Consider the reaction below at 298K,
`C("graphite")+2H_(2)(g) to CH_(4) (G)`
`triangle_(f) H^(@) ("Kj/mol") -74.9`
`S_(m)^(@) ("J/mol-K")+5.6 +130.7 +186.3`
Which statement below is correct?

Options

A. `triangle_(r)G^(@)` is -50.8 kJ and the reaction is driven by enthalpy only
B. `triangle_(r)G^(@)` is -50.8 kJ and the reaction is driven by entropy only
C. `triangle_(r)G^(@)` is +50.8 kJ and the reaction is driven by both enthalpy and entropy
D. `triangle_(r)G^(@)`is -50,8 kJ and the reaction is driven by both enthalpy and entropy

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the reaction and identify given values
The reaction is:
\[ C(\text{graphite}) + 2H_2(g) \rightarrow CH_4(g) \]

Given values:
- Standard enthalpy of formation, \( \Delta_f H^\circ = -74.9 \, \text{kJ/mol} \)
- Standard entropies:
- \( S^\circ(C) = 5.6 \, \text{J/mol-K} \)
- \( S^\circ(H_2) = 130.7 \, \text{J/mol-K} \) (for 1 mole of \( H_2 \))
- \( S^\circ(CH_4) = 186.3 \, \text{J/mol-K} \)

### Step 2: Calculate the standard entropy change (\( \Delta S^\circ \))
Using the formula for entropy change:
\[ \Delta S^\circ = S^\circ(\text{products}) - S^\circ(\text{reactants}) \]
Substituting the values:
\[ \Delta S^\circ = S^\circ(CH_4) - [S^\circ(C) + 2 \times S^\circ(H_2)] \]
\[ \Delta S^\circ = 186.3 - [5.6 + 2 \times 130.7] \]
Calculating the reactants' entropy:
\[ \Delta S^\circ = 186.3 - [5.6 + 261.4] \]
\[ \Delta S^\circ = 186.3 - 267 = -80.7 \, \text{J/mol-K} \]

### Step 3: Convert \( \Delta S^\circ \) to kJ
Since \( \Delta S^\circ \) is in J/mol-K, we convert it to kJ:
\[ \Delta S^\circ = -80.7 \, \text{J/mol-K} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = -0.0807 \, \text{kJ/mol-K} \]

### Step 4: Calculate the standard Gibbs free energy change (\( \Delta G^\circ \))
Using the Gibbs free energy equation:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Substituting the values:
\[ \Delta G^\circ = -74.9 \, \text{kJ/mol} - 298 \, \text{K} \times (-0.0807 \, \text{kJ/mol-K}) \]
Calculating the second term:
\[ 298 \times -0.0807 = -24.0736 \, \text{kJ/mol} \]
Now substituting back:
\[ \Delta G^\circ = -74.9 + 24.0736 = -50.8264 \, \text{kJ/mol} \]

### Step 5: Round off the value
Rounding off gives:
\[ \Delta G^\circ \approx -50.83 \, \text{kJ/mol} \]

### Step 6: Analyze the options
We need to determine which statement is correct based on our calculated value of \( \Delta G^\circ \).

### Conclusion
The correct statement is that the standard Gibbs free energy is approximately \( -50.8 \, \text{kJ/mol} \) and the reaction is driven by both enthalpy and entropy.

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