Given the following thermochemical data at 298 K and 1 barΔHvap ∘CH3OH=40.0kJmol−1ΔfH∘:H(g)=220kJmol−1O(g)=250kJmol−1C(g)=710kJmol−1Bond dissociation energy(C−H)=420kJmol−1(C−O)=350kJmol−1(O−H)=465kJmol−1 The ΔfH∘ of liquid methyl alcohol in kJmol−1 is

ΔfH∘=ΔfH∘(C,g)+4ΔfH∘(H,g)+ΔfH∘(O,g)−[3(BE)C−H+(BE)C−O+(BE)O−H=(710+4×220+250)−(30×420+350+465)=−235kJCH3OH(g)→CH3OH(l);ΔH∘=−40kJ Thus, ΔfH∘CH3OH,l=−235−40=−275kJmol−1 Note ΔfH∘, based on (BE) value requires all species in gaseous state thus, 40 kJ of heat is released which converted CH3OH (g) into CH3OH(l).