`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 reaction where nitrogen monoxide (NO) decomposes into nitrogen gas (N₂) and oxygen gas (O₂), we can use the relationship between the standard Gibbs free energy change (\( \Delta G^\circ \)) and the equilibrium constant (\( K_p \)).

### Step-by-Step Solution:

1. **Write the Reaction**:
The decomposition of NO can be represented as:
\[ 2 \text{NO} \rightarrow \text{N}_2 + \text{O}_2 \]

2. **Identify Given Values**:
We are given:
- \( \Delta G^\circ = -87 \, \text{kJ/mol} \)
- Temperature \( T = 25^\circ C = 298 \, \text{K} \)

3. **Convert \( \Delta G^\circ \) to Joules**:
Since the universal gas constant \( R \) is typically expressed in J/(mol·K), we convert \( \Delta G^\circ \) from kJ to J:
\[ \Delta G^\circ = -87 \times 1000 = -87000 \, \text{J/mol} \]

4. **Use the Gibbs Free Energy Equation**:
The relationship between \( \Delta G^\circ \) and \( K_p \) is given by:
\[ \Delta G^\circ = -RT \ln K_p \]
Rearranging this gives:
\[ \ln K_p = -\frac{\Delta G^\circ}{RT} \]

5. **Substitute Values into the Equation**:
The universal gas constant \( R \) is \( 8.314 \, \text{J/(mol·K)} \). Now we can substitute the values:
\[ \ln K_p = -\frac{-87000 \, \text{J/mol}}{(8.314 \, \text{J/(mol·K)})(298 \, \text{K})} \]

6. **Calculate the Denominator**:
Calculate \( R \times T \):
\[ R \times T = 8.314 \times 298 = 2477.572 \, \text{J/mol} \]

7. **Calculate \( \ln K_p \)**:
Now substitute this value back into the equation:
\[ \ln K_p = \frac{87000}{2477.572} \approx 35.05 \]

8. **Convert \( \ln K_p \) to \( K_p \)**:
To find \( K_p \), we exponentiate:
\[ K_p = e^{35.05} \approx 1.73 \times 10^{15} \]

### Final Answer:
The value of \( K_p \) at 25°C for the reaction is approximately:
\[ K_p \approx 1.73 \times 10^{15} \]