Question

Calculate the enthalpy change for the reaction: 2H2(g) + O2(g) --> 2H2O(g) given the following enthalpy changes: H2(g) + 1/2O2(g) --> H2O(g) ΔH1 = -286 kJ/molH2(g) + 1/2Cl2(g) --> HCl(g) ΔH2 = -92 kJ/mol Cl2(g) --> 2Cl(g) ΔH3 = +242 kJ/molHCl(g) + 1/2O2(g) --> H2O(g) + Cl2(g) ΔH4 = -184 kJ/mol

Answer 1

To calculate the enthalpy change for the reaction 2H2 g  + O2 g  --> 2H2O g , we can use Hess's Law. Hess's Law states that the enthalpy change of a reaction is the same, whether it occurs in one step or several steps. We can manipulate the given reactions to create the desired reaction and then add the enthalpy changes.First, we need to manipulate the given reactions:1. H2 g  + 1/2O2 g  --> H2O g  H1 = -286 kJ/mol    This reaction is already in the correct form 2. H2 g  + 1/2Cl2 g  --> HCl g  H2 = -92 kJ/mol   HCl g  + 1/2O2 g  --> H2O g  + Cl2 g  H4 = -184 kJ/mol    Add these two reactions to eliminate HCl and Cl2    H2 g  + 1/2O2 g  --> H2O g  H2+4 =  -92  +  -184  = -276 kJ/molNow we have two reactions:1. H2 g  + 1/2O2 g  --> H2O g  H1 = -286 kJ/mol2. H2 g  + 1/2O2 g  --> H2O g  H2+4 = -276 kJ/molSince we need 2 moles of H2O in the final reaction, we can add these two reactions:1. H2 g  + 1/2O2 g  --> H2O g  H1 = -286 kJ/mol2. H2 g  + 1/2O2 g  --> H2O g  H2+4 = -276 kJ/mol---------------------------------------------------   2H2 g  + O2 g  --> 2H2O g  H =  -286  +  -276  = -562 kJ/molSo, the enthalpy change for the reaction 2H2 g  + O2 g  --> 2H2O g  is -562 kJ/mol.