Calculate the standard free energy change for the formation of methane at `300K`:
`C("graphite") +2H_(2) (g) rarr CH_(4)(g)`
The following data are given:
`Delta_(f)H^(Theta) (kJ mol^(-1)): CH_(4)(g) =- 74.81`
`Delta_(f)S^(Theta)(JK^(-1) mol^(-1)): C("graphite") = 5.70, H_(2)(g) = 130.7 CH_(4)(g) = 186.3`

Options

A. `Delta_(r)G^(@)` is -50.8 kJ and the reactionis driven by enthalpy only
B. `Delta_(r)G^(@)` is`-50.8 kJ` and the reaction is driven by entropy only
C. `Delta_(r)G^(@)` is `+5.0 kJ` and the reaction is driven by enthalpy and entropy
D. `Delta_(r)G^(@)` is `-50.8 kJ` and the reaction is driven by enthalpy and entropy

Explanation

To calculate the standard free energy change (ΔG°) for the formation of methane (CH₄) at 300 K from graphite (C) and hydrogen gas (H₂), we will use the following thermodynamic relationship:

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

### Step 1: Calculate ΔH° (Standard Enthalpy Change)

The standard enthalpy change for the reaction can be calculated using the formula:

\[ \Delta H^\circ = \Delta H_f^\circ (\text{products}) - \Delta H_f^\circ (\text{reactants}) \]

For the reaction:

\[ C(\text{graphite}) + 2H_2(g) \rightarrow CH_4(g) \]

We have:

- ΔH_f°(CH₄) = -74.81 kJ/mol
- ΔH_f°(C) = 0 kJ/mol (standard state)
- ΔH_f°(H₂) = 0 kJ/mol (standard state)

Thus, the calculation becomes:

\[ \Delta H^\circ = (-74.81) - (0 + 0) = -74.81 \text{ kJ/mol} \]

### Step 2: Calculate ΔS° (Standard Entropy Change)

Next, we calculate the standard entropy change using the formula:

\[ \Delta S^\circ = S^\circ (\text{products}) - S^\circ (\text{reactants}) \]

Given the entropy values:

- S°(CH₄) = 186.3 J/(K·mol)
- S°(C) = 5.70 J/(K·mol)
- S°(H₂) = 130.7 J/(K·mol)

The calculation becomes:

\[ \Delta S^\circ = 186.3 - (5.70 + 2 \times 130.7) \]

Calculating the reactants' entropy:

\[ \Delta S^\circ = 186.3 - (5.70 + 261.4) = 186.3 - 267.1 = -80.8 \text{ J/(K·mol)} \]

To convert this to kJ/(K·mol):

\[ \Delta S^\circ = -80.8 \times 10^{-3} \text{ kJ/(K·mol)} \]

### Step 3: Calculate ΔG°

Now we can substitute ΔH° and ΔS° into the Gibbs free energy equation:

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

Substituting the values:

\[ \Delta G^\circ = -74.81 \text{ kJ/mol} - (300 \text{ K}) \times (-80.8 \times 10^{-3} \text{ kJ/(K·mol)}) \]

Calculating the second term:

\[ \Delta G^\circ = -74.81 + 24.24 = -50.57 \text{ kJ/mol} \]

### Final Answer

Thus, the standard free energy change for the formation of methane at 300 K is:

\[ \Delta G^\circ \approx -50.57 \text{ kJ/mol} \]