Enthalpy of combustion of propane, graphite and H2 at 298K are C3H8(g)+5O2(g)→3CO2(g)+4H2O(I),∆H1=−2220‌kJ‌mol−1 C(‌ graphite ‌)+O2(g)→CO2(g),‌‌∆H2=−393.5‌kJ‌mol−1 H2(g)+‌
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O2(g)→H2O(I),‌‌∆H3=−285.8‌kJ‌mol−1 The desired reaction is 3C‌ (graphite) ‌+4H2(g)→C3H8(g) ∆Hf=3∆H2+4∆H3−∆H1 =3(−393.5)+4(−285.8)−(−2220) =−103.7‌kJ‌mol−1 |∆Hf|≃104‌kJ‌mol−1