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What is the standard enthalpy change for the reaction between 1 mol of ethanoic acid and 1 mol of sodium hydroxide in aqueous solution, given that the reaction produces 1 mol of sodium ethanoate and 1 mol of water?

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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 .

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