`DeltaS^(@)` wil be highest for the reaction

Options

A. `Ca(s)+1/2O_(2)(g)toCaO(s)`
B. `CaCO_(3)(s)toCaO(s)+CO_(2)(g)`
C. `C(s)+O_(2)(g)toCO_(2)(g)`
D. `N_(2)(g)+O_(2)(g)to2NO(g)`

Explanation

To determine which reaction has the highest standard change in entropy (ΔS°), we need to analyze the gaseous moles on both sides of each reaction. The formula for calculating ΔS° is:

\[ \Delta S° = S_{products} - S_{reactants} \]

Where \( S \) refers to the sum of the gaseous moles. The greater the difference between the gaseous moles on the product side and the reactant side, the higher the ΔS° value.

Let's analyze each option step by step:

### Step 1: Analyze Option A
- **Reaction**: \( \text{Solid} \rightarrow \text{Gas} \)
- **Gaseous moles**:
- Products: 0 (since there are no gaseous products)
- Reactants: 1 (solid does not count)
- **Calculation**:
\[ \Delta S° = 0 - 1 = -1 \]

### Step 2: Analyze Option B
- **Reaction**: \( \text{Solid} + \text{Solid} \rightarrow \text{Gas} \)
- **Gaseous moles**:
- Products: 1 (1 mole of gas)
- Reactants: 0 (both are solids)
- **Calculation**:
\[ \Delta S° = 1 - 0 = 1 \]

### Step 3: Analyze Option C
- **Reaction**: \( \text{Gas} \rightarrow \text{Gas} \)
- **Gaseous moles**:
- Products: 1 (1 mole of gas)
- Reactants: 1 (1 mole of gas)
- **Calculation**:
\[ \Delta S° = 1 - 1 = 0 \]

### Step 4: Analyze Option D
- **Reaction**: \( \text{Gas} + \text{Gas} \rightarrow \text{Gas} \)
- **Gaseous moles**:
- Products: 1 (1 mole of gas)
- Reactants: 2 (2 moles of gas)
- **Calculation**:
\[ \Delta S° = 1 - 2 = -1 \]

### Step 5: Compare the ΔS° values
- Option A: ΔS° = -1
- Option B: ΔS° = 1
- Option C: ΔS° = 0
- Option D: ΔS° = -1

### Conclusion
The highest ΔS° value is found in **Option B**, which has a ΔS° of 1. Therefore, the reaction with the highest standard change in entropy is Option B.