Question

Select the most appropriate option.

Bond enthalpies of H–H, Cl–Cl, and H–Cl bonds are 434 kJ mol^–1, 242 kJ mol^–1, and 431 kJ mol^–1, respectively. Enthalpy of formation of HCl is _______.

पर्याय

• 245 kJ mol^–1

• –93 kJ mol^–1

• –245 kJ mol^–1

• 93 kJ mol^–1

Answer 1

Bond enthalpies of H–H, Cl–Cl, and H–Cl bonds are 434 kJ mol^–1, 242 kJ mol^–1, and 431 kJ mol^–1, respectively. Enthalpy of formation of HCl is `bb(underline(–93  kJ  mol^(–1))`.

Explanation:

Δ_rH° = ∑ ΔH° (reactant bonds) – ∑ΔH° (products bonds)

\[\ce{H2_{(g)} + Cl2_{(g)} → 2HCl_{(g)}}\] 

∴ Δ_rH° = [1 mol × 434 kJ mol^–1 + 1 mol × 242 kJ mol^–1] – [2 mol × 431 kJ mol^–1]

= –186 kJ

∴ \[\ce{H2_{(g)} + Cl2_{(g)} → 2HCl_{(g)}, ΔrH° = -186 kJ}\]

For enthalpy of formation of HCl, the reaction is

\[\ce{1/2 H2_{(g)} + 1/2 Cl2_{(g)} -> HCl_{(g)}}\],

Δ_rH° =`(- 186  "kJ")/(2  "mol")`= –93 kJ mol^–1