Calculate the enthalpy of formation of carbon disulfide given that the enthalpy of combustion of it is 110.2 kJ mol-1 and those of sulfur and carbon are 297.4 kJ and 394.5 kJ/g atoms respectively.
Our aim is C(s) + 2S(s) → CS2(l); ΔH =? Given (i) CS2(l) + 3O2(g) → CO2(g) + 2SO2(g); ΔH = – 110.2 kJ mol-1 (ii) S(s) + O2(g) → SO2(g); ΔH = – 297.4 kj mol-1 (iii) C(s) + O2(g) → CO2(g); ΔH= – 394.5 kj mol-1 Add (iii) + 2(ii) and subtract (i), it gives, on rearranging C(s) + 2S(s) → CS2(l); ΔH = (- 394.5) +- 2(- 297.4) – (- 110.2) = -879.1 kj mol-1 Thus the enthalpy of formation of CS2 = – 879.1 kJ mol-1
