Question

Calculate the ∆Hº (on a per mole basis) for the HCl/NaOH neutralization. You may assume that the specific heat and density of the resulting solution to be the same as water. Be sure to use the combined total volumes of HCl and NAOH when calculating the total mass for this reaction. Remember to take into account the moles of limiting reagent when calculating the ∆Hº on a per mole basis.

Exact Concentrations: HCl = .9637 NaOH = 1.016

Trial 1 : temp of HCl = 19 degrees C Temp of NaOH = 19 degrees C constant temp of HCl & NaOH = 24.5 degrees C

Trial 2 : temp of HCl = 20 degrees C Temp of NaOH = 20 degrees C constant temp of HCl & NaOH = 24 degrees C

Answer 1

Solution:

Lets assume volume of both NaOH and HCl = 1.0 L

Calculation of moles of NaOH

n NaOH = molarity x volume in L

= 1.016 M x 1.0 L = 1.016 mol

n HCl = 0.9637 M x 1.0 L = 0.9637 mol

Limiting reactant:

Neutralization reaction

NaOH (aq) + HCl (aq) --- > NaCl (aq) + H2O (l)

Mol ratio of NaOH : HCl is 1 : 1

So moles of HCl are limiting reactant.

Calculation of average Delta T

Trial 1

Delta T = 24.5 – 19 deg C = 5.5 deg C

Trial 2

Delta T = 24 -19 deg C = 5 deg C

Average Delta T = 5.5 + 5.0 / 2

= 10.5 /2 = 5.25 deg C

Lets assume density of the solution = 1.0 g/mL

Heat capacity = 4.184 J/g C

Lets calcuatle q (heat )

q = m C Delta T

m is mass of the solution

C is heat capacity

Mass of solution = 2.0 L (both volume HCl and NaoH ) x (1000 mL / 1 L) x 1g / mL

= 2000 g

Lets plug this value in order to get q

q = 2000 g x 4.184 J/gC x 5.25 deg C

= 43932 J

Delta H rxn = - q / n

Here n is (mol HCl)

Delta H rxn = -43932 J/0.9637 mol

= -45586.8 J/mol

Delta Hrxn = -45586.8 J/mol