The `K_p` value for the equilibrium `H_(2)(g)+I_(2)(s) Leftrightarrow 2HI (g)` is 871 at `25^@C`. If the vapour pressure of iodine is `4 xx 10^(-4)` atm, calculate the equilibrium constant in of partial pressures at the same temperature for the reaction, `H_(2) (g)+I_(2) (s) Leftrightarrow 2HI(g)`.

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_p \) for the reaction \( H_2(g) + I_2(s) \leftrightarrow 2HI(g) \) given the vapor pressure of iodine and the provided \( K_p \) value.

### Step-by-step Solution:

1. **Understand the Reaction and Given Data**:
- The reaction is:
\[ H_2(g) + I_2(s) \leftrightarrow 2HI(g) \]
- The given \( K_p \) at 25°C is 871.
- The vapor pressure of iodine \( I_2 \) is \( 4 \times 10^{-4} \) atm.

2. **Write the Expression for \( K_p \)**:
- The expression for \( K_p \) for the reaction is:
\[ K_p = \frac{(P_{HI})^2}{P_{H_2} \cdot P_{I_2}} \]
- Since \( I_2 \) is a solid, its activity is 1, and we can ignore it in the equilibrium expression.

3. **Rearranging the Expression**:
- Since \( P_{I_2} \) is given as a vapor pressure, we can substitute it into the equation:
\[ K_p = \frac{(P_{HI})^2}{P_{H_2} \cdot (4 \times 10^{-4})} \]
- Rearranging gives:
\[ (P_{HI})^2 = K_p \cdot P_{H_2} \cdot (4 \times 10^{-4}) \]

4. **Substituting the Known Values**:
- We know \( K_p = 871 \):
\[ (P_{HI})^2 = 871 \cdot P_{H_2} \cdot (4 \times 10^{-4}) \]

5. **Finding the Value of \( P_{H_2} \)**:
- To find \( P_{H_2} \), we need to isolate it:
\[ P_{H_2} = \frac{(P_{HI})^2}{871 \cdot (4 \times 10^{-4})} \]

6. **Calculating the Pressure**:
- We can calculate \( P_{H_2} \) using the values:
\[ P_{H_2} = \frac{871 \cdot (4 \times 10^{-4})}{871} = 4 \times 10^{-4} \]
- Now, substituting back into the equation:
\[ (P_{HI})^2 = 871 \cdot (4 \times 10^{-4})^2 \]
- Calculate \( (4 \times 10^{-4})^2 = 16 \times 10^{-8} \):
\[ (P_{HI})^2 = 871 \cdot 16 \times 10^{-8} = 13936 \times 10^{-8} = 0.0013936 \]
- Thus, \( P_{HI} = \sqrt{0.0013936} \approx 0.0373 \) atm.

7. **Final Calculation of \( K_p \)**:
- Finally, we can find the new \( K_p \):
\[ K_p' = \frac{(P_{HI})^2}{P_{H_2} \cdot (4 \times 10^{-4})} \]
- Substituting the values:
\[ K_p' = \frac{0.0013936}{(4 \times 10^{-4}) \cdot (4 \times 10^{-4})} = \frac{0.0013936}{16 \times 10^{-8}} = 871 \]

### Conclusion:
The equilibrium constant in terms of partial pressures at the same temperature for the reaction is \( K_p' = 871 \).