Question

Calculate the enthalpy change for the following reaction using Hess's Law calculations:2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)Given the following information:2H2(g) + O2(g) → 2H2O(l) ΔH1 = -484 kJ/mol2S(s) + 3O2(g) → 2SO3(g) ΔH2 = -791 kJ/mol2H2(g) + S(s) → 2H2S(g) ΔH3 = -36 kJ/mol

Answer 1

To solve this problem using Hess's Law, we need to manipulate the given reactions in such a way that when they are added together, they will result in the desired reaction. First, we need to reverse the third reaction and multiply it by 2:2H2S g   4H2 g  + 2S s  H3' = +72 kJ/molNow, we need to subtract 2SO3 g  from the second reaction to get 2SO2 g  in the products:2SO3 g   2S s  + 3O2 g  H2' = +791 kJ/mol2SO3 g   2SO2 g  + O2 g  H4 = unknownSubtracting the second equation from the first:2S s  + 2O2 g   2SO2 g  H4 = -198 kJ/molNow, we can add the manipulated reactions together:2H2 g  + O2 g   2H2O l  H1 = -484 kJ/mol2H2S g   4H2 g  + 2S s  H3' = +72 kJ/mol2S s  + 2O2 g   2SO2 g  H4 = -198 kJ/mol---------------------------------------------2H2S g  + 3O2 g   2H2O l  + 2SO2 g  H = ?Adding the enthalpy changes:H = H1 + H3' + H4H = -484 kJ/mol + 72 kJ/mol - 198 kJ/molH = -610 kJ/molSo, the enthalpy change for the given reaction is -610 kJ/mol.