For the reaction
`H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l),DeltaH=-285.8 kJ mol^(-1)`
`DeltaS=-0.163 kJ mol^(-1) K^(-1)`. What is the value of free energy change at `27^(@)C` for the reaction

Options

A. `-236.9 kJ mol^(-1)`
B. `-281.4 kJ mol^(-1)`
C. `-334.7 kJ mol^(-1)`
D. `+334.7 kJ mol^(-1)`

Explanation

To find the free energy change (ΔG) for the reaction at 27°C, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

### Step 1: Convert the temperature to Kelvin
First, we need to convert the temperature from Celsius to Kelvin. The conversion formula is:

\[ T(K) = T(°C) + 273.15 \]

For 27°C:

\[ T = 27 + 273.15 = 300.15 \, K \]

### Step 2: Substitute the values into the Gibbs free energy equation
We have the following values:
- ΔH = -285.8 kJ/mol
- ΔS = -0.163 kJ/(mol·K)
- T = 300.15 K

Now we can substitute these values into the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

\[ \Delta G = -285.8 \, \text{kJ/mol} - (300.15 \, K \times -0.163 \, \text{kJ/(mol·K)}) \]

### Step 3: Calculate TΔS
Now we calculate TΔS:

\[ T \Delta S = 300.15 \, K \times -0.163 \, \text{kJ/(mol·K)} = -48.96 \, \text{kJ/mol} \]

### Step 4: Substitute TΔS back into the equation
Now we can substitute TΔS back into the equation for ΔG:

\[ \Delta G = -285.8 \, \text{kJ/mol} - (-48.96 \, \text{kJ/mol}) \]

\[ \Delta G = -285.8 \, \text{kJ/mol} + 48.96 \, \text{kJ/mol} \]

### Step 5: Final calculation
Now we perform the final calculation:

\[ \Delta G = -236.84 \, \text{kJ/mol} \]

### Conclusion
Thus, the value of free energy change (ΔG) at 27°C for the reaction is approximately:

\[ \Delta G \approx -236.84 \, \text{kJ/mol} \]