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),(DeltaG_(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 calculate the standard entropy \( S_{298}^\circ [H_2O(g)] \) for water vapor from the given reaction and data, we will follow these steps:

### Step 1: Write the Reaction and Identify Given Data
The reaction is:
\[ H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g) \]
From the table, we have the following data:
- \( \Delta H_{298}^\circ \) (kcal/mole):
- CO(g): -26.42
- CO2(g): -94.05
- H2O(g): -57.8
- H2(g): 0
- \( \Delta G_{298}^\circ \) (kcal/mole):
- CO(g): -32.79
- CO2(g): -94.24
- H2O(g): -54.64
- H2(g): 0
- \( S_{298}^\circ \) (cal/(k mol)):
- CO(g): 47.3
- CO2(g): 51.1
- H2(g): 31.2
- H2O(g): ?

### Step 2: Calculate \( \Delta S_{298}^\circ \) for the Reaction
Using the formula for the change in entropy for the reaction:
\[ \Delta S_{298}^\circ = S_{products} - S_{reactants} \]
Substituting the known values:
\[ \Delta S_{298}^\circ = [S_{H_2} + S_{CO_2}] - [S_{H_2O} + S_{CO}] \]
This can be written as:
\[ \Delta S_{298}^\circ = [31.2 + 51.1] - [S_{H_2O} + 47.3] \]
\[ \Delta S_{298}^\circ = 82.3 - [S_{H_2O} + 47.3] \]

### Step 3: Calculate \( \Delta G_{298}^\circ \) and \( \Delta H_{298}^\circ \)
Using the Gibbs free energy equation:
\[ \Delta G_{298}^\circ = \Delta H_{298}^\circ - T \Delta S_{298}^\circ \]
First, calculate \( \Delta G_{298}^\circ \):
\[ \Delta G_{298}^\circ = [0 - (-94.24)] - [-57.8 - (-26.42)] \]
\[ \Delta G_{298}^\circ = 94.24 - (-31.38) = 94.24 + 31.38 = 125.62 \text{ kcal/mole} \]

Now calculate \( \Delta H_{298}^\circ \):
\[ \Delta H_{298}^\circ = [-94.05 + 0] - [-57.8 - (-26.42)] \]
\[ \Delta H_{298}^\circ = -94.05 + 31.38 = -62.67 \text{ kcal/mole} \]

### Step 4: Substitute into Gibbs Free Energy Equation
Now we can substitute into the Gibbs free energy equation:
\[ -6.81 = -62.67 - 298 \Delta S_{298}^\circ \]
Rearranging gives:
\[ 298 \Delta S_{298}^\circ = -62.67 + 6.81 \]
\[ 298 \Delta S_{298}^\circ = -55.86 \]
\[ \Delta S_{298}^\circ = \frac{-55.86}{298} = -0.187 \text{ kcal/(K mole)} \]

### Step 5: Solve for \( S_{298}^\circ [H_2O(g)] \)
Now substitute back to find \( S_{H_2O} \):
\[ \Delta S_{298}^\circ = 82.3 - [S_{H_2O} + 47.3] \]
Setting this equal to the previously calculated \( \Delta S_{298}^\circ \):
\[ -0.187 = 82.3 - [S_{H_2O} + 47.3] \]
Rearranging gives:
\[ S_{H_2O} + 47.3 = 82.3 + 0.187 \]
\[ S_{H_2O} + 47.3 = 82.487 \]
\[ S_{H_2O} = 82.487 - 47.3 = 35.187 \text{ cal/(K mole)} \]

### Final Answer
Thus, the standard entropy \( S_{298}^\circ [H_2O(g)] \) is approximately:
\[ S_{298}^\circ [H_2O(g)] \approx 35.19 \text{ cal/(K mole)} \]