Akhmalazmi86@blogspot.com
Form 5 Chapter 3
CHAPTER 4 : THERMOCHEMISTRY
ANALYSIS OF PAST YEAR QUESTIONS FROM 2003 – 2008
Year
Paper
No.
Type of
question
Question
No
2003
P2
S
2004
P3
E
6
S
P2
E
S
4
2005
P3
E
S
P2
E
S
2006
P3
E
S
3
5
P2
E
S
2007
P3
E
S
1
P2
E
S
2008
P3
E
S
1
P2
E
S
P3
E
S
E
6
1. SPM 2003/P2/Q6
A student carried out an experiment to determine the heat of displacement for the reaction
between copper and silver nitrate solution. In this experiment, excess copper powder was added
to 100 cm3 of silver nitrate solution 0.5 mol dm-3. The heat of displacement in this experiment was
-105 kJ mol-1.
[Specific heat capacity of the solution is 4.2 Jg-1 C-1, and the density of the solution is 1 g cm-3]
(a) What is meant by heat of displacement ?
…………………………………………………………………………………………………………
[1 mark]
(b) Besides the data given above, state one other piece of data that is needed to calculate the
heat of displacement.
(c)
…………………………………………………………………………………………………………
[1 mark]
State one precaution that must be taken while carrying out the experiment.
…………………………………………………………………………………………………………
[1 mark]
State one observation of the experiment.
(d) (i)
(ii)
…………………………………………………………………………………………………….
.
[1 mark]
State the reason for the observation in (d) (i)
…………………………………………………………………………………………………….
.
[1 mark
]
1
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
(e) Based on the information from this experiment, calculate
(i)
the number of moles of silver ions reacted.
[1 mark]
(ii)
the amount of heat released.
[1 mark]
(iii)
the change in temperature.
[1 mark]
(f)
Draw an energy level diagram for the reaction in this experiment.
[2 marks
]
2
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
(g) The experiment is repeated using 100 cm3 of 1.0 mol dm-3 silver nitrate solution and excess
copper powder. Calculate the temperature change in this experiment. Explain why this
change of temperature is different from that in (e) (iii).
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
[3 marks]
2. SPM 2004/P2/Q4
Figure 4 show the set-up of the apparatus of an experiment to determine the heat of
precipitation. 25.0 cm3 of 0.5 mol dm-3 silver nitrate solution is reacted with 25 cm3 of 0.5 mol dm3
sodium chloride solution. As a result there is a change in temperature of the mixture and a
white precipitate is formed.
Before
reaction
After
reaction
Figure 4
3
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
(b) Why is a polystyrene cup used in the experiment ?
…………………………………………………………………………………………………………
[1 mark]
(i)
(ii)
Based on the change of temperature in the experiment, state the type of reaction that
occurred.
…………………………………………………………………………………………………….
[1 mark]
How is the total energy of the product different from the total energy of the reactants?
…………………………………………………………………………………………………….
[1 mark]
(c)
State one step that should be taken while adding the two solutions in order to get a more
accurate result.
…………………………………………………………………………………………………………
[1 mark]
(d) The ionic equation for the precipitation reaction of silver chloride is :
Ag+(aq) + Cl-(aq)  AgCl(s)
(i)
What is the number of moles of Ag+ ions that reacted with Cl- ions ?
[1 mark]
(ii)
Calculate the heat change of the precipitation reaction that has taken place. Use the
information that the specific heat capacity of water is 4.2 J g-1 C-1 and density water is
1 g cm-3.
[2 marks]
4
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
(iii) Calculate the heat of precipitation for this reaction.
[2 marks]
(e) The calculated value of the heat of precipitation for this reaction is less than the actual
value.
Give a reason.
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
[1 mark]
3. SPM 2005/P2/Q3
A pupil carried out an experiment to determine the value of heat of displacement. Figure 3 shows
the set up of the apparatus used in the experiment.
Zinc Powder
Glass cup
Copper (II) sulphate
Solution
Figure 3
The following data was obtained :
Initial temperature of copper(II) sulphate solution, 1 = 28 oC
Highest temperature of the mixture of product, 2
= 48 oC
(a) Complete the ionic equation for the reaction that occurred.
Zn + Cu2+ 
………………………………………………………………………………………….
[1 mark]
5
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
(b) In this experiment, excess zinc is added to 100 cm3 of 0.5 mol dm-3 copper(II) sulphate
solution.
Given that the specific heat capacity of the solution is 4.2 J g-1 oC-1 and the density of the
solution is 1.0 g cm-3.
