Answer:
<em>1. H₂(g) + Cl₂(g) → 2HCl(g) </em>
- <u>ΔS is too close to zero to decide</u>
<em>2. COCl₂(g) → CO(g) + Cl₂(g) </em>
<em><u></u></em>
<em>3. 2HBr(g) + Cl₂(g) → 2HCl(g) + Br₂(g) </em>
- <u>ΔS is too close to zero to decide</u>
<em>4. 2H₂S(g) + 3O₂(g) → 2H₂O(g) + 2SO₂(g) </em>
<em>5. CaCO₃(s) → CaO(s) + CO₂(g) </em>
Explanation:
You can <em>predict whether ΔS is greater than zero, less than zero, or too close to zero</em> 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 <em>ΔS > 0.</em>
- When the number of molecules in gas phase decreases, the entropy of the products is less than the entropy of the reactants and <em>ΔS < 0.</em>
- 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 <em>ΔS ≈ 0.</em>
<em>1. H₂(g) + Cl₂(g) → 2HCl(g) </em>
<em></em>
- 2 gas molecules → 2 gas molecules ⇒ ΔS is too close to 0 to decide
<em>2. COCl₂(g) → CO(g) + Cl₂(g) </em>
- 1 gas molecule → 2 gas molecules ⇒ ΔS > 0
<em>3. 2HBr(g) + Cl₂(g) → 2HCl(g) + Br₂(g) </em>
- 3 gas molecules → 3 gas molecules ⇒ ΔS is too close to zero to decide
<em>4. 2H₂S(g) + 3O₂(g) → 2H₂O(g) + 2SO₂(g) </em>
- 5 gas molecules → 4 gas molecules ⇒ ΔS < 0
<em>5. CaCO</em><em>₃</em><em>(s) → CaO(s) + CO₂(g) </em>
- 1 solid unit → 1 solid unit and 1 gas molecule ⇒ ΔS > 0