Question

Predict whether S for each reaction would be greater than zero, less than zero, or too close to zero to decide. Clear All H2(g) + Cl2(g)2HCl(g) COCl2(g)CO(g) + Cl2(g) 2HBr(g) + Cl2(g)2HCl(g) + Br2(g) 2H2S(g) + 3O2(g)2H2O(g) + 2SO2(g) CaCO3(s)CaO(s) + CO2(g) S > 0 S < 0 too close to decide

Answer 1

1. H₂(g) + Cl₂(g) → 2HCl(g)

• ΔS is too close to zero to decide

2. COCl₂(g) → CO(g) + Cl₂(g)

• ΔS > 0

3. 2HBr(g) + Cl₂(g) → 2HCl(g) + Br₂(g)

• ΔS is too close to zero to decide

4. 2H₂S(g) + 3O₂(g) → 2H₂O(g) + 2SO₂(g)

• ΔS < 0

5. CaCO₃(s) → CaO(s) + CO₂(g)

• ΔS > 0

Step-by-step explanation:

You can predict whether ΔS is greater than zero, less than zero, or too close to zero by looking at the physical states of the reactant and product compounds.

• When the number of molecules in gas phase increases, the entropy of the products is greater than the entropy of the reactants and ΔS > 0.

• When the number of molecules in gas phase decreases, the entropy of the products is less than the entropy of the reactants and ΔS < 0.

• When the number of molecules in gas phase does not change, the entropy of the products is similar to the entropy of the reactants and ΔS ≈ 0.

1. H₂(g) + Cl₂(g) → 2HCl(g)

• 2 gas molecules → 2 gas molecules ⇒ ΔS is too close to 0 to decide

2. COCl₂(g) → CO(g) + Cl₂(g)

• 1 gas molecule → 2 gas molecules ⇒ ΔS > 0

3. 2HBr(g) + Cl₂(g) → 2HCl(g) + Br₂(g)

• 3 gas molecules → 3 gas molecules ⇒ ΔS is too close to zero to decide

4. 2H₂S(g) + 3O₂(g) → 2H₂O(g) + 2SO₂(g)

• 5 gas molecules → 4 gas molecules ⇒ ΔS < 0

5. CaCO₃(s) → CaO(s) + CO₂(g)

• 1 solid unit → 1 solid unit and 1 gas molecule ⇒ ΔS > 0