Calculate H for the process Zns Ss 2O g ZnSO s 298 24 from
Calculate H°   for the process Zn(s) + S(s) + 2O (g)  ZnSO (s) 298 24
 from the following information:
 Zn(s) + S(s)  ZnS(s)   H°   = 206.0 kJ 298
 ZnS(s) + 2O (g)  ZnSO (s)   H°   = 776.8 kJ
Solution
1) Zn + S ---> ZnS , ,,, dH1 = -260 kJ
2) ZnS + 2 O ---> ZnSO ,,,,, dH2 = -776.8 kJ
3) Zn + S + 2 O ---> ZnSO ,,,,,, dH3 = ???
it can be derived that
equation 3 = equation 1 + equation 2
now according to Hess`s law
dH3 = dH1 + dH2
dH3 = -206 kJ + (-776.8 kJ)
dH3 = -982.8 kJ
so
the required dHo value is -982.8 kJ

