The bond dissociation energy of gaseous H2, Cl2 and HCl are 104, 58 and 103 kcal mol–1 respectively. The enthalpy of formation for HCl gas will be
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Given
H2 (g) → 2H (g); ΔH = 104 kcal ...(1)
Cl2 (g) → 2Cl(g) ; ΔH = 58 kcal ...(2)
HCl (g) → H(g) + Cl(g) ; ΔH = 103 kcal ...(3)
Heat of formation for HCl
H2 (g) + Cl2 (g) → HCl (g) ; ΔH = ?
Divide equation (1) and (2) by 2, and then add
H2 (g) + Cl2 (g) → H(g) + Cl(g) ; ΔH = 81 kcal ...(4)
Subtracting equation (3) from equation (4)
HCl (g) → H(g) + Cl(g) ; ΔH = 103 kcal ...(3)
- – – –
__________________________________________________
H2 (g) + Cl2 (g) → HCl(g) ; ΔH = – 22.0 kcal
∴ Enthalpy of formation of HCl gas = – 22.0 kcal
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