`K_(C) " for " H_(2) (g)+ 1//2 O_(2)(g) rArr H_(2) `O (l)at 500 K is `2.4 xx 10^(47), K_(p) `of the reaction is .

Options

A. High P
B. High T
C. Presence of Pt
D. Addition of He (g)

Answer

To find the equilibrium constant \( K_p \) for the reaction \( H_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons H_2O(l) \) at 500 K, we can use the relationship between \( K_c \) and \( K_p \).

### Step-by-Step Solution:

1. **Identify the given data**:
- The equilibrium constant \( K_c \) is given as \( 2.4 \times 10^{47} \).
- The temperature \( T \) is \( 500 \, K \).
- The universal gas constant \( R \) is \( 8.314 \, \text{J/(K·mol)} \).

2. **Write the relationship between \( K_p \) and \( K_c \)**:
\[ K_p = K_c \cdot R T^{\Delta n} \]
where \( \Delta n \) is the change in the number of moles of gas during the reaction.

3. **Calculate \( \Delta n \)**:
- Count the moles of gaseous products and reactants:
- Products: \( H_2O(l) \) (liquid, not counted in \( \Delta n \)).
- Reactants: \( H_2(g) \) and \( \frac{1}{2} O_2(g) \).
- Total moles of gaseous reactants = \( 1 + \frac{1}{2} = 1.5 \).
- Total moles of gaseous products = \( 0 \) (since \( H_2O \) is liquid).
- Therefore, \( \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 0 - 1.5 = -1.5 \).

4. **Substitute the values into the equation**:
\[ K_p = K_c \cdot R T^{\Delta n} \]
\[ K_p = (2.4 \times 10^{47}) \cdot (8.314) \cdot (500)^{-1.5} \]

5. **Calculate \( (500)^{-1.5} \)**:
\[ (500)^{-1.5} = \frac{1}{500^{1.5}} = \frac{1}{500 \cdot \sqrt{500}} \approx \frac{1}{11180.34} \approx 8.94 \times 10^{-5} \]

6. **Calculate \( K_p \)**:
\[ K_p = (2.4 \times 10^{47}) \cdot (8.314) \cdot (8.94 \times 10^{-5}) \]
\[ K_p \approx (2.4 \times 10^{47}) \cdot (8.314 \times 8.94 \times 10^{-5}) \approx (2.4 \times 10^{47}) \cdot (0.000207) \approx 9.54 \times 10^{40} \]

7. **Final Result**:
\[ K_p \approx 9.54 \times 10^{40} \]