H₂S(g)+ 3/2 O₂(g)→ H₂O (L)+ SO₂(g)Δ_rH° = -562

question 83

Multiple Choice

H₂S(g) + 3/2 O₂(g) → H₂O (l) + SO₂(g) Δ_rH° = -562.3 kJ/mol
1/8 S₈(s) + O₂(g) → SO₂(g) Δ_rH° = -297.0 kJ/mol
H₂(g) + 1/2O₂ → H₂O (l) Δ_rH° = -266.9 kJ/mol
Compute Δ_rH° for H₂(g) + 1/8 S₈ (s) → H₂S(g) in kJ/mol.

• A)562.3 kJ/mol
• B)276.6 kJ/mol
• C)265.3 kJ/mol
• D)20.5 kJ/mol
• E)-1.6 kJ/mol

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