Given `Delta_(r)S^(@)=-266` and the listed `[S_(m)^(@)value]` Calculate `S^(@) "for" Fe_(3)O_(4)(s)`
`4Fe_(3)O_(4)(s)[...] +O_(2)(g)[205] to6Fe_(2)O_(3)(s)[87]`

Options

A. calculate `S^(@)` for `Fe_(3)O_(4)(s)` : `4Fe_(3)O_(4)(s)[......]+O_(2)(g)[205]rarr6Fe_(2)O_(3)(s)[87]`
B. `+111.1`
C. `+122.4`
D. `145.75`

Explanation

To calculate the standard entropy \( S^\circ \) for \( \text{Fe}_3\text{O}_4(s) \), we can use the given standard change in entropy \( \Delta_r S^\circ \) for the reaction and the standard entropies of the products and reactants involved in the reaction.

### Step-by-Step Solution:

1. **Write the reaction**:
The reaction given is:
\[ 4 \text{Fe}_3\text{O}_4(s) + \text{O}_2(g) \rightarrow 6 \text{Fe}_2\text{O}_3(s) \]

2. **Identify the known values**:
We have:
- \( \Delta_r S^\circ = -266 \, \text{J/K} \)
- \( S^\circ \) for \( \text{Fe}_2\text{O}_3(s) = 87 \, \text{J/K} \)

3. **Set up the equation for the reaction**:
The standard entropy change for the reaction can be expressed as:
\[ \Delta_r S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants}) \]
This can be written as:
\[ \Delta_r S^\circ = 6 S^\circ(\text{Fe}_2\text{O}_3) - 4 S^\circ(\text{Fe}_3\text{O}_4) - S^\circ(\text{O}_2) \]

4. **Substitute known values into the equation**:
Plugging in the values we know:
\[ -266 = 6(87) - 4 S^\circ(\text{Fe}_3\text{O}_4) - 205 \]

5. **Calculate the right side**:
Calculate \( 6(87) \):
\[ 6(87) = 522 \]
Now substitute this back into the equation:
\[ -266 = 522 - 4 S^\circ(\text{Fe}_3\text{O}_4) - 205 \]

6. **Simplify the equation**:
Combine the constants on the right side:
\[ -266 = 522 - 205 - 4 S^\circ(\text{Fe}_3\text{O}_4) \]
\[ -266 = 317 - 4 S^\circ(\text{Fe}_3\text{O}_4) \]

7. **Rearrange to solve for \( S^\circ(\text{Fe}_3\text{O}_4) \)**:
Move \( 317 \) to the left side:
\[ -266 - 317 = -4 S^\circ(\text{Fe}_3\text{O}_4) \]
\[ -583 = -4 S^\circ(\text{Fe}_3\text{O}_4) \]

8. **Divide by -4**:
\[ S^\circ(\text{Fe}_3\text{O}_4) = \frac{583}{4} = 145.75 \, \text{J/K} \]

### Final Answer:
Thus, the standard entropy \( S^\circ \) for \( \text{Fe}_3\text{O}_4(s) \) is:
\[ S^\circ(\text{Fe}_3\text{O}_4) = 145.75 \, \text{J/K} \]

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