Question

Calculate the enthalpy change for the process
$CCl_4(g) → C(g) + 4Cl(g)$
and calculate bond enthalpy of $C-Cl$ in $CCl_4(g)$.
$∆_{vap}H^Θ(CCl_4) = 30.5\ kJ mol^{-1}$.
$∆_fH^Θ(CCl_4)= –135.5\ kJ mol^{-1}$.
$∆_aH^Θ (C)= 715.0\ kJ mol^{-1}$, where $∆_aH^Θ$ is enthalpy of atomisation.
$∆_aH^Θ (Cl_2) = 242\ kJ mol^{-1}$.

Answer 1

The chemical equations implying to the given values of enthalpies are:
(i) $CCl_4(l) → CCl_4(g)$ ; $∆_{vap}H^Θ(CCl_4) = 30.5\ kJ mol^{–1}$
(ii) $C(s) → C(g)$ ; $∆_aH^Θ(C) = 715.0\ kJ mol^{–1}$
(iii) $Cl_2(g) → 2Cl(g)$ ; $∆_aH^Θ(Cl_2) = 242\ kJ mol^{–1}$
(iv) $C + 4Cl → CCl_4(g)$ ; $∆_fH^Θ(CCl_4) = –135.5\ kJ mol^{–1}$
Enthalpy change for the given process $CCl_4(g) → C(g) + 4 Cl(g)$ can be calculated using the following algebraic calculations as:
Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)
$∆H = ∆_aH^Θ(C) + 2∆_aH^Θ (Cl_2) – ∆_{vap}H^Θ – ∆_fH$
= $(715.0\ kJ mol^{–1}) + 2(242\ kJ mol^{–1}) – (30.5 \ J mol^{–1}) – (–135.5\ kJ mol^{–1})$
$∆H = 1304 \ kJ mol^{–1}$
Bond enthalpy of $C–Cl$ bond in $CCl_4 (g)$
= $\frac {1304}{4} kJ mol^{–1}$
= $326\ kJ mol^{–1}$