Question

. Calculate ∆rH for the following reaction: CH4(g) + 3Cl2 (g) → CHCl3(l) + 3HCl (g)

Answer 1

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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)

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