Calculate the standard enthalpy change for the following reaction: 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) Given the following information: - The standard enthalpy of formation of NaOH(aq) is -469.20 kJ/mol - The standard enthalpy of formation of H2SO4(aq) is -814.50 kJ/mol - The standard enthalpy of formation of Na2SO4(aq) is -1388.10 kJ/mol Note: Make sure to balance the equation and use Hess's Law if necessary.

Question

Calculate the standard enthalpy change for the following reaction: 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) Given the following information: - The standard enthalpy of formation of NaOH(aq) is -469.20 kJ/mol - The standard enthalpy of formation of H2SO4(aq) is -814.50 kJ/mol - The standard enthalpy of formation of Na2SO4(aq) is -1388.10 kJ/mol Note: Make sure to balance the equation and use Hess's

Answer 1

To calculate the standard enthalpy change for the reaction, we can use Hess's Law. According to Hess's Law, the enthalpy change for a reaction is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.For the given reaction:2NaOH aq  + H2SO4 aq   Na2SO4 aq  + 2H2O l We have the following standard enthalpies of formation:Hf NaOH  = -469.20 kJ/molHf H2SO4  = -814.50 kJ/molHf Na2SO4  = -1388.10 kJ/molThe standard enthalpy of formation of H2O l  is -285.83 kJ/mol.Now, we can calculate the standard enthalpy change for the reaction  H :H = [Hf Na2SO4  + 2 * Hf H2O ] - [2 * Hf NaOH  + Hf H2SO4 ]H = [ -1388.10  + 2 *  -285.83 ] - [2 *  -469.20  +  -814.50 ]H =  -1388.10 - 571.66  -  -938.40 - 814.50 H = -1959.76 + 1752.90H = -206.86 kJTherefore, the standard enthalpy change for the given reaction is -206.86 kJ.