Softcopy for Teachers FORM 5

MODULE • Chemistry Form 5

Energy Changes in Chemical Reactions
Perubahan Tenaga dalam Tindak Balas Kimia

Almost all chemical reactions During chemical reaction, heat energy is either released to the surrounding or absorbed from the
involve energy change. Why is surrounding.
there energy change? Semasa tindak balas kimia, tenaga haba dibebaskan ke persekitaran atau diserap dari persekitaran.
Hampir semua tindak balas kimia
melibatkan perubahan tenaga.
Mengapakah ada perubahan
tenaga?

What are the two types of energy Two types of reactions that occur are:
changes in chemical reaction? Dua jenis tindak balas yang berlaku adalah:
Apakah dua jenis perubahan
tenaga dalam tindak balas kimia? (a) exothermic reaction / tindak balas eksotermik

(b) endothermic reaction / tindak balas endotermik

How do we know energy changes Measure the temperature change in the surrounding.
during chemical reaction? Ukur perubahan suhu persekitaran.
Bagaimanakah kita mengetahui
perubahan tenaga semasa tindak Remark / Catatan:
balas kimia? Other possible ways to detect energy change in a reaction quickly is by touching.
Cara lain yang untuk mengesan perubahan tenaga dalam suatu tindak balas ialah dengan meyentuh.

Define exothermic reaction. It is a chemical reaction that releases heat to the surrounding.
Nyatakan maksud tindak balas
eksotermik. Tindak balas kimia yang membebaskan haba ke persekitaran.

How do we know if a reaction is The temperature in the surrounding rises .
exothermic? / Bagaimanakah kita Suhu persekitaran meningkat .
mengetahui jika tindak balas
adalah eksotermik?

Examples of exothermic reactions. Reaction between zinc and copper(II) sulphate / Tindak balas antara zink dengan kuprum(II) sulfat
Contoh tindak balas eksotermik.

Temperature increases 34 35 36 37 38 39 40 41 42 43
Suhu meningkat

Excess zinc powder
Serbuk zink berlebihan

Heat / Haba Copper(II) sulphate solution
Larutan kuprum(II) sulfat
Heat / Haba

Heat energy is released to the surroundings U
Tenaga haba dibebaskan kepada persekitaran N
I
Give other examples of chemical Type of reaction Example of chemical equation for the reaction T
equations which are exothermic Jenis tindak balas Contoh persamaan kimia untuk tindak balas
reaction. 4
Berikan contoh lain persamaan Neutralisation / Peneutralan 2KOH + H2SO4 → K2SO4 + 2H2O
kimia yang merupakan tindak
balas eksotermik. Reaction between acids and metals Mg + 2HCl → MgCl2 + H2
Tindak balas antara asid dengan logam

Reaction between acids and metal carbonate CaCO3 + 2HCl → CaCl2 + H2O + CO2
Tindak balas antara asid dengan karbonat
logam

Combustion of alcohol / Pembakaran alkohol C2H5OH + 3O2 → 2CO2 + 3H2O

Remark / Catatan:
– * Neutralisation reaction, reaction between metal with acid, reaction between carbonate and acids were studied in topic

“Acid and base” (Form 4).
* Tindak balas peneutralan, tindak balas antara logam dengan asid, tindak balas antara karbonat dan asid dipelajari

dalam tajuk “Asid dan Bes” (Tingkatan 4)
– *Combustion of alcohols was studied in topic “Carbon Compound” (Form 5).
*Pembakaran alkohol dipelajari dalam tajuk “Sebatian Karbon” (Tingkatan 5).

149 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Can physical changes also be Yes / Ya
exothermic? Give examples.
Bolehkah perubahan fizikal juga Physical changes Example
menjadi eksotermik? Berikan Perubahan fizikal Contoh
contoh.
Dissolving sodium hydroxide/potassium hydroxide in water
Define endothermic reaction.
Nyatakan maksud tindak balas Melarutkan natrium hidroksida/kalium hidroksida dalam air
endotermik.
How do we know if a reaction is NaOH(s/p) H2O Na+ (aq/ak) + OH–(aq/ak)
endothermic? / Bagaimanakah kita
mengetahui jika suatu tindak Dissolving substance in water Dissolving anhydrous salt such as copper(II) sulphate in water
balas adalah endotermik? Melarutkan bahan dalam air
Examples of endothermic Melarutkan garam kontang seperti kuprum(II) sulfat dalam air
reaction. Change of state of matter
Contoh tindak balas endotermik. Perubahan keadaan jirim CuSO4(s/p) H2O Cu2+(aq/ak) + SO42–(aq/ak)

Adding water to concentrated acid

Menambahkan air kepada asid pekat

H2SO4(aq/ak) H2O 2H+ (aq/ak) + SO42–(aq/ak)

– Condensation / Kondensasi
– Freezing / Pembekuan
– Sublimation (gas changes to solid)
Pemejalwapan (gas bertukar kepada pepejal)

Remark / Catatan:
1 Not all dissolving of substances in water are exothermic. Some can be endothermic.
2 Change of state of matter was studied in topic “ Atomic Structure” (Form 4)
1 Tidak semua melarutkan bahan dalam air adalah eksotermik. Sebahagian adalah endotermik.
2 Perubahan keadaan jirim dipelajari dalam topik “Struktur Atom” (Tingkatan 4)

It is a chemical reaction that absorbs heat energy from the surrounding.

Tindak balas kimia yang menyerap tenaga haba dari persekitaran.

The temperature in the surrounding decreases .
Suhu persekitaran menurun .

Dissolving ammonium nitrate in water / Melarutkan ammonium nitrat dalam air

Temperature decreases 34 35 36 37 38 39 40 41 42 43
Suhu menurun

Ammonium nitrate
Ammonium nitrat

U Heat / Haba Water / Air
N Heat / Haba
I
T Heat energy is absorbed from the surroundings
Tenaga haba diserap dari persekitaran
4 Give other examples of chemical
equations which are endothermic Type of reaction Example of chemical equation for the reaction
reaction: Jenis tindak balas Contoh persamaan kimia tindak balas
Berikan contoh lain persamaan
kimia yang merupakan tindak Thermal decomposition of nitrate salts reaction. 2Cu(NO3)2 Δ 2CuO + 4NO2 + O2
balas endotermik: Penguraian terma tindak balas garam nitrat.
ZnCO3 Δ ZnO + CO2
© Nilam Publication Sdn Bhd Thermal decomposition of carbonate salts
reaction. HCl + KHCO3 KCl + CO2 + H2O
Penguraian terma tindak balas garam karbonat.
6CO2 + 6H2O UV C6H12O6 + 6O2
Reaction of acids with hydrogen carbonate.
Tindak balas asid dengan hidrogen karbonat.

Photosynthesis
Fotosintesis

Remark / Catatan:
Decomposition of nitrate and carbonate salts was studied in topic “Salt” (Form 4).
Penguraian garam nitrat dan karbonat dipelajari dalam tajuk “Garam” (Tingkatan 4).

150

MODULE • Chemistry Form 5

Can physical changes also be Yes / Ya Example
endothermic? Give examples. Physical changes Contoh
Bolehkah perubahan fizikal juga Perubahan fizikal
menjadi endotermik? Berikan Dissolving ammonium salts/nitrate salts in water
contoh. Dissolving substance in water Melarutkan garam ammonium/garam nitrat dalam air
Melarutkan bahan dalam air
H2O NH4+ (aq/ak) + NO3– (aq/ak)
Heating of hydrated salt NH4NO3(s/p)
Pemanasan garam terhidrat
KNO3(s/p) H2O K+ (aq/ak) + NO3– (aq/ak)

Decomposition of hydrated salt to anhydrous salt and water

Penguraian garam terhidrat kepada garam kontang dan air

CuSO4.5H2O(s/p) D CuSO4(s/p) + 5H2O(l/ce)

(blue/biru) (white/putih)

Change of state of matter – Melting
Perubahan keadaan jirim Peleburan
– Boiling/evaporation
Pendidihan/penyejatan
– Sublimation (solid changes to gas)
Pemejalwapan (pepejal bertukar kepada gas)

Remark / Catatan:
Change of state of matter is in the topic “Atomic Structure”( Form 4)
Perubahan keadaan jirim dalam tajuk “Stuktur Atom” (Tingkatan 4)

Energy change during formation and breaking of bonds
Perubahan tenaga semasa pembentukan dan pemutusan ikatan

The energy changes are caused (i) Breaking of bonds in the reactant
by two processes taking place Pemutusan ikatan dalam bahan tindak balas
during chemical reaction (ii) Formation of bonds in the products
when reactants change to Pembentukan ikatan dalam hasil
products. What are these two
processes? Heat energy is absorbed to break the bonds in the reactants. U
Perubahan tenaga disebabkan Tenaga haba diserap untuk memecahkan ikatan dalam bahan tindak balas. N
oleh dua proses yang berlaku I
semasa tindak balas kimia apabila Heat energy is released during bond formation in the products. T
bahan tindak balas berubah Tenaga haba dibebaskan semasa pembentukan ikatan dalam hasil tindak balas.
kepada hasil. Apakah dua proses 4
ini?
It depends on the strength of bonds:
How is breaking of bonds in the Ia bergantung kepada kekuatan ikatan:
reactants occur? (i) More energy is absorbed to break a strong bond compared to a weak bond.
Bagaimanakah pemutusan ikatan Lebih tenaga diserap untuk memutuskan ikatan yang kuat berbanding ikatan lemah.
dalam bahan tindak balas (ii) More energy is released when a strong bond is formed compared to a weak bond.
berlaku? Lebih tenaga dibebaskan apabila ikatan yang kuat terbentuk berbanding ikatan yang lemah.

How is formation of bond in the Heat of reaction, ∆H is the difference between heat energy absorbed and heat energy released when
product occur? 1 mole of reactant react or 1 mole of product is formed.
Bagaimana pembentukan ikatan Haba tindak balas, ∆H ialah perbezaan antara tenaga haba yang dibebaskan dan tenaga haba yang
dalam hasil berlaku? diserap apabila 1 mol bahan bertindak balas atau 1 mol hasil terbentuk.

What is the factor that affect the
quantity of heat absorbed or
released during breaking and
formation of bonds?
Apakah faktor yang
mempengaruhi kuantiti haba yang
diserap atau dibebaskan semasa
pemutusan dan pembentukan
ikatan?

How heat of reaction, ΔH is
obtained?
Bagaimana haba tindak balas, ΔH
diperolehi?

151 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Example of reaction: / Contoh tindak balas: 2H2 + O2 → 2H2O, ∆H = –486 kJ Bonds are formed in the products:
Ikatan terbentuk dalam hasil tindak balas:
Bonds are broken in the reactants:
Ikatan diputuskan dalam bahan tindak balas:

Heat is absorbed Heat is released
Haba diserap Haba dibebaskan

(+1 370 kJ heat energy absorbed) (–1 856 kJ heat energy released)
(+1 370 kJ tenaga haba diserap) (–1 856 kJ tenaga haba dibebaskan)

➢ The heat released from bond formation is higher than heat absorbed for bond breaking.

Tenaga haba yang dibebaskan dari pembentukan ikatan lebih tinggi daripada haba yang diserap semasa pemutusan ikatan.

➢ A negative sign for ∆H shows that heat is released .

Tanda negatif untuk ∆H menunjukkan bahawa haba dibebaskan .

Energy change in exothermic reaction / Perubahan tenaga dalam tindak balas eksotermik

Energy profile diagram for Energy / Tenaga
exothermic reactions.
Rajah profil tenaga untuk tindak Heat energy
balas eksotermik. is absorbed
(+ve)
Tenaga haba Heat energy is
diserap (+if) released (–ve)
Tenaga haba
Reactants dibebaskan
Bahan tindak (–if)
balas Products
Hasil tindak
∆H is balas
negative
∆H
adalah
negatif

Remark / Catatan:
Heat energy absorbed for bond breaking is equal to the activation energy, Ea (topic“Rate of Reaction”).
Tenaga haba diserap untuk memutuskan ikatan adalah sama dengan tenaga pengaktifan, Ea (tajuk “Kadar Tindak
Balas”).

Compare the quantity of heat The quantity of heat energy absorbed for bonds breaking in the reactants is lower than
energy absorbed and released in heat energy released for the formation of bonds in the products.
U the reaction.
N
I Bandingkan kuantiti tenaga haba Kuantiti tenaga haba yang diserap untuk pemecahan ikatan dalam bahan tindak balas adalah lebih
yang diserap dan dibebaskan
dalam tindak balas tersebut. rendah daripada tenaga haba yang dibebaskan untuk pembentukan ikatan dalam hasil

T tindak balas.
4 Compare strength of bonds in the
reactants and products. Weak bonds are broken and strong bonds are formed.
Bandingkan kekuatan ikatan Ikatan lemah dipecahkan dan ikatan kuat dibentuk.

dalam bahan tindak balas dan

hasil tindak balas.

What is the sign of ΔH? The sign of ∆H is negative . / Tanda bagi ∆H adalah negatif .
Apakah tanda ΔH?

Why is the sign of ΔH negative? A negative sign for ∆H shows that heat is released to the surrounding.
Mengapa tanda ΔH adalah ke persekitaran.
negatif? Tanda negatif ∆H menunjukkan haba dibebaskan

Why is the temperature Heat energy is released to the surroundings, temperature of the surroundings
increases?
Mengapakah suhu meningkat? increases . (Surroundings include the reaction solution, container and the air).

Tenaga haba dibebaskan ke persekitaran, suhu persekitaran naik . (Persekitaran

termasuklah larutan bahan tindak balas, bekas dan udara)

© Nilam Publication Sdn Bhd 152

MODULE • Chemistry Form 5

What is the energy change in the Energy change: / Perubahan tenaga: energy → Heat energy
reaction? Chemical → Tenaga haba
Apakah perubahan tenaga dalam
tindak balas? Tenaga kimia

Compare the total energy content Total energy content of the products is less than total energy of the reactants.
of the reactants and products. Jumlah kandungan tenaga hasil kurang daripada jumlah kandungan tenaga bahan tindak balas.
Bandingkan jumlah kandungan
tenaga bahan dan hasil tindak Energy / Tenaga
balas. Reactants
Bahan tindak balas
Draw the energy level diagram for
exothermic reaction.
Lukis rajah aras tenaga untuk
tindak balas eksotermik.

∆ H is negative (heat is released)
∆ H adalah negatif (haba dibebaskan)
Products / Hasil tindak balas

Energy change in endothermic reaction: Energy
Perubahan tenaga dalam tindak balas endotermik: Tenaga

Energy profile diagram for
endothermic reactions.
Rajah profil tenaga untuk tindak
balas endotermik.

Heat energy Heat energy is released (–ve)
is absorbed Tenaga haba dibebaskan (–if)
(+ve) Products
Tenaga haba Hasil tindak
diserap (+if) balas

Reactants ∆H is positive
Bahan tindak ∆H adalah positif
balas

Compare the quantity of heat The quantity of heat energy absorbed for bonds breaking in the reactants is higher than U
energy absorbed and released in heat energy released from the formation of bonds in the products. N
the reaction. I
Bandingkan kuantiti tenaga haba Kuantiti tenaga haba diserap untuk pemecahan ikatan dalam bahan tindak balas lebih tinggi T
yang diserap dan dibebaskan daripada tenaga haba dibebaskan dari pembentukan ikatan dalam hasil tindak balas.
dalam tindak balas tersebut. 4
Strong bonds are broken and weak bonds are formed.
Compare strength of bonds in the I katan kuat dipecahkan dan ikatan lemah dibentuk.
reactants and products.
Bandingkan kekuatan ikatan Heat of reaction, ∆H is the difference between heat energy absorbed and heat energy released.
dalam bahan tindak balas dan Haba tindak balas, ∆H adalah perbezaan antara tenaga haba yang diserap dengan tenaga haba yang
hasil tindak balas. dibebaskan.

How heat of reaction, ΔH is The sign of ∆H is positive . / Tanda untuk ∆H adalah positif .
obtained?
Bagaimana haba tindak balas, ΔH A positive sign for ∆H shows that heat is absorbed from the surrounding.
diperolehi? Tanda positif ∆H menunjukkan haba diserap dari persekitaran.

What is the sign of ΔH?
Apakah tanda bagi ΔH?

Why is the sign of ΔH positive?
Mengapa tanda ΔH adalah positif?

153 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Why is the temperature Heat is absorbed from the surroundings, temperature of the surrounding decreases (Surrounding
decreases? include the reaction solution, container and the air). / Haba diserap dari persekitaran, suhu
Mengapakah suhu menurun? persekitaran menurun (Persekitaran termasuklah larutan bahan tindak balas, bekas dan udara).

What is the energy change in the Energy change: / Perubahan tenaga: energy → Chemical energy
reaction? Heat
Apakah perubahan tenaga dalam
tindak balas tersebut? Tenaga haba → Tenaga kimia

Compare the total energy content Total energy content of the products is more than total energy of the reactants.
of the reactants and products. Jumlah kandungan tenaga hasil tindak balas lebih daripada jumlah kandungan tenaga bahan tindak balas.
Bandingkan jumlah kandungan
tenaga bahan dan hasil tindak Energy / Tenaga
balas.
Products / Hasil tindak balas
Draw the energy level diagram for
exothermic reaction.
Lukis rajah aras tenaga untuk
tindak balas eksotermik.

