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Class 11
Chemistry
Thermodynamics
calculate the enthalpy change for the process c...

Question:
Calculate the enthalpy change for the process

CCl₄(g) → C(g) + 4 Cl(g)

and calculate bond enthalpy of C – Cl in CCl₄(g).

Δ_vapH⁰(CCl₄) = 30.5 kJ mol^–1.

Δ_fH⁰ (CCl₄) = –135.5 kJ mol^–1.

Δ_aH⁰ (C) = 715.0 kJ mol^–1 , where Δ_aH⁰ is enthalpy of atomisation

Δ_aH⁰ (Cl₂) = 242 kJ mol^–1

Answer:

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

(i) CCl_4(l) → CCL_4(g) Δ_vapH^θ = 30.5 kJ mol^-1

(ii) C_(s) → C_(g) Δ_aH^θ = 715.0 kJ mol^-1

(iii) Cl_2(g) → 2Cl_(g) Δ_aH^θ = 242 kJ mol^-1

(iv) C_(g) + 4Cl_(g) → CCl_4(g) Δ_fH = -135.5 kJ mol^-1

Enthalpy change for the given process C_(g) + 4Cl_(g) → CCl_4(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) = 326 kJ mol^-1

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