Calculate standard entropy change in the reaction
`Fe_(2)O_(3)(s)+3H_(2)(g) rarr 2Fe(s)+3H_(2)O(l)`
Given `: S_(m_(0))(Fe_(2)O_(3).S)=87.4,S_(m)^(@)(Fe,S)=27.3`
`S_(m)^(@)(H_(2),g)=130.7,S_(m)^(@)(H_(2)O,l)=69.9JK^(-1)mol^(-1)`

Options

A. `-212.5 JK^(-1)mol^(-1)`
B. `-215.2JK^(-1)mol^(-1)`
C. `-120.9JK^(-1)mol^(-1)`
D. None of these

Explanation

To calculate the standard entropy change (ΔS°) for the reaction:

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

we will use the formula:

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

where \( \nu_i \) is the stoichiometric coefficient and \( S_m^\circ \) is the standard molar entropy of the species involved.

### Step 1: Identify the standard molar entropies

Given:
- \( S_m^\circ(\text{Fe}_2\text{O}_3) = 87.4 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S_m^\circ(\text{Fe}) = 27.3 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S_m^\circ(\text{H}_2) = 130.7 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S_m^\circ(\text{H}_2\text{O}) = 69.9 \, \text{J K}^{-1} \text{mol}^{-1} \)

### Step 2: Calculate the entropy for products

For the products:
- For \( 2\text{Fe} \):
\[ 2 \times S_m^\circ(\text{Fe}) = 2 \times 27.3 = 54.6 \, \text{J K}^{-1} \text{mol}^{-1} \]
- For \( 3\text{H}_2\text{O} \):
\[ 3 \times S_m^\circ(\text{H}_2\text{O}) = 3 \times 69.9 = 209.7 \, \text{J K}^{-1} \text{mol}^{-1} \]

Total for products:
\[ S_m^\circ(\text{products}) = 54.6 + 209.7 = 264.3 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 3: Calculate the entropy for reactants

For the reactants:
- For \( \text{Fe}_2\text{O}_3 \):
\[ S_m^\circ(\text{Fe}_2\text{O}_3) = 87.4 \, \text{J K}^{-1} \text{mol}^{-1} \]
- For \( 3\text{H}_2 \):
\[ 3 \times S_m^\circ(\text{H}_2) = 3 \times 130.7 = 392.1 \, \text{J K}^{-1} \text{mol}^{-1} \]

Total for reactants:
\[ S_m^\circ(\text{reactants}) = 87.4 + 392.1 = 479.5 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 4: Calculate the standard entropy change

Now we can calculate the standard entropy change:
\[ \Delta S^\circ = S_m^\circ(\text{products}) - S_m^\circ(\text{reactants}) \]
\[ \Delta S^\circ = 264.3 - 479.5 = -215.2 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Final Result
The standard entropy change for the reaction is:
\[ \Delta S^\circ \approx -215.2 \, \text{J K}^{-1} \text{mol}^{-1} \]