From the given table answer the following question:
`{:(,CO(g),CO_(2)(g),H_(2)O(g),H_(2)(g)),(DeltaH_(298)^(@)(-"KCal"//"mole"),-26.42,-94.05,-57.8,0),(DeltaH_(298)^(@)(-"KCal"//"mole"),-32.79,-94.24,-54.64,0),(S_(298)^(@)(-"Cal"//"k mol"),47.3,51.1,?,31.2):}`
Reaction:`H_(2)O(g) + CO(g) hArr H_(2)(g)+CO_(2)(g)`
Calculate `S_(298)^(@) [H_(2)O(g)]`

Options

A. `-119.47` Cal/k mole
B. `+119.47` Cal/k mole
C. `-45.13` Cal/k mole
D. `+45.13` Cal/k mole

Explanation

To solve the problem, we need to calculate the standard entropy \( S_{298}^\circ \) for \( H_2O(g) \) using the provided data and the reaction:

\[ H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g) \]

### Step 1: Write the reaction and identify the components
The reaction is:
\[ H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g) \]

### Step 2: Calculate \( \Delta H^\circ \) for the reaction
Using the standard enthalpy values from the table:
- \( \Delta H^\circ \) for \( H_2O(g) = -57.8 \, \text{kcal/mole} \)
- \( \Delta H^\circ \) for \( CO(g) = -26.42 \, \text{kcal/mole} \)
- \( \Delta H^\circ \) for \( H_2(g) = 0 \, \text{kcal/mole} \)
- \( \Delta H^\circ \) for \( CO_2(g) = -94.05 \, \text{kcal/mole} \)

Using the formula:
\[ \Delta H^\circ = \sum H^\circ_{\text{products}} - \sum H^\circ_{\text{reactants}} \]

Calculating:
\[ \Delta H^\circ = [0 + (-94.05)] - [(-57.8) + (-26.42)] \]
\[ \Delta H^\circ = -94.05 + 57.8 + 26.42 = -10.83 \, \text{kcal/mole} \]

### Step 3: Calculate \( \Delta G^\circ \) for the reaction
Using the standard Gibbs free energy values from the table:
- \( G^\circ \) for \( H_2O(g) = -54.64 \, \text{kcal/mole} \)
- \( G^\circ \) for \( CO(g) = -32.79 \, \text{kcal/mole} \)
- \( G^\circ \) for \( H_2(g) = 0 \, \text{kcal/mole} \)
- \( G^\circ \) for \( CO_2(g) = -94.24 \, \text{kcal/mole} \)

Calculating:
\[ \Delta G^\circ = [0 + (-94.24)] - [(-54.64) + (-32.79)] \]
\[ \Delta G^\circ = -94.24 + 54.64 + 32.79 = -6.81 \, \text{kcal/mole} \]

### Step 4: Use the Gibbs free energy equation to find \( \Delta S^\circ \)
Using the relationship:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Rearranging gives:
\[ \Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T} \]
Substituting the values:
\[ \Delta S^\circ = \frac{-10.83 - (-6.81)}{298} \]
\[ \Delta S^\circ = \frac{-10.83 + 6.81}{298} = \frac{-4.02}{298} \approx -0.0135 \, \text{kcal/K} \]
Converting to cal/K:
\[ \Delta S^\circ \approx -13.5 \, \text{cal/K} \]

### Step 5: Calculate \( S_{298}^\circ[H_2O(g)] \)
Using the formula for entropy change:
\[ \Delta S^\circ = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} \]
Where:
- \( S^\circ_{H_2} = 31.2 \, \text{cal/K} \)
- \( S^\circ_{CO} = 47.3 \, \text{cal/K} \)
- \( S^\circ_{CO_2} = 51.1 \, \text{cal/K} \)

Substituting:
\[ -13.5 = [31.2 + 51.1] - [S^\circ_{H_2O} + 47.3] \]
\[ -13.5 = 82.3 - [S^\circ_{H_2O} + 47.3] \]
Rearranging gives:
\[ S^\circ_{H_2O} = 82.3 + 13.5 - 47.3 \]
\[ S^\circ_{H_2O} = 48.5 \, \text{cal/K} \]

### Final Answer
The standard entropy \( S_{298}^\circ[H_2O(g)] \) is approximately \( 48.5 \, \text{cal/K} \).