Question

Calculate the standard enthalpy change (∆H°) for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) given the following information:- Balanced chemical equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)- ∆H°f of NaCl(aq) = -407.3 kJ/mol- ∆H°f of H2O(l) = -285.8 kJ/mol- Heat capacity of the system, Cp = 4.18 J/g.K- Mass of the HCl solution used = 50.0 g- Concentration of HCl solution = 0.100 M- Volume of NaOH solution used = 100.0 mL- Concentration of NaOH solution = 0.200 MAssume that the specific heat of the solution is equal to that of water at 4.18 J/g.K and that the density of both solutions is equal to 1.00 g/mL.

Answer 1

First, we need to determine the limiting reactant. To do this, we'll find the moles of HCl and NaOH in the reaction.Moles of HCl =  concentration of HCl    volume of HCl solution Moles of HCl =  0.100 mol/L    50.0 g / 1.00 g/mL    1 L / 1000 mL  = 0.00500 molMoles of NaOH =  concentration of NaOH    volume of NaOH solution Moles of NaOH =  0.200 mol/L    100.0 mL / 1000 mL/L  = 0.0200 molSince the balanced chemical equation has a 1:1 ratio of HCl to NaOH, HCl is the limiting reactant.Next, we'll use Hess's Law to find the standard enthalpy change for the reaction:H = [Hf NaCl  + Hf H2O ] - [Hf HCl  + Hf NaOH ]Since HCl and NaOH are strong acids and bases, we can assume their standard enthalpies of formation are approximately zero. Therefore, the equation simplifies to:H = [Hf NaCl  + Hf H2O ]Now, we can plug in the given values for the standard enthalpies of formation:H = [ -407.3 kJ/mol  +  -285.8 kJ/mol ] = -693.1 kJ/molThe standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide is -693.1 kJ/mol.