∆H is positive (heat is absorbed)
∆H adalah positif (haba diserap)
Reactants
Bahan tindak balas

Application of knowledge of exothermic and endothermic reactions in everyday life.
Aplikasi tindak balas eksotermik dan endotermik dalam kehidupan seharian.

I Hot packs:
Pek panas:

State the type of reaction take Contain chemicals that release heat, application of exothermic reaction. eksotermik .
place in the hot packs. Mengandungi bahan kimia yang membebaskan haba, aplikasi bagi tindak balas
Nyatakan jenis tindak balas yang
berlaku di dalam pek panas.

How is the structure of the hot It is a plastic bag containing separate compartments of water and anhydrous calcium chloride.
packs? / Bagaimanakah struktur Ia adalah beg plastik yang mengandungi ruang berasingan air dan kalsium klorida kontang.
pek panas?

Explain how the reaction take The anhydrous calcium chloride dissolves in water to release heat, thus causing the temperature

place in the hot pack. to increase . / Kalsium klorida kontang larut dalam air dan membebaskan haba yang seterusnya

U Terangkan bagaimana tindak menyebabkan suhu meningkat .
N balas berlaku dalam pek panas. CaCl2(s/p) H2O Ca2+(aq/ak) + 2Cl–(aq/ak)

TI ∆ H = –83 kJ mol–1

4 What are other substances that Other substances that can be used in a hot pack are anhydrous magnesium sulphate, anhydrous
can be used in the hot packs? copper(II) sulphate and calcium oxide.
Apakah bahan lain yang boleh Bahan lain yang boleh digunakan dalam pek panas adalah magnesium sulfat kontang, kuprum(II) sulfat
kontang dan kalsium oksida.
digunakan dalam pek panas?

How is the structure of reusable A reusable hot pack uses supersaturated solution of sodium ethanoate crystallisation and resolution.
hot packs? Pek panas yang boleh dipakai semula menggunakan larutan tepu natrium etanoat yang akan menghablur.
Bagaimanakah struktur pek panas
yang boleh diguna semula?

State the uses of hot packs. – Mountain climbers use hot packs to keep their hands and feet warm.
Nyatakan kegunaan pek panas. Pendaki gunung menggunakan pek panas untuk menghangatkan tangan

dan kaki.
– Heat can increase blood flow and help restore movement to injured tissue.
Haba dapat meningkatkan pengaliran darah dan membantu mengembalikan

pergerakan tisu yang tercedera.
– Heat can also reduce joint stiffness, pain, and muscle spasm.
Haba juga mengurangkan sengal sendi, kesakitan dan kejang otot.

© Nilam Publication Sdn Bhd 154

MODULE • Chemistry Form 5

II Cold packs:
Pek sejuk:

State the type of reaction take Contain chemicals that absorb heat, application of endothermic reaction. .
place in the cold packs. Mengandungi bahan kimia yang menyerap haba, aplikasi bagi tindak balas endotermik
Nyatakan jenis tindak balas
yang berlaku dalam pek sejuk.

How is the structure of the cold It is a plastic bag containing separate compartments of water and solid ammonium nitrate.
packs? Ia adalah beg plastik yang mengandungi ruang yang berasingan air dan pepejal ammonium nitrat.
Bagaimanakah struktur pek
sejuk? The solid ammonium nitrate dissolves in water to absorb heat, thus causing the temperature to

Explain how the reaction take decrease .
place in the cold pack.
Terangkan bagaimana tindak Pepejal ammonium nitrat larut dalam air menyerap haba yang seterusnya menyebabkan suhu
balas berlaku dalam pek sejuk.
menurun . H2O NH4+(aq/ak) + NO3– (aq/ak) ∆H = + 26 kJ mol–1
What are other substances can be NH4NO3(s/p)
used in the cold packs?
Apakah bahan lain yang boleh Other substances that can be used in a cold pack are ammonium chloride, potassium nitrate and sodium
digunakan dalam pek sejuk? thiosulphate.
Bahan lain yang boleh digunakan dalam pek sejuk adalah ammonium klorida, kalium nitrat dan natrium
State the uses of cold packs. tiosulfat.
Nyatakan kegunaan pek sejuk.
– To reduce body temperature for a patient who has fever.
Untuk mengurangkan suhu badan pesakit yang demam.
– To ease pain from both acute and chronic injuries, such as sprains and
arthritis.
Untuk melegakan kesakitan kecederaan akut dan kronik, seperti terseliuh
dan artritis.

Exercise / Latihan

1 The following is the thermochemical equation of neutralisation:
Berikut ialah persamaan termokimia peneutralan:

HCl + NaOH NaCl + H2O            ∆H = –57 kJ mol–1 U
N
Remark: / Catatan: I
T
Thermochemical equation is an equation that shows the heat of reaction together with the balanced equation.
4
Persamaan termokimia ialah satu persamaan yang menunjukkan haba tindak balas bersama dengan persamaan seimbang.

Construct energy Energy / Tenaga
level diagram for the HCl + NaOH
thermochemical equation. ∆H = –57 kJ mol–1
Bina rajah aras tenaga
untuk persamaan termokimia
tersebut.

NaCl + H2O

155 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Give four statements to (i) The reaction between hydrochloric acid and sodium hydroxide is exothermic .
interpret the energy level .
diagram that you have (ii) During reaction, the temperature of the mixture increases
constructed.
Berikan empat pernyataan (iii) When one mole of hydrochloric acid reacts with one mole of sodium hydroxide to produce
untuk mentafsir rajah aras one mole of sodium chloride and one mole of water, the quantity of heat released is
tenaga yang telah anda bina.
57 kJ.
(iv) The total energy of 1 mole of hydrochloric acid and 1 mole of sodium hydroxide is

more than the total energy of 1 mole of sodium chloride and 1 mole of water.
The difference in energy is 57 kJ.

(i) Tindak balas antara asid hidroklorik dengan natrium hidroksida adalah
.
eksotermik .

(ii) Semasa tindak balas, suhu campuran meningkat

(iii) Apabila satu mol asid hidroklorik bertindak balas dengan satu mol natrium hidroksida

menghasilkan satu mol natrium klorida dan satu mol air, kuantiti haba yang
dibebaskan ialah 57 kJ.

(iv) Jumlah tenaga bagi 1 mol asid hidroklorik dan 1 mol natrium hidroksida lebih
daripada jumlah tenaga 1 mol natrium klorida dan 1 mol air. Perbezaan tenaga

adalah 57 kJ.

2 The following is the thermochemical equation for the dissolving of sodium nitrate in water:
Berikut ialah persamaan termokimia bagi melarutkan natrium nitrat dalam air:

NH4NO3(s/p) H2O NH4+ (aq/ak) + NO3–(aq/ak)            ∆H = +26 kJ mol–1

Construct energy Energy / Tenaga
level diagram for the
thermochemical equation. NH4+ (aq/ak) + NO3– (aq/ak)
Bina rajah aras tenaga
untuk persamaan termokimia ∆H = +26 kJ mol–1
tersebut.

NH4NO3(s/p)

U Give four statements to (i) Dissolving ammonium nitrate in water is endothermic .
N interpret the energy level
I diagram that you have (ii) Temperature of the solution decreases .
T
4 constructed. (iii) When one mole of ammonium nitrate dissolves in water, the quantity of heat
Berikan empat pernyataan
absorbed is 26 kJ.

untuk mentafsir rajah aras (iv) The total energy of 1 mole of solid ammonium nitrate is less than the total

tenaga yang telah anda bina. energy of ammonium nitrate solution. The difference in energy is 26 kJ.

(i) Melarutkan ammonium nitrat dalam air adalah endotermik .

(ii) Suhu larutan menurun .

(iii) Apabila satu mol ammonium nitrat larut dalam air, kuantiti haba yang diserap

ialah 26 kJ.

(iv) Jumlah tenaga bagi 1 mol pepejal ammonium nitrat adalah kurang daripada

jumlah tenaga 1 mol larutan ammonium nitrat. Perbezaan tenaga adalah 26 kJ.

© Nilam Publication Sdn Bhd 156

MODULE • Chemistry Form 5

3 The diagram below shows the energy level of reaction I and II.
Gambar rajah di bawah menunjukkan aras tenaga bagi tindak balas I dan II.

Energy / Tenaga Energy / Tenaga

2NO2(g/g) Zn + CuSO4

∆H = +66 kJ mol–1 ∆H = –210 kJ mol–1
N2(g/g) + 2O2(g/g)
ZnSO4 + Cu
Reaction I / Tindak balas I Reaction II / Tindak balas II

Based on the diagram above, (i) Reaction I is endothermic while reaction II is exothermic .
compare the energy level (ii) Heat is absorbed from the surrounding in reaction I while heat is released to the
diagram in reaction I and
reaction II. surrounding in reaction II.
Berdasarkan gambar rajah di (iii) The total energy content of 1 mole of nitrogen gas and 2 moles of oxygen gas is lower
atas, bandingkan rajah aras
tenaga tindak balas I dan than the total energy content of 2 moles of nitrogen dioxide in reaction I. The total energy
tindak balas II. of the content of 1 mole of zinc and 1 mole of copper(II) sulphate is higher than the

total energy content of 1 mole of zinc sulphate and 1 mole of copper in reaction II.
(iv) The quantity of heat absorbed during reaction I is 66 kJ (heat of reaction is

+66 kJ mol–1) while the quantity of heat released during reaction II is 210 kJ

(heat of reaction is –210 kJ mol–1).

(i) Tindak balas I adalah endotermik manakala tindak balas II adalah

eksotermik .

(ii) Haba diserap dari persekitaran dalam tindak balas I manakala haba dibebaskan ke

persekitaran dalam tindak balas II.
(iii) Jumlah kandungan tenaga 1 mol gas nitrogen dan 2 mol gas oksigen lebih rendah

daripada jumlah kandungan tenaga 2 mol nitrogen dioksida dalam tindak balas
I. Jumlah kandungan tenaga 1 mol zink dan 1 mol kuprum(II) sulfat lebih tinggi

daripada jumlah tenaga 1 mol zink sulfat dan 1 mol kuprum dalam tindak balas II. U
(iv) Kuantiti haba yang diserap semasa tindak balas I adalah 66 kJ (haba tindak N
I
balas ialah +66 kJ mol–1) manakala kuantiti haba yang dibebaskan semasa tindak T
balas II adalah 210 kJ
4
(haba tindak balas ialah –210 kJ mol–1).

Endothermic and
Exothermic

Endotermik dan Eksotermik

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157 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

The Heat of Reaction (∆H) / Haba Tindak Balas (∆H)

What is heat of reaction, ΔH? The amount of heat energy released or absorbed during chemical reaction.
Apakah haba tindak balas, ΔH? Jumlah tenaga haba dibebaskan atau diserap semasa tindak balas kimia.

Different types of chemical Or / Atau
reactions will give different value
of heat of reaction. What are the The difference between total energy content of the reactants and the products.
heat of reactions for specific Perbezaan antara jumlah kandungan tenaga bahan tindak balas dan hasil.
reactions in this chapter?
Jenis tindak balas kimia yang Type of reaction Specific heat of reaction Remark
berbeza akan memberikan nilai Jenis tindak balas Haba tentu tindak balas Catatan
haba tindak balas yang berbeza.
Apakah haba tindak balas untuk Precipitation Heat of precipitation Precipitation reaction in topic “Salt” (Form 4)
tindak balas tertentu dalam bab Pemendakan Haba pemendakan Tindak balas pemendakan dalam tajuk “Garam”
ini? (Tingkatan 4)

Displacement Heat of displacement Displacement reaction in topic “Electrochemistry”
Penyesaran Haba penyesaran (Form 4)
Tindak balas penyesaran dalam tajuk
“Elektrokimia” (Tingkatan 4)

Neutralisation Heat of neutralisation Neutralisation reaction in topic “Acid and Base”
Peneutralan Haba peneutralan (Form 4)
Tindak balas peneutralan dalam tajuk “Asid dan
Bes” (Tingkatan 4)

Combustion Heat of combustion Combustion of alcohol in topic “Carbon
Pembakaran Haba pembakaran Compound” (Form 5)
Pembakaran alkohol dalam tajuk “Sebatian
Karbon” (Tingkatan 5)

Complete the following table.
Lengkapkan jadual di bawah.

Heat of reaction Definition Example
Haba tindak balas Definisi Contoh

Heat of Heat of precipitation is heat change Thermochemical equation:

precipitation when 1 mole of precipitate is Persamaan termokimia:

Haba pemendakan formed from its ions in aqueous Pb(NO3)2(aq/ak) + Na2SO4(aq/ak) → PbSO4(s/p) + 2NaNO3(aq/ak)
solution.
Haba pemendakan ialah perubahan ∆H = –50.4 kJ mol–1

haba apabila 1 mol mendakan

terbentuk dari ion-ionnya dalam Ionic equation:

larutan akueus. Persamaan ion:
U Remark / Catatan:
N Precipitation reaction occurs when two Pb2+ + SO42– → PbSO4
I solutions containing cations and anions
T of insoluble salts are added together. • 50.4 kJ heat energy is released when 1 mole of lead(II) ions reacted with
This reaction is used to prepare any
insoluble salt. 1 mole of sulphate ions to form 1 mole of lead(II) sulphate .
50.4 kJ tenaga haba dibebaskan apabila 1 mol ion plumbum(II) bertindak
4 Tindak balas pemendakan berlaku
apabila dua larutan mengandungi kation balas dengan 1 mol ion sulfat untuk membentuk 1 mol plumbum(II) sulfat .

dan anion garam tak terlarutkan

dicampur bersama. Tindak balas ini Energy level diagram:
digunakan untuk menyediakan garam Gambar rajah aras tenaga:
tak terlarutkan.

Energy / Tenaga

Pb2+(aq/ak) + SO42–(aq/ak)

ΔH = –50.4 kJ mol–1

PbSO4(s/p)

© Nilam Publication Sdn Bhd 158

MODULE • Chemistry Form 5

Heat of reaction Definition Example
Haba tindak balas Definisi Contoh

Heat of Heat of displacement is heat Thermochemical equation: / Persamaan termokimia:
displacement change when 1 mole of a metal is Zn(s/p) + CuSO4(aq/ak) → ZnSO4(aq/ak) + Cu(s/p)
Haba penyesaran displaced from its solution by a ∆H = –217 kJ mol–1
more electropositive metal.
Haba penyesaran ialah perubahan Ionic equation: / Persamaan ion:
haba apabila 1 mol logam
disesarkan dari larutan garamnya Cu2+ + Zn → Zn2+ + Cu
oleh logam yang lebih
elektropositif. • 217 kJ heat energy is released when 1 mole of copper is displaced
from copper(II) sulphate solution by zinc .
Remark / Catatan:
Displacement reaction occurs when a 217 kJ tenaga haba dibebaskan apabila 1 mol kuprum disesarkan
metal which is situated at a higher dari larutan kuprum(II) sulfat oleh zink.
position (higher tendency to release
electron) in the electrochemical series Energy level diagram: / Gambar rajah aras tenaga:
displaces a metal below it from its salt Energy / Tenaga
solution. Cu2+(ak/aq) + Zn(s/p)
Tindak balas penyesaran berlaku
apabila logam yang berada di ∆H = –217 kJ mol–1
kedudukan yang lebih tinggi (lebih
cenderung melepaskan elektron) dalam Zn2+(ak/aq) + Cu(s/p)
siri elektrokimia menyesar logam di
bawahnya dari larutan garamnya.

