Heat of precipitation

Module PdPR

Chemistry Form 5

Thermochemistry

Heat of Precipitation

Prepared by Translated by
Pn. Wong Choy Wan Pn. Choong Peit Chun
Guru Cemerlang Kimia SMK Buntong, Ipoh, Perak.
SMK Buntong, Ipoh, Perak.

Let’s learn about HEAT OF
PRECIPITATION. Try to read and
understand the notes given by the teacher. If
you have a problem, you can tell the teacher
via Whatsapp. Teacher is always there with
you. Do remember you have to finish this
module within a week. Send the answers of
your assessment to the teacher by
Date: __________________
Time : ________________

Learning Theme: Heat At the end of a. State the meaning of heat of
area the module, precipitation.
11.0 pupils are
Content Thermochemistry able to: b. determining the
Standard endothermic
and exothermic heat of
Learning precipitation for magnesium
Standard carbonate (MgCO3) and
silver chloride (AgCl)

11.2 Heat of reaction c. Write thermochemical
equation for
11.2.1 Determine heat of precipitation precipitation reaction
through activity
d. construction of energy level
diagram

e. Solve numeric problems
involving the heat of
precipitation

Recall back and write your answer
in a notebook.

1. Does magnesium carbonate soluble in water?
2. Suggest two types of substances which are needed to prepare

magnesium carbonate.
3. Write the chemical equation for the preparation of magnesium

carbonate.
4. Sketch a labeled picture to show the steps of preparation of

magnesium carbonate.

Heat of precipitation How to determine the heat of
precipitation for the formation
Definition:
The heat change of silver chloride?
when one mole
of precipitate is Scan the QR code for experimental steps to
formed from its determine the heat of precipitation of silver
ions in aqueous chloride.
solution under
standard 12
conditions.

TP 3

Comprehension test

Describe an experiment to determine the heat of
precipitation for magnesium carbonate.

Your answer should include the following:

 Materials and apparatus required
 Procedure
 Precautions
 Observation and tabulation of data
 Steps to calculate the heat of precipitation

Solving numeric problems
involving the heat of precipitation

Assumptions:

1. The solution is dilute. It has the same density as water, which
is 1.0 g cm-3.

2. The solution has the same specific heat capacity as water,
which is 4.2 J g-1 °C-1

3. No heat is absorbed by the apparatus of the experiment.

A. If Ө, temperature change in the solution is given.

Question requirements:

Number Heat change ∆H
of mole

• • mcϴ ϴ
1000
1000

m = mass of the solution, g  mcϴ / 1000 to change the unit into

c = specific heat capacity of the solution, kilojoule, kJ.
4.2 J g-1 °C-1
 This can be done in step 2.
ϴ = temperature change in the solution,  Put a ‘+’ or ‘-’ sign in front of the value.
°C  Temperature rises, ‘-‘ sign
 Temperature drops, ‘+‘ sign
During the calculation, the unit for the

answer obtained is the Joule, J.
g x J g-1 °C-1 x °C

B. For the question which ∆H is given

find from energy level diagrams or thermochemical equations
Question requirements: Calculate the change in temperature, °C

Number of mole, n Heat change Temperature change

n x ∆H n x ΔH = mcӨ
1000 Ө = n x ΔH

• During the calculation, the unit for the answer mc

obtained is kilojoule, kJ.  Ensure that
• mol x kJ mol-1 n x ∆H is in the
• Change into the unit J (x 1000) to continue unit of J

with the following steps.

Scan the QR code for
an example of heat of
precipitation
calculation

Example 1

The figure shows the arrangement of the apparatus for determining the heat of
precipitation of silver chloride.

Thermometer

Polystyrene
cups

25 cm3 potassium chloride 25 cm3 silver nitrate solution, 1.0 Mixed solution
solution, 1.0 mol dm-3 mol dm-3

The initial temperature readings of potassium chloride solution and silver nitrate solution as
well as the highest temperature of the mixed solution are recorded in the table below.

