CHEMISTRY PROBLEM

Question

Given that

Δ𝐻∘f[Cl(g)]=121.3 kJ⋅mol−1

Δ𝐻∘f[C(g)]=716.7 kJ⋅mol−1

Δ𝐻∘f[CCl4(g)]=−95.7 kJ⋅mol−1

calculate the average molar bond enthalpy of the carbon–chlorine bond in a CCl4 molecule.

Answer 1

C(g) + 4Cl(g) ==> CCl₄(g) ∆H = -95.7 kJ/mol

Cl₂(g) ==> 2Cl(g) ∆H = 121.3 kJ/mol

C(s) ==> C(g) ∆H = 716.7 kJ/mol

Reactants = 4 x 121.3 + 716.7 = 485.2 + 716.7 = 1201.9 kJ/mol

Products = -95.7 kJ/mol

Products - Reactants = Reaction

-95.7 - 1201.9 = -1297.6 kJ/mol

Since there are 4 C-Cl bonds in CCl₄, we divide by 4 to get the enthalpy of the C-Cl bond

Enthalpy of the C-Cl bond = 1297.6 kJ/mol ÷ 4 = 324.4 kJ/mol