Heat of Heat of neutralisation is heat Thermochemical equation: / Persamaan termokimia:
neutralisation released when 1 mole of water is
Haba peneutralan formed from neutralisation of acid KOH(aq/ak) + HNO3(aq/ak) → KNO3(aq/ak) + H2O(l/ce)
with an alkali.
Haba peneutralan ialah ∆H = –57 kJ mol–1
perubahan haba yang dibebaskan
apabila 1 mol air terbentuk dari Ionic equation: / Persamaan ion:
peneutralan asid dan alkali.
H+ + OH– → H2O
Remark / Catatan:
Neutralisation is the reaction between • 57 kJ heat energy is released when 1 mol of water formed from
an acid and a base to form only salt and neutralisation of potassium hydroxide with nitric acid .
water.
Peneutralan ialah tindak balas antara 57 kJ haba dibebaskan apabila 1 mol air terbentuk dari peneutralan
asid dan bes menghasilkan garam dan kalium hidroksida dengan asid nitrik .
air sahaja.
Energy level diagram: / Gambar rajah aras tenaga:

Energy / Tenaga

H+(aq/ak) + OH–(aq/ak)

ΔH = –57 kJ mol–1 U
H2O(l/ce) N
I
Heat of combustion Heat of combustion is heat Thermochemical equation: / Persamaan termokimia: T
Haba pembakaran released when 1 mole of fuel is
burnt completely in excess C2H5OH + 3O2 → 2CO2 + 3H2O 4
oxygen.
Haba pembakaran ialah haba yang ∆H = –1 366 kJ mol–1
dibebaskan apabila 1 mol bahan
api terbakar lengkap dalam • 1 366 kJ heat energy is released when one mole of ethanol is burnt
oksigen berlebihan. completely in excess oxygen . / 1 366 kJ tenaga haba dibebaskan

Remark / Catatan: apabila 1 mol etanol dibakar lengkap dalam oksigen berlebihan .
Combustion is a reaction when a
substance burns completely in the Energy level diagram: / Gambar rajah aras tenaga:
excess oxygen. Energy / Tenaga
Tindak balas pembakaran adalah C2H5OH + 3O2
tindak balas yang berlaku apabila
bahan terbakar lengkap dalam ∆H = –1 366 kJ mol–1
oksigen berlebihan.
2CO2 + 3H2O

159 © Nilam Publication Sdn Bhd

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© Nilam Publication Sdn Bhd Activity/Experiment to Determine Heat of Reaction MODULE • Chemistry Form 5
Aktiviti/Eksperimen untuk Menentukan Haba Tindak Balas

Heat of reaction (DH) is the energy change when one mole of reactant reacts or when one mole of product is formed.
Haba tindak balas (DH) ialah perbezaan haba apabila 1 mol bahan bertindak balas atau apabila 1 mol hasil tindak balas dihasilkan.

B

CALCULATION: / PENGIRAAN:

Heat of reaction, / Haba tindak balas, Heat released/Heat absorbed/ Heat change / Haba dibebaskan/ Haba diserap/Perubahan haba,
mcq
B DH = ± X H = mcq
B B
B

160 ∆H = Heat of precipitation: Heat change ∆H = Heat of displacement: Heat change ∆H = Heat of neutralisation: Heat released ∆H = H eat of combustion: Heat released when 1 mole
when 1 mol of precipitate is formed when 1 mole of metal is displaced from when 1 mole of water is formed from of fuel is burnt completely in excess oxygen.
from its ions in aqueous solution. its solution by a more electropositive
metal. / Haba penyesaran: Perbezaan neutralisation of acid with an alkali. Haba pembakaran: Haba yang terbebas apabila
Haba pemendakan: Perubahan haba apabila 1 mol logam disesarkan 1 mol bahan api terbakar lengkap dalam oksigen
haba apabila 1 mol mendakan daripada larutannya oleh logam yang Haba peneutralan: Haba yang terbebas berlebihan.
dihasilkan daripada ion-ionnya di lebih elektropositif. apabila 1 mol air dihasilkan daripada
dalam larutan akueus. peneutralan asid dengan suatu alkali. m = v olume of water in copper can
m = volume of salt solution isi padu air di dalam tin kuprum
m = t otal volume of salt solution isi padu larutan garam m = t otal volume of acid and alkali
jumlah isi padu larutan garam X = number of moles of metals displaced jumlah isi padu asid dan alkali X = number of moles of alcohol (from mass fuel in the
lamp)
X = n umber of moles of precipitate (from the balanced equation) X = number of moles of water (from the
(from the balanced equation) bilangan mol logam yang disesarkan balanced equation) bilangan mol alkohol (dari perbezaan jisim pelita)

bilangan mol mendakan (dari B (dari persamaan seimbang) bilangan mol air (dari persamaan B
seimbang)
B persamaan seimbang)
B


Wind shield Thermometer
Pengadang Termometer
angin
34 35 36 37 38 39 40 41 42 43
34 35 36 37 38 39 40 41 42 43

34 35 36 37 38 39 40 41 42 43
34 35 36 37 38 39 40 41 42 43
Sodium chloride solution Zinc powder Hydrochloric acid Water Copper Can
Larutan natrium klorida Serbuk zink Asid hidroklorik Air Tin kuprum
Silver nitrate solution Copper(II) sulphate solution Fuel
Larutan argentum nitrat Larutan kuprum(II) sulfat Sodium hydroxide Bahan api
solution
Material: / Bahan: Material: / Bahan: Larutan natrium Manipulated variable:
0.5 mol dm–3 of sodium chloride solution, 0.5 mol dm–3 of copper(II) sulphate, zinc hidroksida Pemboleh ubah dimanipulasikan:
0.5 mol dm–3 of silver nitrate solution powder / Larutan kuprum(II) sulfat Type of alcohols / Jenis alkohol
Larutan natrium klorida 0.5 mol dm–3, 0.5 mol dm–3, serbuk zink Manipulated variable: / Pemboleh ubah
larutan argentum nitrat 0.5 mol dm–3 Apparatus: / Radas: dimanipulasikan: Responding variable:
Polystyrene cup, measuring cylinder, Type of acid / Jenis asid Pemboleh ubah bergerak balas:
Apparatus: / Radas: thermometer Responding variable: Heat of combustion / Haba pembakaran
Polystyrene cup, measuring cylinder, Bekas polistirena, silinder penyukat, Pemboleh ubah bergerak balas:
thermometer / Bekas polistirena, silinder termometer Heat of neutralisation / Haba peneutralan Constant variable: / Pemboleh ubah dimalarkan:
penyukat, termometer Constant variable: / Pemboleh ubah Volume of water, type of metal can
Procedure: / Prosedur: dimalarkan:
Procedure: / Prosedur: 1 25 cm3 of 0.5 mol dm–3 copper(II) Volume and concentration of acid and alkali, Isi padu air, jenis bekas logam
1 25 cm3 of 0.5 mol dm–3 silver nitrate
sulphate solution is measured with type of alkali / Isi padu dan kepekatan asid dan Hypothesis: / Hipotesis:
solution is measured with measuring alkali, jenis alkali The higher the number of carbon and hydrogen atom per
measuring cylinder and poured into a Hypothesis: / Hipotesis:
cylinder and poured into a polystyrene Reaction between hydrochloric acid and sodium molecule alcohols, the higher the heat of combustion.
polystyrene cup. hydroxide has higher heat of neutralisation than
cup. Semakin bertambah bilangan atom karbon dan hidrogen

161 2 25 cm3 of 0.5 mol dm–3 sodium 2 A thermometer is placed into the reaction between ethanoic acid and sodium setiap molekul alkohol, semakin bertambah haba
chloride solution is measured with solution and the initial temperature T1
hydroxide / Tindak balas asid hidroklorik pembakaran.
another measuring cylinder and of the solution is recorded.
dengan natrium hidroksida menghasilkan Material: / Bahan:
poured into another polystyrene cup. 3 Half spatula of zinc powder is quickly Methanol, ethanol, propanol, butanol
and carefully added into the copper(II) haba peneutralan yang lebih tinggi dari haba
3 A thermometer is placed into each Metanol, etanol, propanol, butanol
solution and the initial temperature T1 sulphate solution. peneutralan antara asid etanoik dengan
Apparatus: / Radas:
and T2 of each solution is recorded. 4 The reaction mixture is stirred with the natrium hidroksida. Lamp, copper can, thermometer, wind shield, wooden
thermometer and the highest
4 The sodium chloride solution is poured Material: / Bahan block, measuring cylinder, tripod stand
quickly and carefully into the silver temperature, T2 is recorded. 2 mol dm–3 of hydrochloric acid, 2 mol dm–3
ethanoic acid, 2 mol dm–3 of sodium Pelita, bekas kuprum, termometer, pengadang, bongkah
nitrate solution. 1 25 cm3 larutan kuprum(II) sulfat kayu, silinder penyukat, tungku kaki tiga
berkepekatan 0.5 mol dm–3 disukat hydroxide solution
5 The reaction of mixture is stirred with dengan silinder penyukat dan dituang Asid hidroklorik 2 mol dm–3, asid etanoik Procedure: / Prosedur:
the thermometer and the highest ke dalam cawan polistirena. 2 mol dm–3, larutan natrium hidroksida 2 mol dm–3 1 100 cm3 of water is measured with measuring cylinder
Apparatus: / Radas:
temperature, T3 is recorded. 2 Termometer diletakkan di dalam Polystyrene cup, measuring cylinder, and poured into the copper can.
larutan tersebut dan suhu awal T1
1 25 cm3 larutan argentum nitrat dicatat. thermometer / Bekas polistirena, silinder 2 A thermometer is placed into the water and the initial
berkepekatan 0.5 mol dm–3 disukat penyukat, termometer temperature, T1 is recorded.
dengan silinder penyukat dan dituang 3 Separuh spatula serbuk zink Procedure: / Prosedur:
ke dalam cawan polistirena. ditambahkan dengan cepat dan 1 50 cm3 of 2 mol dm–3 sodium hydroxide 3 The copper can is placed on a tripod stand.
cermat ke dalam larutan kuprum(II) 4 A lamp is filled with methanol. The lamp is weighed
2 25 cm3 larutan natrium klorida sulfat. solution is measured with measuring
berkepekatan 0.5 mol dm–3 disukat and the initial mass, m1 is recorded.
dengan silinder penyukat dan dituang 4 Campuran tindak balas dikacau cylinder and poured into a polystyrene cup.
ke dalam cawan polistirena yang lain. dengan termometer dan suhu tertinggi, 2 50 cm3 of 2 mol dm–3 hydrochloric acid is 5 A wind shield is placed as shown in the diagram to
T2 dicatat. minimise heat loss to the moving air in the surrounding.
3 Termometer diletakkan di dalam setiap measured with another measuring cylinder
larutan dan suhu awal setiap larutan T1 6 The lamp is placed near the base of the copper can to
dan T2 dicatat. and poured into another polystyrene cup. maximise the heat transfer and the wick is lighted.

4 Larutan natrium klorida dituang 3 A thermometer is placed into each solution 7 The water is stirred continuously with the thermometer
dengan cepat dan cermat ke dalam and the initial temperature of sodium until its temperature increased by 30°C, the flame is
larutan argentum nitrat.
hydroxide solution and hydrochloric acid, put off and the highest temperature, T2 reached by the
5 Campuran tindak balas dikacau
dengan termometer dan suhu tertinggi, T1 and T2 are recorded. water is recorded.
T3 dicatat.
4 Hydrochloric acid is poured quickly and 8 The final mass of the lamp, m2 and its content is
© Nilam Publication Sdn Bhd carefully into the sodium hydroxide solution. weighed immediately and recorded. MODULE • Chemistry Form 5

5 The reaction mixture is stirred with the 9 Steps 1 to 8 are repeated with ethanol, propanol and
thermometer and the highest temperature, butanol.

T3 is recorded. 1 100 cm3 air disukat dengan silinder penyukat dan
dituang ke dalam bekas kuprum.
6 Steps 1 to 5 are repeated using sodium
hydroxide solution and ethanoic acid. 2 Termometer diletakkan di dalam air dan suhu awal, T1
dicatat.
1 50 cm3 larutan natrium hidroksida 2 mol dm–3
disukat dengan silinder penyukat dan 3 Bekas kuprum diletakkan di atas tungku kaki tiga.
4 Pelita diisikan dengan metanol dan ditimbang. Jisim
dituang ke dalam cawan polistirena.
2 50 cm3 asid hidroklorik berkepekatan 2 mol dm–3 awalnya, m1 dicatat.
5 Penghadang angin diletakkan seperti ditunjukkan
disukat dengan silinder penyukat yang lain dan
dituang ke dalam cawan polistirena yang lain. dalam rajah untuk mengurangkan kehilangan haba ke
3 Termometer diletakkan di dalam setiap larutan udara persekitarannya.
dan suhu awal larutan natrium hidroksida dan 6 Pelita diletak dekat dengan bekas kuprum untuk
asid hidroklorik, T1 dan T2 dicatat. memaksimakan pemindahan haba. Sumbu pelita
4 Asid hidroklorik dituang dengan cepat dan tersebut dinyalakan.
cermat ke dalam larutan natrium hidroksida. 7 Air tersebut dikacau berterusan dengan termometer
5 Campuran tindak balas dikacau dengan sehingga suhunya meningkat sebanyak 30°C, api
termometer dan suhu tertinggi, T3 dicatat. dipadamkan dan suhu tertinggi, T2 dicapai oleh air dicatat.
6 Langkah 1 hingga 5 diulang dengan 8 Jisim terakhir pelita m2 dan kandungannya segera
menggunakan larutan natrium hidroksida ditimbang dan dicatat.
dan asid etanoik. 9 Langkah 1 hingga 8 diulangi dengan menggunakan
etanol, propanol dan butanol.

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MODULE • Chemistry Form 5

Calculating heat of reaction, / Pengiraan haba tindak balas, ∆H:

What are the quantities needed to (i) Mass of substance (m in grams)
calculate heat change / heat Jisim bahan (m dalam gram)
absorbed / heat given out or heat (ii) Specific heat capacity of a substance (c in J g–1 °C–1)
release, H in a substance? Muatan haba tentu bahan (c dalam J g–1 °C–1)
Apakah kuantiti yang diperlukan (iii) Temperature change (θ°C)
untuk mengira perubahan haba / Perubahan suhu (θ°C)
haba yang diserap atau haba yang
dibebaskan, H dalam bahan?

What are the assumptions made For a chemical reaction that occurs in an aqueous solution (precipitation, displacement of metal and
to get mass, specific capacity and neutralisation), assumptions are made during the calculation of heat of reaction:
temperature of the substance? Untuk tindak balas kimia yang berlaku dalam larutan akueus (pemendakan, penyesaran logam dan
Apakah andaian yang dibuat peneutralan) anggapan dibuat semasa pengiraan haba tindak balas:
untuk mendapatkan jisim, muatan (i) Density of aqueous solution is equal to the density of water = 1 g cm–3, for example:
tentu dan suhu bahan tersebut? Ketumpatan larutan akueus sama dengan ketumpatan air = 1 g cm–3, contoh:

• 1 cm3 of aqueous solution has a mass of 1 g
1 cm3 larutan akueus mempunyai jisim 1 g
• 2 cm3 of aqueous solution has a mass of 2 g
2 cm3 larutan akueus mempunyai jisim 2 g
• m cm3 of aqueous solution has a mass of m g
m cm3 larutan akueus mempunyai jisim m g
(ii) Specific heat capacity of solution = Specific heat capacity of water = 4.2 J g–1 °C–1
Muatan haba tentu bahan larutan = Muatan haba tentu bahan air = 4.2 J g–1 °C–1
(iii) No heat lost to the surroundings:
Tiada haba hilang ke persekitaran:
– all heat released in an exothermic reaction = heat absorbed by the solution (temperature increases)
semua haba dibebaskan dalam satu tindak balas eksotermik = haba yang diserap oleh larutan (suhu

meningkat)
– all heat absorbed in endothermic reaction = heat lost by the solution (temperature decreases)
semua haba diserap dalam satu tindak balas endotermik = haba yang hilang oleh larutan (suhu

menurun)

How to calculate heat change / The heat change, H in a reaction can be calculated with the following formula
heat absorbed / heat given out or Perubahan haba, H dalam tindak balas boleh dikira dengan formula berikut:
heat released, H in a reaction?
Bagaimana mengira perubahan Heat change (H) / Perubahan haba (H) = mcθ J
haba / haba yang diserap / haba
yang dikeluarkan atau haba where / di mana
dibebaskan, H dalam tindak m = mass of the solution in gram
balas? jisim larutan dalam gram
c = specific heat capacity of solution in J g–1 °C–1
muatan haba tentu larutan dalam J g–1 °C–1
θ = temperature change in °C
perubahan suhu dalam °C

How to calculate heat of reaction, (i) Heat of reaction (∆H) is the energy change when one mole of reactant reacts or when one mole of

ΔH in a reaction? product is formed.

4U Bagaimanakah cara mengira haba Haba tindak balas (∆H) ialah perubahan tenaga apabila satu mol bahan bertindak balas atau satu mol
tindak balas, ΔH dalam tindak
N balas? hasil terbentuk.
I
T (ii) X mol of reactant/product absorbs/releases H J of heat energy

X mol bahan/hasil menyerap/membebaskan H J tenaga haba H
X
1 mol of reactant/ product absorbs/releases / 1 mol of bahan/hasil menyerap/membebaskan J mol–1

⇒ ∆H (heat of reaction) / ∆H (Haba tindak balas) = +/–  HJ ,
X mol

X = number of moles of reactant/product / bilangan mol bahan/hasil

Remark / Catatan: negative for exothermic reaction (temperature increases).
(i) The sign of ∆H is negatif untuk tindak balas eksotermik (suhu menaik).
Tanda ∆H adalah positive for endothermic reaction (temperature decreases).
(ii) The sign of ∆H is positif untuk tindak balas endotermik (suhu menurun).
Tanda ∆H adalah

What is the unit heat of reaction? The unit for heat of reaction is kJ mol–1.
Apakah unit haba bagi tindak Unit untuk haba tindak balas ialah kJ mol–1.
balas?