(i)
(ii)
Calculate the change of heat capacity of the experiment.
Use the formula, H = mc
[2 marks]
Calculate the heat of displacement in the experiment.
The number of moles of copper(II)
sulphate that reacted
= ………………………………………………......
Heat of displacement
(c)
= …………………………………………………...
[2 marks]
Draw the energy level diagram for the reaction.
[2 marks]
(d) It was found that the heat of displacement value in (b)(ii) is not the same as the actual value.
Suggest one step that must be taken to get a more accurate value.
…………………………………………………………………………………………………………
[1 mark]
(e) Based on the experiment, what is meant by the heat of displacement ?
…………………………………………………………………………………………………………
[1 mark]
(f)
The pupil repeats the experiment, replacing the metal zinc with metal X.
The following equation shows the reaction and the value of heat of displacement of metal
iron and metal X.
6
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
Equation I :
Fe (s) + CuSO4 (aq)  FeSO4 (aq) + Cu(s), H = -150 kJ mol-1
Equation II :
X(s) + CuSO4 (aq)  XSO4 (aq) + Cu(s),
H = - 100 kJ mol-1
Predict the metal X.
Choose from this list : Aluminium, magnesium and tin.
…………………………………………………………………………………………………………
[1 mark]
4. SPM 2005/P2/Q5
(a) What is the meaning of the heat of combustion of an alcohol?
…………………………………………………………………………………………………………
[1 mark]
(b) Table 5 shows the heat of combustion of three types of alcohol.
The number of carbon atoms and the attractive force between molecules are among the
factors that affect the value of heat of combustion.
Name of alcohol
Methanol
Ethanol
Propanol
Molecular formula
CH3OH
C2H5OH
C3H7OH
Heat of combustion /kJ
mol-1
725
1376
2015
Table 5
(i)
Use data from Table 5 to draw the graph of the heat of combustion against number of
carbon atoms on the graph paper provided.
[2 marks]
7
Akhmalazmi86@blogspot.com
(ii)
Form 5 Chapter 3
Based on the graph in (b)(i), as the number of carbon atoms increases so does the
value of the heat of combustion.
Explain why.
……………………………………………………………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………………………………
[2 marks]
8
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
(iii) Calculate the heat released when 2.3g of ethanol is completely burnt in air.
Given that the relative atomic mass of C=12, H=1, O=16.
Use the formula :
Heat released = Number of moles X Heat of combustion.
[2 marks]
(c)
Methanol and ethanol do not have isomers.
Propanol has two isomers.
Draw the structures of the two isomers of propanol.
[2 marks]
(d) Table 5.2 shows the freezing and the boiling points of mercury, methanol, ethanol
and butanol.
Substance
Mercury
Methanol
Ethanol
Butanol
Freezing point /0C
-39
-97
-117
-90
Boiling point /0C
357
64
79
117
Table 5.2
A thermometer may contain mercury or an alcohol.
A mercury thermometer is not suitable to measure the temperature at around -100 0C.
Name a suitable alcohol that can be used in a thermometer to measure the temperature at
around –100 0C.
Give one reason for your choice.
Name of alcohol
: ………………………………………………………………………....
Reason
: ………………………………………………………………………....
………………………………………………………………………....
………………………………………………………………………....
[2 marks]
9
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
5. SPM 2006/P3/Q1
Diagram 1.1 shows two experiments to determine the heat of neutralisation.
Experiment I
Reaction between 25 cm3 of sodium hydroxide solution, NaOH, 2.0 mol dm-3 and 25 cm3 of
ethanoic acid, CH3COOH, 2.0 mol dm-3.
Initial temperature of mixture
: ……………… °C
Highest temperature of mixture
: ……………… °C
Change in temperature
: ……………… °C
Experiment II
Reaction between 25 cm3 of sodium hydroxide solution, NaOH, 2.0 mol dm-3 and 25 cm3 of
hydrochloric acid, HCl, 2.0 mol dm-3.
Initial temperature of mixture
: T1 °C
Highest temperature of mixture
: T2 °C
Change in temperature
: T3 °C
Diagram 1.1
10
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
(a) Write the initial and the highest temperature of the mixture and change in temperature for
Experiment I in Diagram 1.1.
[3 marks]
(b) Construct a table that can be used to record the data from both experiments.
[3 marks]
(c)
State one hypothesis for both experiments.
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
[3 marks]
(d) Based on the temperature in Experiment I, predict the predict the change in temperature in
Experiment II.
