Enthalpy change of neutralisation.

Assumptions:

The density of acids is equal to the density of water and amounts to 1.00 g cm-3

Key variables:

m – mass of a substance in grams

s – specific heat capacity in J g-1 K-1

∆T - the amount by which the temperature is increased in K

Planning (b)

Requirements:

• 1 burette (25 ml)
• 2 beakers
• 3 calibrated flasks (500 ml)
• 1 plastic bottle (1500 ml)
• phenolphthalein

Procedure:

We were provided with 2 mol dm-3 hydrochloric acid (HCl), 2 mol dm-3 nitric acid (HNO3), 2 mol dm-3 potassium hydroxide (KOH), 2 mol dm-3 sodium hydroxide (NaOH) and 4 mol dm-3 sodium hydroxide (NaOH).

1. We measured 30 cm3 of approximately of 2 mol dm-3 nitric acid into the beaker.
2. We took the temperature of the nitric acid and recorded it in table 1.
3. We measured 30 cm3 of approximately of 2 mol dm-3 sodium hydroxide into the beaker.
4. We took the temperature of the sodium hydroxide and recorded it in table 1.
5. Subsequently we added the NaOH to the HNO3 and stirred the mixture carefully with the thermometer.
6. While mixing we recorded the maximum temperature of the solution.
7. We repeated it 5 times with different sets of acids and hydroxides.

1. hydrochloric acid and sodium hydroxide
2. hydrochloric acid and potassium hydroxide
3. nitric acid and potassium hydroxide
4. nitric acid and sodium hydroxide
5. hydrochloric acid and 4 mol dm-3 sodium hydroxide

Data Collection

a)

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

H+ + Cl- + Na+ + OH- → Na+ + Cl- + H2O

H+ + OH-→ H2O

Table 1.

b)

HCl (aq) + KOH (aq) → KCl (aq) + H2O (l)

Table 2.

c)

HNO3 (aq) + KOH (aq) → KNO3 (aq) + H2O (l)

Table 3.

d) 

HNO3 (aq) + NaOH (aq) → NaNO3 (aq) + H2O (l)

Table 4.

e)

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

Table 5.

Data Processing and Presentation

Heat required = ms∆T

m =d V

n = c V

∆T = Tmix - (T1 + T2)

∆H = heat required * 1/n

s = 4.18 J g-1 K-1

The amount of heat required to heat the water can be calculated as follows (we assume that the heat energy required to change the temperature of the other substances present may be ignored):

a) HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

V = 60 cm3

d = 1.00 g cm-3

m = d V = 60 cm3 * 1.00 g cm-3 = 60 g

∆T = Tmix - ½(T1 + T2) = 31.0 oC - 20.0 oC = 11.0 oC

heat required = ms∆T = 60.0 g * 4.18 J g-1 K-1  * 11.0 oC = 2758 J = 2.758 kJ

nHCl = c V = 2 mol dm-3  * 0.3 dm-3  = 0.06 moles

nNaOH = c V = 2 mol dm-3  * 0.3 dm-3  = 0.06 moles

∆H = heat required * 1/n = 2.758 kJ * 1/0.06 moles = 45.97 kJ mol-1

∆H = - 45.97 kJ mol-1

b) HCl (aq) + KOH (aq) → KCl (aq) + H2O (l)

V = 60 cm3

d = 1.00 g cm-3

m = d V = 60 cm3 * 1.00 g cm-3 = 60 g

∆T = Tmix - ½(T1 + T2) = 32.0 oC - 20.5 oC = 10.5 oC

heat required = ms∆T = 60.0 g * 4.18 J g-1 K-1  * 10.5 oC = 2633.40 J = 2.633 kJ

nHCl = c V = 2 mol dm-3  * 0.3 dm-3  = 0.06 moles

nKOH = c V = 2 mol dm-3  * 0.3 dm-3  = 0.06 moles

∆H = heat required * 1/n = 2.633 kJ * 1/0.06 moles = 43.88 kJ mol-1

∆H = - 43.88 kJ mol-1

c) HNO3 (aq) + KOH (aq) → KNO3 (aq) + H2O (l)

V = 60 cm3

d = 1.00 g cm-3

m = d V = 60 cm3 * 1.00 g cm-3 = 60 g

∆T = Tmix - ½(T1 + T2) = 33.0 oC - 21.75 oC = 11.25 oC

heat required = ms∆T = 60.0 g * 4.18 J g-1 K-1  * 11.25 oC = 2821.50 J = 2.822 kJ

nHCl = c V = 2 mol dm-3  * 0.3 dm-3  = 0.06 moles

nKOH = c V = 2 mol dm-3  * 0.3 dm-3  = 0.06 moles

∆H = heat required * 1/n = 2.822 kJ * 1/0.06 moles = 47.03 kJ mol-1

∆H = - 47.03 kJ mol-1

d) HNO3 (aq) + NaOH (aq) → NaNO3 (aq) + H2O (l)

V = 60 cm3

d = 1.00 g cm-3

m = d V = 60 cm3 * 1.00 g cm-3 = 60 g

∆T = Tmix - ½(T1 + T2) = 33.0 oC - 22.75 oC = 10.25 oC

heat required = ms∆T = 60.0 g * 4.18 J g-1 K-1  * 10.25 oC = 2570.70 J = 2.571 kJ

nHCl = c V = 2 mol dm-3  * 0.3 dm-3  = 0.06 moles

nKOH = c V = 2 mol dm-3  * 0.3 dm-3  = 0.06 moles

∆H = heat required * 1/n = 2.751 kJ * 1/0.06 moles = 45.85 kJ mol-1

∆H = - 45.85 kJ mol-1

e) HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

V = 60 cm3

d = 1.00 g cm-3

m = d V = 60 cm3 * 1.00 g cm-3 = 60 g

∆T = Tmix - ½(T1 + T2) = 33.0 oC - 21.5 oC = 11.5 oC

heat required = ms∆T = 60.0 g * 4.18 J g-1 K-1  * 11.5 oC = 2884.20 J = 2.884 kJ

nHCl = c V = 2 mol dm-3  * 0.3 dm-3  = 0.06 moles

nKOH = c V = 2 mol dm-3  * 0.3 dm-3  = 0.06 moles

∆H = heat required * 1/n = 2.884 kJ * 1/0.06 moles = 48.07 kJ mol-1

∆H = - 48.07 kJ mol-1

Conclusion and Evaluation

As we can see from the results above, the prediction made at the very beginning of this lab was correct. Neither type of acid or base nor the concentration of acid does not have influence on the enthalpy of neutralisation.

Hence we may assume that the enthalpy of neutralisation is equal to the enthalpy change for

H+ + OH-→ H2O.

The enthalpy change for this reaction, however, is –57.9 kJ mol-1. The differences between my results and the theoretical value may come from the fact that the measurements were not very accurate. The temperatures of the acids, bases and mixtures might have been influenced by cool beakers. Therefore the temperatures were a bit lower than they should have been. If the ∆T was higher by 3oC, the enthalpy of neutralisation would be almost the same as in the sources.

I do not know how to improve the experiment so that data gathered will be similar to theoretical values. I reckon in classroom conditions such mistake is not a serious one.

Sources:

1. Green J, Damji S. 2001. Chemistry. Second edition. IBID Press, Victioria, Australia.
2. http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_neutralisation

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