Given the following thermochemical equations S (rhombic) + O2(g) = SO2 (g) Δ H = -297.5 KJ/Mol S (monoclinic) + O2 (g) = SO2 (g) Δ H = -300 KJ/Mol Calculate ΔH for the transformation of 1 gram atom of rhombic sulphur into monoclinic sulphur.

Answer 1

S (rhombic) + O2(g) = SO2 (g).............. Δ H = -297.5 KJ/Mol
S (monoclinic) + O2 (g) = SO2 (g) .........Δ H = -300 KJ/Mol
Write eqn 1 as is; reverse eqn 2 and add to eqn 1. Cancel items that appear on both sides which should cancel SO2(g) and O2(g) and leave you with the eqn you want or S(rhombic) ==> S(monoclinic) dH = ?
Calculate dH by adding dH from eqn 1 to dH of reversed equation 2.
Remember to change the sign of dH for eqn 2 when it is reversed.