1. The document discusses various types of energy changes that occur during chemical reactions including exothermic reactions, endothermic reactions, and energy level diagrams. 2. It provides examples of heat changes associated with different types of chemical reactions such as heat of precipitation, heat of displacement, heat of neutralization, and heat of combustion. 3. The key concepts covered are how to calculate heat changes using thermochemical equations and energy level diagrams, and how temperature changes can indicate whether a reaction is exothermic or endothermic.

2. THERMOCHEMISTRY
Exothermic
reaction
Endothermic
reaction
Different energy
level diagrams
Heat of reaction
Heat of
precipitation
Heat of
displacement
Heat of
neutralisation
Heat of
combustion

3. Exothermic
reaction
- is a chemical reaction that
releases energy to the
surrounding in which the
energy of products is less
than the energy of the
reactants
Endothermic
reaction
- is a chemical reaction that
absorbs energy from the
surrounding in which the
energy of the product is
more than the energy of
reactants
Energy level
diagram
-is a graph that shows energy
change of a chemical reaction

4. Energy changes in chemical reaction
Exothermic Endothermic
Energy is given off
during a chemical
reaction
The surrounding
will become hot
The T of the reaction
mixture will rise
Energy is absorbed
from the surrounding
during a chemical
reaction
The surrounding
will become cold
The T of the reaction
mixture will fall

5. Example of exothermic and
endothermic reaction
Exothermic Endothermic
-photosynthesis
-dissolving ammonium
salts in water
-frying an egg
-decomposition of metals
carbonates
-decomposition of metals
nitrate
-Respiration
-neutralisation
-Reaction acids + metals
-reaction acids + metal
carbonates
-burning of fuels
-dissolving NaOH in water
-Adding H2O to
concentrated acids
-reaction reactive metals
+ H2O
-oxidations of metals
-Haber process

6. Energy level diagrams
Chemical reactions
Certain quantity of heat
is given off or absorbed
(ΔH)
(ΔH) is the difference between
the energy of the reactants and
the energy of the products
(ΔH) = Hproducts – Hreactants
Exothermic reaction
Energy is given off
Hproducts < Hreactants
(ΔH) has a negative sign
Energy
reactants
products
(ΔH) -negative
sign

7. (ΔH) is the difference between
the energy of the reactants and
the energy of the products
(ΔH) = Hproducts – Hreactants
Endothermic reaction
Energy is absorbed
Hproducts > Hreactants
(ΔH) has a positive sign
Energy
reactants
products
(ΔH) positive
sign

8. Interpreting energy level diagrams
Eg.
Energy
Zn + 2HCl
ZnCl2 + H2
ΔH = -126 kJ
Base on the energy level
diagram ,we can say :
The reaction between Zn and HCl to
form ZnCl2 and H2 is exothermic
When one mole of Zn reacts with
two moles of HCl to form one mole
of ZnCl2 and H2, the quantity of heat
released is 126 kJ.
The total energy of one mole of Zn
and two moles of HCl is more than
the total energy of one mole of
ZnCl2 and 1 mole of H2. The
difference in energy is 126 kJ
The temperature of the reaction
mixture will rise

9. Energy
N2 + 2O2
2 NO2
ΔH =+66 kJ
Base on the energy level
diagram ,we can say :
Exercise
The reaction between N2 and O2 to
form NO2 is endothermic
When one mole of N2 gas reacts
with two moles of O2 gas to form
two moles of NO2, the quantity of
heat absorbed is 66 kJ.
The total energy of two moles of
NO2 gas is more than the total
energy of one mole of N2 gas and
two moles of O2. The difference in
energy is 66 kJ
The temperature of the reaction
mixture will fall

10. Energy change and chemical bonds
Refer to your text book –pg 147
Type of
reaction
Exothermic
Endothermic
Energy change Sign of ΔH
Energy absorbed for
bond breaking < energy
released during bond
formation
Energy absorbed for
bond breaking > energy
released during bond
formation
negative
positive

11. Heat of reaction, ΔH is the energy change
of a chemical reaction, that is the
difference between the energy of the
reactants and the energy of the products
The heat of reaction, ΔH is
written at the end of a
chemical equation .
H2 + Cl2  2 HCl ΔH = -184kJ
N2 + O2  2 NO ΔH = +225kJ
These equations called a
thermochemical equation

