Question

Calculate the heat capacity of the calorimeter system using equation 3 and the molar enthalpy of neutralization ΔHneut (provided on p. 2).

Equation 3:

ΔHtransformation = −ΔHcalorimeter
=−CP,cal(Tf−Ti) (3)

ΔHneut: is −56.02 kJ per mole

Limiting reagent: 0.00112 moles of NaOH

Tf = 29.0 C

Ti = 22.50 C

Mass of dry calorimeter assembly: 6.36 g

Mass of calorimeter after HCl added: 54.05 g

Mass of calorimeter with NaOH solution: 103.77 g

Answer 1

NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H₂O(l) ; ΔH = -56.02 kJ/mole

Molar mass of HCl = 36.5 g/mole

Thus, moles of HCl in (54.05 - 6.36)g of it = mass/molar mass = 47.69/36.5 = 1.307

Molar mass of NaOH = 40 g/mole

moles of NaOH in (103.77 - 54.05)g of it = 49.72/40 = 1.243

Now, as per the balanced reaction , NaOH and HCl reacts in the molar ratio of 1:1

Thus, NaOH is limiting reagent

Now, heat released on the neutralisation reaction = moles reacting*ΔH = 1.243*56.02 = 69.63 kJ

Now, heat released in the reaction = heat capacity of calorimeter*rise in temperature

or, 69630 = heat capacity*(29 - 22.5)

or, heat capacity of calorimeter = 10.713 J/⁰C