Question

In determining molar enthalpies of neutralization, the contribution of qcal to the total heat evolved is often neglected, and Equation 1 is simplified to ΔH = −ΔT × (heat capacity of contents). Recalculate ΔHneut for determination 1 (table below) using this simplification and compare your answer with the ΔHneut your obtained in this experiment. Does it seem reasonable to neglect the contribution of qcal?

I calculated ΔHneut to be -5.0418 E-5

Chemical system:

HCl + NaOH

	Determination 1	Determination 2
Molarity of acid, M	1.00	1.00
Molarity of NaOH, M	1.05	1.05
Volume of acid, mL	49.9	51.1
Volume of NaOH, mL	49.8	50.0
Temperature of acid: Ti, 0C	21.7	22.6
Max temperature: Tf, 0C	29.7	30.3

Answer 1

Determination 1: Heat gained by acid solution = Q = m * s * (T₂-T₁)

m = mass of solution = volume * density = 49.9 ml* 1(g/ ml) = 49.9 grams

s = heat capacity of water = 4.18 J / g C, T2 = 29.7 C, T1 = 21.7 C

So, Q = 49.9 grams * 4.18 J/ g C * (29.7-21.7) C = 1669 Joules

Now 49.9 ml of 1 M acid gets neutralised by 49.8 ml of 1.05 M base

49.9 ml of 1 M acid contains (49.9*1)/1000 mole = 0.0499 mole of acid

49.8 ml of 1.05 M base contains (49.9*1.05)/1000 mole = 0.0523 mole of base

So, 0.0499 mole of acid in neutralised by 0.0499 mole of base and the neutralisation process releases 1669 Joules of heat.

So, heat released per mole of neutralisation is 1669/ 0.0499 Joules = 33446 J = 33.446 KJ heat

But actual heat of neutralistaion of HCl and NaOH is 55 KJ/ mole which greater than calculated value.

It is because we have not included heat gained by calorimeter (Qcal). So, it is not reasonable in this case to neglect Qcal.