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Determine the enthalpy of hydrolysis for sodium acetate (NaC2H3O2) given that the enthalpy of neutralization for acetic acid (HC2H3O2) and sodium hydroxide (NaOH) is -55.9 kJ/mol and the heat of formation for NaC2H3O2 is -382.7 kJ/mol.

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To determine the enthalpy of hydrolysis for sodium acetate  NaC2H3O2 , we need to consider the reactions involved in the process. The hydrolysis of sodium acetate can be represented by the following equation:NaC2H3O2  aq  + H2O  l   HC2H3O2  aq  + NaOH  aq The enthalpy change for this reaction  H_hydrolysis  can be calculated using Hess's Law, which states that the enthalpy change for a reaction is the same, whether it occurs in one step or several steps. We can use the given enthalpy values for the neutralization of acetic acid and sodium hydroxide and the heat of formation for sodium acetate to find the enthalpy change for the hydrolysis reaction.First, let's consider the neutralization reaction:HC2H3O2  aq  + NaOH  aq   NaC2H3O2  aq  + H2O  l       H_neutralization = -55.9 kJ/molNow, we need to reverse this reaction to get the formation reaction for sodium acetate:NaC2H3O2  aq  + H2O  l   HC2H3O2  aq  + NaOH  aq       H_formation = - -55.9 kJ/mol  = 55.9 kJ/molNow, we can use the heat of formation for sodium acetate  -382.7 kJ/mol  and the enthalpy change for the formation reaction  55.9 kJ/mol  to find the enthalpy change for the hydrolysis reaction:H_hydrolysis = H_formation - H_neutralizationH_hydrolysis =  -382.7 kJ/mol  -  55.9 kJ/mol H_hydrolysis = -438.6 kJ/molTherefore, the enthalpy of hydrolysis for sodium acetate  NaC2H3O2  is -438.6 kJ/mol.

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