12. 1. HEAT OF PRECIPITATION
Definition – the energy change when one
mole of precipitate is formed
from its ions
The heat of reaction can be calculated by
using the formula :
Energy change = mcθ
In which, m is the mass of the aqueous
reaction mixture
C is the specific heat capacity of
the aqueous reaction mixture
θ is the change in T

13. Assumptions in this calculation :
1. ρ of the aqueous reaction mixture is
1 g cm-3 ,that is the ρ of water.
2. c of the aqueous reaction mixture is
the same as the specific heat
capacity of water .
The value is 4.2 J g-1 0C-1
3. No heat is lost to or absorbed from
the surroundings
4. No heat is absorbed by the
apparatus of the experiment

14. Determining the heat of
precipitation of AgCl
thermometer
Plastic cup
50 cm3 of
0.5 mol dm-3
AgNO3 solution
50 cm3 of
0.5 mol dm-3
NaCl solution
cover
Reacting
mixture

15. Results
Initial temperature of AgNO3 / 0C
Initial temperature of NaCl / 0C
Average temperature of the
mixture / 0C
Highest temperature of the
mixture/ 0C
Increase in temperature / 0C
28.0
28.0
28.0
32.0
32.0 - 28.0 = 4.0

16. Step 2
Moles of Ag+ = MV/1000
= 0.5x50/1000
= 0.025 mol
Moles of Cl- = MV/1000
= 0.5x50/1000
= 0.025 mol
Calculation :
Step 1
Heat given out
in the reaction = mcθ
= (50 + 50)gx4.2Jg-1 0C-1x40C
= 1680 J
Ag+ + Cl-  AgCl
Based on the equation,
1 mole of Ag+will react
with 1 mole of Cl- to
form 1 mole of AgCl.
Therefore , 0.025 mole
of AgCl will formed in
this reaction

17. Step 3
Precipitation of 0.025 mole of AgCl gives out 1680 J
:. Precipitation of 1 mole of AgCl gives out
1680J x 1 mole = 67,200J
0.025 mole
The heat of precipitation of AgCl = -67.2kJ mol-1
(ΔHprecipitation)
Energy
Ag+ + Cl-
AgCl
ΔH = -67.2 kJ mol-1

18. Discussion :
1. The value of heat of precipitation obtained
in the experiment is less than the
theoretical because some heat is lost to
the surroundings .
2. To reduced heat lost to the surroundings:
i) use a plastic cup (good insulator)
ii) mix the AgNO3 solution and NaCl
solution quickly.

19. Exercise 1 – to find the ΔH
100 cm3 of 1 mol dm-3 lead(II) nitrate solution
was mixed with 100 cm3 of 1 mol dm-3
potassium sulphate solution. The temperature
of the reaction rose from 270C to 320C .
Calculate the heat of precipitation of lead(II)
sulphate .(c=4.2 J g-1 0C-1 , density of solution
, 1 gcm-3 )
Solution :
1.Moles of Pb2+ = 1x100/1000 = 0.1 mol
Moles of SO4
2- = 1x100/1000 = 0.1 mol
Pb2+ + SO4
2-  PbSO4
- 0.1 mol of Pb2+ react with 0.1 mol of SO4
2-
to form 0.1 mol PbSO4

20. 2. Energy
change = mcθ
= (100 + 100)gx4.2Jg-1 0C-1x(32-27)0C
= 4200J
3.
Precipitation of 0.1 mole of PbSO4 gives out 4200 J
:. Precipitation of 1 mole of PbSO4 gives out
4200J x 1 mole = 42,000J
0.1 mole
The heat of precipitation of PbSO4 = -42kJ mol-1
(ΔHprecipitation)
Pb2+ + SO4
2-  PbSO4 ΔH= -42kJ
Exercise – pg 83 (cerdik publication)

