Determine the ∆��rxn for the neutralization and write a thermochemical equation for the reaction.

Question

A chemist mixes 25.0 ml of 1.0 mol/L diprotic acid H2X (aq) with 50 ml.0 of 1.0 mol/L NaOH(aq) in a

coffee cup calorimeter. Both solutions are initially at 25.0℃. The temperature stabilizes to 33.9 ℃.

Determine the ∆��rxn for the neutralization and write a thermochemical equation for the reaction.

Answer 1

First, let's look at the balanced equation for the reaction:

H₂X + 2NaOH ==> 2H₂O + Na₂X

Next, we'll determine which, if either reactant, is limiting:

moles H2X = 25.0 ml x 1 L / 1000 ml x 1.0 mol / L = 0.025 mols H2X

mols NaOH = 50 ml x 1 L / 1000 ml x 1.0 mol / L = 0.05 mols NaOH

Since the mol ratio in the balanced equation shows 2 mols NaOH needed per 1 mol H2X, neither is limiting.

Next, we can use the equation q = mC∆T to find the heat generated by this reaction. We know that heat is generated (given off) because the temperature rises from 25º to 33.9º (an exothermic reaction).

q = heat = ?

m = mass = 25.0 ml + 50.0 ml = 75.0 ml x 1 g / ml = 75 g (assuming a density of the reaction soln of 1g/ml)

C = 4.184 J/gº (this is the specific heat for water)

∆T = 33.9º - 25º = 8.9º

Solving for q: q = (75 g)(4.184 J/gº)(8.9º)

q = 2793 J

This will be heat generated from 0.025 mols H2X, or from 0.05 mols NaOH. Thus, we can write the ∆Hrxn using units of kJ / mol as follows:

∆Hrxn = 2793 J / 0.025 mols x 1 kJ / 1000 J = 112 kJ / mol H₂X

The thermochemical reaction would be:

H₂X + 2NaOH ==> 2H₂O + Na₂X + heat

H₂X + 2NaOH ==> 2H₂O + Na₂X + 112 kJ