Question

Calculate the enthalpy change for the following reaction, using Hess's Law:2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)Given the following enthalpy changes: 2H2(g) + O2(g) → 2H2O(l)       ΔH1 = -484 kJ2SO2(g) + O2(g) → 2SO3(g)     ΔH2 = -198 kJ2H2(g) + S(s) → H2S(g)          ΔH3 = +25kJ

Answer 1

To solve this problem using Hess's Law, we need to manipulate the given reactions and their enthalpy changes to match the target reaction. Target reaction: 2H2S g  + 3O2 g   2H2O l  + 2SO2 g 1. Reverse the third reaction and multiply by 2:   2H2S g   4H2 g  + 2S s           H3' = -50 kJ2. Add the first reaction to the modified third reaction:   4H2 g  + 2S s  + 2H2 g  + O2 g   4H2O l  + 2S s  + 2SO2 g    H = H1 + H3' = -484 kJ - 50 kJ = -534 kJ3. Now we have the reaction:   4H2 g  + 2S s  + O2 g   4H2O l  + 2SO2 g 4. Divide the entire reaction by 2 to match the target reaction:   2H2S g  + 3O2 g   2H2O l  + 2SO2 g    H = -534 kJ / 2 = -267 kJThe enthalpy change for the target reaction is -267 kJ.