`K_(p)`has the value of `10^(-6) atm^(3) and 10^(-4)atm^(3)` at 298 K and 323 K respectiely for the reaction
`CuSO_(4).3H_(2)O(s)hArrCuSO_(4)(s)+3H_(2)O(g)`
`Delta_(r)H^(@)` for the reaction is :

Options

A. `7.7KJ//mol`
B. `-147.41KJ//mol`
C. `147.41KJ//mol`
D. none of these

Answer

To solve for the standard enthalpy change (ΔH°) for the reaction given the equilibrium constants (Kp) at two different temperatures, we can use the van 't Hoff equation:

\[ \ln \left(\frac{K_{p2}}{K_{p1}}\right) = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

### Step-by-Step Solution:

1. **Identify the given values:**
- \( K_{p1} = 10^{-6} \, \text{atm}^3 \) at \( T_1 = 298 \, \text{K} \)
- \( K_{p2} = 10^{-4} \, \text{atm}^3 \) at \( T_2 = 323 \, \text{K} \)
- Gas constant \( R = 8.314 \, \text{J/(K·mol)} \)

2. **Substitute the values into the van 't Hoff equation:**
\[ \ln \left(\frac{K_{p2}}{K_{p1}}\right) = \ln \left(\frac{10^{-4}}{10^{-6}}\right) = \ln(10^{2}) = 2 \ln(10) \]
Since \( \ln(10) \approx 2.303 \):
\[ \ln(10^2) = 2 \cdot 2.303 = 4.606 \]

3. **Calculate the temperature difference:**
\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{298} - \frac{1}{323} \]
To compute this:
\[ \frac{1}{298} \approx 0.003356 \, \text{K}^{-1} \]
\[ \frac{1}{323} \approx 0.003096 \, \text{K}^{-1} \]
\[ \frac{1}{298} - \frac{1}{323} \approx 0.003356 - 0.003096 = 0.000260 \, \text{K}^{-1} \]

4. **Rearranging the van 't Hoff equation to solve for ΔH°:**
\[ 4.606 = -\frac{\Delta H^{\circ}}{8.314} \cdot 0.000260 \]
Rearranging gives:
\[ \Delta H^{\circ} = -\frac{4.606 \cdot 8.314}{0.000260} \]

5. **Calculate ΔH°:**
\[ \Delta H^{\circ} = -\frac{38.308 \, \text{J}}{0.000260} \approx -147439.12 \, \text{J/mol} \]
Converting to kilojoules:
\[ \Delta H^{\circ} \approx -147.44 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta H^{\circ} \approx 147.44 \, \text{kJ/mol} \]