To calculate the standard enthalpy change for the reaction between 1 mol of ethanoic acid and 1 mol of sodium hydroxide in aqueous solution, we need to use the following formula:H_reaction = H_f products - H_f reactants where H_f is the standard enthalpy of formation for each compound.The reaction is as follows:CH3COOH aq + NaOH aq CH3COONa aq + H2O l We need the standard enthalpies of formation H_f for each compound:H_f CH3COOH = -484.5 kJ/molH_f NaOH = -425.6 kJ/molH_f CH3COONa = -379.2 kJ/molH_f H2O = -285.8 kJ/molNow, we can plug these values into the formula:H_reaction = [ -379.2 + -285.8 ] - [ -484.5 + -425.6 ]H_reaction = -664.0 - -910.1 H_reaction = 246.1 kJ/molThe standard enthalpy change for the reaction between 1 mol of ethanoic acid and 1 mol of sodium hydroxide in aqueous solution is 246.1 kJ/mol exothermic, since the value is positive .