Calculate the enthalpy of formation (in kcal/mol) of HBr(g) from the following data: H 2 (g) + Cl 2 (g) → 2HCl(g); ΔH = -44.0 kcal HCl(g) + aq → HCl(aq); ΔH = -17.0 kcal HBr(g) + aq → HBr(aq); ΔH = -2

Question

Calculate the enthalpy of formation (in kcal/mol) of HBr(g) from the following data:

H₂(g) + Cl₂(g) → 2HCl(g); ΔH = -44.0 kcal

HCl(g) + aq → HCl(aq); ΔH = -17.0 kcal

HBr(g) + aq → HBr(aq); ΔH = -20.0 kcal

KOH(aq) + HCl(aq) → KCl(aq); ΔH = -13.5 kcal

KOH(aq) + HBr(aq) → KBr(aq); ΔH = -13.5 kcal

Cl₂(g) + 2KBr(aq) → 2KCl(aq) + Br₂(l); ΔH = -23.0 kcal

TestHub · Accepted Answer

Target: H 2 (g) + Br 2 (l) → 2HBr(g). Add: H 2 + Cl 2 → 2HCl(g) (-44.0); 2HCl(g) → 2HCl(aq) (-34.0); 2HCl(aq) + 2KOH → 2KCl(aq) (-27.0); 2KCl(aq) + Br 2 → Cl 2 + 2KBr(aq) (+23.0); 2KBr(aq) → 2KOH + 2HBr(aq) (+27.0); 2HBr(aq) → 2HBr(g) (+40.0). Sum = -15.0 kcal for 2 mol HBr, so Δ f H(HBr) = -7.5 kcal/mol. This is option (b).

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