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 of calculating the standard free energy change (ΔG°) for the dissociation of water vapor at 1000 K, we will follow these steps:

### Step 1: Determine the degree of dissociation (α)
The degree of dissociation is given as \(3 \times 10^{-5}\%\). To convert this percentage into a fraction, we divide by 100:

\[ \alpha = \frac{3 \times 10^{-5}}{100} = 3 \times 10^{-7} \]

### Step 2: Write the balanced chemical equation
The dissociation of water vapor can be represented by the following balanced equation:

\[ 2 \text{H}_2\text{O} (g) \rightleftharpoons 2 \text{H}_2 (g) + \text{O}_2 (g) \]

### Step 3: Calculate the initial and equilibrium concentrations
Assuming we start with 1 mole of water vapor:

- Initial moles of H2O = 1
- Moles of H2O dissociated = \(2 \times \alpha = 2 \times 3 \times 10^{-7} = 6 \times 10^{-7}\)
- At equilibrium:
- Moles of H2O = \(1 - 6 \times 10^{-7} \approx 1\) (since \(6 \times 10^{-7}\) is very small)
- Moles of H2 = \(2 \times \alpha = 6 \times 10^{-7}\)
- Moles of O2 = \(\alpha = 3 \times 10^{-7}\)

### Step 4: Calculate the partial pressures at equilibrium
Using the ideal gas law, we can express the partial pressures:

- \(P_{\text{H}_2} = 2 \times \alpha \times 1 \text{ atm} = 6 \times 10^{-7} \text{ atm}\)
- \(P_{\text{O}_2} = \alpha \times 1 \text{ atm} = 3 \times 10^{-7} \text{ atm}\)
- \(P_{\text{H}_2\text{O}} \approx 1 \text{ atm}\)

### Step 5: Calculate the equilibrium constant (Kp)
The equilibrium constant \(K_p\) for the reaction is given by:

\[ K_p = \frac{(P_{\text{H}_2})^2 (P_{\text{O}_2})}{(P_{\text{H}_2\text{O}})^2} \]

Substituting the values:

\[ K_p = \frac{(6 \times 10^{-7})^2 (3 \times 10^{-7})}{(1)^2} = \frac{36 \times 10^{-14} \times 3 \times 10^{-7}}{1} = 108 \times 10^{-21} = 1.08 \times 10^{-19} \]

### Step 6: Calculate the standard free energy change (ΔG°)
The relationship between ΔG° and Kp is given by the equation:

\[ \Delta G° = -RT \ln K_p \]

Where:
- \(R = 8.314 \, \text{J/(mol K)}\)
- \(T = 1000 \, \text{K}\)

Calculating ΔG°:

\[ \Delta G° = - (8.314 \, \text{J/(mol K)})(1000 \, \text{K}) \ln(1.08 \times 10^{-19}) \]

Calculating \(\ln(1.08 \times 10^{-19})\):

\[ \ln(1.08 \times 10^{-19}) \approx -43.2 \]

Now substituting back:

\[ \Delta G° = - (8.314)(1000)(-43.2) \approx 360,000 \, \text{J} \approx 360 \, \text{kJ} \]

### Final Step: Convert to kJ
Since we need the answer in kJ:

\[ \Delta G° \approx 360 \, \text{kJ} \]

### Conclusion
The standard free energy change of the system at standard state is approximately **360 kJ**.

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