© Nilam Publication Sdn Bhd 162

MODULE • Chemistry Form 5

Steps to calculate heat of reaction, Step 1 / Langkah 1: Step 2 / Langkah 2: Step 3 / Langkah 3:
ΔH. Calculate the number Calculate the *heat
Langkah-langkah untuk mengira of moles of reactants change of the Calculate *heat of
haba tindak balas, ΔH. or products, X reaction, H
Hitung bilangan mol Hitung *perubahan reaction, ΔH
bahan atau hasil haba tindak balas, H
tindak balas, X Hitung *haba tindak
H = mcθ
balas, ΔH

ΔH = +/– H J
X

Remark / Catatan:
For specific reaction in this chapter: / Bagi tindak balas tertentu dalam bab ini:
1 *Heat change, H is heat released for exothermic reaction or heat absorbed for endothermic reaction.
*Perubahan haba, H ialah haba yang dibebaskan untuk tindak balas eksotermik atau haba yang diserap untuk

tindak balas endotermik.
2 *Heat of reaction, ΔH is / * Haba tindak balas, ΔH ialah

(i) Heat of precipitation for precipitation reaction / Haba pemendakan bagi tindak balas pemendakan
(ii) Heat of displacement for displacement reaction / Haba penyesaran bagi tindak balas penyesaran
(iii) Heat of neutralisation for neutralisation reaction / Haba peneutralan bagi tindak balas peneutralan
(iv) Heat of combustion for combustion reaction / Haba pembakaran bagi tindak balas pembakaran

Example: / Contoh:
60 cm3 of 0.25 mol dm–3 silver nitrate solution reacts with 60 cm3 of 0.25 mol dm–3 potassium bromide solution with an average
temperature of 29°C. A yellow precipitate was formed and the highest temperature reached is 32°C. Determine the heat of reaction,
∆H. / 60 cm3 larutan argentum nitrat 0.25 mol dm–3 bertindak balas dengan 60 cm3 larutan kalium bromida 0.25 mol dm–3 dengan
suhu purata 29°C. Mendakan kuning terbentuk dan suhu tertinggi dicapai ialah 32°C. Tentukan haba tindak balas, ∆H.

Steps / Langkah-langkah Calculation / Pengiraan
Step 1: / Langkah 1:
Calculate the number of moles of silver AgNO3(aq/ak) + KBr (aq/ak) → AgBr(s/p) + KNO3(aq/ak)
bromide precipitated (X)
Hitung bilangan mol mendakan argentum or / atau Ag+ + Br– → AgBr
bromida (X) Number of moles of Ag+ / Bilangan mol Ag+

Step 2: / Langkah 2: = 60 dm3 × 0.25 mol dm–3 = 0.015 mol
Calculate the heat released, H
Hitung haba yang dibebaskan, H 1 000

Step 3: / Langkah 3: Number of moles of Br– / Bilangan mol Br –
Calculate the heat of precipitation, ΔH
Hitung haba pemendakan, ΔH = 60 dm3 × 0.25 mol dm–3 = 0.015 mol

1 000

From the equation: / Daripada persamaan:
1 mole of Ag+ ions reacts with 1 mole of Br– ions to form 1 mole of AgBr
1 mol ion Ag+ bertindak balas dengan 1 mol ion Br – membentuk 1 mol AgBr

0.015 mole of Ag+ ions reacts with 0.015 mole Br– ions to form 0.015 mole AgBr

0.015 mol ion Ag+ bertindak balas dengan 0.015 mol ion Br – membentuk 0.015 mol AgBr

X = 0.015

H = mcθ J U
H = 120 g × 4.2 J g–1 °C–1 × 3°C N
I
= 1 512 J T

Remark / Catatan: 4
1 Mass of the solution, m / Jisim larutan, m
= (60 cm3 + 60 cm3) × 1 g cm–3
= 120 g
2 Temperature change, q / Perubahan suhu, q
= (32 – 29)°C
= 3°C

∆H = – H (negative because heat is released to the surrounding or temperature increases)
X (negatif sebab haba dibebaskan ke persekitaran atau suhu menaik)

= – 1 512 = –100.8 kJ mol–1
0.015 mol

Draw the energy level diagram: Energy / Tenaga
Lukis rajah aras tenaga: AgNO3(aq/ak) + KBr(aq/ak)

ΔH = –100.8 kJ mol–1

AgBr(s/p) + KNO3(aq/ak)

163 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Numerical Problems Involving Heat of Displacement
Pengiraan Melibatkan Haba Penyesaran

Example 1 / Contoh 1:

Excess of zinc powder is added to 50 cm3 of 0.1 mol dm–3 copper(II) sulphate solution. The temperature of reaction mixture
increases by 5°C. Calculate the heat of displacement of copper by zinc from copper(II) sulphate solution.
[Specific heat capacity of solution = 4.2 J g–1 o C–1, density of solution = 1 g cm–3]
Serbuk zink berlebihan ditambah kepada 50 cm3 larutan kuprum(II) sulfat 0.1 mol dm–3. Suhu campuran tindak balas meningkat
sebanyak 5°C. Hitungkan haba penyesaran kuprum oleh zink dari larutan kuprum(II) sulfat.
[Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3]

Step 1: Calculate number of mole of copper displaced Excess / Lebih 50 cm3 ? mol
Langkah 1: Hitung bilangan mol kuprum yang 0.1 mol dm–3 ZnSO4 + Cu
disesarkan Zn +
CuSO4 →
Step 2: Calculate the heat released, H
Langkah 2: Hitung haba dibebaskan, H Number of moles of CuSO4 / Bilangan mol CuSO4

= 50 × 0.1 = 0.005 mol
1 000

From the equation, / Dari persamaan,
1 mol CuSO4 : 1 mol Cu
0.005 mol CuSO4 : 0.005 mol Cu

Heat released in the experiment,
Haba dibebaskan dalam eksperimen,
H = 50 × 4.2 × 5 J
= 1 050 J

Step 3: Calculate the heat of reaction (ΔH) Heat of displacement, / Haba penyesaran
Langkah 3: Hitung haba tindak balas (ΔH)
∆H = –  1 050 J
0.005 mol

= –210 kJ mol–1

Example 2 / Contoh 2:

The following is the thermochemical equation for a reaction.

Berikut adalah persamaan termokimia untuk suatu tindak balas.

Zn + CuSO4 → ZnSO4 + Cu ∆H = –210 kJ mol–1

U Calculate the heat released when 50 cm3 of 1.0 mol dm–3 copper(II) sulphate solution reacts with excess zinc.
N
Hitungkan haba yang dibebaskan apabila 50 cm3 larutan kuprum(II) sulfat 1.0 mol dm–3 bertindak balas dengan zink berlebihan.

I Step 1: Calculate number of mole of copper displaced Number of moles of CuSO4 / Bilangan mol CuSO4

T Langkah 1: Hitung bilangan mol kuprum yang = 50 × 1
1 000
4 disesarkan
= 0.05 mol

From the equation, / Dari persamaan
1 mol CuSO4 : 1 mol Cu
0.05 mol CuSO4 : 0.05 mol Cu

Step 2: Calculate the heat released, H H
Langkah 2: Hitung haba dibebaskan, H ∆H = X

H
210 kJ mol–1 = 0.05
H = 210 kJ mol–1 × 0.05 mol
= 10.5 kJ

© Nilam Publication Sdn Bhd 164

MODULE • Chemistry Form 5

Exercise / Latihan

The thermochemical ionic equation below represents the reaction between magnesium powder and iron(II) sulphate solution.
Persamaan ion termokimia di bawah mewakili tindak balas antara serbuk magnesium dengan larutan ferum(II) sulfat.

Mg(s/p) + Fe2+(aq/ak) → Mg2+(aq/ak) + Fe(s/p) ∆H = –189 kJ mol–1

Calculate the increase in temperature when excess magnesium powder is added into 80 cm3 of 0.4 mol dm–3 iron(II) sulphate
solution. [Specific heat capacity of solution = 4.2 J g–1 °C–1, density of solution = 1 g cm–3]
Hitungkan kenaikan suhu apabila serbuk magnesium berlebihan ditambah kepada 80 cm3 larutan ferum(II) sulfat 0.4 mol dm–3.
[Muatan haba tentu bahan larutan = 4.2 J g –1 °C–1, ketumpatan larutan = 1 g cm–3]

Step 1: Calculate number of mole of iron displaced Number of moles of FeSO4 / Bilangan mol FeSO4
Langkah 1: Hitung bilangan mol ferum yang disesarkan
= 80 × 0.4
1 000

= 0.032 mol

From the equation, / Dari persamaan
1 mol Fe 2+ : 1 mol Fe
0.032 mol Fe 2+ : 0.032 mol Fe

Step 2: Calculate the heat released, H H
Langkah 2: Hitung haba dibebaskan, H ∆H = X

189 kJ mol–1 = H ; H = H eat released in the experiment
0.032 mol Haba dibebaskan dalam eksperimen

H = 189 kJ mol–1 × 0.032 mol
= 6.048 kJ = 6 048 J

Step 3: Calculate the increase in temperature, θ 6 048 J = mcθ = 80 × 4.2 × θ
Langkah 3: Hitung peningkatan suhu, θ θ = 18°C

Numerical Problems Involving Heat of Precipitation U
Pengiraan Melibatkan Haba Pemendakan N
I
T

4

1 When 25 cm3 of 0.25 mol dm–3 silver nitrate solution is added into 25 cm3 of 0.25 mol dm–3 sodium chloride solution, the
temperature of the mixture rises by 3°C. What is the quantity of heat released in this experiment?

[Specific heat capacity of a solution = 4.2 J g–1 °C–1]
Apabila 25 cm3 larutan argentum nitrat 0.25 mol dm–3 ditambah kepada 25 cm3 larutan natrium klorida 0.25 mol dm–3, suhu

campuran tindak balas naik sebanyak 3ºC. Berapa kuantiti haba yang dibebaskan dalam eksperimen ini?
[Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1]

Answer: / Jawapan:

Heat released in the experiment, / Haba dibebaskan dalam eksperimen,
H = 50 g × 4.2 J g–1 ºC–1 × 3ºC

= 630 J

165 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

2 The thermochemical ionic equation below represents the reaction between lead(II) nitrate solution and potassium sulphate

solution.

Persamaan ion termokimia di bawah mewakili tindak balas antara larutan plumbum(II) nitrat dengan larutan kalium sulfat.

Pb2+ + SO42– → PbSO4 ∆H = –50.4 kJ mol–1

Calculate the increase in temperature when 25 cm3 of 1 mol dm–3 of lead(II) nitrate solution is added into 25 cm3 of 1 mol dm–3

of potassium sulphate solution. [Specific heat capacity of solution = 4.2 J g–1 °C–1, density of solution = 1 g cm–3]

Hitungkan kenaikan suhu apabila 25 cm3 larutan plumbum(II) nitrat 1 mol dm–3 ditambah kepada 25 cm3 larutan kalium sulfat

1 mol dm–3. [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3]

Step 1: Calculate number of mole of lead(II) sulphate Number of moles of Pb2+ = 25 × 1 = 0.025 mol,
precipitate formed 1 000
Langkah 1: Hitung bilangan mol mendakan Bilangan mol Pb2+
plumbum(II) sulfat yang terbentuk
Number of moles of SO42– = 25 × 1 = 0.025 mol
Step 2: Calculate the heat released, H Bilangan mol SO42– 1 000
Langkah 2: Hitung haba dibebaskan, H
From the equation: / Dari persamaan
• 1 mol of Pb2+ ions reacts with 1 mol of SO42– ions to form 1 mole of

PbSO4
1 mol ion Pb2+ bertindak balas dengan 1 mol ion SO42– membentuk

1 mol PbSO4
• 0.025 mole of Pb2+ ions reacts with 0.025 mole SO42– ions to form

0.025 mole of PbSO4
0.025 mol ion Pb2+ bertindak balas dengan 0.025 mol ion SO42–

membentuk 0.025 mol PbSO4

X = 0.025 mol
H

50.4 kJ mol–1 = 0.025 mol
Heat released / Haba dibebaskan, H = 1.26 kJ = 1 260 J

Step 3: Calculate the increase in temperature, θ 1 260 J = mcθ = 50 g × 4.2 J g–1 ºC–1 × θ
Langkah 3: Hitung peningkatan suhu, θ θ = 6 ºC

Numerical Problems Involving Heat of Neutralisation
Pengiraan Melibatkan Haba Peneutralan

1 100 cm3 of 2.0 mol dm–3 sodium hydroxide solution is added into 100 cm3 of 2.0 mol dm–3 ethanoic acid. The initial temperature

for both solutions is 28.0ºC and the highest temperature is 41.0ºC. Calculate heat of neutralisation. / 100 cm3 larutan natrium

hidroksida 2.0 mol dm–3 ditambah kepada 100 cm3 asid etanoik 2.0 mol dm–3. Suhu awal kedua-dua larutan ialah 28.0ºC dan

suhu tertinggi ialah 41.0ºC. Hitungkan haba peneutralan.
U [Specific heat capacity of a solution = 4.2 J g–1 °C–1] / [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1]
N
I Step 1: Calculate number of mole of water formed Number of mole of NaOH / Bilangan mol NaOH

T Langkah 1: Hitung bilangan mol air yang terbentuk = 100 × 2 = 0.2 mol
1 000
4
Number of mole of CH3COOH / Bilangan mol CH3COOH

= 100 × 2 = 0.2 mol
1 000

CH3COOH + NaOH → CH3COONa + H2O
From the equation / Daripada persamaan:
1 mol CH3COOH : 1 mol NaOH : 1 mol H2O
0.2 mol CH3COOH : 0.2 mol NaOH : 0.2 mol H2O

Step 2: Calculate the heat released, H Heat released / Haba dibebaskan,
Langkah 2: Hitung haba dibebaskan, H H = (100 + 100) g × 4.2 J g–1 °C–1 × (41 – 28) °C–1
= 10 920 J = 10.92 kJ

Step 3: Calculate the heat of neutralisation (ΔH) Heat of neutralisation / Haba peneutralan,
Langkah 3: Hitung haba peneutralan (ΔH) 10.92 kJ

ΔH = 0.2 mol = –54.6 kJ mol–1

© Nilam Publication Sdn Bhd 166

MODULE • Chemistry Form 5

2 The reaction between 25 cm3 of hydrochloric acid and 25 cm3 of sodium hydroxide solution releases the heat of 2 100 J. What
is the temperature change of the mixture?

Tindak balas antara 25 cm3 asid hidroklorik dan 25 cm3 larutan natrium hidroksida membebaskan haba sebanyak 2 100 J.
Apakah perubahan suhu campuran tindak balas?

[Specific heat capacity of a solution = 4.2 J g–1 °C–1] / [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1]

Answer: / Jawapan:
Heat released / Haba dibebaskan = mcθ = 2 100 J
(25 + 25) g × 4.2 J g–1 °C–1 × θ = 2 100 J
θ = 10°C

Comparing the Heat of Neutralisation / Membandingkan Haba Peneutralan

What is heat of neutralisation? Heat of neutralisation is heat released when one mole of hydrogen ions from acid reacts with one mole of
Apakah haba peneutralan? hydroxide ions from alkali to produce one mole of water:
Haba peneutralan ialah haba yang dibebaskan apabila satu mol ion hidrogen dari asid bertindak balas
dengan satu mol ion hidroksida dari alkali menghasilkan satu mol air:

H+(aq/ak) + OH–(aq/ak) → H2O ∆ H = –57 kJ mol–1

Explain why the value of heat of – Hydrochloric acid and nitric acid are strong monoprotic acids. One mole hydrochloric acid or nitric acid
neutralisation between sodium ionise completely in water to produce one mole of hydrogen ions.
hydroxide solution / potassium
hydroxide solution with – Sodium hydroxide and potassium hydroxide are strong alkali. One mole sodium hydroxide or potassium
hydrochloric acid / nitric acid is hydroxide ionise completely in water to produce one mole of hydroxide ions.
57 kJ mol–1.
Terangkan mengapa nilai haba – Heat of neutralisation of sodium hydroxide solution/potassium hydroxide solution with hydrochloric
peneutralan antara larutan natrium acid/nitric acid is –57 kJ mol–1 because all the reactions produce one mol of water.
hidroksida / larutan kalium
hidroksida dengan asid – Asid hidroklorik dan asid nitrik adalah asid monoprotik kuat. Satu mol asid hidroklorik atau asid nitrik
hidroklorik / asid nitrik adalah mengion sepenuhnya dalam air untuk menghasilkan satu mol ion hidrogen.
57 kJ mol–1.
– Natrium hidroksida dan kalium hidroksida adalah alkali kuat. Satu mol natrium hidroksida atau kalium
hidroksida mengion sepenuhnya dalam air untuk menghasilkan satu mol ion hidroksida.

– Haba peneutralan bagi larutan natrium hidroksida/kalium hidroksida dengan asid hidroklorik/asid nitrik
ialah –57 kJ mol–1 kerana semua tindak balas menghasilkan satu mol air.