21. Exercise 2 – to find θ
The heat of of precipitation of calcium carbonate
Is 12 kJ mol-1. 50 cm3 of 2 mol dm-3 calcium
chloride solution and 50 cm3 of 2 mol dm-3 sodium
carbonate solution are mixed together .Calculate
the temperature change of the reaction mixture .
(c=4.2 J g-1 0C-1 , density of solution, 1 gcm-3 )
Exercise 3-to find the volume of solution
The heat of of precipitation of iron(III) hydroxide
Is -10 kJ mol-1. When a 2 mol dm-3 sodium hydroxide
solution was mixed together with a solution of
iron(III) chloride , 200 J of heat was released .
Calculate the volume of sodium hydroxide solution
that was used .
Ans: 2.860C , 30cm3

22. 2. HEAT OF DISPLACEMENT
Definition – Heat released when one
mole of metal is displaced
from its salt solution by a
more electropositive metal

23. thermometer
Plastic
cup
Mg
powder
50 cm3 of 0.2
mol dm-3 CuSO4
solution
Reacting
mixture
cover

24. Results
Initial temperature of CuSO4
solution / 0C
Highest temperature of the
mixture/ 0C
Increase in temperature / 0C
28.0
37.0
37.0 - 28.0 = 9.0

25. Step 2
Moles of Cu2+ = MV/1000
= 0.2x50/1000
= 0.01 mol
CuSO4 + Mg  MgSO4 + Cu
Based on the equation,
1 mole of CuSO4 will
produced 1 mole of Cu
Calculation :
Step 1
Heat given out
in the reaction = mcθ
= (50)gx4.2Jg-1 0C-1x (37.0-28.0)0C
= 1,890 J
Therefore , 0.01 mole
of Cu2+ will produced
0.01 mole of Cu in
this reaction

26. Step 3
Displacement of 0.01 mole of Cu gives out 1,890 J
:. Displacement of 1 mole of Cu gives out
1,890J x 1 mole = 189,000J
0.01 mole
The heat of diplacement of Cu = -189 kJ mol-1
(ΔHdisplacement)
Energy
Cu2+ +Mg
Mg2+ + Cu
ΔH = -189 kJ mol-1

27. Discussion :
1. The value of heat of displacement obtained
in the experiment is less than the
theoretical because some heat is lost to
the surroundings .
2. To reduced heat lost to the surroundings:
i) use a plastic cup (good insulator)
ii) the metals are added quickly to the
solution.
iii) metals in the powder form are used, so
that the reaction take a shorter time to
complete .

28. Exercise 1 – to find the ΔH
Excess zinc powder is added to 100 cm3 of
1 mol dm-3 iron(III) sulphate solution. The
temperature of the reaction mixture rise by 9.60C .
Calculate the heat of displacement of iron from its
salts solution by zinc . (c=4.2 J g-1 0C-1 , density of
solution, 1 gcm-3 )
Answer = -20.16 kJ mol-1
Exercise – pg 31 (cerdik publication)

29. Relating the volume and the concentration
of solution with temperature change
When excess zinc is added to 50 cm3 of 1.0 mol dm-3
lead(II) nitrate solution,the temperature change
is θ0C. What is the temperature change when
excess zinc is added to :
a)100cm3 of 1.0 mol dm-3 lead(II) nitrate solution ?
b)50 cm3 of 2.0 mol dm-3 lead(II) nitrate solution?
c)100 cm3 of 0.5 mol dm-3 lead(II) nitrate solution?
Try your best !!!

30. Answer :
Solution original Part(a) Part(b) Part(c)
Concentration/
mol dm-3
Volume/ cm3
No.of moles=
MV/1000
Change in T/
0C
1 1 2 0.5
50 100 50 100
1x50
1000
= 0.05
1x100
1000
= 0.1
2x50
1000
= 0.1
0.5x100
1000
= 0.05
θ θ 2θ θ/2

31. Part A
-the no. of moles of Pb(NO3)2 is twice
that of the original solution .
(0.1 mole compared to 0.05 mole)
The heat given off is twice as much
But this heat is distributed over a volume
of solution that is twice as much
(100 cm3 compared to 50 cm3)
The temperature rise is still θ0C

32. Part B
-the no. of moles of Pb(NO3)2 is twice
that of the original solution .
(0.1 mole compared to 0.05 mole)
The heat given off is twice as much
But this heat is distributed over a volume
that is the same
( 50 cm3)
The temperature rise is 2θ 0C

33. Part C
-the no. of moles of Pb(NO3)2 is the
same as that of the original solution .
(0.05 mole)
The heat given off is the same as that
of the original solution.
But this heat is distributed over a volume
that is twice as much
( 100 cm3 compared to 50 cm3)
The temperature rise is θ/2 0C

34. 3. HEAT OF NEUTRALISATION
Definition – the heat
released when one
mole of water is formed
from the neutralisation
between one mole of
hydrogen ions from an acid
and one mole of hydroxide
ions from an alkali .

