3.2 HEAT OF REACTION

3.2 HEAT OF REACTIONS

3.2.1 determine heat of precipitation through activity.
3.2.2 determine heat of displacement through activity.
3.2.3 compare heat of neutralisation through experiments for reactions
between item:
3.2.4
(a) strong acid and strong alkali,
(b) weak acid and strong alkali,
(c) strong acid and weak alkali,
(d) weak acid and weak alkali.
compare heat of combustion for various types of alcohol through
experiment.

1. The heat of reaction can be
determined through experiments by
establishing the temperature change
when the reaction occurs.

2. The value of temperature change
obtained is used to calculate heat of
reaction.



Assumptions a during calculations:

Heat of precipitation

1. The thermochemical equation for the formation of barium
sulphate, BaSO4 precipitate is:

2. Based on the thermochemical equation, 42 kJ of heat is
released when 1 mole of barium sulphate, BaSO4 precipitate is
formed. Therefore, the heat of precipitation of barium sulphate,
BaSO4 is -42 kJ mol−1.

Write the balanced of chemical equation and ionic equation

Chemical equation Ionic equation
a) Pb(NO3)2 + Na2SO4 → Pb2+ + SO42- →

b) BaCl2 + Na2SO4 →

c) AgNO3 + KCl →

d) Pb(NO3)2 + 2KI →















1 : Solve the question

25.0 cm3 of 2.0 mol dm-3 barium chloride, BaCl2 solution is added to the 25.0 cm3 of
2.0 mol dm-3 potassium sulphate, K2SO4 solution. Calculate the heat of precipitation of
this reaction, when the temperature increased 10.0 0C during experiment

2 : Solve the question

50 cm3 of 0.5 moldm-3 silver nitrate, AgNO3 solution at 29.5 0C is added to 50 cm3 of
0.5 mol dm-3 potassium chloride, KCl solution which is at a temperature of 28.5 0C.
The mixture is stirred and the highest temperature reached is 32.0 0C. Calculate the
heat of precipitation for silver chloride









Result : Reaction between silver nitrate solution and sodium chloride solution

Measurement Temperature ( °C )
Initial temperature of silver nitrate solution 28.1
Initial temperature of sodium Chloride 28.3
solution
Average initial temperature mixture 31.0
Highest temperature mixture
Temperature change

3 : Solve the question

Diagram 34 shows the set-up of the apparatus used in an experiment to determine
the heat of precipitation.
25.0 cm3 of 0.5 moldm-3 of barium chloride solution is added with 25.0 cm3 of 0.5
moldm-3 of sodium sulphate solution.

The following data in was obtained.

a) State the meaning of heat of precipitation in this experiment ?
is the heat released when 1 mole of barium sulphate is formed from the reaction between sodium
sulphate solution and barium chloride solution
b) State one observation of the experiment.
white precipitate produced (due to formation of BaSO4) // Thermometer reading increases (because
temperature increase) // polystyrene cup become hotter

c) Write the ionic equation for the reaction that occurred.

d) Calculate:
i) The change of heat in the experiment.
[Specific heat capacity of water = 4.2 J g-1 oC-1; density of solution = 1 g cm-3]

ii) The heat of precipitation in the experiment.

Heat of displacement

1. The following thermochemical equation represents a displacement reaction.

2. Based on the thermochemical equation, 250 kJ of heat is released when one mole
of copper, Cu is displaced from copper(II) sulphate, CuSO4 solution by iron, Fe.
3. Therefore, the heat of displacement of copper from copper(II) sulphate, CuSO4
solution by iron, Fe is -250 kJ mol−1.







1 : Solve the question

When excess magnesium powder is added to 50 cm3 of 0.2 mol dm-3 iron (II) nitrate solution,
the temperature increases from 30.5 0C to 40.0 0C. What is the heat of displacement of iron?
Specific heat capacity of solution = 4.2 J g-1 oC-1; density of solution =1 g cm-3]

2 : Solve the question

The thermochemical equation represents the reaction between iron metal and copper(II)
sulphate solution.
Fe + CuSO4 → FeSO4 + Cu H = -150 kJ mol-1
What is the increasing in temperature of the mixture when iron powder is added to 200 cm3 of
0.5 mol dm-3 copper(II) sulphate solution.

