If `Delta G_(298K)` for the reaction `2H_(2(g. 1atm)) + O_(2(g. 1atm)) rarr 2H_(2)O_((g.1atm))` is `-240kJ`, what will be `Delta G_(298K)` for the reaction `H_(2)O_((g.0.2 atm)) rarr H_(2(g.4atm)) + (1)/(2) O_(2(g.0.26atm))`

Options

A. 245.7 kJ
B. 239.3 kJ
C. 125.7 kJ
D. `-125.7 kJ`

Answer

To solve the problem, we need to determine the Gibbs free energy change (ΔG) for the reaction:

\[ \text{H}_2\text{O(g, 0.2 atm)} \rightarrow \text{H}_2(g, 4 \text{ atm}) + \frac{1}{2} \text{O}_2(g, 0.26 \text{ atm}) \]

Given that the ΔG at standard conditions (1 atm) for the formation of water is -240 kJ for the reaction:

\[ 2 \text{H}_2(g, 1 \text{ atm}) + \text{O}_2(g, 1 \text{ atm}) \rightarrow 2 \text{H}_2\text{O(g, 1 atm)} \]

### Step-by-Step Solution:

**Step 1: Calculate ΔG for the formation of 1 mole of water.**
- The given ΔG for the formation of 2 moles of water is -240 kJ.
- Therefore, for 1 mole of water, ΔG is:
\[ \Delta G_{\text{formation}} = \frac{-240 \text{ kJ}}{2} = -120 \text{ kJ} \]

**Step 2: Determine ΔG for the decomposition of 1 mole of water.**
- The decomposition reaction is the reverse of the formation reaction, so:
\[ \Delta G_{\text{decomposition}} = +120 \text{ kJ} \]

**Step 3: Use the reaction quotient (Q) to adjust ΔG.**
- The reaction quotient \( Q \) for the decomposition reaction is given by:
\[ Q = \frac{P_{\text{H}_2} \times (P_{\text{O}_2})^{1/2}}{P_{\text{H}_2\text{O}}} \]
- Plugging in the pressures:
\[ Q = \frac{(4 \text{ atm}) \times (0.26 \text{ atm})^{1/2}}{0.2 \text{ atm}} = \frac{4 \times 0.51}{0.2} = \frac{2.04}{0.2} = 10.2 \]

**Step 4: Calculate the change in Gibbs free energy using the formula.**
- The formula for ΔG in terms of Q is:
\[ \Delta G = \Delta G^\circ + RT \ln Q \]
- Where \( R = 8.314 \text{ J/(mol K)} \) and \( T = 298 \text{ K} \).
- Convert ΔG° to Joules:
\[ \Delta G^\circ = 120 \text{ kJ} = 120,000 \text{ J} \]
- Now calculate \( RT \ln Q \):
\[ RT = 8.314 \times 298 = 2477.572 \text{ J} \]
\[ \ln Q = \ln(10.2) \approx 2.31 \]
\[ RT \ln Q = 2477.572 \times 2.31 \approx 5711.5 \text{ J} \]

**Step 5: Combine the values to find ΔG.**
- Now substitute back into the ΔG equation:
\[ \Delta G = 120,000 \text{ J} + 5711.5 \text{ J} = 125,711.5 \text{ J} \approx 125.7 \text{ kJ} \]

### Final Answer:
\[ \Delta G_{298K} \approx 125.7 \text{ kJ} \]