`DeltaG^(@)` for the converstion of NO to `N_(2)` and `O_(2)` is `-"87 kJ/mole"` at `25^(@)C`. The value of `K_(p)` at this temperature is

Options

A. 4.9
B. `1.8xx10^(15)`
C. `5.7xx10^(-16)`
D. 35

Answer

To find the value of \( K_p \) for the conversion of \( 2NO \) to \( N_2 + O_2 \) given that \( \Delta G^\circ \) is \(-87 \, \text{kJ/mol}\) at \( 25^\circ C \), we can follow these steps:

### Step-by-Step Solution:

1. **Convert \( \Delta G^\circ \) to Joules**:
\[ \Delta G^\circ = -87 \, \text{kJ/mol} = -87 \times 10^3 \, \text{J/mol} = -87000 \, \text{J/mol} \]

**Hint**: Remember to convert kilojoules to joules by multiplying by \( 1000 \).

2. **Use the relationship between \( \Delta G^\circ \) and \( K_c \)**:
The equation relating \( \Delta G^\circ \) to the equilibrium constant \( K_c \) is:
\[ \Delta G^\circ = -2.303 \, R \, T \, \log K_c \]
where:
- \( R = 8.314 \, \text{J/(K·mol)} \)
- \( T = 25^\circ C = 298 \, \text{K} \)

3. **Rearranging the equation to solve for \( \log K_c \)**:
\[ \log K_c = -\frac{\Delta G^\circ}{2.303 \, R \, T} \]

4. **Substituting the values into the equation**:
\[ \log K_c = -\frac{-87000}{2.303 \times 8.314 \times 298} \]

5. **Calculating the denominator**:
\[ 2.303 \times 8.314 \times 298 \approx 5730.8 \]

6. **Calculating \( \log K_c \)**:
\[ \log K_c \approx \frac{87000}{5730.8} \approx 15.19 \]

7. **Finding \( K_c \)**:
\[ K_c = 10^{15.19} \approx 1.58 \times 10^{15} \]

8. **Relating \( K_p \) and \( K_c \)**:
The relationship between \( K_p \) and \( K_c \) is given by:
\[ K_p = K_c \cdot R^{\Delta n} \cdot T^{\Delta n} \]
where \( \Delta n \) is the change in moles of gas:
- Products: \( N_2 + O_2 \) = 2 moles
- Reactants: \( 2NO \) = 2 moles
- Thus, \( \Delta n = 2 - 2 = 0 \)

9. **Calculating \( K_p \)**:
Since \( \Delta n = 0 \):
\[ K_p = K_c \cdot R^0 \cdot T^0 = K_c \]
Therefore, \( K_p \approx 1.58 \times 10^{15} \).

### Final Answer:
\[ K_p \approx 1.58 \times 10^{15} \]