Question

Calculate the enthalpy change for the process

\[\ce{CCl_4 (g) → C(g) + 4Cl(g)}\]

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:

1) \[\ce{CCl_{4(l)} -> CCl_{4(g)}}\]  Δ_vapH^θ = 30.5 kJ mol^–1

2) \[\ce{C_{(s)} -> C_{(g)}}\] _aH^θ = 715.0 kJ mol^–1

3) \[\ce{Cl_{2(g)}  -> 2Cl_{(g)}}\] Δ_aH^θ = 242 kJ mol^–1

4) \[\ce{C_{(g)} +  4Cl_{(g)} -> CCl_{4(g)}}\] Δ_fH = - 135.5 kJ mol^–1

Enthalpy change for the given process \[\ce{CCl_{4(g)} -> C_{(g)} + 4Cl_{(g)}}\]   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)}

`= 1304/4 " kJ mol"^(-1)`

= 326 kJ mol^-1