please help me... You are given the following standard enthalpies of formationat
ID: 679625 • Letter: P
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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 reactionExplanation / 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 /molRelated Questions
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