35. Determining the heat of
neutralisation
thermometer
Plastic cup
50 cm3 of
2.0 mol dm-3
NaOH solution
50 cm3 of
2.0 mol dm-3
HCl acid
cover
Reacting
mixture

36. Results
Initial temperature of NaOH / 0C
Initial temperature of HCl / 0C
Average temperature of acid
and alkali / 0C
Highest temperature of the
mixture/ 0C
Increase in temperature / 0C
28.0
28.0
28.0
40.0
40.0 - 28.0 = 12.0

37. Step 2
Moles of H+ = MV/1000
= 2.0x50/1000
= 0.1 mol
Moles of OH- = MV/1000
= 2.0x50/1000
=0.1 mol
Calculation :
Step 1
Heat given out
in the reaction = mcθ
= (50 + 50)gx4.2Jg-1 0C-1x120C
= 5040J
H+ + OH-  H2O
Based on the equation,
1 mole of H+will react
with 1 mole of OH- to
form 1 mole of H2O.
Therefore , 0.1 mole
of H2O will formed in
this reaction

38. Step 3
Formation of 0.1 mole of H2O gives out 5040J
:. Formation of 1 mole of H2O gives out
5040J x 1 mole =50400 J
0.1 mole
The heat of neutralisation = - 50.4kJ mol-1
Between NaOH and HCl
(ΔHneutralisation)
Energy
H+ + OH-
H2O
ΔH = -50.4 kJ mol-1

39. Discussion :
1.Value of heat of neutralisation between a
strong acid and a strong alkali is constant ,
that is -57.3 kJ mol-1
2. The value of heat of neutralisation obtained
in the experiment is less than the
theoretical because some heat is lost to
the surroundings .
3. To reduced heat lost to the surroundings:
i) use a plastic cup (good insulator)
ii) mix the acid and the alkali quickly

40. 4.Value of heat of neutralisation between a
strong acid and a strong alkali is constant ,
that is -57.3 kJ mol-1
.Is it the same if we use :
i) a strong acid and a weak alkali ??
ii) a weak acid and a strong alkali ??
iii) a weak acid and a weak alkali ??
Alert !!!
i) Ethanoic acid is a weak acid and ammonia
solution is a weak alkali ,they both dissociate
partially in water. Most of them still exist as
molecules.
ii) Some of the heat given out during the
neutralisation reaction is used to dissociate
the weak acid or the weak alkali completely in
water . That is why the heat of neutralisation
involving weak acid or weak alkali is less than
-57.3 kJ mol-1

41. Exercise 1 – to find the ΔH
When 100 cm3 of 2 mol dm-3 dilute nitric acid is
added to 100 cm3 of 2 mol dm-3 sodium hydroxide ,
the temperature of the reaction mixture rises from
270C to 40.650C . Calculate the heat of neutralisation
(c=4.2 J g-1 0C-1 , density of solution, 1 gcm-3 )
Answer = -57.33 kJ mol-1
Exercise – pg 219 (cerdik publication)

42. Exercise 2 – to find θ
The thermochemical equation for the reaction
between nitric acid and sodium hydroxide
solution is shown below :
HNO3 + NaOH  NaNO3 + H2O ΔH = -57.3 kJ mol-1
When 250 cm3 of 1.0 mol dm-3 nitric acid is added
to 200 cm3 of 2.0 mol dm-3 sodium hydroxide
solution, what is the change in temperature ?
(c = 4.2 J g-1 0C-1 , density of water = 1g cm-3)
Ans: 7.60C

43. Exercise 3 – calculation involving a
dibasic acid
200 cm3 of 2 mol dm-3 sodium hydroxide solution
was added to a fixed volume of 1 mol dm-3 dilute
sulphuric acid . Calculate :
1) the quantity of heat given off
2) the volume of dilute sulphuric acid used .
(c = 4.2 J g-1 0C-1 , density of water = 1g cm-3 ,
ΔH = -57.3 kJ mol-1 )
Ans: 22920J , 200cm3

