What will be `DeltaG` for the reaction at `25^(@)C` when partial pressures of reactants `H_(2), CO_(2),H_(2)O and CO` are 10, 20, 0.02 and 0.01 atm respectively. (Given : `G_(H_(2)O)^(@)=-"228.58 kJ, "G_(CO)^(@)=-"137,15 kJ, "G_(CO)^(@)=-"394.37 kJ"`.)

Options

A. `+5.61kJ`
B. `-5.61 kJ`
C. `7.09 kJ`
D. `-8.13kJ`

Explanation

To find the value of ΔG for the reaction at 25°C, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction can be represented as:
\[ \text{H}_2(g) + \text{CO}_2(g) \rightleftharpoons \text{H}_2\text{O}(g) + \text{CO}(g) \]

### Step 2: Calculate ΔG° for the reaction
Using the standard Gibbs free energy of formation values provided:
- \( G^\circ_{H_2O} = -228.58 \, \text{kJ} \)
- \( G^\circ_{CO} = -137.15 \, \text{kJ} \)
- \( G^\circ_{H_2} = 0 \, \text{kJ} \) (as it is a standard state)
- \( G^\circ_{CO_2} = -394.37 \, \text{kJ} \)

The formula for ΔG° is:
\[ \Delta G^\circ = \sum G^\circ_{\text{products}} - \sum G^\circ_{\text{reactants}} \]

Substituting the values:
\[ \Delta G^\circ = [G^\circ_{H_2O} + G^\circ_{CO}] - [G^\circ_{H_2} + G^\circ_{CO_2}] \]
\[ \Delta G^\circ = [-228.58 + (-137.15)] - [0 + (-394.37)] \]
\[ \Delta G^\circ = -365.73 + 394.37 = 28.64 \, \text{kJ} \]

### Step 3: Calculate the reaction quotient Q
The reaction quotient Q is given by:
\[ Q = \frac{P_{H_2O} \cdot P_{CO}}{P_{H_2} \cdot P_{CO_2}} \]
Substituting the given partial pressures:
- \( P_{H_2} = 10 \, \text{atm} \)
- \( P_{CO_2} = 20 \, \text{atm} \)
- \( P_{H_2O} = 0.02 \, \text{atm} \)
- \( P_{CO} = 0.01 \, \text{atm} \)

Calculating Q:
\[ Q = \frac{(0.02)(0.01)}{(10)(20)} = \frac{0.0002}{200} = 1 \times 10^{-6} \]

### Step 4: Calculate ΔG using the formula
The formula for ΔG is:
\[ \Delta G = \Delta G^\circ + RT \ln Q \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \) (or \( 0.0831 \, \text{kJ/(mol K)} \))
- \( T = 298 \, \text{K} \)

Calculating \( RT \ln Q \):
\[ RT = 0.0831 \times 298 \approx 24.79 \, \text{kJ/mol} \]
\[ \ln Q = \ln(1 \times 10^{-6}) \approx -13.8155 \]
\[ RT \ln Q \approx 24.79 \times (-13.8155) \approx -342.5 \, \text{kJ} \]

Now substituting back into the ΔG equation:
\[ \Delta G = 28.64 + (-342.5) = -313.86 \, \text{kJ} \]

### Final Answer
Thus, the value of ΔG for the reaction at 25°C is:
\[ \Delta G \approx -313.86 \, \text{kJ} \]