The free energy of formation of NO is `78 kJ mol^(-1)` at the temperature of an authomobile engine `(1000 K)`. What is the equilibrium constant for this reaction at `1000 K`?
`1/2 N_(2)(g)+1/2 O_(2)(g) hArr NO(g)`

Options

A. `8.4xx10^(-5)`
B. `7.1xx10^(-9)`
C. `4.2xx10^(-10)`
D. `1.7xx10^(-19)`

Explanation

To find the equilibrium constant (K) for the reaction at 1000 K, we can use the relationship between the Gibbs free energy change (ΔG°) and the equilibrium constant (K). The formula is:

\[ \Delta G° = -RT \ln K \]

Where:
- ΔG° is the standard Gibbs free energy change (in joules per mole),
- R is the universal gas constant (8.314 J/(mol·K)),
- T is the temperature in Kelvin,
- K is the equilibrium constant.

### Step-by-Step Solution:

1. **Convert ΔG° from kJ/mol to J/mol**:
Given that ΔG° for the formation of NO is 78 kJ/mol, we convert this to joules:
\[ \Delta G° = 78 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 78000 \, \text{J/mol} \]

2. **Substitute values into the equation**:
We have ΔG° = 78000 J/mol, R = 8.314 J/(mol·K), and T = 1000 K. Plugging these values into the equation:
\[ 78000 = - (8.314) \times (1000) \ln K \]

3. **Calculate the right-hand side**:
Calculate \( - (8.314) \times (1000) \):
\[ - (8.314 \times 1000) = -8314 \]
So the equation becomes:
\[ 78000 = -8314 \ln K \]

4. **Rearrange to solve for ln K**:
Divide both sides by -8314:
\[ \ln K = \frac{-78000}{8314} \]

5. **Calculate the value of ln K**:
\[ \ln K \approx -9.39 \]

6. **Exponentiate to find K**:
To find K, we exponentiate both sides:
\[ K = e^{-9.39} \]
Using a calculator, we find:
\[ K \approx 8.4 \times 10^{-5} \]

### Final Answer:
The equilibrium constant \( K \) for the reaction at 1000 K is approximately \( 8.4 \times 10^{-5} \).

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