Calculate the enthalpy change for the process

CCl₄(g) → C(g) + 4 Cl(g)

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

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, where is enthalpy of atomisation

(Cl₂) = 242 kJ mol^–1

Given:

Δ_vapH^° (CCl₄) = 30.5 kJ mol^–1

Δ_fH^° (CCl₄) = -135.5 kJ mol^–1

Δ_aH^° (C) = 715.0 kJ mol^–1

Δ_aH^° (Cl₂) =242 kJ mol^–1

We can write,

CCl₄(l) → CCl₄(g); ΔH = 30.5 kJ mol^–1 (1)

C(s) + 2Cl₂(g) → CCl₄(l); ΔH = -135.5 kJ mol^–1 (2)

C(s) → C(g); ΔH = 715.0 kJ mol^–1 (3)

Cl₂(g) → 2Cl(g); ΔH =242 kJ mol^–1 (4)

To get the required reaction, CCl₄(g) → C(g) + 4 Cl(g)

We follow the steps given below:

Step 1: Add t the both reaction (1) and (2), we get

C(s) + 2Cl₂(g) → CCl₄(l); ΔH = -135.5 kJ mol^–1

CCl₄(l) → CCl₄(g); ΔH = 30.5 kJ mol^–1

C(s) + 2Cl₂(g) → CCl₄(g); ΔH = -105 kJ mol^–1 (5)

Step 2: Add the both reaction (3) and (4), we get

C(s) → C(g); ΔH = 715.0 kJ mol^–1

2Cl₂ (g) → 4Cl (g); ΔH = 484 kJ mol^–1 [ 2×reaction 4]

C(s) + 2Cl₂(g) → C(g) + 4Cl(g); ΔH = 1199 kJ mol^–1 (6)

Step 3: Subtract the both reaction (5) and (6), we get

C(s) + 2Cl₂(g) → C(g) + 4Cl(g); ΔH = 1199 kJ mol^–1

CCl₄(g) → C(s) + 2Cl₂(g); ΔH = -105 kJ mol^–1

CCl₄(g) → C(g) + 4 Cl(g); ΔH = 1304 kJ mol^–1

(formation of required reaction)

Thus, the enthalpy change for the process is 1304 kJ mol^–1

⇒Calculation the bond enthalpy of C-Cl in CCl₄(g)

Bond enthalpy of C-Cl bond in CCl₄ is equal to one fourth of the energy of dissociation of CCl₄

∴ Bond enthalpy of C – Cl in CCl₄(g) =

As ΔH =1304 kJ mol^–1

= 326 kJ mol^-1

Note: Bond enthalpy is defined as the amount of energy required to break one mole of bond of a particular type between the atoms in the gaseous state, i.e., to separate the atoms in the gaseous state under 1 am pressure and the specified temperature is called bond enthalpy.