Thermochemisty Q

Question

In one experiment, a student placed 50.0 mL of 1.00 M acetic acid (HC₂H₃O₂) at 25.5 °C in a coffee cup calorimeter. To this was added 50.0 mL of 1.50 M NaOH solution. The mixture was stirred, and the temperature quickly increased by +6.6 °C

The net ionic equation is:

What is the ∆Hº ?

Answer 1

The molecular equation is HC₂H₃O₂ + NaOH ==> NaC₂H₃O₂ + H₂O

The complete ionic equation is HC₂H₃O₂(aq) + Na⁺(aq) + OH^-(aq) ==> Na⁺(aq) + C₂H₃O₂^-(aq) + H₂O(l)

The NET IONIC equation is HC₂H₃O₂(aq) + OH^-(aq) ==> C₂H₃O₂^-(aq) + H₂O(l)

∆H = mC∆T

∆H = heat of reaction = ?

m = mass of water = 50.0 ml + 50.0 ml = 100.0 ml = 100.0 g (assuming density of 1g/ml)

C = specific heat for water = 4.184 J/g/deg

∆T = change in temperature = 6.6 degrees

∆H = (100.0 g)(4.184 J/g/deg)(6.6 deg) = 2761 J (This is the ∆H reaction)

If you want the molar ∆H, which is usually what is asked for, then we need to find the moles of acid and base and correct for that.

moles acetic acid = 0.050 L x 1.00 mol/L = 0.050 moles

moles NaOH = 0.05 L x 1.00 mol/L = 0.050 moles

Molar ∆H = 2761 J/0.05 moles = 55,220 J/mole = 55.22 kJ/mole