Which of the following has `DeltaS^(@)` greater than zero

Options

A. `CaO_(s) + CO_(2(g)) iff CaCO_(3(s))`
B. `NaCl_(aq) iff NaCl_(s)`
C. `NaNO_3(s) iff Na^(aq)^(+) NO_(aq)^(-)`
D. `N_(2(g)) + 3H_(2(g)) iff 2NH_(3(g))`

Explanation

To determine which of the given options has a change in entropy (ΔS°) greater than zero, we need to analyze the states of the reactants and products in each option. The general principle is that the entropy of gases is higher than that of liquids, which is higher than that of solids. Therefore, a reaction that produces more gaseous products or converts solids into liquids or gases will typically have a positive ΔS°.

### Step-by-Step Solution:

1. **Understanding Entropy (ΔS°)**:
- Entropy is a measure of disorder or randomness in a system.
- The more disordered the system, the higher the entropy.

2. **Analyzing Each Option**:
- **Option A**: Reactants include CO₂ (gas) and products are solids.
- Since the product is a solid and the reactant is a gas, the entropy decreases. Thus, ΔS° < 0.

- **Option B**: Reactants are aqueous solutions and products are solids.
- The transition from a liquid (aqueous) to a solid decreases randomness. Thus, ΔS° < 0.

- **Option C**: Reactants are solids and products are aqueous solutions.
- The transition from solid to aqueous increases randomness. Thus, ΔS° > 0.

- **Option D**: Both reactants and products are gases, but the number of gas molecules decreases from 4 to 2.
- Fewer gas molecules mean less randomness. Thus, ΔS° < 0.

3. **Conclusion**:
- The only option where ΔS° is greater than zero is **Option C**, where the reaction involves the conversion of a solid to an aqueous state.

### Final Answer:
**Option C** has ΔS° greater than zero.