At 1000K, water vapour at 1 atm has been found to be dissociated into hydrogen and oxygen to the extent of `3xx10^(-5)%`. The std free energy change of the system at standard state, assuming ideal behaviour in kJ will be

Options

A. 313.11 kJ
B. 434.88 kJ
C. 380.44 kJ
D. 1023.56 kJ

Explanation

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) for the dissociation of water vapor into hydrogen and oxygen at 1000 K, given that the extent of dissociation is \(3 \times 10^{-5}\%\).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The dissociation of water vapor can be represented by the following chemical equation:
\[ 2 H_2O(g) \rightleftharpoons 2 H_2(g) + O_2(g) \]

2. **Convert Extent of Dissociation to Alpha**:
The extent of dissociation is given as \(3 \times 10^{-5}\%\). To convert this percentage to a fraction (alpha), we divide by 100:
\[ \alpha = \frac{3 \times 10^{-5}}{100} = 3 \times 10^{-7} \]

3. **Calculate Partial Pressures at Equilibrium**:
- Initial pressure of water vapor, \(P_{H_2O} = 1 \, \text{atm}\).
- At equilibrium:
- Pressure of \(H_2\) = \(\alpha = 3 \times 10^{-7} \, \text{atm}\)
- Pressure of \(O_2\) = \(\frac{\alpha}{2} = \frac{3 \times 10^{-7}}{2} = 1.5 \times 10^{-7} \, \text{atm}\)
- Pressure of \(H_2O\) = \(1 - \alpha \approx 1 \, \text{atm}\) (since \(\alpha\) is very small)

4. **Calculate the Equilibrium Constant \(K_p\)**:
The expression for \(K_p\) for the reaction is:
\[ K_p = \frac{(P_{H_2})^2 (P_{O_2})}{(P_{H_2O})^2} \]
Substituting the values:
\[ K_p = \frac{(3 \times 10^{-7})^2 (1.5 \times 10^{-7})}{(1)^2} \]
\[ K_p = \frac{9 \times 10^{-14} \cdot 1.5 \times 10^{-7}}{1} = 1.35 \times 10^{-20} \, \text{atm}^2 \]

5. **Calculate Standard Gibbs Free Energy Change (\(ΔG°\))**:
The formula for calculating \(ΔG°\) is:
\[ ΔG° = -2.303 \cdot R \cdot T \cdot \log(K_p) \]
Where:
- \(R = 0.082 \, \text{L atm/(K mol)}\)
- \(T = 1000 \, \text{K}\)
- \(K_p = 1.35 \times 10^{-20}\)

Now substituting the values:
\[ ΔG° = -2.303 \cdot 0.082 \cdot 1000 \cdot \log(1.35 \times 10^{-20}) \]

First, calculate \(\log(1.35 \times 10^{-20})\):
\[ \log(1.35 \times 10^{-20}) \approx -20 + \log(1.35) \approx -20 + 0.130 = -19.87 \]

Now substituting this back into the equation:
\[ ΔG° = -2.303 \cdot 0.082 \cdot 1000 \cdot (-19.87) \]
\[ ΔG° \approx 2.303 \cdot 0.082 \cdot 1000 \cdot 19.87 \]
\[ ΔG° \approx 380.447 \, \text{kJ} \]

### Final Answer:
The standard free energy change of the system at standard state is approximately:
\[ ΔG° \approx 380.447 \, \text{kJ} \]