Question

calculate the heats of combustion for the following reactions using the standard entalpies
- C2H4(g) + 3O2(g) --> 2CO2(g) + 2H2O(l)
-2H2S(g) + 3O2(g) --> 2H2O(l) + 2SO2(g)

Answer 1

For a reaction,ΔH^o=ΣΔH_f^o(products)-ΣΔH_f^o(reactants)

Look up thermodynamic data:

ΔH_f^o(C₂H₄)=+52.47kJ/mol;ΔH_f^o(O₂)=0;ΔH_f^o(CO₂)=-393.51kJ/mol;ΔH_f^o(H₂Oliquid)=-285.83kJ/mol;

ΔH_f^o(H₂S)=-20.63kJ/mol;ΔH_f^o(SO₂)=-296.83kJ/mol

For reaction1,ΔH^o=2ΔH_f^o(CO₂)+2ΔH_f^o(H₂O)-ΔH_f^o(C₂H₄)=(2*-393.51kJ/mol)+(2*-285.83kJ/mol)-(+52.47kJ/mol)=-1411.15kJ/mol;

So combustion heat=1411.15kJ/mol.

For reaction2,ΔH^o=2ΔH_f^o(H₂O)+2ΔH_f^o(SO₂)-2ΔH^o_f(H2S)=(2*-285.83kJ/mol)+(2*-296.83kJ/mol)-(2*-20.63kJ/mol)=-1124.06kJ/mol;

It's for every 2 moles of H2S, so combustion heat=1124.06/2=562.03kJ/mol.