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
[3 marks]
(e) Why must the initial temperature and the highest temperature be recorded in these
experiments?
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
[3 marks]
(f)
How can the value of the changes in temperature be obtained?
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
[3 marks]
(g) State three observations that you could obtain in Experiment I other than the change in
temperature.
1…………………………………………………………………………………………………………
2…………………………………………………………………………………………………………
11
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
3…………………………………………………………………………………………………………
[3 marks]
(h) State three constant variables in this experiment.
1…………………………………………………………………………………………………………
2…………………………………………………………………………………………………………
3…………………………………………………………………………………………………………
[3 marks]
(i)
Diagram 1.2 shows the calculation to determine the heat of neutralization for the reaction in
Experiment I and Experiment II.
Experiment I
Experiment II
Heat release
= mc
Heat release
= mc
= 50 g x 4.2 4.2 J g-1 oC-1 x ………… oC = 50 g x 4.2 4.2 J g-1 oC-1 x T3 oC
= xJ
= yJ
Heat of neutralization
Heat of neutralization
=
=
x kJ
_________________________________
Number of mole of water produced
y kJ
_________________________________
Number of mole of water produced
Diagram 1.2
Based on Diagram 1.2 :
(i) Give the operational definition for the heat of neutralization.
…………………………………………………………………………………………………….
…………………………………………………………………………………………………….
[3 marks]
(ii)
It was found that the value of y is greater than the value of x.
Explain why.
…………………………………………………………………………………………………….
…………………………………………………………………………………………………….
[3 marks]
(j)
The experiment is repeated using the methanoic acid.
The values of the heat of neutralization of these acids are given in Table 1.
Complete Table 1 by classifying the acids as strong acid or weak acid.
12
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
Name of acid
Ethanoic acid
Heat of neutralization/kJ mol-1
-50.3
Hydrochloric acid
-57.2
Mathanoic acid
-50.5
Type of acid
Table 1
[3 marks]
6. SPM 2007/P3/Q1
Diagram 1.1 shows the apparatus set-up for experiments I, II, III and IV.
The magnification of the thermometers shows the readings of the initial temperature and the
highest or lowest temperatures in each experiment.
(a) (i)
Record the temperature readings in the spaces provided in Diagram 1.1.
[3 marks]
Experiment I
Experiment II
Experiment III
13
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
Experiment IV
Diagram 1.1
(ii)
Construct a table to show all the data in each of these experiments.
[3 marks]
(iii) Classify the reactions in these experiments as either exothermic reactions or
endothermic reactions.
Exothermic reaction
Endothermic reaction
[3 marks]
(b) A student repeated Experiment I several times.
(i) State three things must be kept constant in these experiments.
1…………………………………………………………………………………………………
2…………………………………………………………………………………………………
3…………………………………………………………………………………………………
[3 marks]
14
Akhmalazmi86@blogspot.com
(ii)
Form 5 Chapter 3
State the hypothesis for Experiment I.
…………………………………………………………………………………………………….
…………………………………………………………………………………………………….
[3 marks]
(c)
Based on Experiment II :
(i) Sate the temperature change and give two reasons for the change.
Temperature change :
…………………………………………………………………………………………………….
Reason1 :
…………………………………………………………………………………………………….
Reason2 :
…………………………………………………………………………………………………….
[3 marks]
(ii)
Sate the operational definition for the reaction that takes place.
…………………………………………………………………………………………………….
…………………………………………………………………………………………………….
[3 marks]
(d) The reaction in Experiment III is a neutralization reaction.
Other acids can be substituted for hydrochloric acid.
These acids have the same volume and concentration as the hydrochloric acid in
Experiment III.
Predict the temperature in the neutralization reactions of these acids.
1. Sulphuric acid : …………………………………………….. oC
2. Nitric acid
: …………………………………………….. oC
3. Ethanoic acid
: …………………………………………….. oC
[3 marks]
15
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
(e) Diagram 1.2 shows some observations in experiment IV.
Diagram 1.2
(i)
State three observations shown in Diagram 1.2.
1…………………………………………………………………………………………………
2…………………………………………………………………………………………………
(ii)
3…………………………………………………………………………………………………
[3 marks]
The following chemical equation represents the reaction in Experiment IV.
HCl(aq) + NaHCO3(aq) → NaCl(aq) + CO2(g) + H2O(l)
Based on the chemical equation, and the answer in 1(e)(i), what inference can be
made from Experiment IV?
…………………………………………………………………………………………………….
…………………………………………………………………………………………………….