The initial temperature of the potassium chloride solution/ °C 28.0
The initial temperature of the silver nitrate solution/ °C 29.0
The highest temperature of the mixed solution/ °C 34.0

a. Write a chemical equation for this precipitation reaction.
b. Calculate the heat of precipitation of silver chloride.
c. Draw an energy level diagram for this reaction.
d. State one reason why that polystyrene cups are used.
e. Suggest a step to determine the heat value of precipitation more

accurately.
f. If a solution of potassium chloride is replaced with a solution of sodium

chloride of the same concentration, predict the heat of precipitation. Explain
why.

Answers

a. KCl + AgNO3  AgCl + KNO3

b. Step 1:

Number of mole KCl = Number of mole AgNO3
= MV / 1000

= (1.0)(25) / 1000

= 0.025 mol

From the chemical equation, 1 mol KCl produces 1 mol AgCl

Thus 0.025 mol KCl produces 0.025 mol AgCl

Step 2: = mcϴ Average initial temperature
Heat change = (25+25)(4.2)(5.5) = (28.0 + 29.0) / 2
= 1155 J = 28.5 °C
= 1.155 kJ
ϴ = the highest temperature – initial
temperature

= 34.0 – 28.5
= 5.5 °C

Step 3:

When 0.025 mol AgCl is produced, 1.155 kJ heat is released.
Heat released when 1 mol AgCl is produced
= 1.155 kJ / 0.025 mol
= 46.2 kJ mol-1

Heat of precipitation , ∆H = - [ . . ] • As the temperature rises, put a ‘-’ sign in
front of the value.
= - 46.2 kJ mol-1
• This indicates that the AgCl precipitation
reaction is an exothermic reaction.

c. Balanced chemical
equation can be
Heat replaced by ionic
KCl + AgNO3 equation.

∆H = - 46.2 kJ mol-1

AgCl + KNO3

d. Polystyrene cups are thermal insulators. It reduces the release of
the thermal energy to the environment.

e. Cover the polystyrene cup with a lid.
f. The heat precipitation value is the same. This is because

potassium ions and nitrate ions are observer ions. These ions are
not involved in the formation of argentum chloride precipitates.
Only argentum ions and chloride ions are involved in the formation
of the precipitate.

Prevent the release of heat
energy to the surrounding -
WRONG

Example 2

In an experiment to determine the heat of precipitation of silver chloride, 50 cm3 of
silver nitrate solution, 1.0 mol dm-3 was mixed into 50 cm3 sodium chloride solution,
1.0 mol dm-3.
The thermochemical equation for the precipitation reaction of silver chloride is as
follow.

Ag+ + Cl-  AgCl ∆H = -58.8 kJ mol-1

a. Calculate the temperature change that occurs in this reaction.
b. If the experiment repeated using 50 cm3 silver nitrate solution, 0.5 mol dm-3 and

50 cm3 sodium chloride solution, 0.5 mol dm-3, predict the temperature change for
this reaction. Explain.

Answers

a. Step 1:

Number of mole NaCl = Number of mole AgNO3
= MV / 1000

= (1.0)(50) / 1000

= 0.05 mol

From the chemical equation, 1 mole KCl produces 1 mole AgCl

Thus 0.05 mole KCl produces 0.05 mole AgCl

Sutep 2: = n x ∆H Step 3:
Heat change = 0.05 mol x 58.8 kJ mol-1 mcϴ = n x ∆H
= 2.94 kJ
= 2940 J (50+50)(4.2)ϴ = 2940 J

ϴ =
.

= 7.0 °C

b. The temperature change is 3.5 °C.
The concentration of the solution used is half.
The number of moles of sodium chloride solution and silver nitrate solution
was half that of the original experiment.
The number of moles of silver chloride precipitate produced was half that of
the original experiment.

It is time to test on
you. What is your
level of mastery?

TemDpuorahtimona:sa:
30 minit

Objective questions TP 2

1. The thermochemical equation below shows the reaction between lead
(II) ions, Pb2+ dan chloride ions, Cl-.

Pb2+ + Cl-  PbCl2 ∆H = -58 kJ mol-1

Which of the following statements is correct about this reaction?

A Heat is absorbed from the environment.
B The temperature of the mixture solution increases.
C This reaction is an endothermic reaction.
D 58 kJ of heat energy is absorbed when 1 mole of lead (II) chloride

is formed.

