In: Chemistry
Acetic Acid |
NaOH |
Ti°C |
Tf°C |
|
Trial 1 |
5mL |
5mL |
23.5 |
28.6 |
Trial 2 |
5mL |
5mL |
23.8 |
29.3 |
Write a balanced equation chemical equation for the neutralization reaction.
Using the density of the salt solution, calculate the mass of the salt solution for each trial.
Calculate the heat (qrxn) for each trial
Calculate the average(qrxn).
From the balanced equation, calculate the moles of water formed in the reaction.
Calculate Enthalpy of Neutralization (△H neutralization) in kJ/mol.
Calculate % error by comparing the calculated value to the actual value.
-density of sodium acetate solution= 1.53g/mL
-Enthalpy of Neutralization for acetic acid with NaOH is 55.2kJ
a) The balanced chemical equation for the reaction between acetic acid (CH3COOH) and NaOH is
CH3COOH (aq) + NaOH (aq) -------> CH3COONa (aq) + H2O (l)
b) Total volume of the solution for each trial = (5 + 5) mL = 10 mL.
Density of sodium acetate (CH3COONa) is 1.53 g/mL.
Mass of CH3COONa = (volume of salt solution)*(density of CH3COONa) = (10 mL)*(1.53 g/mL) = 15.3 g.
c) Assume the specific heat of the salt solution to be the same as the specific heat of water, i.e, C = 4.18 J/g.⁰C (you haven’t provided a different value).
Trail 1: qrxn = (mass of CH3COONa)*(specific heat of CH3COONa)*(change in temperature) = (15.3 g)*(4.18 J/g.⁰C)*(28.6 – 23.5)⁰C = 326.1654 J (ans).
Trial 2: qrxn = (mass of CH3COONa)*(specific heat of CH3COONa)*(change in temperature) = (15.3 g)*(4.18 J/g.⁰C)*(29.3 – 23.8)⁰C = 351.7470 J (ans).
d) Average qrxn = ½*[(qrxn, Trial 1) + (qrxn, Trial 2)] = ½*(326.1654 + 351.7470) J = 338.9562 J (ans).
e) As per the balanced stoichiometric equation in part (a) above,
1 mole CH3COONa = 1 mole H2O
We have formed 15.3 g CH3COONa.
Molar mass of CH3COONa = (2*12.01 + 3*1.008 + 2*15.9994 + 1*22.9897) g/mol = 82.0325 g/mol.
Moles of CH3COONa = moles of H2O = (15.3 g)/(82.0325 g/mol) = 0.1865 mole.
f) Enthalpy of neutralization = (Average qrxn)/(moles of H2O) = (338.9562 J)/(0.1865 mole) = 1817.4595 J/mol = (1817.4595 J/mol)*(1 kJ/1000 J) = 1.8174595 kJ/mol ≈ 1.82 kJ/mol (ans).