[3 marks]
(iii) Sketch a graph to show the change in the volume of carbon dioxide gas produced
against time.
[3 marks]
16
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
7. SPM 2008/P2/Q6
The thermochemical equation for the neutralization reaction between nitric acid and sodium
hydroxide solution is given below.
HNO3 + NaOH  NaNO3 + H2O , ∆H = -57.3 kJ mol-1
(a) State the meaning of heat of neutralisation.
…………………………………………………………………………………………………………
[1 mark]
(b) Based on the given thermochemical equation, state one observation when dilute nitric acid
is added to sodium hydroxide solution.
Explain your answer.
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
(c)
…………………………………………………………………………………………………………
[2 marks]
In an experiment, 100 cm3 of 2 mol dm-3 nitric acid solution was added to 100 cm3 of 2 mol
dm-3 sodium hydroxide solution.
[Specific heat capacity of solution = 4.2 J g-1 oC-1;
Density of solution = 1 g cm-3]
Calculate
(i) The heat energy released in this experiment,
[2 marks]
(ii)
The temperature change in this experiment.
[2 marks]
(d) Draw the energy level diagram for the reaction between nitric acid and sodium hydroxide.
[2 marks]
17
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
(e) Nitric acid and ethanoic acid both react with sodium hydroxide by a neutralisation reaction.
HNO3 + NaOH  NaNO3 + H2O , ∆H = -57.3 kJ mol-1
CH3COOH + NaOH  CH3COONa + H2O , ∆H = -55.2 kJ mol-1
Explain why the heat of neutralisation for each reaction is slightly different.
………………………………………………………………………………………………………..
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
[2 marks]
18
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
CHAPTER 4 : THERMOCHEMISTRY
1. SPM 2003/P2/Q6
(a)
Heat change when 1 mol of metal is displaced from its salt solution by a more electropositive
metal.
(b)
Initial temperature and highest temperature.
(c)
1. Stir the mixture.
2. Add the two solutions as quickly as possible.
3. Use polystyrene or plastic cup
(any one)
(d)
(i)
1. Grey solid is deposited
2. Colourless solution turns blue
3. The thermometer reading rises or the container becomes hot or warm.
(any one)
(ii)
(e)
1. Silver metal is produced
2. copper(II) ion is produced
3. exothermic reaction/ heat is released to the surroundings
(i)
= 0.5 x 100
1000
= 0.05 mol
(ii)
= 0.05 x 105 kJ
= 5250 J
(iii)
Ө = 5.25 x 1000
100 x 4.2
= 12.5 oC
(f)
Energy
+
Cu + Ag
-1
∆H = -105 kJ mol
2+
Cu
(g)
+ Ag
1. Mol of Ag+ = 1 x 100 = 0.1 mol
1000
19
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
= 25 oC
2. Heat change, Ө = 0.1 x 10500
100 x 4.2
3. Number of mol of Ag+ is double or concentration of silver nitrate is double.
2. SPM 2004/P2/Q4
(a) To reduce the heat loss to the surroundings.
(b) (i)
(ii)
(c)
Exothermic reaction
Total energy of products is less than total energy of reactants
Mix the solutions quickly and stir the reaction mixture.
(d) (i)
(ii)
Number of moles Ag+ = 25 x 0.5
100
= 0.0125 mol
The heat change = mcө
= 50 x 4.2 x (31.5-29.0)
= 525 J
(iii) 0.0125 mol of Ag+ ions that reacted with Cl- ions released 525 J
1 mol of Ag+ ions that reacted with Cl- ions released =
525
J
0.0125
= 42000 J
Heat of precipitation = -42 kJmol-1
(e) Heat is released to surroundings.
3. SPM 2005/P2/Q3
(a) Zn2+ + Cu
(b) (i)
(ii)
∆H = 100 x 4.2 x 20 = 8400 J
Number of moles CuSO4 reacted = 0.5 x 100
1000
Heat of displacement
=
= 0.05 mol
___ mcө_______
Number of moles
= 8400
0.05
= -168 000 J mol-1
= -168 kJ mol-1
20
Akhmalazmi86@blogspot.com
(c)
Form 5 Chapter 3
Energy
2+
Zn
+ Cu
-1
∆H = -168 kJ mol
2+
Zn
+ Cu
1. Use a plastic / polystyrene cup
2. add the zinc powder quickly.
3. stir the solution
(d)
(any one)
(e) The heat released when 1 mole of copper is displaced from its solution.
