For a reaction: `X rarr Y+Z`
Absolute entropies are `X="120 J K"^(-1)"mol"^(-1),`
`Y="213.8 J K"^(-1)"mol"^(-1) and Z="197.9 J K"^(-1)"mol"^(-1)`.
What will be the entropy change at 298 K and 1 atm?

Options

A. `"291.7 J K"^(-1)`
B. `"255 J K"^(-1)`
C. `213.8 J K"^(-1)`
D. `"257.3 J K"^(-1)`

Explanation

To calculate the entropy change (ΔS) for the reaction \( X \rightarrow Y + Z \), we will follow these steps:

### Step 1: Identify the absolute entropies of the reactants and products.
- Given:
- \( S_X = 120 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S_Y = 213.8 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S_Z = 197.9 \, \text{J K}^{-1} \text{mol}^{-1} \)

### Step 2: Calculate the total entropy of the products.
- The products of the reaction are \( Y \) and \( Z \).
- Total entropy of products \( S_{\text{products}} = S_Y + S_Z \)
- Calculation:
\[ S_{\text{products}} = 213.8 + 197.9 = 411.7 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 3: Calculate the total entropy of the reactants.
- The reactant of the reaction is \( X \).
- Total entropy of reactants \( S_{\text{reactants}} = S_X \)
- Thus, \( S_{\text{reactants}} = 120 \, \text{J K}^{-1} \text{mol}^{-1} \)

### Step 4: Calculate the entropy change (ΔS) for the reaction.
- The entropy change is given by the formula:
\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]
- Substituting the values we calculated:
\[ \Delta S = 411.7 - 120 = 291.7 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Final Answer:
The entropy change (ΔS) at 298 K and 1 atm for the reaction \( X \rightarrow Y + Z \) is \( 291.7 \, \text{J K}^{-1} \text{mol}^{-1} \).

---