3 : Solve the question
Diagram 14 shows the set-up of apparatus of an experiment to determine the heat of
displacement for the reaction between copper(II) sulphate solution and excess zinc
powder.

a) Write the ionic equation for the reaction in this exsperiment.

b) State one observation from the experiment. Zn + CuSO4 ---> ZnSO4 + Cu

Zinc powder dissolve // brown solid deposited // thermometer reading increases // intensity of blue
colour decrease (based on balance equation.... by comparing reactants and products)

c) Based on the experiment, calculate:

i) heat released ii) number of moles of copper that is formed

iii) heat of displacement for the reaction

d) Why zinc powder is used in the experiment?
to increase rate of reaction

e) Draw the energy level diagram for the reaction

f) The experiment is repeated by replacing the zinc powder with magnesium powder.
i) Predict the heat of displacement that will be obtained.

Higher than heat of displacement when Zinc reacts with CuSO4 solution

ii) Explain
Distance between Magnesium and Copper is further than Zinc and Copper in Electrochemical series







Procedure:

1. Measure 25 cm3 of 0.5 mol dm−3 copper(II) sulphate, CuSO4 solution and pour it
into a polystyrene cup.
2. Put a thermometer into the solution and leave aside for two minutes.
3. Record the initial temperature of the solution in a table..
4. Add one spatula of magnesium powder, Mg quickly and carefully into the
polystyrene cup..
5. Close the polystyrene cup and stir the mixture using the thermometer.
6. Record the highest temperature of the mixture.
7. Repeat steps 1 to 6 by using zinc powder, Zn to replace magnesium, Mg

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Result : Reaction between zinc, Zn and Magnesium , Mg with copper(II) sulphate,
CuSO4 solution respectively

Measurement (°C ) Magnesium Zinc
28.0
Initial temperature of copper(II) sulphate solution 28.0 33.0

Highest temperature mixture 39.0

Temperature change

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Discussion

1. (a)

(b) ΔH of copper by Mg = - 92.4 KJ/mol (show ur working solution properly)
ΔH of copper by Zn = - 42.0 KJ/mol (show ur working solution properly)

(c) Magnesium is more electropositive than zinc. The distance between Mg and
Cu in Electrochemical series is further than the distance between Zn and Cu.
So, heat of displacement of copper by Magnesium is higher compare by Zinc.

(d)

2. (e) To make sure all copper(II) sulphate solution reacts completely with
Magnesium or Zinc powder

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Discussion
3. Brown solid deposited // Zn powder or Magnesium powder dissolves // intensity of
blue colour of CuSO4 decrease // polystyrene cup becomes hotter
4. Heat of displacement is : the heat released when thermometer reading increases
when 1 mol of Copper is displaced from copper(II) sulphate solution when Zinc /
Magnesium powder is added into the solution.

5. Yes. Heat of displacement only involves copper(II) ions which are present in
copper(II) nitrate solution.

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### Calculation guide

Temperature Relationship Concentration Volume of solution

No change No change Changing the

Change, n If volume
times Change n No change

times

3 : Solve the question

Excess copper powder is added to 100 cm3
0.5 mol/dm3 silver nitrate solution
Excess copper powder is added to 400 cm3
0.5 mol/dm3 silver nitrate solution

1. What is the temperature change in experiment II ? Derive your answer



4 : Solve the question

1. What is the temperature change in experiment II ? Derive your answer



Heat of neutralisation

Example:
Themochemical equation shows the neutralisation reaction between hydrochloric
acid, HCl and sodium hydroxide, NaOH solution :

The thermochemical equation shows 57 kJ heat is released when one mole of water,
H2O is formed from the neutralisation reaction between 1 mole of hydrochloric acid,
HCl (1 mole of hydrogen ions, H+) and 1 mole of sodium hydroxide, NaOH (1 mole of
hydroxide ions, OH−)