Using listed informations, calculate `Delta_(r)G^(@)` (in kJ/mol) at `27^(@)C`
`Co_(3)O_(4)(s)+4CO(g)rarr3Co(s)+4CO_(2)(g)`
Given : At 300 K `DeltaH_(f)^(@)(kJ//mol)-891,-110.5,0.0,-393.5`
`S^(@)(J//K-mol)102.5, 197.7, 30.0, 213.7`

Options

A. `-214.8`
B. `-195.0`
C. `-200.3`
D. -256.45

Explanation

To calculate the standard Gibbs free energy change (ΔrG°) for the reaction at 27°C (300 K), we can follow these steps:

### Step 1: Write the Reaction
The reaction given is:
\[ \text{Co}_3\text{O}_4(s) + 4\text{CO}(g) \rightarrow 3\text{Co}(s) + 4\text{CO}_2(g) \]

### Step 2: Identify ΔH°f and S° Values
We are provided with the following standard enthalpy of formation (ΔH°f) and standard entropy (S°) values:
- For Co₃O₄: ΔH°f = -891 kJ/mol, S° = 102.5 J/K·mol
- For CO: ΔH°f = -110.5 kJ/mol, S° = 197.7 J/K·mol
- For Co: ΔH°f = 0.0 kJ/mol, S° = 30.0 J/K·mol
- For CO₂: ΔH°f = -393.5 kJ/mol, S° = 213.7 J/K·mol

### Step 3: Calculate ΔH° of the Reaction
Using the formula:
\[ \Delta H°_{\text{reaction}} = \sum (\Delta H°_f \text{ of products}) - \sum (\Delta H°_f \text{ of reactants}) \]

Calculating for products:
- For 3 Co: \( 3 \times 0.0 = 0 \) kJ
- For 4 CO₂: \( 4 \times (-393.5) = -1574.0 \) kJ

Total for products = \( 0 - 1574.0 = -1574.0 \) kJ

Calculating for reactants:
- For Co₃O₄: \( -891 \) kJ
- For 4 CO: \( 4 \times (-110.5) = -442.0 \) kJ

Total for reactants = \( -891 - 442 = -1333 \) kJ

Now, substituting into the ΔH° equation:
\[ \Delta H°_{\text{reaction}} = -1574.0 - (-1333) = -241.0 \text{ kJ/mol} \]

### Step 4: Calculate ΔS° of the Reaction
Using the formula:
\[ \Delta S°_{\text{reaction}} = \sum (S° \text{ of products}) - \sum (S° \text{ of reactants}) \]

Calculating for products:
- For 3 Co: \( 3 \times 30.0 = 90.0 \) J/K·mol
- For 4 CO₂: \( 4 \times 213.7 = 854.8 \) J/K·mol

Total for products = \( 90.0 + 854.8 = 944.8 \) J/K·mol

Calculating for reactants:
- For Co₃O₄: \( 102.5 \) J/K·mol
- For 4 CO: \( 4 \times 197.7 = 790.8 \) J/K·mol

Total for reactants = \( 102.5 + 790.8 = 893.3 \) J/K·mol

Now, substituting into the ΔS° equation:
\[ \Delta S°_{\text{reaction}} = 944.8 - 893.3 = 51.5 \text{ J/K·mol} \]

### Step 5: Calculate ΔG° of the Reaction
Using the Gibbs free energy equation:
\[ \Delta G° = \Delta H° - T \Delta S° \]

Where:
- T = 300 K
- Convert ΔS° to kJ: \( 51.5 \text{ J/K·mol} = 0.0515 \text{ kJ/K·mol} \)

Now substituting the values:
\[ \Delta G° = -241.0 \text{ kJ/mol} - (300 \text{ K} \times 0.0515 \text{ kJ/K·mol}) \]
\[ \Delta G° = -241.0 - 15.45 = -256.45 \text{ kJ/mol} \]

### Final Answer
Thus, the standard Gibbs free energy change (ΔrG°) at 27°C is:
\[ \Delta rG° = -256.45 \text{ kJ/mol} \]

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