For which of these reactions will there be `Delta S` positive?

Options

A. `H_(2)O(g) rightarrow H_(2)O(l)`
B. `H_(2)(g) + I_(2)(g) rightarrow 2HI(g)`
C. `CaCO_(3) (s) rightarrow CaO(s) + CO_(2)(g) `
D. `N_(2)(g) + 3H_(2)(g) rightarrow 2NH_(3)(g)`

Answer

To determine which of the given reactions will have a positive change in entropy (ΔS > 0), we need to analyze the number of moles of gaseous products and reactants in each reaction. A positive ΔS indicates an increase in disorder or randomness in the system, which typically occurs when the number of gaseous product moles is greater than the number of gaseous reactant moles.

### Step-by-Step Solution:

1. **Analyze the first reaction:**
- Reaction: H2O(g) → H2O(l)
- In this reaction, water vapor (gas) is converted to liquid water.
- **Moles of reactants:** 1 (H2O(g))
- **Moles of products:** 1 (H2O(l))
- Since there is no increase in the number of moles, ΔS is negative.

2. **Analyze the second reaction:**
- Reaction: H2(g) + I2(g) → 2 HI(g)
- Here, 1 mole of H2 and 1 mole of I2 (total 2 moles of reactants) produce 2 moles of HI.
- **Moles of reactants:** 2 (H2 + I2)
- **Moles of products:** 2 (2 HI)
- The number of moles remains the same, so ΔS is zero.

3. **Analyze the third reaction:**
- Reaction: CaCO3(s) → CaO(s) + CO2(g)
- In this reaction, solid calcium carbonate decomposes into solid calcium oxide and gaseous carbon dioxide.
- **Moles of reactants:** 1 (CaCO3)
- **Moles of products:** 1 (CaO) + 1 (CO2) = 2
- The number of gaseous moles increases (from 0 to 1), so ΔS is positive.

4. **Analyze the fourth reaction:**
- Reaction: 2 H2(g) + 3 N2(g) → 2 NH3(g)
- Here, 2 moles of H2 and 3 moles of N2 (total 5 moles of reactants) produce 2 moles of NH3.
- **Moles of reactants:** 5 (2 H2 + 3 N2)
- **Moles of products:** 2 (2 NH3)
- The number of moles decreases, so ΔS is negative.

### Conclusion:
The only reaction that results in a positive change in entropy (ΔS > 0) is the third reaction:
**CaCO3(s) → CaO(s) + CO2(g)**.