Given
(i) `2Fe_(2)O_(3)(s) rarr 4Fe(s)+3O_(2)(g)`,
`DeltaG^(@)=+1487.0 kj mol^(-1)`
(ii) `2CI(g)+O_(2) rarr 2CO_(2)(g)`,
`Delta_(r )G^(@)=-514.4 kj mol^(-1)`
Free energy change `Delta_(r)G^(@)` for the reaction
`2Fe_(2)O_(3)(s)+6CO(g) rarr 4Fe(s)+6cO_(2)(g)` will be :

Options

A. `-1 1 2 .4 K J mol^(-1)`
B. `-5 6 . 2 K J mol^(-1)`
C. `-1 6 8 . 2 K J mol^(-1)`
D. `- 2 0 8 .0 K J mol^(-1)`

Explanation

To find the free energy change (ΔG°) for the reaction:

\[ 2Fe_{2}O_{3}(s) + 6CO(g) \rightarrow 4Fe(s) + 6CO_{2}(g) \]

we will use the provided reactions and their corresponding ΔG° values.

### Step 1: Identify the given reactions and their ΔG° values

1. **Reaction 1:**
\[ 2Fe_{2}O_{3}(s) \rightarrow 4Fe(s) + 3O_{2}(g) \quad \Delta G^{\circ}_{1} = +1487.0 \, \text{kJ/mol} \]

2. **Reaction 2:**
\[ 2CO(g) + O_{2}(g) \rightarrow 2CO_{2}(g) \quad \Delta G^{\circ}_{2} = -514.4 \, \text{kJ/mol} \]

### Step 2: Adjust Reaction 2 for the required stoichiometry

We need 6 moles of CO in the reaction. Therefore, we will multiply Reaction 2 by 3:

\[ 3 \times (2CO(g) + O_{2}(g) \rightarrow 2CO_{2}(g)) \]

This gives us:

\[ 6CO(g) + 3O_{2}(g) \rightarrow 6CO_{2}(g) \quad \Delta G^{\circ}_{2} = 3 \times (-514.4) = -1543.2 \, \text{kJ/mol} \]

### Step 3: Combine the adjusted Reaction 2 with Reaction 1

Now, we will add Reaction 1 and the adjusted Reaction 2:

1. **Reaction 1:**
\[ 2Fe_{2}O_{3}(s) \rightarrow 4Fe(s) + 3O_{2}(g) \quad \Delta G^{\circ}_{1} = +1487.0 \, \text{kJ/mol} \]

2. **Adjusted Reaction 2:**
\[ 6CO(g) + 3O_{2}(g) \rightarrow 6CO_{2}(g) \quad \Delta G^{\circ}_{2} = -1543.2 \, \text{kJ/mol} \]

When we add these reactions, the \(3O_{2}(g)\) cancels out:

\[ 2Fe_{2}O_{3}(s) + 6CO(g) \rightarrow 4Fe(s) + 6CO_{2}(g) \]

### Step 4: Calculate the overall ΔG°

Now we can calculate the overall ΔG° for the reaction:

\[ \Delta G^{\circ}_{\text{total}} = \Delta G^{\circ}_{1} + \Delta G^{\circ}_{2} \]

Substituting the values:

\[ \Delta G^{\circ}_{\text{total}} = 1487.0 \, \text{kJ/mol} - 1543.2 \, \text{kJ/mol} \]

Calculating this gives:

\[ \Delta G^{\circ}_{\text{total}} = -56.2 \, \text{kJ/mol} \]

### Final Answer

The free energy change \( \Delta G^{\circ} \) for the reaction \( 2Fe_{2}O_{3}(s) + 6CO(g) \rightarrow 4Fe(s) + 6CO_{2}(g) \) is:

\[ \Delta G^{\circ} = -56.2 \, \text{kJ/mol} \]