44. 4. HEAT OF COMBUSTION
Definition – the heat released when one
mole of a substance is burnt
completely in an excess of
oxygen

45. Determining the heat of
combustion
thermometer
water
ethanol
windshield
Copper can
Spirit lamp
Wooden block

46. Results
Initial temperature of 200g
water /0C
Highest temperature of 200
g water / 0C
Increase in the temperature
/ 0C
Mass of lamp and ethanol
before burning / g
Mass of lamp and ethanol
after burning / g
Mass of ethanol burnt / g
28.0
58.0
58.0 – 28.0 = 30.0
245.85
245.85 - 244.95 = 0.90
244.95

47. Step 2
Mass of ethanol
= 245.85 - 244.95
= 0.90g
Moles of ethanol
= 0.9 /molar mass
=0.9g/2(12)+5(1)+16(1)+1(1)gmol-1
=0.0196 mol
Calculation :
Step 1
Heat given out
in the reaction = mcθ
= (200)gx4.2Jg-1 0C-1x300C
= 25200J

48. Step 3
0.0196 mole of C2H5OH gives out 25200 J
:. 1 mole of C2H5OH gives out
25200J x 1 mole =1286000 J
0.0196 mole
The heat of combustion = - 1286 kJ mol-1
of ethanol
(ΔHcombustion)
Energy
C2H5OH + 3O2
2CO2 + 3H2O
ΔH = -1286 kJ mol-1

49. Exercise 1 – to find the ΔH
1.6 g methanol is burnt completely in excess oxygen
.The heat given off is used up to heat up 300 cm3 of
Water . The temperature of the water rises by 28.80C.
Calculate the heat of combustion of methanol
(Ar H = 1 , C=12, O=16 , c=4.2 J g-1 0C-1 , density
of water, 1 gcm-3 )
Answer = -725.76 kJ mol-1
Exercise – pg 141 (cerdik publication)

50. Exercise 2 – to find θ
The thermochemical equation for the combustion
of glucose is shown below :
C6H12O6 + 6 O2  6 CO2 + 6 H2O ΔH = -2400 kJ mol-1
36 g of glucose is burnt completely to heat up
800 cm3 of water. Calculate the rise in
temperature of the water.
(Mr of glucose = 180 , c = 4.2 J g-1 0C-1 , density
of water = 1g cm-3)
Ans: 142.90C

51. Exercise 3 – to find mass of alcohol burnt
Complete combustion of 1 mole of butan-1-ol
produces 2678 kJ of heat . Calculate the mass of
butan-1-ol needed to burn completely in excess
oxygen in order to raise the temperature of
500 cm3 of water by 350C .
(c = 4.2 J g-1 0C-1 , density of water = 1g cm-3)
Ans: 2.03g

52. Relationship between relative molecular
mass and heat of combustion of alcohols
Relative molecular mass
of alcohol increases
Heat of combustion of
alcohol increases
Refer your revision book

53. FUEL VALUE
Definition :
-the amount of heat energy given out when
one gram of the fuel is completely burnt
in excess of oxygen .
-the unit for fuel value if kJ g-1
-a fuel with a high fuel value releases a lot
of heat per gram when it burns.

54. Exercise 4 – to find mass of alcohol burnt
The fuel value of charcoal is 35 kJ g-1 . How much
charcoal must be burnt to boil 1.8 dm3 of water ?
(c = 4.2 J g-1 0C-1 , density of water = 1g cm-3 ,
room temperature of water ,270C )
Ans: 15.77g

55. End of the topic – please come and see
me if you have any problems
Success is simply the natural outcome of
our directed intentions and actions .
SUCCESS DOES NOT COME TO YOU .
YOU GO TO IT.
SUCCESS IS A RESULT , NOT A GOAL

56. Never doubt
You have great
potential within you.
Know yourself and
appreciate all the special
qualities you have.
Never doubt your abilities.
Concentrate on whatever
make you happy and
build your spirit up.
Always aim higher than
you believe you can reach.
Be persistent and consistent.
Do not be afraid to walk
on new path if your
heart lead you to.
Believe in yourself-
that you can ultimately
achieve what you want in life .