Question

Calculate the standard enthalpy change for the reaction: 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)Given the following data:ΔHf°(NaOH) = -469.14 kJ/molΔHf°(H2SO4) = -814.00 kJ/molΔHf°(Na2SO4) = -1388.40 kJ/molΔHf°(H2O(l)) = -285.83 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following formula:H reaction  =  Hf products  -  Hf reactants For the reactants:2 moles of NaOH: 2   -469.14 kJ/mol  = -938.28 kJ1 mole of H2SO4: 1   -814.00 kJ/mol  = -814.00 kJFor the products:1 mole of Na2SO4: 1   -1388.40 kJ/mol  = -1388.40 kJ2 moles of H2O l : 2   -285.83 kJ/mol  = -571.66 kJNow, we can plug these values into the formula:H reaction  =  -1388.40 kJ +  -571.66 kJ   -  -938.28 kJ +  -814.00 kJ  H reaction  =  -1960.06 kJ  -  -1752.28 kJ H reaction  = 207.78 kJThe standard enthalpy change for the reaction is 207.78 kJ.