Chloroform, CHCl3, is formed by the following reaction: CH4(g) + 3 Cl2(g) → 3 HCl(g) + CHCl3 (g) Determine the enthalpy change for this reaction (ΔH°rxn)

Question

Chloroform, CHCl₃, is formed by the following reaction: 

CH₄(g) + 3 Cl₂(g) → 3 HCl(g) + CHCl₃ (g)

Determine the enthalpy change for this reaction (ΔH°_rxn), using the enthalpy of formation of CHCl₃ (g), ΔH°_f = – 103.1 kJ/mol, and the following:

                               CH₄(g) + 2 O₂(g) → 2 H₂O(l) + CO₂(g)       ΔH°_rxn = – 890.4 kJ/mol

                               2 HCl (g) → H₂ (g) + Cl₂(g)                            ΔH°_rxn = + 184.6 kJ/mol

                               C (graphite) + O₂(g) → CO₂(g)                    ΔH°_rxn = – 393.5 kJ/mol

                               H₂ (g) + ½ O₂(g) → H₂O(l)                             ΔH°_rxn = – 285.8 kJ/mol

I made the equation:

C(graphite) + 1/2H₂(g) + 3/2Cl₂(g) -->CHCl₃(g) ΔH°_f = – 103.1 kJ/mol

based on the information given.

I then used this equation (along with the other equations) to obtain the target equation of: CH₄(g) + 3 Cl₂(g) → 3 HCl(g) + CHCl₃ (g), by flipping some of the equations and multiplying some of them by numbers.

When doing this I got: -103.1 kJ/mol + (-890.4 kJ/mol) + (-276.9 kJ/mol) + (+393.5 kJ/mol) + (+571.6 kJ/mol) = -305.3 kJ/mol.

However, the answer choices are not in kJ/mol, they are only in kJ. These are the answer choices provided:

a) -305.3 kJ

b) -103.1 kJ

c) +305.3 kJ

d) 145.5 kJ

e) -145.5 kJ

How do I get from kJ/mol to just kJ?

Answer 1

The values are given as kJ/mole but then you multiply by the number of moles in the balanced equation, giving you kJ. The moles cancel. So, if you did the arithmetic correctly, the answer should be -305.3 kJ.