Question

Calculate the enthalpy change for the process

CCl_4(g) → C_(g) + 4Cl_(g)

and calculate bond enthalpy of C–Cl in CCl_4(g).

Δ_vapH^θ (CCl₄) = 30.5 kJ mol^–1.

Δ_fH^θ (CCl₄) = –135.5 kJ mol^–1.

Δ_aH^θ (C) = 715.0 kJ mol^–1, where Δ_aH^θ is enthalpy of atomisation

Δ_aH^θ (Cl₂) = 242 kJ mol^–1

Answer 1

The chemical equations implying to the given values of enthalpies are:

Δ_vapH^θ = 30.5 kJ mol^–1

Δ_aH^θ = 715.0 kJ mol^–1

Δ_aH^θ = 242 kJ mol^–1

Δ_fH = –135.5 kJ mol^–1

Enthalpy change for the given process can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

ΔH = Δ_aH^θ(C) + 2Δ_aH^θ (Cl₂) – Δ_vapH^θ – Δ_fH

= (715.0 kJ mol^–1) + 2(242 kJ mol^–1) – (30.5 kJ mol^–1) – (–135.5 kJ mol^–1)

ΔH = 1304 kJ mol^–1

Bond enthalpy of C–Cl bond in CCl_{4 (g)}

= 326 kJ mol^–1