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

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

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

Ibelieve it's celsius. hope this .

Bthats what im doing to and i figured it out it is b