Question
Class 11ChemistryThermodynamics

Calculate the enthalpy change for the process:

CCl4(g) → C(g) + 4Cl(g)

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

Given:

ΔvapH°(CCl4) = 30.5 kJ mol−1

ΔfH°(CCl4) = −135.5 kJ mol−1

ΔaH°(C) = 715.0 kJ mol−1 (enthalpy of atomization)

ΔaH°(Cl2) = 242 kJ mol−1

Verified Answer

The given data imply as under:

(i) CCl4(l) → CCl4(g), ΔH = 30.5 kJ mol−1

(ii) C(s) + 2Cl2(g) → CCl4(l), ΔH = −135.5 kJ mol−1

(iii) C(s) → C(g), ΔH = 715.0 kJ mol−1

(iv) Cl2(g) → 2Cl(g), ΔH = 242 kJ mol−1

Aim: CCl4(g) → C(g) + 4Cl(g), ΔH = ?

Eqn. (iii) + 2 × eqn. (iv) −eqn. (i) −eqn. (ii) gives the required equation:

ΔH = 715.0 + 2(242) − 30.5 − (−135.5) kJ mol−1

= 1304 kJ mol−1

Bond enthalpy of C–Cl in CCl4 (average value) = 1304/4 = 326 kJ mol−1