please help me... You are given the following standard enthalpies of formationat

Question

please help me... You are given the following standard enthalpies of formationat 25 0C. NH4OH(aq)          -80.29 kJ/mol H+(aq)                          0 kJ/mol NH4+(aq)              -132.5kJ/mol H2O(l)                  -285.85 kJ/mol a) Calculate the standard enthalpy of neutralization ofNH4OH(aq) NH4OH(aq) + H+(aq)   -> NH4+(aq) +H2O(l) b) Using the value of -56.2 kJ/mol as the standard enthalpychange for the reaction H+ (aq) + OH- (aq) -> H2O(l) calculate the standard enthalpy change for the reaction NH4OH(aq)                                       ->   NH4+(aq) +OH-(aq) please help me... You are given the following standard enthalpies of formationat 25 0C. NH4OH(aq)          -80.29 kJ/mol H+(aq)                          0 kJ/mol NH4+(aq)              -132.5kJ/mol H2O(l)                  -285.85 kJ/mol a) Calculate the standard enthalpy of neutralization ofNH4OH(aq) NH4OH(aq) + H+(aq)   -> NH4+(aq) +H2O(l) b) Using the value of -56.2 kJ/mol as the standard enthalpychange for the reaction H+ (aq) + OH- (aq) -> H2O(l) calculate the standard enthalpy change for the reaction

Explanation / Answer

      a)  NH4OH(aq) + H+(aq)   -> NH4+(aq) +H2O(l)              Horxn = Hofproducts - Hof reactants                           = [Hof NH4+ +Hof H2O ] - [Hof NH4OH +Hof H+ ]                           = (-132.5 kJ/mol + -285.85kJ/mol) - (-80.29 kJ /mol +0)                            =-338.06 kJ /mol      b)   H+ (aq) +OH- (aq) -> H2O (l)          Horxn = HofH2O - [Hof OH- + Hof H+]         -56.2 kJ /mol =(-285.85 kJ/mol ) - (Hof OH- + 0)                   Hof OH- = -229.65 kJ /mol          c)NH4OH(aq) ->  NH4+(aq) + OH-(aq)                      Horxn   = Hof NH4+ +Hof OH- - (Hof NH4OH)                         = -132.5 kJ /mol -229.65 kJ /mol - (-80.29 kJ /mol )                          = -281.86 kJ /mol

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