Question

Determine the standard enthalpy change of the reaction between 50.0 mL of 0.200 M hydrochloric acid (HCl) and 50.0 mL of 0.100 M sodium hydroxide (NaOH) at 25°C. The reaction is a neutralization reaction between HCl and NaOH, and the products are water and sodium chloride (NaCl). The molar enthalpy of formation of NaCl is -411.2 kJ/mol, and the specific heat capacity of the aqueous solution is 4.18 J/g°C.

Answer 1

First, we need to determine the limiting reactant in the reaction between HCl and NaOH. The balanced chemical equation for the reaction is:HCl aq  + NaOH aq   NaCl aq  + H2O l Moles of HCl =  volume in L    concentration in mol/L Moles of HCl =  50.0 mL   1 L / 1000 mL    0.200 mol/L = 0.0100 molMoles of NaOH =  volume in L    concentration in mol/L Moles of NaOH =  50.0 mL   1 L / 1000 mL    0.100 mol/L = 0.00500 molSince there are fewer moles of NaOH, it is the limiting reactant.Now, we can calculate the enthalpy change of the reaction using the molar enthalpy of formation of NaCl.H_reaction = moles of NaOH  H_formation of NaClH_reaction = 0.00500 mol   -411.2 kJ/mol  = -2.056 kJThe standard enthalpy change of the reaction between 50.0 mL of 0.200 M hydrochloric acid and 50.0 mL of 0.100 M sodium hydroxide at 25C is -2.056 kJ.