Question

Calculate the standard enthalpy change for the neutralization reaction between 50.0 mL of 0.100 M hydrochloric acid and 75.0 mL of 0.080 M sodium hydroxide. Given: the specific heat capacity of the solution is 4.18 J/g°C and the density of the solution is 1.00 g/mL. The balanced chemical equation for the reaction is HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l).

Answer 1

To calculate the standard enthalpy change for the neutralization reaction, we first need to determine the limiting reactant and the amount of heat released during the reaction.1. Determine the limiting reactant:Moles of HCl =  volume in L    concentration in mol/L  =  50.0 mL  1 L/1000 mL   0.100 M = 0.00500 molMoles of NaOH =  75.0 mL  1 L/1000 mL   0.080 M = 0.00600 molSince there is a 1:1 ratio between HCl and NaOH in the balanced equation, HCl is the limiting reactant.2. Determine the heat released during the reaction:The standard enthalpy of neutralization for a strong acid and strong base is approximately -55.8 kJ/mol. Since HCl is the limiting reactant, the heat released during the reaction can be calculated as follows:q = moles of HCl   -55.8 kJ/mol  = 0.00500 mol   -55.8 kJ/mol  = -0.279 kJ3. Calculate the standard enthalpy change:The standard enthalpy change for the reaction can be calculated using the heat released and the moles of the limiting reactant:H = q / moles of limiting reactant = -0.279 kJ / 0.00500 mol = -55.8 kJ/molTherefore, the standard enthalpy change for the neutralization reaction between 50.0 mL of 0.100 M hydrochloric acid and 75.0 mL of 0.080 M sodium hydroxide is -55.8 kJ/mol.