Question

Consider the reaction CH4(g) + X2(g) → CH3X(g) + HX(g), where X is a halogen-like atom (Group 7A). Calculate ∆H for this reaction if bond energy tables give the following values: C H : 416 kJ/mol X X : 230 kJ/mol H X : 277 kJ/mol C X : 222 kJ/mol Answer in units of kJ/mol.

Answer 1

ΔH of reaction is -147 kJ/mol

Step-by-step explanation:

For the reaction:

CH₄(g) + X₂(g) → CH₃X(g) + HX(g)

It is possible to obtain ΔHr of reaction from the sum of ΔH of products minus ΔH of reactants:

ΔHr = ΔH CH₃X + ΔH HX - (ΔH CH₄ + ΔH X₂) (1)

The ΔH of each compound is obtained from bond energies thus:

ΔH CH₃X = 3×C-H + C-X = 3×416 kJ/mol + 222 kJ/mol = 1470 kJ/mol

ΔH HX = H-X = 277 kJ/mol

ΔH CH₄ = 4×C-H = 4×416kJ/mol = 1664 kJ/mol

ΔH X₂ = X-X = 230 kJ/mol

Replacing in (1):

ΔHr = 1470 kJ/mol + 277kJ/mol - (1664 kJ/mol + 230 kJ/mol) = -147 kJ/mol

I hope it helps!