HCl + KOH → KCl + H2O H+ + OH– → H2O, ∆H = –57 kJ mol–1
HCl + NaOH → NaCl + H2O
HNO3 + KOH → KNO3 + H2O
HNO3 + NaOH → NaNO3 + H2O

⇒ 1 mol of hydrogen ions react with 1 mol of hydroxide ions to form 1 mol of water to release 57 kJ of U
heat energy. N
I
1 mol ion hidrogen bertindak balas dengan ion hidroksida membentuk 1 mol air dan membebaskan T
57 kJ tenaga haba.
4
What is the value of heat of – Thermochemical equation for the neutralisation between sodium hydroxide with sulphuric acid (diprotic
neutralisation of sulphuric acid acid):
(strong diprotic acid) with strong
alkali? Explain. Persamaan termokimia untuk peneutralan antara natrium hidroksida dengan asid sulfurik (asid diprotik):
Apakah nilai haba peneutralan 2NaOH + H2SO4 → Na2SO4 + 2H2O
asid sulfurik (asid diprotik kuat)
dengan alkali kuat? Terangkan. – 2 mol hydroxide ions react with 2 mol of hydrogen ions to form 2 mol H2O. Heat released is
2 × 57 kJ = 114 kJ.

2 mol ion hidroksida bertindak balas dengan 2 mol ion hidrogen membentuk 2 mol H2O. Haba yang
dibebaskan ialah 2 × 57 kJ = 114 kJ.
2H+ + 2OH– → 2H2O,  ΔH = –114 kJ

– Heat of neutralisation of sulphuric acid with sodium hydroxide remains at –57 kJ mol–1 because the
definition for heat of neutralisation is heat released for the formation of one mol of water.

Haba peneutralan bagi asid sulfurik dengan natrium hidroksida masih –57 kJ mol–1 kerana maksud
haba peneutralan adalah haba yang dibebaskan bagi pembentukan satu mol air.
H+ + OH– → H2O,  ΔH = –57 kJ mol–1

167 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Explain why the heat of Magnitude of heat of neutralisation for a weak acid with a strong alkali or strong acid with weak alkali is
neutralisation between weak acid less than 57 kJ mol–1.
and strong alkali is less than Magnitud haba peneutralan untuk asid lemah dengan alkali kuat atau asid kuat dengan alkali lemah
–57 kJ mol–1. adalah kurang daripada 57 kJ mol–1.
Terangkan mengapa haba
peneutralan antara asid lemah dan Example: / Contoh: ∆H = –55 kJ mol–1
alkali kuat kurang daripada ∆H = –12 kJ mol–1
–57 kJ mol–1. N aOH + CH3COOH → CH3COONa + H2O

NaOH + HCN → NaCN + H2O

Explanation: / Penerangan:

(i) Weak acids ionise partially in water to produce hydrogen ions.

Asid lemah mengion separa dalam air menghasilkan ion hidrogen .

CH3COOH CH3COO– + H+

(ii) Some of the particles still remain in the form of molecules .

Sebahagian zarah masih kekal dalam bentuk molekul .

(iii) Heat energy is absorbed to ionise molecules of the weak acid that have not been ionised so that

they ionise completely.

Tenaga haba diserap untuk mengionkan molekul asid lemah yang masih belum mengion supaya

mengion sepenuhnya.

(iv) Part of the heat that is released is used/absorbed to ionise the molecules of weak acid that has not

been ionised.

Sebahagian haba yang dibebaskan digunakan/diserap untuk mengionkan molekul asid lemah

yang masih belum mengion.

Calculation guide: / Panduan pengiraan:

# Calculation guide I: / Panduan pengiraan I:
If any reaction is repeated by changing the volume without changing the concentration, change in temperature is the same.
Jika sebarang tindak balas diulangi dengan menukarkan isi padu tanpa menukar kepekatan, perubahan suhu adalah sama.

Example 1: / Contoh 1:
• Reaction I: / Tindak balas I:
50 cm3 of 2 mol dm–3 hydrochloric acid is added to 50 cm3 of 2 mol dm–3 potassium hydroxide solution. The temperature

rises by 13°C. / 50 cm3 asid hidroklorik 2 mol dm–3 ditambah dengan 50 cm3 larutan kalium hidroksida 2 mol dm–3. Suhu naik
sebanyak 13°C.
• Reaction II: / Tindak balas II:
100 cm3 of 2 mol dm–3 hydrochloric acid is added to 100 cm3 of 2 mol dm–3 potassium hydroxide solution. What is the
temperature change in this reaction? / 100 cm3 asid hidroklorik 2 mol dm–3 ditambah dengan 100 cm3 larutan kalium hidroksida
2 mol dm–3. Apakah perubahan suhu dalam tindak balas ini?

U Answer: / Jawapan: where / di mana X = Number of moles of water / Bilangan mol air
N H H = Heat change (heat released in the reaction)
I
T ∆H = X


4 Perubahan haba (haba dibebaskan dalam tindak balas)

= mcq
HCl + KOH → KCl + H2O

1 mol 1 mol 1 mol

From the equation / Daripada persamaan:

Reaction I: / Tindak balas I: 1 mol HCl : 1 mol KOH : 1 mol H2O

0.1 mol HCl : 0.1 mol HCl : 0.1 mol H2O

Reaction II: / Tindak balas II: 1 mol HCl : 1 mol KOH : 1 mol H2O

0.2 mol HCl : 0.2 mol HCl : 0.2 mol H2O

Reaction I: / Tindak balas I: ∆H = 100 × 4.2 × 13 J = 54 600 J
0.1

Reaction II: / Tindak balas II: 54 600 J = 600 × 4.2 × T , where / di mana T = t emperature change in reaction II
0.2 perubahan suhu dalam tindak balas II

T = 13ºC

© Nilam Publication Sdn Bhd 168

MODULE • Chemistry Form 5

Explanation:
– When Reaction II is repeated by doubled volume of acid and alkali, the number of moles of water produced in Reaction II is

double, hence amount of heat energy released is double.
– The amount of heat energy which is doubled is used to increase total volume of solution which is also double.
– Therefore, the increase in temperature remains the same.

Penerangan:
– Apabila Tindak balas II diulang dengan menggandakan isi padu asid dan alkali, bilangan mol air yang dihasilkan dalam

Tindak balas II adalah dua kali ganda, oleh itu jumlah tenaga haba yang dibebaskan adalah dua kali ganda.
– Jumlah tenaga haba yang berganda digunakan untuk meningkatkan jumlah isi padu larutan yang juga dua kali ganda.
– Oleh itu, peningkatan suhu kekal sama.

# Calculation guide II: / Panduan pengiraan II:
If the reaction is repeated by changing the concentration of the solution by n times without changing the volume, the
temperature change is n times. / Jika sebarang tindak balas diulangi dengan menukarkan kepekatan larutan sebanyak n kali
tanpa menukar isi padu, perubahan suhu adalah n kali.

Example 2: / Contoh 2:
• Reaction I: / Tindak balas I:
50 cm3 of 0.2 mol dm–3 lead(II) nitrate is added to 50 cm3 of 0.2 mol dm–3 sodium carbonate solution. The temperature of the

mixture rises by 2.4°C. / 50 cm3 larutan plumbum(II) nitrat 2 mol dm–3 ditambah dengan 50 cm3 larutan natrium karbonat
0.2 mol dm–3. Suhu naik sebanyak 2.4°C.
• Reaction II: / Tindak balas II:
50 cm3 of 0.6 mol dm–3 lead(II) nitrate solution is added to 50 cm3 of 0.6 mol dm–3 sodium carbonate solution. What is the
temperature rise in this experiment? / 50 cm3 larutan plumbum(II) nitrat 0.6 mol dm–3 ditambah dengan 50 cm3 larutan natrium
karbonat 0.6 mol dm–3. Apakah kenaikan suhu dalam eksperimen ini?

Answer: / Jawapan:
H

∆H = X where / di mana ∆H = Heat of precipitation of lead(II) carbonate
Haba pemendakan plumbum(II) karbonat

X = Number of moles of lead(II) carbonate precipitated

Bilangan mol mendakan plumbum(II) karbonat
H = Heat change / Perubahan haba

= mcθ

Ionic equation for both reactions: / Persamaan ion untuk kedua-dua tindak balas:

Pb2+ + CO32– → PbCO3

Reaction I: / Tindak balas I: ∆H = 100 × 4.2 × 2.4 J = 100 800 J mol–1
0.01

Reaction II: / Tindak balas II: 100 800 J = 100 ×0 .40.32 × T , where / di mana T = temperature change in reaction II U
perubahan suhu dalam tindak balas N
II I
T
T = 7.2°C (The temperature changes is 3 times more than reaction I)
4
(Perubahan suhu adalah 3 kali lebih daripada tindak balas I)

Explanation:
– When Reaction II is repeated by increasing the concentration of lead(II) nitrate solution and sodium carbonate solution by

3 times, number of moles lead(II) carbonate produced is also increased by 3 times. Hence, the amount of heat energy
released is increased 3 times.
– The heat energy is used to increase the same total volume of solution.
– Therefore, the increase in temperature is 3 times.

Penerangan:
– Apabila Tindak balas II diulang dengan meningkatkan kepekatan larutan plumbum(II) nitrat dan larutan natrium karbonat

sebanyak 3 kali, bilangan mol plumbum(II) karbonat yang dihasilkan juga meningkat sebanyak 3 kali ganda. Oleh itu,
jumlah tenaga haba yang dilepaskan meningkat 3 kali ganda.
– Tenaga haba digunakan untuk meningkatkan jumlah larutan yang sama.
– Oleh itu, kenaikan suhu adalah 3 kali ganda.

169 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Numerical Problems Involving Heat of Combustion
Pengiraan Melibatkan Haba Pembakaran

1 Methanol burns in oxygen in a reaction as shown in the thermochemical equation below.

Metanol terbakar dalam oksigen seperti persamaan termokimia di bawah.

CH3OH(s/p) + 3  O2(g/g) → CO2 (g/g) + 2H2O ∆H = –725 kJ mol–1
2

What is the mass of methanol that must be burnt completely to produce 145 kJ of heat? [Relative atomic mass: C, 12; O, 16]

Apakah jisim metanol yang perlu dibakar lengkap untuk menghasilkan 145 kJ haba? [Jisim atom relatif: C, 12; O, 16]

Calculate number of mole of methanol ∆H = H ,  X = Number of moles of methanol / Bilangan mol metanol
Kira bilangan mol metanol X

725 kJ mol–1 = 145 kJ
X

Number of moles of methanol / Bilangan mol metanol

= 145 kJ
725 kJ mol–1

= 0.2 mol

Calculate mass of methanol Mass of methanol / Jisim metanol
Hitung jisim metanol = 0.2 × [12 × 1 + 4 × 1 + 16]
= 6.4 g

2 22 g of butanol is burnt completely in excess of oxygen. The heat released is used to heat up 500 cm3 of water from 27.5°C to
55.8°C. Calculate the heat of combustion of butanol.

22 g butanol terbakar lengkap dalam oksigen berlebihan. Haba yang dibebaskan memanaskan 500 cm3 air dari 27.5°C ke
55.8°C. Hitungkan haba pembakaran butanol.

[Specific heat capacity of a solution = 4.2 J g–1 °C–1, relative atomic mass: H, 1; C, 12; O, 16 ]
[Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, jisim atom relatif: H, 1; C, 12; O, 16]

Calculate number of mole of butanol Number of moles of butanol / Bilangan mol butanol
Kira bilangan mol butanol
= 22 g
U 74 g mol–1
N
I H = 500 cm3 × 4.2 J g–1 °C–1 × 28.3°C
T = 59 430 J
= 59.43 kJ
4 Calculate heat released, H
Hitung haba dibebaskan, H

Calculate the heat of combustion, (ΔH) ∆H = 59.43 kJ
Hitung haba pembakaran, (ΔH) 22 mol
74

= 199.9 kJ mol–1

© Nilam Publication Sdn Bhd 170

MODULE • Chemistry Form 5

Comparing the Heat of Combustion of Various Fuels
Membandingkan Haba Pembakaran Pelbagai Bahan Api

Define heat of combustion. Heat of combustion is heat energy released when 1 mole of fuel is burnt completely in excess oxygen.
Nyatakan maksud haba Haba pembakaran ialah tenaga haba yang dibebaskan apabila 1 mol bahan api dibakar lengkap dalam oksigen
pembakaran. berlebihan.

What is the difference in heat of The higher the number of carbon and hydrogen atoms per molecule of alcohols, the higher the heat
combustion between various energy released by the combustion of 1 mole of alcohols.
alcohols? Semakin tinggi bilangan atom karbon dan hidrogen dalam setiap molekul alkohol, semakin banyak tenaga
Apakah perbezaan haba haba dibebaskan dari pembakaran 1 mol alkohol.
pembakaran di antara pelbagai
alkohol?

Example: / Contoh:
The diagram below shows the graph of heat of combustion of alcohols against number of carbon atom per molecule.
Rajah di bawah menunjukkan graf haba pembakaran melawan bilangan atom karbon dalam setiap molekul alkohol.

Heat of combustion of alcohol (kJ mol–1)
Haba pembakaran alkohol (kJ mol–1)

3 000
2 000

1 000

0 12 34 Number of carbon atom per molecule
Bilangan atom karbon per molekul

State relationship between the number of carbon atom per molecule of alcohol with the heat of combustion. Explain.
Nyatakan hubungan antara bilangan atom karbon per molekul alkohol dengan haba pembakaran. Terangkan.

Answer: / Jawapan:
– When the number of carbon atom per molecule of alcohol increases, the heat combustion increases.

– When the number of carbon atom per molecule of alcohol increases , the number of carbon dioxide and water molecules produced

as products increases .
– More bonds between atoms in carbon dioxide and water molecules are formed, more heat is released.
– Apabila bilangan atom karbon dalam setiap molekul alkohol bertambah, haba pembakaran bertambah.

– Apabila bilangan atom karbon dalam setiap molekul alkohol bertambah , bilangan molekul karbon dioksida dan air yang
dihasilkan bertambah .

– Lebih banyak ikatan antara atom dalam molekul air dan karbon dioksida terbentuk, lebih banyak haba dibebaskan.

Fuel value / Nilai bahan api Fuels are substances that burn in the air to produce heat energy. U
Bahan api ialah sebatian yang terbakar dalam udara untuk menghasilkan tenaga haba. N
What are fuels? I
Apakah bahan api? Fuel value is the amount of heat released when 1 g of fuel burns completely, the unit is kJ g–1. T
Nilai bahan api adalah jumlah haba yang dibebaskan apabila 1 g bahan api terbakar lengkap, unitnya
What is fuel value? adalah kJ g–1. 4
Apakah nilai bahan api?
Fuel value is used to compare the cost of energy for various fuel. A fuel with high fuel value can supply
What are the application of fuel more energy. / Nilai bahan api digunakan untuk membandingkan kos tenaga pelbagai bahan api. Bahan
value? api dengan nilai bahan api yang tinggi boleh membekalkan lebih tenaga.
Apakah aplikasi nilai bahan api?
Example: / Contoh:

Fuel / Bahan api Fuel value/ kJ g–1 / Nilai bahan api/ kJ g–1
Methanol / Metanol 23
Charcoal / Arang kayu 35
Crude oil / Minyak mentah 45
Kerosene / Kerosin 37
Petrol / Petrol 34
Natural gas / Gas asli 50

171 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

What are the aspects to be – Fuel value of the fuel. / Nilai bahan api bagi bahan api.
considered when choosing a fuel – Cost of energy/fuel. / Harga tenaga/bahan api.
in industry? – Availability and sources of the fuel.
Apakah aspek yang harus Sumber bahan api.
dipertimbangkan ketika memilih – Effect of the fuel to the environment.
bahan api dalam industri? Kesan bahan api kepada persekitaran.

What is the major sources of Fossil fuels such as coal, petroleum and natural gas
energy? Bahan api fosil seperti arang batu, petroleum dan gas asli
Apakah sumber utama tenaga?
Fossil fuels are non-renewable source of energy.
Why fossil fuel is eventually will Bahan api fosil adalah sumber tenaga yang tidak boleh diperbaharui.
be used up?
Mengapa bahan api fosil akhirnya Other sources of energy are the sun, biomass, water and radioactive substances.
akan habis digunakan? Sumber tenaga yang lain adalah matahari, biojisim, air dan bahan radioaktif.

State other sources of energy.
Nyatakan sumber tenaga lain.

Structured Questions / Soalan Struktur

1 The diagram below shows the apparatus set-up for an experiment to determine the heat of displacement of silver.
Rajah di bawah menunjukkan susunan radas untuk eksperimen menentukan haba penyesaran argentum.

34 35 36 37 38 39 40 41 42 43 Excess of copper powder
Serbuk kuprum berlebihan

Plastic cup 100 cm3 of 0.5 mol dm–3 silver nitrate solution
Cawan plastik 100 cm3 larutan argentum nitrat 0.5 mol dm–3

The following data was obtained: / Berikut adalah data yang diperoleh:

Initial temperature of silver nitrate solution / Suhu awal larutan argentum nitrat = 28.0ºC
H ighest temperature of the mixture of product / Suhu tertinggi campuran hasil tindak balas = 40.5ºC

[Given specific heat capacity of solution = 4.2 J g–1 °C–1, density of solution = 1 g cm–3]
[Muatan haba tentu larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3]
U
N (a) What is meant by the ‘heat of displacement’ in the experiment?
I Apakah yang dimaksudkan dengan ‘haba penyesaran’ dalam eksperimen itu?
T Heat released when one mole of silver is displaced from silver nitrate solution by copper.

