Question

Calculate the standard enthalpy change for the reaction: 2H2(g) + O2(g) --> 2H2O(l)Given the following enthalpies of formation: ∆Hf°(H2O, l) = -285.8 kJ/mol ∆Hf°(H2, g) = -241.8 kJ/mol ∆Hf°(O2, g) = 0 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following formula:H reaction  =  [Hf products ] -  [Hf reactants ]For this reaction, we have:H reaction  = [2 * Hf H2O, l ] - [2 * Hf H2, g  + Hf O2, g ]Substitute the given values:H reaction  = [2 *  -285.8 kJ/mol ] - [2 *  -241.8 kJ/mol  + 0 kJ/mol]H reaction  =  -571.6 kJ/mol  -  -483.6 kJ/mol H reaction  = -88 kJ/molThe standard enthalpy change for the reaction is -88 kJ/mol.