(f) Tin (Sn)
4. SPM 2005/P2/Q5
(a) The heat released when 1 mole of alcohol is completely burnt in excess oxygen.
(b) (i)
(ii)
1.
2.
all points are transferred correctly
draw a straight line
The greater the number of carbon dioxide molecules, more products are formed which
causes more heat to be released during the formation of bonds.
(iii) Relative molecular mass of ethanol
= (12 x 2) + (1 x 6) + 16 = 46
Number of moles ethanol = 2.3 = 0.05 mol
46
Heat released = 0.05 x 1376
= 68.8 kJ
= 68 800 J
(c)
(d) - Ethanol
- The freezing point of ethanol is -117 oC, which is lower than -100 oC.
5. SPM 2006/P3/Q1
(a) Initial temperature of mixture
: 28.0 oC
Highest temperature of mixture : 40.0 oC
21
Akhmalazmi86@blogspot.com
Change in temperature
Form 5 Chapter 3
: 12.0 oC
(b)
Experiment
Initial temperature of
mixture/ oC
Highest l temperature of
mixture/ oC
Change in temperature/ oC
(c)
Experiment I
28.0
Experiment II
T1
40.0
T2
12.0
T3
Strong acid produces higher heat of neutralization than weak acid.
(d) 12.5 oC - 15.0 oC
(e) To enable us to obtain the change in temperature for both experiments.
(f)
Change in temperature = Highest temperature of mixture - Initial temperature of mixture
(g) 1. A colourless mixture of solution is obtained.
2. The vinegar smell of ethanoic acid disappears.
3. The polysterene cup becomes hot.
4. Thermometer reading is rises
(h) 1. The volumes of the acid and the alkali.
2. The concentrations of the acid and the alkali.
3. The type of cup used in the experiment.
(i)
(i)
The heat of neutralization is defined as the amount of heat released when 1 mole of
water is produced.
(ii)
Experiment II uses a strong acid whereas Experiment I uses a weak acid.
(j)
Name of acid
Ethanoic acid
Hydrochloric acid
Methanoic acid
Type of acid
Weak acid
Strong acid
Weak acid
6. SPM 2007/P3/Q1
(a) (i)
Experiment I
Experiment II
Experiment III
Experiment IV
Initial temperature
(oC)
28.0
29.0
27.0
30.0
Highest
temperature (oC)
36.0
25.0
32.0
27.0
Initial temperature
(oC)
Highest
temperature (oC)
(ii)
Experiment
22
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
I
II
III
IV
(iii)
(b) (i)
(c)
28.0
29.0
27.0
30.0
Exothermic reaction
Experiment I
Experiment III
36.0
25.0
32.0
27.0
Endothermic reaction
Experiment II
Experiment IV
1. The mass of sodium hydroxide.
2. the volume of water in the cup.
3. The size of the polystyrene cup.
(ii)
The reaction between sodium hydroxide and water is an exothermic reaction.
(i)
Temperature change = 4 oC
Reason 1 :
Heat energy is absorbed by the reactants from the surroundings.
Reason 2 :
The energy of the products is more than the energy of the reactants.
(ii)
The decrease in temperature shows that endothermic reaction happens where heat
energy is absorbed from the surroundings.
(d) 1. 37 oC
2. 32 oC
3. 30 oC
(e) (i)
(ii)
1. Final temperature is lower than the initial temperature.
2. The temperature reading decreases.
3. Bubbles of gas are released.
Heat energy is absorbed when hydrochloric acid reacts with sodium hydrogen
carbonate to produce sodium chloride, carbon dioxide and water.
(iii)
Volume of carbon
dioxide gas,/cm3
Time /minute
23
Akhmalazmi86@blogspot.com
Form 5 Chapter 3
7. SPM 2008/P2/Q6
(a) Heat change when 1 mole of hydrogen ions reacts with 1 mole of hydroxide ions to form
1 mole of water.
(b) Observation : the mixture becomes hot or temperature increase
Explanation : the reaction is exothermic
(c)
(i)
No. of moles of NaOH = 100 x 2 = 0.2 mol
1000
Energy released = 0.2 x 57.3
= 11.46 kJ
(ii)
(d)
Temperature change = 11.46 x 1000
200 x 4.2
= 13.6 oC
Energy
NaOH + HNO3
-1
∆H = -57.3 kJ mol
Na NO3 + H2O
(e)
1. Ethanoic acid is a weak acid which partially ionize in water, nitric acid is strong acid
that ionize completely in water.
2. energy is used to ionize/dissociate weak acid.
24