4 Haba dibebaskan apabila satu mol argentum disesarkan dari larutan argentum nitrat oleh kuprum.
(b) State three observations in the experiment and the reason for each observation.
Nyatakan tiga pemerhatian dalam eksperimen itu dan berikan sebab untuk setiap pemerhatian.

(i) Grey solid is deposited because silver metal is displaced by copper from silver nitrate solution / Pepejal kelabu berkilat

terenap kerana logam argentum disesar oleh kuprum dari larutan argentum nitrat.

(ii) Colourless solution turns blue because copper(II) ion is produced / Larutan tanpa warna menjadi biru kerana kuprum(II)

ion dihasilkan

(iii) The thermometer reading rises or the container becomes hot or warm because heat is released to the surroundings/the

reaction is exothermic / Bacaan termometer meningkat atau bekas menjadi panas kerana haba dibebaskan ke

persekitaran/ tindak balas adalah eksotermik

© Nilam Publication Sdn Bhd 172

MODULE • Chemistry Form 5

(c) Why is a plastic cup used in the experiment?
Mengapakah cawan plastik digunakan dalam eksperimen itu?

To reduce heat loss to the surrounding. / Untuk mengurangkan kehilangan haba ke persekitaran

(d) Write the ionic equation for the reaction.
Tulis persamaan ion untuk tindak balas itu.

Cu + 2Ag+ → Cu2+ + 2Ag

(e) Based on the information given in the experiment, calculate:
Berdasarkan maklumat yang diberi, hitungkan:
(i) change in temperature / perubahan suhu

θ = 40.5 – 28.0 = 12.5°C

(ii) the heat given out in the experiment / haba yang dibebaskan dalam eksperimen
H = (100)(4.2)(12.5) = 5 250 J

(iii) the heat of displacement of silver / haba penyesaran argentum

Number of moles of AgNO3 / Bilangan mol AgNO3 = 100 × 0.5 = 0.05 mol
1 000

From the equation, / Dari persamaan,

2 mol of AgNO3 produce 2 mol of Ag / 2 mol AgNO3 menghasilkan 2 mol Ag
0.05 mol of AgNO3 produce 0.05 mol of Ag / 0.05 mol AgNO3 menghasilkan 0.05 mol Ag

Heat of displacement of silver / Haba penyesaran argentum = –5 250 J
0.05 mol

= –105 kJ mol–1

(f) (i) 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.

Eksperimen itu diulangi menggunakan 100 cm3 larutan argentum nitrat 1.0 mol dm–3 dan serbuk kuprum yang
berlebihan. Hitungkan perubahan suhu dalam eksperimen ini.

Number of moles of Ag+ / Bilangan mol Ag+ = 1 × 100 = 0.1 mol,
1 000
U
Number of moles of Ag displaced / Bilangan mol Ag disesarkan = 0.1 mol N
I
Temperature change, / Perubahan suhu, θ = 0.1 × 105 000 = 25 ºC T
100 × 4.2
4

(ii) Explain why this change of temperature is different from that in (e)(i).
Terangkan mengapa perubahan suhu berbeza dengan (e)(i).

The number of mol of silver displaced is doubled, hence amount of heat energy released is also doubled. The amount of
heat energy which is doubled is used to increase the temperature of the same volume of solution. The increase in
temperature of the solution is also doubled. / Bilangan mol argentum disesar adalah dua kali ganda, maka jumlah
tenaga haba dibebaskan juga dua kali ganda. Jumlah tenaga haba yang dua kali ganda digunakan untuk meningkatkan
suhu larutan yang sama isi padunya. Kenaikan suhu larutan juga menjadi dua kali ganda.

173 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

2 Experiment I is carried out to determine the heat of neutralisation between strong acid and strong alkali. 50 cm3 of 0.5 mol dm–3
sodium hydroxide solution is poured into a plastic cup and the initial temperature is recorded. 50 cm3 of 0.5 mol dm–3 nitric
acid is then poured into the cup containing the sodium hydroxide solution. The mixture is stirred and heat produced raises the
temperature by 3°C. [Specific heat capacity of the solution = 4.2 J g–1 °C–1]

Eksperimen I dijalankan untuk menentukan haba peneutralan antara asid kuat dengan alkali kuat. 50 cm3 larutan natrium
hidroksida 0.5 mol dm–3 dituangkan dalam cawan plastik dan suhu awal dicatat. 50 cm3 asid nitrik 0.5 mol dm–3 kemudian
dituangkan ke dalam cawan mengandungi larutan natrium hidroksida. Campuran tindak balas dikacau dan haba yang terbebas
menaikkan suhu sebanyak 3°C. [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1]

Plastic cup 34 35 36 37 38 39 40 41 42 43 50 cm3 of 0.5 mol dm–3 sodium hydroxide solution
Cawan plastik and 50 cm3 of 0.5 mol dm–3 nitric acid
50 cm3 larutan natrium hidroksida 0.5 mol dm–3
dan 50 cm3 asid nitrik 0.5 mol dm–3

Experiment I / Eksperimen I

(a) What is meant by ‘heat of neutralisation’ in the experiment?
Apakah yang dimaksudkan ‘haba peneutralan’ dalam eksperimen ini?

Heat released when 1 mole of water is formed from the neutralisation between nitric acid and sodium hydroxide solution.

Haba yang dibebaskan apabila 1 mol air terbentuk dari tindak balas antara asid nitrik dan larutan natrium hidroksida.

(b) Calculate / Hitungkan
(i) the number of moles of sodium hydroxide that reacts with nitric acid.
bilangan mol natrium hidroksida yang bertindak balas dengan asid nitrik.

50 × 0.5
Number of moles / Bilangan mol = 1 000

= 0.025 mol

(ii) the heat released in the experiment.

haba yang dibebaskan dalam tindak balas itu.

Heat released / Haba dibebaskan = Heat changed / Perubahan haba

= 100 × 4.2 × 3

U = 1 260 J
N
I (iii) the heat of neutralisation for the reaction.
T haba peneutralan bagi tindak balas.
4 NaOH + HNO3 → NaNO3 + H2O

0.025 mol 0.025 mol 0.025 mol

0.025 mole of NaOH reacts with 0.025 mole of HNO3 to form 0.025 mole of H2O

0.025 mol NaOH bertindak balas dengan 0.025 mol HNO3 membentuk 0.025 mol H2O

The heat change is 1 260 J / Haba dibebaskan ialah 1 260 J
1 260 J

Heat of neutralisation / Haba peneutralan = ∆H = – 0.025 mol

= –50.4 kJ mol–1

(c) Write the thermochemical equation for the reaction in the experiment.
Tulis persamaan termokimia untuk tindak balas dalam eksperimen.

NaOH + HNO3 → NaNO3 + H2O ∆H = –50.4 kJ mol–1

© Nilam Publication Sdn Bhd 174

MODULE • Chemistry Form 5

(d) Experiment II is carried out under the same conditions as experiment I, whereby a 50 cm3 of 1 mol dm–3 ethanoic
acid is added to 50 cm3 of 1 mol dm–3 sodium hydroxide solution. The temperature of the mixture increased by 5.5°C.
Eksperimen II dijalankan dalam keadaan yang sama dengan eksperimen I di mana 50 cm3 asid etanoik 1 mol dm–3 ditambah
kepada 50 cm3 larutan natrium hidroksida 1 mol dm–3. Suhu campuran meningkat sebanyak 5.5ºC.

Plastic cup 34 35 36 37 38 39 40 41 42 43 50 cm3 of 1 mol dm–3 sodium hydroxide solution
Cawan plastik and 50 cm3 of 1 mol dm–3 ethanoic acid
50 cm3 larutan natrium hidroksida 1 mol dm–3
dan 50 cm3 asid etanoik 1 mol dm–3

Experiment II / Eksperimen II

(i) Calculate the number of moles of sodium hydroxide used.

Hitungkan bilangan mol natrium hidroksida digunakan.

Number of moles of sodium hydroxide / Bilangan mol natrium hidroksida = MV = 1 × (50)
1 000 1 000

= 0.05 mol

(ii) Calculate the heat of neutralisation for the reaction between ethanoic acid and sodium hydroxide solution.

[Specific capacity for all solutions is 4.2 J g–1 °C–1 and the density of all solutions is 1.0 g cm–3]

Hitungkan haba peneutralan bagi tindak balas antara asid etanoik dengan larutan natrium hidroksida.

[Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3]
Heat change / Haba dibebaskan = mcθ
= (50 + 50) × 4.2 × 5.5 = 2 310 J
2 310 J
Heat of neutralisation / Haba peneutralan = – 0.05 mol = – 46 200 J mol–1 = 46.2 kJ mol–1

(e) Compare the heat of neutralisation for Experiment I and Experiment II. Explain your answer. U
Bandingkan haba peneutralan dalam Eksperimen I dan Eksperimen II. Terangkan jawapan anda. N
I
The heat of neutralisation for Experiment I is higher than Experiment II. Nitric acid is a strong acid which ionises completely T

in water. Ethanoic acid is a weak acid which ionises partially in water, some of the ethanoic acid still remain in the form of 4

molecules. Some of heat released in Experiment II during neutralisation is absorbed to ionise the molecules of ethanoic acid.

Haba peneutralan dalam Eksperimen I lebih tinggi daripada Eksperimen II. Asid nitrik adalah asid kuat mengion lengkap

dalam air. Asid etanoik adalah asid lemah yang mengion separa dalam air, sebahagian asid etanoik wujud dalam bentuk

molekul. Sebahagian haba yang dibebaskan dalam Eksperimen II semasa peneutralan diserap untuk mengionkan molekul asid

etanoik yang belum mengion.

3 An experiment was carried out to determine the heat of precipitation for the reaction between lead(II) nitrate and potassium
sulphate. 50.0 cm3 of 0.5 mol dm–3 lead(II) nitrate solution was added to 50.0 cm3 of 0.5 mol dm–3 of potassium sulphate solution
in a plastic cup.

Satu eksperimen dijalankan untuk menentukan haba pemendakan antara plumbum(II) nitrat dan kalium sulfat. 50.0 cm3 larutan
plumbum(II) nitrat 0.5 mol dm–3 ditambahkan kepada 50.0 cm3 larutan kalium sulfat 0.5 mol dm–3 di dalam cawan plastik.

The thermochemical equation for the reaction is shown below: / Persamaan termokimia untuk tindak balas seperti berikut:
Pb(NO3)2 + K2SO4 → PbSO4 + 2KNO3 ∆H = –50.4 kJ mol–1

[Specific heat capacity of the solution = 4.2 J g–1 °C–1, density solution = 1 g cm–3]
[Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3]

(a) What is meant by ‘heat of precepitation’ in the experiment?
Apakah yang dimaksudkan dengan ‘haba pemendakan’ dalam eksperimen itu?

Heat is released when 1 mole of lead(II) sulphate is precipitated from mixing the aqueous solution of the Pb2+ ions and SO42–

ions. / Haba yang dibebaskan apabila 1 mol plumbum(II) sulfat termendak dari larutan akueus yang mengandungi ion Pb2+

dan ion SO42–.

175 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

(b) State one observation in the experiment.
Nyatakan satu pemerhatian dalam eksperimen.

White precipitate is formed. / Mendakan putih terbentuk.

(c) Calculate / Hitungkan

(i) number of moles of lead(II) nitrate

Bilangan mol plumbum(II) nitrat
50 × 0.5

Number of moles / Bilangan mol = 1 000 = 0.025 mol

(ii) Heat change in the experiment.

Perubahan haba dalam eksperimen.

Pb(NO3)2 + K2SO4 → PbSO4 + 2KNO3 ∆H = –50.4 kJ mol–1
Number of moles of PbSO4 / Bilangan mol PbSO4 = 0.025 mol
1 mole of lead(II) sulphate is precipitated, heat released is 50.4 kJ

1 mol plumbum(II) sulfat termendak, haba terbebas ialah 50.4 kJ

0.025 mol of lead(II) sulphate, heat released is / 0.025 mol of plumbum(II) sulfat termendak, haba terbebas ialah

= 50.4 × 0.025 = 1.26 kJ
or / atau H

50.4 kJ = 0.025

H = 50.4 × 0.025 = 1.26 kJ

(iii) Temperature change / Perubahan suhu

1 260 J = 100 × 4.2 × θ

θ= 1 260 = 3°C
100 × 4.2

(d) Construct energy level diagram for the reaction.
Lukis gambar rajah aras tenaga untuk tindak balas tersebut.

Energy / Tenaga
Pb(NO3)2 + K2SO4

∆H = –50.4 kJ mol–1
PbSO4 + 2KNO3

(e) Write an ionic equation for the above reaction.

U Tulis persamaan ion untuk tindak balas di atas.
N
I Pb2+ + SO42– → PbSO4

T
4 (f) The experiment is repeated by using 50.0 cm3 of 0.5 mol dm-3 lead(II) ethanoate and 50.0 cm3 of 0.5 mol dm–3 sodium
sulphate solution. What is the change in temperature for the reaction? Explain your answer.
Eksperimen diulangi dengan menggunakan 50.0 cm3 plumbum(II) etanoat 0.5 mol dm–3 dan 50.0 cm3 larutan natrium

sulfat 0.5 mol dm–3. Apakah perubahan suhu untuk tindak balas itu? Terangkan jawapan anda.

3°C. The precipitation of lead(II) sulphate only involves Pb2+ ions and SO42– ions.

3°C. Pemendakan plumbum(II) sulfat hanya melibatkan ion Pb2+ dan ion SO42–.

(g) Why is a plastic cup used in this experiment?
Mengapakah cawan plastik digunakan dalam eksperimen ini?

Plastic is a good heat insulator // to reduce heat loss to the surrounding.

Plastik adalah penebat haba yang baik // untuk mengurangkan kehilangan haba ke persekitaran.

© Nilam Publication Sdn Bhd 176

MODULE • Chemistry Form 5

(h) In another experiment where calcium chloride solution is reacted with sodium carbonate solution, the temperature of the
mixture decreases. The temperature change is recorded and ∆H is calculated.

Dalam eksperimen lain, larutan kalsium klorida ditindak balaskan dengan larutan natrium karbonat, suhu campuran
tindak balas berkurang. Perubahan suhu direkod dan ∆H dihitung.

(i) Write a balanced equation for the reaction above.
Tulis persamaan seimbang untuk tindak balas di atas.

CaCl2 + Na2CO3 → CaCO3 + 2NaCl

(ii) Construct an energy level diagram for the reaction.
Lukis gambar rajah aras tenaga untuk tindak balas itu.

Energy / Tenaga

CaCO3 + 2NaCl

CaCl2 + Na2CO3

4 The apparatus set-up below was used to determine the heat of combustion of butanol.
Susunan radas di bawah telah digunakan untuk menentukan haba pembakaran butanol.

Thermometer 34 35 36 37 38 39 40 41 42 43
Termometer

Metal can / Tin logam

Water / Air

Lamp + Butanol
Pelita + Butanol

The results are as follows: / Keputusan adalah seperti berikut: U
Initial mass of lamp + butanol / Jisim awal pelita + butanol = 502.28 g N
Final mass of lamp + butanol / Jisim akhir pelita + butanol = 500.00 g I
Initial temperature of water / Suhu awal air = 29°C T
Highest temperature of water / Suhu tertinggi air = 59°C
Volume of water / Isi padu air = 500 cm3 4
[Specific heat capacity of water = 4.2 J g–1 °C–1] / [Muatan haba tentu air = 4.2 J g–1 °C–1]
© Nilam Publication Sdn Bhd
(a) Write the equation for the combustion of butanol, C4H9OH.
Tulis persamaan untuk pembakaran butanol, C4H9OH.

C4H9OH + 6O2 → 4CO2 + 5H2O

(b) Calculate the heat energy change for the combustion of butanol in the above experiment.
Hitungkan perubahan haba untuk pembakaran butanol di dalam eksperimen di atas.

Heat change, / Perubahan haba, H = 500 × 4.2 × 30
= 63 000 J/63 kJ

(c) Calculate the number of moles of butanol that was burnt.
Hitungkan bilangan mol butanol yang telah terbakar.
[Relative atomic mass: / Jisim atom relatif: C = 12, H = 1]

Relative molecular mass / Jisim molekul realtif = 4(12) + 10(1) + 16

= 74

Number of moles / Bilangan mol = 2.28
74

= 0.03 mol

177

MODULE • Chemistry Form 5

(d) Calculate the heat of combustion for butanol. / Hitungkan haba pembakaran butanol.
0.03 mol of butanol releases 63 kJ of heat energy / 0.03 mol butanol membebaskan 63 kJ
63 kJ
1 mol of butanol releases / 1 mol butanol membebaskan = 0.03 mol = 2 100 kJ of heat energy / tenaga haba
DH = –2 100 kJ mol–1

(e) Give two precautionary steps that should be taken when conducting the experiment above.
Berikan dua langkah berjaga-jaga yang harus diambil semasa menjalankan eksperimen di atas.

