#### 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_{vap}*H*^{θ} (CCl_{4}) = 30.5 kJ mol^{–1}.

Δ_{f}*H*^{θ} (CCl_{4}) = –135.5 kJ mol^{–1}.

Δ_{a}*H*^{θ} (C) = 715.0 kJ mol^{–1}, where Δ_{a}*H*^{θ} is enthalpy of atomisation

Δ_{a}*H*^{θ} (Cl_{2}) = 242 kJ mol^{–1}

#### Solution

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

1)`"CCl"_(4(l)) -> "CCl"_(4(g))` Δ_{vap}*H*^{θ} = 30.5 kJ mol^{–1}

2)` C_((s)) -> C_(g)` _{a}*H*^{θ} = 715.0 kJ mol^{–1}

3) `Cl_(2(g)) -> 2Cl_(g)` Δ_{a}*H*^{θ} = 242 kJ mol^{–1}

4) `C_(g) + 4Cl_(g) -> "CCl"_(4(g))` Δ_{f}*H* = –135.5 kJ mol^{–1}

Enthalpy change for the given process `"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)

Δ*H* = Δ_{a}*H*^{θ}(C) + 2Δ_{a}*H*^{θ} (Cl_{2}) – Δ_{vap}*H*^{θ} – Δ_{f}*H*

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

∴ Δ*H* = 1304 kJ mol^{–1}

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

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

= 326 kJ mol^{–1}