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), usingthe following:
enthalpy of formation of CHCl3 (g), ΔH°f = – 103.1 kJ/mol
                                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
 Â
+ 145.4 kJ
– 305.3 kJ
– 145.4 kJ
+ 305.3 kJ
– 103.1 kJ
