If `DeltaG = -177`K cal for `" "(1)2Fe(s)+(3)/(2)O_(2)(g) rarr Fe_(2)O_(3)(s)`
and `DeltaG =- 19` K cal for `" "(2)4Fe_(2)O_(3)(s)+ Fe(s) rarr 3Fe_(3)O_(4)(s)`
What is the Gibbs free energy of formation fo `Fe_(3)O_(4)(s)`?

Options

A. `+229.6` Kcal/mol
B. `-242.3` Kcal/mol
C. `-727` Kcal/mol
D. `-229.6` Kcal/mol

Explanation

To find the Gibbs free energy of formation for \( \text{Fe}_3\text{O}_4(s) \), we will use the provided reactions and their corresponding Gibbs free energy changes.

### Step 1: Write down the given reactions and their Gibbs free energy changes.

1. \( 2 \text{Fe}(s) + \frac{3}{2} \text{O}_2(g) \rightarrow \text{Fe}_2\text{O}_3(s) \)
\( \Delta G_1 = -177 \, \text{kcal} \)

2. \( 4 \text{Fe}_2\text{O}_3(s) + \text{Fe}(s) \rightarrow 3 \text{Fe}_3\text{O}_4(s) \)
\( \Delta G_2 = -19 \, \text{kcal} \)

### Step 2: Adjust the first reaction to match the stoichiometry needed for the second reaction.

Since the second reaction involves \( \text{Fe}_2\text{O}_3 \) in a quantity of 4, we will multiply the first reaction by 2 to match the stoichiometry:

\[ 4 \text{Fe}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{Fe}_2\text{O}_3(s) \]
The Gibbs free energy change for this adjusted reaction will be:
\[ \Delta G_1' = 2 \times (-177 \, \text{kcal}) = -354 \, \text{kcal} \]

### Step 3: Add the adjusted first reaction to the second reaction.

Now, we add the adjusted first reaction to the second reaction:

\[ 4 \text{Fe}(s) + 3 \text{O}_2(g) + 4 \text{Fe}_2\text{O}_3(s) + \text{Fe}(s) \rightarrow 2 \text{Fe}_2\text{O}_3(s) + 3 \text{Fe}_3\text{O}_4(s) \]

When we combine these reactions, the \( 2 \text{Fe}_2\text{O}_3(s) \) cancels out:

\[ 5 \text{Fe}(s) + 3 \text{O}_2(g) \rightarrow 3 \text{Fe}_3\text{O}_4(s) \]

### Step 4: Calculate the total Gibbs free energy change for the formation of \( \text{Fe}_3\text{O}_4 \).

The total Gibbs free energy change for the formation of \( 3 \text{Fe}_3\text{O}_4(s) \) is:

\[ \Delta G_{\text{total}} = \Delta G_1' + \Delta G_2 = -354 \, \text{kcal} + (-19 \, \text{kcal}) = -373 \, \text{kcal} \]

### Step 5: Calculate the Gibbs free energy of formation for 1 mole of \( \text{Fe}_3\text{O}_4 \).

Since the total Gibbs free energy change calculated is for the formation of \( 3 \text{Fe}_3\text{O}_4(s) \), we need to divide by 3 to find the Gibbs free energy of formation for 1 mole:

\[ \Delta G_f (\text{Fe}_3\text{O}_4) = \frac{-373 \, \text{kcal}}{3} = -124.33 \, \text{kcal} \]

### Final Answer

The Gibbs free energy of formation for \( \text{Fe}_3\text{O}_4(s) \) is approximately \( -124.33 \, \text{kcal} \).