For the water gas reaction
`C(s)+H_(2)O(g)hArrCO(g)+H_(2)(g)`
The standard Gibb's energy of energy of reaction (at 1000K) is `-8.1KJmol^(-1)`. Value of equilibrium constant is -

Options

A. 2.6
B. 6.2
C. 8.2
D. 10

Explanation

To find the equilibrium constant (K) for the water gas reaction given the standard Gibbs energy change (ΔG°) at 1000 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between ΔG° and K**:
The relationship between the standard Gibbs energy change (ΔG°) and the equilibrium constant (K) is given by the equation:
\[ \Delta G^\circ = -2.303RT \log K \]

2. **Rearrange the equation to solve for log K**:
We can rearrange the equation to isolate log K:
\[ \log K = \frac{-\Delta G^\circ}{2.303RT} \]

3. **Substitute the known values**:
We know:
- ΔG° = -8.1 kJ/mol = -8.1 × 10³ J/mol (convert kJ to J)
- R (gas constant) = 8.314 J/(K·mol)
- T (temperature) = 1000 K

Substituting these values into the equation:
\[ \log K = \frac{-(-8.1 \times 10^3)}{2.303 \times 8.314 \times 1000} \]

4. **Calculate the denominator**:
First, calculate the denominator:
\[ 2.303 \times 8.314 \times 1000 = 19115.982 \]

5. **Calculate log K**:
Now substitute back into the equation:
\[ \log K = \frac{8.1 \times 10^3}{19115.982} \approx 0.423 \]

6. **Find K**:
To find K, we use the antilogarithm:
\[ K = 10^{0.423} \approx 2.63 \]

7. **Round the answer**:
Rounding to one decimal place, we get:
\[ K \approx 2.6 \]

### Final Answer:
The value of the equilibrium constant (K) is approximately **2.6**.