons A(g) \) using the equilibrium constants at two different temperatures, we can follow these steps:

### Step-by-Step Solution

1. **Identify Given Data**:
- Equilibrium constant \( K_1 = 1 \) at temperature \( T_1 = 27^\circ C = 300 \, K \)
- Equilibrium constant \( K_2 = 4 \) at temperature \( T_2 = 47^\circ C = 320 \, K \)
- Gas constant \( R = 2 \, \text{cal/mol-K} \)

2. **Use the Van't Hoff Equation**:
The equation relating the equilibrium constants at two temperatures is:
\[ \log\left(\frac{K_2}{K_1}\right) = \frac{\Delta_r H^\circ}{2.303 R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

3. **Substitute Known Values**:
Substitute \( K_1 \), \( K_2 \), \( T_1 \), \( T_2 \), and \( R \) into the equation:
\[ \log\left(\frac{4}{1}\right) = \frac{\Delta_r H^\circ}{2.303 \times 2} \left(\frac{1}{300} - \frac{1}{320}\right) \]

4. **Calculate the Left Side**:
\[ \log(4) = 0.602 \]

5. **Calculate the Right Side**:
First, calculate \( \left(\frac{1}{300} - \frac{1}{320}\right) \):
\[ \frac{1}{300} - \frac{1}{320} = \frac{320 - 300}{300 \times 320} = \frac{20}{96000} = \frac{1}{4800} \]

Now substitute this into the equation:
\[ 0.602 = \frac{\Delta_r H^\circ}{2.303 \times 2} \times \frac{1}{4800} \]

6. **Rearranging the Equation**:
Rearranging to solve for \( \Delta_r H^\circ \):
\[ \Delta_r H^\circ = 0.602 \times 2.303 \times 2 \times 4800 \]

7. **Calculate \( \Delta_r H^\circ \)**:
\[ \Delta_r H^\circ = 0.602 \times 2.303 \times 2 \times 4800 \approx 13309.49 \, \text{cal/mol} \]

8. **Convert to Kilocalories**:
\[ \Delta_r H^\circ \approx 13.31 \, \text{kcal/mol} \]

9. **Determine the Enthalpy Change for the Reverse Reaction**:
Since the reaction \( A(g) \rightleftharpoons B(g) + C(g) \) is the reverse of the original reaction, the enthalpy change will be negative:
\[ \Delta_r H^\circ \text{ for } B(g) + C(g) \rightleftharpoons A(g) = -13.31 \, \text{kcal/mol} \]

### Final Answer:
The enthalpy change for the reaction \( B(g) + C(g) \rightleftharpoons A(g) \) is:
\[ \Delta_r H^\circ = -13.31 \, \text{kcal/mol} \]