Question

A student reaceted 50.0 ml of 1.05 N sodium hydroxide with 50.0 ml of 1.00 N sulfuric acid in a coffee cup calorimeter. From the temperature vs time plot, the average Ti was 25.7 celsius and the Tf of the reaction mixture was 31.2 Celsius. (ignore calorimeter constant).

What is the enthalpy of neutralization?

Answer 1

Coffee cup calorimeter: is a constant pressure calorimeter. In this case, the heat that is measured is equivalent to the change in enthalpy

NaOH taken = 50 mL (1.05 N) = 52.5 mmol (mol = Normality x volume /acidity factor)

H₂SO₄ taken = 50 mL (1N) = 25 mmol ((mol = Normality x volume /basicity factor)

Total volume of the solution after mixing = 100 mL

Change in temperature, ∆T = Tf-Ti = 31.2 – 25.7 = 5.5

Specific heat capacity of water, c = 4.186 J/g ^oC

The neutralization reaction is:

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + H₂O(l) + heat

Heat released during the reaction will be absorbed by Na₂SO₄ solution.

Assumptions are:

1. No energy is lost to the surroundings (coffe cup calorimeter)
2. Density of the solution and the specific heat capacity of the solution are approximately those of water – 1g/mL and 4.186 J/g^oC

If q (solution) = energy released in the reaction = energy absorbed by the solution, then

q = m. c. ∆T    (where m = mass of resultant solution = 100 g)

= 100 g x 4.186 J/g^oC x 5.5 ^oC = 2.302 kJ

At constant pressure, enthalpy of the reaction is given by ∆H = q/v, (where v = moles of the limiting reactant; in this case H2SO4 = 0.025 mol)

∆H = q/v = 2.302 kJ / 0.025 mol = 92.08 kJ/mol

Enthalpy of neutralization, ∆H = 92.08 kJ/mol