• Use a wind shield / Gunakan pengadang angin

• Make sure the flame touches the bottom of the metal can / Pastikan nyalaan pelita menyentuh bahagaian bawah bekas logam.

• Stir the water in the metal can continuously (any 2) / Kacau air dalam bekas logam secara berterusan (mana-mana 2)

(f) The theoretical value for the heat of combustion of butanol is –2 877 kJ. Explain why the experimental value for the heat
of combustion of butanol is lower than the theoretical value.

Nilai teori untuk haba pembakaran butanol ialah –2 877 kJ. Terangkan mengapa nilai dari eksperimen untuk haba
pembakaran butanol adalah lebih rendah daripada nilai teori.
Heat is lost to the surrounding. Incomplete combustion of butanol. Heat from the flame during the burning of butanol is

absorbed by the tin/heats the tin. / Haba hilang ke persekitaran. Pembakaran butanol yang tidak lengkap. Haba dari nyalaan

pembakaran butanol diserap oleh bekas logam/memanaskan bekas logam.

(g) The table below shows the molecular formula and heat of combustion of three types of alcohol.
Jadual di bawah menunjukkan formula molekul dan haba pembakaran untuk tiga jenis alkohol.

Alcohol Molecular formula Heat of combustion/ kJ mol–1
Alkohol Formula molekul Haba pembakaran/ kJ mol–1

Methanol / Metanol CH3OH 725
1 376
Ethanol / Etanol C2H5OH 2 015

Propan-1-ol / Propan-1-ol C3H7OH

Explain why there are differences in the value of heat of combustion of the alcohols in the table.
Terangkan mengapa terdapat perbezaan pada nilai haba pembakaran alkohol dalam jadual di atas.

As the number of carbon and hydrogen atoms per molecule increases, the value of heat combustion increases. The higher

U the number of carbon and hydrogen atoms per molecule, the more carbon dioxide and water molecules products will be formed.
N
I More bonds in the product are formed, more heat is released. / Apabila bilangan atom karbon dalam setiap molekul alkohol

T bertambah, nilai haba pembakaran bertambah. Semakin bertambah bilangan atom karbon dan hidrogen dalam setiap molekul
4 alkohol, bilangan molekul karbon dioksida dan air sebagai hasil juga bertambah. Semakin banyak ikatan dalam hasil
terbentuk, semakin banyak haba dibebaskan.

Additional Questions

Soalan Tambahan

© Nilam Publication Sdn Bhd 178

Examples / Contoh Concept Map / Peta Konsep UNIT
Types of chemicals used / Jenis bahan kimia digunakan
5
Prevent food from Enhance natural flavour/Produce Add or restore the colour Improves the Thicken the food Prevent oxidation
being spoilt artificial flavour/Sweeten the food in a food / Menambah texture of food Memekatkan of food
Menghalang Meningkatkan rasa asal/ Menghasilkan atau mengembalikan Memperbaiki makanan Mencegah
makanan dari rosak perasa tiruan/ memaniskan makanan warna dalam makanan tekstur makanan pengoksidaan

Preservative / Pengawet Flavouring / Perasa Colouring / Pewarna Stabiliser / Penstabil Thickener / Pemekat Antioxidant / Antioksidan

1  Cleaning agent Type / Jenis Effects of food additives on MODULE • Chemistry Form 5
Agen pencuci 3  Food Additive / Bahan tambah makanan health / Kesan bahan tambah
makanan kepada kesihatan CHEMICAL FOR CONSUMERS
Meaning / Maksud CHEMICAL FOR CONSUMERS / BAHAN KIMIA UNTUK PENGGUNA
BAHAN KIMIA UNTUK PENGGUNA
(i)  Soap / Sabun (ii)  Detergent / Detergen (i)  Traditional medicine 2  Medicine / Ubat
Ubatan tradisional

Describe Explain Compare & Explain (ii)  Modern medicine / Ubatan moden Correct usage & side effect
Jelaskan Terangkan Banding & Terangkan Penggunaan yang betul &
kesan sampingan
179 Preparation Cleansing Effectiveness in
Penyediaan action soft water and Psychotherapeutic
hard water Psikoteraputik
Tindakan Keberkesanan Analgesic Antibiotic Function / Fungsi
pembersihan dalam air lembut Analgesik Antibiotik
dan air liat
Function / Fungsi Function / Fungsi

Additive To relieve pain without Used to treat infections Used to treat mental illness
Bahan tambah causing unconsciousness caused by bacteria or fungi Digunakan untuk merawat sakit mental
Untuk melegakan Digunakan untuk merawat
– Biological enzyme kesakitan tanpa jangkitan bakteria dan kulat Type / Jenis
Enzim biologi menyebabkan tidak
– Whitening sedarkan diri Example Stimulants Antidepressant Antipsychotics
Pemutih Contoh Stimulan Antidepresen Antipsikotik
– Fragrances Example  Contoh
Pewangi Example  Contoh
Example   Contoh Example   Contoh

© Nilam Publication Sdn Bhd Paracetamol Aspirin Streptomycin Penicillin Amphetamine Barbiturates & Chloropromazine
Parasetamol Aspirin Streptomisin Penisilin Amfetamin tranquiliser Kloropromazin
Barbiturat &
Learning objective / Objektif pembelajaran trankuliser

1 Analyse soap and detergents 3 Understand the medicine
Menganalisis sabun dan detergen Memahami ubat
2 Evaluate the uses of food additives 4 Appreciate the existence of chemicals
Menilai penggunaan bahan tambah makanan Menghargai kewujudan bahan kimia

U
N
I
T

5

MODULE • Chemistry Form 5

Soap / Sabun

Define soaps. Soaps are sodium or potassium salt of fatty acids.
Nyatakan maksud sabun.
Sabun ialah garam natrium atau kalium bagi asid lemak.

What are fatty acids? Fatty acids are organic acid with long carbon chain CnH2n + 1COOH, n > 10. Fatty acid found naturally as an
Apakah asid lemak? ester with alcohol glycerol (alcohol with 3 OH). Ester of fatty acid with glycerol is fat or oil.
Asid lemak ialah asid organik yang mempunyai rantai karbon panjang CnH2n + 1COOH, n > 10. Asid lemak
boleh didapati secara semula jadi sebagai ester dengan alkohol gliserol. (alkohol dengan 3 OH). Ester
bagi asid lemak ialah lemak atau minyak.

What are the sources of fats and The sources of fats and oils are animal fats and vegetable oils.
oils? / Apakah sumber-sumber sayuran
lemak dan minyak? Sumber lemak dan minyak adalah lemak haiwan dan minyak .

How to prepare soaps from fats Soap are prepared by hydrolysing fats/oils in potassium hydroxide or sodium hydroxide solution,
and oils? the reaction is called saponification . / Sabun boleh disediakan dengan hidrolisis lemak/minyak dalam
Bagaimana untuk menyediakan larutan kalium hidroksida atau natrium hidroksida , tindak balas ini dipanggil saponifikasi .
sabun dari lemak dan minyak?

What is saponification? Saponification is a process involves boiling fats or oils with concentrated sodium hydroxide
Apakah saponifikasi?
solution or potassium hydroxide solution . The products are glycerol and the salts of fatty

acids which are soaps. / Saponifikasi ialah proses yang melibatkan pendidihan lemak atau minyak

dengan larutan natrium hidroksida atau larutan kalium hidroksida yang pekat .

Hasilnya adalah gliserol dan garam dari asid lemak iaitu sabun.

Saponification equation: / Persamaan saponifikasi:

Fats/oil Sodium hydroxide Glycerol Sodium salt of fatty acid
Saponification (ALCOHOL)
(ESTER) (ALKALI) (SOAP)
Lemak/minyak + Natrium hidroksida Saponifikasi + Garam natrium dari asid lemak
Gliserol

(ESTER) (ALKALI) (ALKOHOL) (SABUN)

Complete the following: / Lengkapkan yang berikut: H O
HO H C OH R C O–Na+

H COC R

O O
H C O C R’ +  3NaOH H C OH + R’ C O–Na+

OO

H C O C R” H C OH R” C O–Na+

H Sodium hydroxide H Salt of fatty acid
Natrium hidroksida Garam dari asid lemak
Fat/oil Glycerol
Lemak/minyak Gliserol

U R, R’ and R” are long hydrocarbon chain (alkyl group). R, R’ and R” can be the same or different.
N
R, R’ dan R” adalah rantai hidrokarbon panjang (kumpulan alkil). R, R’ dan R” boleh sama ataupun berbeza.

I The structural formula for soap. O O
T Formula struktur sabun.
5 CH3(CH2)n C O–Na+ or / atau CH3(CH2)n C O–K+
The general formula of soap.
Formula am sabun. RCOO– Na+ or / atau RCOO–K+, where R is alkyl group, CnH2n + 1, n > 10
di mana R ialah kumpulan alkil, CnH2n + 1, n > 10

© Nilam Publication Sdn Bhd 180

MODULE • Chemistry Form 5

Given different formulae of fatty Fatty acid Soap Formula for soap Oil or fat used / Minyak atau
acids, derive the respective Asid lemak Sabun Formula bagi sabun lemak yang digunakan
chemical formulae of the soaps
formed. C11H23COOH Sodium laurate C11H23COONa Coconut oil
Diberikan formula asid lemak yang Lauric acid / Asid laurik Natrium laurat Minyak kelapa
berbeza, terbitkan formula kimia
yang sepadan bagi sabun yang C15H31COOH Sodium palmitate C15H31COONa Palm oil
terbentuk. Minyak sawit
Palmitic acid / Asid palmitik Natrium palmitat

C17H35COOH Sodium stearate C17H35COONa Animal fats
Lemak binatang
Stearic acid / Asid stearik Natrium stearat

Preparing a Sample of Soap in the Laboratory
Menyediakan Sabun dalam Makmal

Materials: Palm oil, sodium chloride, 5 mol dm-3 sodium hydroxide, distilled water
Bahan-bahan: Minyak sawit, natrium klorida, natrium hidroksida 5 mol dm–3, air suling
Apparatus: Beaker, glass rod, wire gauze, tripod stand, Bunsen burner, filter paper, filter funnel, spatula.
Alat radas: Bikar, rod kaca, kasa dawai, tungku kaki tiga, penunu Bunsen, kertas turas, corong penuras, spatula.
Set-up of apparatus: / Susunan alat radas:

Palm oil + concentrated sodium hydroxide
Minyak sawit + natrium hidroksida pekat

Heat
Panaskan

Procedure:

1 10 cm3 of palm oil is measured and poured into a beaker.

2 50 cm3 of 5 mol dm-3 sodium hydroxide solution is measured and added to the palm oil .

3 The mixture is heated while being constantly stirred with a glass rod until it boils.

4 The boiling is continued for about 5 minutes.

5 100 cm3 of distilled water and three spatulas of sodium chloride are added to the mixture.

6 The mixture is boiled again for another five minutes with constant stirring .

7 The mixture is allowed to cool .

8 The cooled mixture is filtered and the residue is rinsed with distilled water. (The residue is soap .)

9 The residue is pressed between a few pieces of filter paper to dry the soap.

10 A small amount of soap is added to 2 cm3 of water in a test tube. The test tube is shaken.

11 The solution formed is touched and felt by fingers.

Prosedur:

1 10 cm3 minyak sawit disukat dan dituang ke dalam bikar.

2 50 cm3 larutan natrium hidroksida 5 mol dm–3 disukat dan ditambah kepada minyak sawit .

3 Campuran tersebut dipanaskan sambil dikacau dengan berterusan menggunakan rod kaca sehingga mendidih.

4 Pendidihan diteruskan selama 5 minit.

5 100 cm3 air suling dan tiga spatula natrium klorida ditambah kepada campuran. U
N
6 Campuran dididihkan sekali lagi selama lima minit sambil dikacau dengan sekata. I
T
7 Campuran tersebut dibiarkan menyejuk .
5
8 Campuran yang telah disejukkan tersebut dituras dan bakinya dibilas dengan air suling. (Baki itu ialah

sabun .)

9 Baki tersebut ditekan di antara beberapa helai kertas turas untuk mengeringkan sabun tersebut.

10 Sedikit sabun ditambahkan kepada 2 cm3 air di dalam tabung uji. Tabung uji digoncangkan.

11 Larutan yang terbentuk disentuh dan dirasa dengan jari.

181 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Observation (a) The residue is a soft white solid.
Pemerhatian
Baki tersebut adalah pepejal putih yang lembut.

(b) Soap feels slippery . / Sabun mempunyai rasa yang licin pada jari.

(c) Soap is foamy when it is shaken in water.

Sabun berbuih apabila digoncangkan dalam air.

The chemical reaction is: / Tindak balas kimia ialah: H C17H35 O
HO H C OH C O–Na+

H C O C C17H35

O H C OH O
C17H35 C O–Na+
H C O C C17H35 H C OH +
O +  3NaOH H
O
H C O C C17H35 Glycerol C17H35 C O–Na+
Gliserol
H Sodium palmitate (soap)
Glyceryl tripalmitate Natrium palmitat (sabun)
Gliserol tripalmitat

What is the purpose of adding The soap can be precipitated by adding sodium chloride. Sodium chloride lowers the solubility of
sodium chloride? soap in water.
Apakah tujuan menambah natrium
klorida? Sabun boleh dimendakkan dengan menambah natrium klorida. Natrium klorida dapat mengurangkan
keterlarutan sabun dalam air.

Conclusion When palm oil is boiled with sodium hydroxide solution, a soap is formed. The reaction is
Kesimpulan called saponification . natrium hidroksida , sabun terbentuk.

Apabila minyak sawit dididihkan dengan larutan
Tindak balas tersebut dipanggil saponifikasi .

Detergent / Detergen

Define detergent. Detergent is cleaning agent that is not soap. Detergents are sodium salts of alkylbenzene sulphonic acid
Nyatakan maksud detergen. or alkyl sulphonic acid.
Detergen ialah agen pembersih yang bukan sabun. Detergen adalah garam natrium dari asid alkilbenzena
What are the two common groups sulfonik atau asid alkil sulfonik.
of detergents?
Apakah dua kumpulan umum Sodium alkylbenzene sulphonate Sodium alkyl sulphate
detergen? Natrium alkilbenzena sulfonat Natrium alkil sulfat

O O

R O S O–Na+ R O S O–Na+

U OO
N
I
T R represents a long chain hydrocarbon / R mewakil rantai hidrokarbon yang panjang

5 Preparation of detergent / Penyediaan detergen

What are the resources to make Detergents are made from synthetic resources such as petroleum fractions and sulphuric acid from
detergents? / Apakah sumber Contact Process. / Detergen dibuat daripada sumber sintetik seperti pembahagian petroleum dan asid
untuk membuat detergen? sulfrufik daripada Proses Sentuh.

© Nilam Publication Sdn Bhd 182

MODULE • Chemistry Form 5

What are the processes involved Detergents are prepared through two processes:
in the preparation of detergent? Detergen dihasilkan melalui dua proses:
Apakah proses yang terlibat (a) Sulphonation / Pensulfonan
dalam penyediaan detergen? (b) Neutralisation / Peneutralan

(a) Preparation of sodium alkyl benzene sulphonate / Penyediaan natrium alkil benzena sulfonat
(i) Sulphonation of alkylbenzene: The alkylbenzene is reacted with concentrated sulphuric acid to form alkylbenzene
sulphonic acid.
Pensulfonan alkilbenzena: Alkilbenzena bertindak balas dengan asid sulfurik pekat untuk membentuk asid alkilbenzena
sulfonik.

CH3 O CH3 O

CH3 (CH2)n C O + HO S OH CH3 (CH2)n C O S OH + H2O

H O HO
Alkylbenzene sulphonic acid
Alkylbenzene Sulphuric acid
Alkilbenzena Asid sulfurik Asid alkilbenzena sulfonik

(ii) Neutralisation: The alkylbenzene sulphonic acid is then neutralised with sodium hydroxide solution to produce
alkylbenzene sulphonate salt, which is detergent .

Peneutralan: asid alkilbenzena sulfonik kemudiannya dineutralkan dengan larutan natrium hidroksida menghasilkan
garam alkilbenzena sulfonat, iaitu detergen .

CH3 O CH3 O

CH3 (CH2)n C O S OH +   NaOH CH3 (CH2)n C O S ONa + H2O

HO HO

Alkylbenzene sulphonic acid Sodium hydroxide Sodium alkylbenzene sulphonate Water

Asid alkilbenzena sulfonik Natrium hidroksida Natrium alkilbenzena sulfonat Air

(b) Preparation of sodium alkyl sulphate / Penyediaan natrium alkil sulfat

(i) Sulphonation of alcohol: The long chain alcohol is reacted with concentrated sulphuric acid to form alkyl sulphonic

acid.

