For the water gas reaction `C(s) + H_(2)O(g) hArr CO(g) + H_(2)(g)`
At `1000K` , the standard Gibbs free energy change of the reaction is `-8.314KJ//mol` . Therefore, at `1000K` the equilibrium constant of the above water gas reaction is

Options

A. 1
B. `10`
C. `(1)/(e)`
D. `2.718`

Explanation

To find the equilibrium constant (K) for the water gas reaction at 1000 K, we can use the relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (K). The equation that relates these quantities is:

\[ \Delta G° = -RT \ln K \]

Where:
- \(\Delta G°\) is the standard Gibbs free energy change (in joules),
- \(R\) 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° to Joules:**
Given that \(\Delta G° = -8.314 \, \text{kJ/mol}\), we need to convert this to joules:
\[ \Delta G° = -8.314 \times 10^3 \, \text{J/mol} = -8314 \, \text{J/mol} \]

2. **Identify the values of R and T:**
- \(R = 8.314 \, \text{J/mol·K}\)
- \(T = 1000 \, \text{K}\)

3. **Substitute the values into the equation:**
Rearranging the Gibbs free energy equation to solve for \(K\):
\[ \ln K = -\frac{\Delta G°}{RT} \]
Substituting the known values:
\[ \ln K = -\frac{-8314}{8.314 \times 1000} \]

4. **Calculate the right-hand side:**
\[ \ln K = \frac{8314}{8314} = 1 \]

5. **Exponentiate to find K:**
To find \(K\), we take the exponential of both sides:
\[ K = e^{\ln K} = e^{1} \approx 2.718 \]

6. **Round the result:**
Rounding this value gives us:
\[ K \approx 2.70 \]

### Final Answer:
The equilibrium constant \(K\) for the water gas reaction at 1000 K is approximately **2.70**.

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