Question

Calculate the standard enthalpy change for the precipitation reaction of copper(II) sulfate and sodium hydroxide, given the following balanced chemical equation:CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq)Assume the reaction takes place under standard conditions and the necessary species are present in their standard states. Round your answer to two decimal places.

Answer 1

To calculate the standard enthalpy change for the reaction, we need the standard enthalpies of formation  Hf  for each species involved in the reaction. The standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states.The standard enthalpies of formation  Hf  for the species involved in the reaction are as follows:CuSO4 aq : -769.98 kJ/molNaOH aq : -469.15 kJ/molCu OH 2 s : -1004.45 kJ/molNa2SO4 aq : -1373.63 kJ/molUsing these values, we can calculate the standard enthalpy change  H  for the reaction using the following equation:H = [  Hf products  -   Hf reactants ]H = [ 1 mol  -1004.45 kJ/mol  +  1 mol  -1373.63 kJ/mol ] - [ 1 mol  -769.98 kJ/mol  +  2 mol  -469.15 kJ/mol ]H =  -1004.45 - 1373.63  -  -769.98 - 2  469.15 H =  -2378.08  -  -1708.28 H = -669.80 kJ/molThe standard enthalpy change for the precipitation reaction of copper II  sulfate and sodium hydroxide is -669.80 kJ/mol.