Question

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

• A. – 44.0 kcal
• B. – 22.0 kcal
• C. 22.0 kcal
• D. 44.0 kcal

Answer 1

Answer: (B)

– 22.0 kcal

Answer 2

Given

H2 (g) $\longrightarrow$ 2H (g);

$\Delta$H = 104 kcal ...(1)

Cl2 (g) $\longrightarrow$ 2Cl(g) ;

$\Delta$H = 58 kcal ...(2)

HCl (g) $\longrightarrow$H(g) + Cl(g) ;

$\Delta$H = 103 kcal ...(3)

Heat of formation for HCl

$\frac{1}{2}$H2 (g) + $\frac{1}{2}$ Cl2 (g) $\longrightarrow$ HCl (g) ;

$\Delta$H = ?

Divide equation (1) and (2) by 2, and then add

$\frac{1}{2}$H2 (g) + $\frac{1}{2}$ Cl2 (g) $\longrightarrow$ H(g) + Cl(g) ;

$\Delta$H = 81 kcal ...(4)

Subtracting equation (3) from equation (4)

HCl (g) $\longrightarrow$ H(g) + Cl(g) ;

$\Delta$H = 103 kcal ...(3)

– – – –

–––––––––––––––––––––––––––––––––––––––––––––––

$\frac{1}{2}$H2 (g) + $\frac{1}{2}$ Cl2 (g) $\longrightarrow$ HCl(g) ;

DH = – 22.0 kcal

$\therefore$ Enthalpy of formation of HCl gas = – 22.0 kcal