2. When 50 cm3 silver nitrate solution, 2.0 mol dm-3 is added to 50 cm3

potassium bromide solution, 2.0 mol dm-3, the temperature of the mixture
increased from 30.0 °C to 42.5 °C. Calculate the heat of precipitation for this
reaction. [Specific heat capacity of solution= 4.2 J g-1 °C-1;

Density of solution = 1.0 g cm-3]

A -26.25 kJ mol-1 TP 3

B -52.5 kJ mol-1
C +26.25 kJ mol-1
D +52.5 kJ mol-1

3. Pb(NO3)2 (aq)+ K2SO4 (aq)  PbSO4(p)+ 2KNO3 (aq) ∆H = -50 kJ mol-1

The above equation shows the precipitation reaction of lead (II) sulphate. If
3.03 g of lead (II) sulphate precipitate is produced, how much heat energy is
released?
[Relative atomic mass: O=16; S=32; Pb=207]

A 0.06 J

B 0.5 J
C 500 J
D 5000 J

TP 3

Subjective questions

1. When 25 cm3 magnesium nitrate solution, Mg(NO3)2 2.0 mol dm-3 with initial
temperature 28 °C is added to 25 cm3 potassium carbonate solution, K2CO3,
2.0 mol dm-3 in a polystyrene cup, the temperature of the mixture is 22.5 °C.

[Specific heat capacity of solution= 4.2 J g-1 °C-1; Density of solution = 1.0 g cm-3]

a. i. Write the chemical equation for the above reaction. TP 2
ii. Write the ionic equation for the above reaction.

b. Is the precipitation of magnesium carbonate, MgCO3 an TP 2
exothermic or endothermic reaction? Explain your answer. TP 3

c. Calculate the heat of precipitation of magnesium carbonate, TP 3
MgCO3.

d. Draw an energy level diagram for the precipitation reaction of
magnesium carbonate, MgCO3.

e. If the experiment is repeated using sodium carbonate solution to

replace the potassium carbonate solution, predict the heat of precipitation

of magnesium carbonate, MgCO3. Explain it.

TP 5

f. How can you reduce heat release to the environment, other than using
polystyrene cups? Sketch a diagram to show your creativity in tackling
this problem.

TP 6

Self- reflection

Tick (/) if you are able to do this.

Define the meaning of heat of precipitation.

Determining the heat of precipitation of endothermic and exothermic
reactions for the precipitation of magnesium carbonate and argentum
chloride.

Write thermochemical equations for precipitation reactions.

Construct an energy level diagram.
Solve calculation problems involving the heat of precipitation.

ANSWERS

Objective questions

1. B 2. B 3. C

Subjective quetions

a. i. Mg(NO3)2 + K2CO3  MgCO3 + 2KNO3
ii. Mg2+ + CO32-  MgCO3.
Endothermic reactions. This is because the temperature
decreases.

c. Number of mole Mg(NO3)2 = Number of mole K2CO3
= MV / 1000
From the chemical equation, = (1.0)(25) / 1000
Maka = 0.025 mol
1 mol Mg(NO3)2 produces 1 mol MgCO3
0.025 mol Mg(NO3)2 produces 0.025 mol MgCO3

Heat change = mcϴ Ө =ϴ2=834–.02–2.258.˚5C
= (25+25)(4.2)(5.5) == 55..55°˚CC
= 1155 J
= 1.155 kJ
When 0.025 mol MgCO3 is produced, 1.155 kJ heat is released.
When releases when 1 mol MgCO3 produced = 1.155 kJ /0.025 mol

= 46.2 kJ mol-1

Heat of precipitation, ∆H = + [1.155 kJ 0.025 mol] • As the temperature decreases,
= + 46.2 kJ mol-1
put a ‘+‘ in front of the value.
• This indicates that the MgCO3
precipitation reaction is an

endothermic reaction.

d. Heat

MgCO3

∆H = + 46.2 kJ mol-1

Mg2+ + CO32-

e. The heat of precipitation does not change = + 46.2 kJ mol-1.
This is because sodium ions and potassium ions are the observer ions in this
reaction.

f. The sketch shows the use of two polystyrene cups or wrapping a polystyrene cup
with cotton. [Accept any reasonable answer]



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