Pensulfonan alkohol: Alkohol rantai panjang ditindak balaskan dengan asid sulfurik pekat untuk membentuk asid

alkil sulfonik. O O

CH3(CH2)nCH2 O H + HO S OH CH3(CH2)nCH2 O S OH +  H2O

Long chain alcohol O O Water
Alkohol rantai panjang Sulphuric acid Alkyl sulphonic acid Air
Asid sulfurik
Asid akil sulfonik

(ii) Neutralisation: The alkyl sulphonic acid is then neutralised with sodium hydroxide solution to produce sodium alkyl

sulphate salt, which is detergent .

Peneutralan: Asid alkil sulfonik kemudiannya dineutralkan dengan larutan natrium hidroksida untuk menghasilkan

garam natrium alkil sulfat, iaitu detergen . U
N
OO I
T
CH3(CH2)nCH2 O S OH +  NaOH CH3(CH2)nCH2 O S ONa + H2O
5
O Sodium hydroxide O Water
Natrium hidroksida Sodium alkyl sulphate Air
Alkyl sulphonic acid
Asid alkil sulfonik Natrium alkil sulfat

183 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

The Cleansing Action of Soap and Detergent
Tindakan Pembersihan Sabun dan Detergen

What are the ions produced when Soap with the general formula RCOO–Na+ / RCOO–K+ ionises in water to produce sodium/
soap ionises in water? mengion dalam air untuk menghasilkan
Apakah ion yang dihasilkan potassium cation, Na+ (or K+) and soap anion, RCOO–.
apabila sabun mengion dalam air?
Sabun mempunyai formula am RCOO–Na+ / RCOO–K+

kation natrium/kalium, Na+ (atau K+) dan anion sabun, RCOO–.

Describe the structure of soap Soap anion is made up of two parts:
anion. Anion sabun terdiri daripada dua bahagian:
Huraikan struktur anion sabun.
O

R C O–

Alkyl Carboxylate ion
Alkil Ion karboksilat

(i) R is a long hydrocarbon chain which are:
• hydrophobic (repelled by water)
• non-polar end (no charge)
• soluble in oil or grease

R ialah rantai hidrokarbon panjang yang bersifat:
• hidrofobik (tak larut dalam air)
• hujung tidak berpolar (tidak bercas)
• larut dalam minyak atau gris

(ii) –COO– is a carboxylate ion which are:
• hydrophilic (soluble in water)
• polar end (negatively charge)
• insoluble in oil or grease

–COO– ialah ion karboksilat yang bersifat:
• hidrofilik (larut dalam air)
• bahagian berpolar (bercas negatif)
• tak larut dalam minyak atau gris

Example:
Contoh:
– Sodium laurate: CH3 – (CH2)14 – COONa in water ionises to: CH3 – (CH2)14 – COO– (soap anion) and Na+.
Natrium laurat: CH3 – (CH2)14 – COONa dalam air mengion kepada: CH3 – (CH2)14 – COO– (anion sabun) dan Na+.
The structural formula of soap anion:
Formula struktur anion sabun:

CH2 CH2 CH2 CH2 CH2 CH2 CH2
CH3 CH2 CH2 CH2 CH2 CH2 CH2 C-O–

O

Hydrocarbon chain or hydrophobic . Carboxylate ion or hydrophilic .
Rantai hidrokarbon atau hidrofobik . Ion karboksilat atau hidrofilik .

U – Simplified representation of soap anion is: Hydrophobic Hydrophilic
N Struktur ringkas anion sabun: Hidrofobik Hidrofilik
I
T

5


© Nilam Publication Sdn Bhd 184

MODULE • Chemistry Form 5

Explain how the hydrophobic and (i) The soap ionises in water to produce free moving soap anions and cations.
hydrophilic parts of soap anions
work together to remove greasy Sabun mengion dalam air menghasilkan anion sabun dan kation yang bebas bergerak.
stains on cloth.
Terangkan bagaimana bahagian (ii) Soap anion reduces the surface tension of water which will increase wetting ability of water.
hidrofobik dan hidrofilik anion
sabun berfungsi bersama untuk Therefore, water wets the dirty cloth.
menghilangkan kotoran
berminyak pada kain. Anion sabun mengurangkan ketegangan permukaan air lalu meningkatkan kebolehan air untuk

membasah. Oleh itu, air membasahi kain kotor.

(iii) The hydrophilic part of the soap anions dissolves in water while the hydrophobic part
dissolves and penetrates into the grease.

Bahagian hidrofilik anion sabun larut di dalam air manakala bahagian hidrofobik larut dalam
gris.

Hydrophobic part Hydrophilic part
Bahagian hidrofobik Bahagian hidrofilik

Grease / Gris

Water Cloth / Kain
Air

(iv) By the movement of water during scrubbing and rubbing, grease is lifted off the surface.

Pergerakan air semasa gosokan dan pengocakan menyebabkan gris tertanggal daripada
permukaan kain.

Hydrophobic part Hydrophilic part
Bahagian hidrofobik Bahagian hidrofilik

Grease / Gris

Water Cloth / Kain
Air

(v) The hydrophilic part of soap anions surround the grease, the grease is suspended in the water.
Bahagian hidrofilik anion sabun mengelilingi gris, gris terapung di dalam air.

(vi) The grease is dispersed into smaller droplets .

Gris berpecah kepada titisan kecil .

(vii) The small droplets do not redeposit on the surface of the cloth due to repulsion between negative
charges on their surface.

Titisan kecil tersebut tidak bergabung semula pada permukaan kain kerana tolakan sesama
cas negatif pada bahagian hidrofilik di permukaan titisan gris.

How Soap Works (viii) The droplets are suspended in water, forming an emulsion .

Bagaimana Sabun Berfungsi Titisan tersebut tersebar sekata dalam air, membentuk emulsi .

https://goo.gl/gJt6FP (ix) Rinsing away the dirty water removes the grease droplets and the surface of the cloth is then cleaned.

Dengan membilas air kotor, titisan gris dapat ditanggalkan dan permukaan kain dapat dibersihkan.

Small droplet of grease
Titisan kecil gris

Cloth / Kain

The cleansing action of detergent / Tindakan pencucian detergen

What are the ions produced when Detergent dissolves in water to form detergent anion and sodium cation . For example the U
detergent ionises in water? ionisation of sodium alkyl sulphate; N
Apakah ion yang dihasilkan I
apabila detergen mengion dalam Detergen melarut dalam air untuk membentuk anion detergen dan kation natrium. T
air? Contohnya pengionan natrium alkil sulfat;
O 5
O CH3 (CH2)n O S O- + Na+

CH3 (CH2)n O S O-Na+ H2O O

O

185 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Describe the structure of The structure of a detergent anion is similar to a soap anion. The detergent anion is also made up of
detergent anion. two parts i.e hydrophobic part and hydrophilic part. / Struktur anion detergen sama dengan anion sabun.
Huraikan struktur anion detergen. Anion detergen juga terdiri daripada dua bahagian iaitu bahagian hidrofobik dan bahagian hidrofilik.
(i) Alkyl sulphate ion: / Ion alkil sulfat:
Explain how the hydrophobic and
hydrophilic parts of detergent O
anions work together to remove
greasy stains on cloth. CH3 (CH2)n O S O–
Terangkan bagaimana bahagian O
anion hidrofobik dan hidrofilik
detergen berfungsi bersama Hydrocarbon chain or hydrophobic . Sulphate ion or hydrophilic .
untuk menghilangkan kesan Rantai hidrokarbon atau hidrofobik . Ion sulfat atau hidrofilik .
berminyak pada kain.
(ii) Alkylbenzene sulphonate ion: / Ion alkilbenzena sulfonat:

O

CH3 (CH2)n O S O-
O

Hydrocarbon chain or hydrophobic . Benzene sulphonate ion or hydrophilic .
Rantai hidrokarbon atau hidrofobik . Ion benzena sulfonat atau hidrofilik .

The cleansing action of detergent is also very similar to soap, the hydrocarbon chain dissolves and
penetrates in grease while the sulphate ion or benzene sulphonate ion remains in water.
Tindakan pembersihan detergen juga sama dengan sabun, rantai hidrokarbon melarut dan
menembusi gris manakala ion sulfat atau ion benzena sulfonat kekal dalam air.

The Effectiveness of the Cleansing of Soap and Detergent
Keberkesanan Pembersihan Sabun dan Detergen

What is hard water? Hard water contains high concentration of calcium ion (Ca2+) and magnesium ion (Mg2+).
Apakah air liat? Air liat mengandungi ion kalsium (Ca2+) dan ion magnesium (Mg2+) dengan kepekatan yang tinggi.

What is the effect of hard water to – The effectiveness of soap is reduced when used in hard water.
the cleansing action of soaps? Keberkesanan sabun berkurang apabila digunakan di dalam air liat.
Explain.
Apakah kesan air liat kepada – Calcium ions and magnesium ions react with soap anions to form insoluble scum.
tindakan pembersihan sabun?
Terangkan. Ion kalsium dan ion magnesium bertindak balas dengan anion sabun untuk membentuk kekat
sabun yang tak larut.

2C17H35COO− (aq/ak) (C17H35COO)2Mg (s/p)↓
Stearic anion/soap anion + Mg2+(aq/ak) →
Anion stearik/anion sabun Insoluble magnesium stearate (scum)

Magnesium stearat tak larut (kekat)

2C17H35COO− (aq/ak) (C17H35COO)2Ca (s/p)↓
Stearic anion/soap anion + Ca2+ (aq/ak) →
Anion stearik/anion sabun Insoluble calcium stearate (scum)

Kalsium stearat tak larut (kekat)

U – Formation of scum reduces the amount of soap available for cleaning.

N Pembentukan kekat mengurangkan jumlah sabun yang diperlukan untuk
I pembersihan.
T
5 What is the effect of hard water to Detergent anions do not form insoluble scum with calcium ions and magnesium ions. This means
the cleansing action of detergent? detergent can act as cleaning agent in hard water. Thus, detergent is more effective than soap in hard
Explain.
Apakah kesan air liat kepada water. / Anion detergen tidak membentuk kekat tak larut dengan ion kalsium dan ion magnesium.

tindakan pembersih detergen? Ini bermakna detergen boleh bertindak sebagai agen pembersih dalam air liat. Maka, detergen lebih

Terangkan. berkesan berbanding sabun dalam air liat.

© Nilam Publication Sdn Bhd 186

Complete the following table: / Lengkapkan jadual berikut: MODULE • Chemistry Form 5

Cleaning agent Soap / Sabun Detergent / Detergen
Agen pembersih Long chain hydrocarbon from petroleum
Rantai hidrokarbon yang panjang daripada petroleum
Sources Animal fats or vegetable oil
Sumber Lemak haiwan atau minyak sayuran R O SO3Na, ROSO3Na

General formula RCOONa
Formula am

The structure of polar O O OO OO
end (Hydrophilic) C O– C O– aSotaru O– O S O–
Struktur hujung berpolar O S O–O orS OO– OO
(Hidrofilik) Carboxylate ion / Ion karboksilat O ataOu
Sulphate ion
Effectiveness Benzene sulphonate ion Ion sulfat
Keberkesanan Ion benzena sulfonat

Effective in soft water only Effective in soft water and hard water
Pembersih yang berkesan dalam air lembut sahaja Berkesan dalam kedua-dua air lembut dan air liat

Formation of scum Forms scum in hard water Do not form scum in hard water
Pembentukan kekat Membentuk kekat dalam air liat Tidak membentuk kekat dalam air liat

pH Slightly alkaline pH value is modified to suit cleaning task

pH Sedikit beralkali Nilai pH boleh diubah mengikut jenis pembersihan

Effect on environment Biodegradable, do not cause pollution Mostly non-biodegradable, cause pollution
Kesan ke atas alam Terbiodegradasi, tidak menyebabkan pencemaran Kebanyakannya tidak terbiodegradasi,
sekitar menyebabkan pencemaran

Why are additives added to Additives are added to enhance its cleaning efficiency. Complete the following table:
detergents? State the function of Bahan-bahan tambah dicampurkan untuk meningkatkan keberkesanan pembersihan. Lengkapkan jadual
the additives. berikut:
Mengapakah bahan-bahan tambah
ditambah kepada detergen? Additive / Bahan tambah Function / Fungsi
Nyatakan fungsi bahan tambah
tersebut. Biological enzyme such as lipase and peptidase Remove protein stains such as blood

Enzim biologi seperti lipase dan peptidase Membuang kotoran-kotoran protein seperti darah

Whitening agent such as sodium perborate and Convert stain to colourless substances
sodium hypochlorite Menukar kotoran kepada sebatian tanpa warna
Agen pemutih seperti natrium perborat dan
natrium hipoklorit

Fragrances Add fragrance to fabrics and detergent
Minyak wangi Menambahkan kewangian fabrik dan detergen

Suspension agent such as To prevent the removed dirt particles redeposited U
Carboxymethylcellulose onto cleaned fabrics N
Agen ampaian seperti Karboksimetilselulosa Untuk mengelakkan zarah kotoran termendap I
semula ke fabrik yang telah dibersihkan T
Anhydrous sodium sulphate and sodium silicate
Natrium sulfat dan natrium silikat kontang Drying agent – keep detergents dry by absorbing 5
water vapour
Agen pengeringan – mengekalkan bahan pencuci
kering dengan menyerap wap air

187 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Experiment to compare the effectiveness of soap and detergent
Eksperimen untuk membandingkan keberkesanan sabun dan detergen
Apparatus: Beaker, measuring cylinder, glass rod
Radas: Bikar, silinder penyukat, rod kaca
Materials: 1 mol dm–3 magnesium sulphate solution, detergent powder, soap powder and cloths with oily stain.
Bahan-bahan: 1 mol dm–3 larutan magnesium sulfat, serbuk detergen, serbuk sabun dan kain dengan kotoran berminyak.

Soap + hard water Cloth with oily stains Detergent + hard water
Sabun + air liat Kain dengan kotoran Detergen + air liat
berminyak

Procedure: / Prosedur:
1 Two beakers are filled with 1 mol dm–3 magnesium sulphate solution until half full.
Dua bikar diisi dengan 1 mol dm–3 magnesium sulfat sehingga separuh penuh.
2 ½ spatula of soap is added to one beaker and ½ spatula of detergent is added to another beaker.
½ spatula serbuk sabun ditambahkan kepada satu bikar dan ½ spatula serbuk detergen ditambahkan kepada bikar yang

lain.
3 The mixtures are stirred with a glass rod. / Campuran tersebut dikacau dengan rod kaca.
4 The dirty clothes are dipped into each of the beakers. / Kain yang kotor direndam di dalam setiap bikar.
5 The dirty clothes in each beaker are washed by scrubbing. / Kain kotor dalam setiap bikar dibasuh dengan gosokan.
6 The cleansing actions of soap and detergent on the dirty clothes are observed and recorded.
Tindakan pembersihan sabun dan detergen ke atas kain-kain kotor tersebut diperhatikan dan direkodkan.

Observation: / Pemerhatian:

Cleaning agent / Agen pembersih Soap / Sabun Detergent / Detergen

Effectiveness Oily stains remained Oily stains removed
Keberkesanan Kotor berminyak kekal Kotor berminyak tanggal

Formation of scum Scum forms No formation of scum
Pembentukan kekat Kekat terbentuk Tiada pembentukan kekat

Conclusion: / Kesimpulan:
Detergent cleans stains more effectively compared to soap in hard water. Detergent is more effective than soap in hard water.
Detergen membersihkan kotoran dengan lebih berkesan berbanding sabun dalam air liat. Detergen lebih berkesan berbanding
sabun dalam air liat.

Exercise / Latihan Substance X
Bahan X
1 The diagrams below show the apparatus set-up for preparing soap.
Rajah di bawah menunjukkan susunan alat radas untuk penyediaan sabun.



U Palm oil + concentrated Soap
N sodium hydroxide Sabun

Minyak sawit + natrium

TI hidroksida pekat
5 Heat Heat
Panaskan Panaskan

(a) (i) What is meant by soap? / Apakah yang dimaksudkan dengan sabun?

Soaps are sodium or potassium salt of long chain fatty acid. / Sabun ialah garam natrium atau kalium bagi asid lemak

yang berantai panjang

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MODULE • Chemistry Form 5

(ii) State the name of the process to prepare soap. / Nyatakan nama proses penyediaan sabun.
Saponification / Saponifikasi

(b) Substance X is added to the soap mixture to complete the process.
Bahan X ditambahkan kepada campuran sabun untuk melengkapkan proses tersebut.
(i) State the name of substance X. / Nyatakan nama bahan X.

Sodium chloride / Natrium klorida
(ii) Why is substance X added to the mixture? / Mengapakah bahan X ditambahkan kepada campuran?

To reduce the solubility of soap in water / to precipitate the soap / Untuk mengurangkan keterlarutan sabun dalam air /

untuk pemendakan sabun

(c) The following equation shows the reaction that takes place in the preparation of soap.
Persamaan berikut menunjukkan tindak balas yang berlaku dalam penyediaan sabun.

Palm oil + Concentrated sodium hydroxide Boil / Pendidihan Sodium palmitate (soap) + Substance Y

Minyak sawit + Natrium hidroksida pekat Natrium palmitat (sabun) + Bahan Y

(i) What is the homologous series of palm oil? / Apakah siri homolog bagi minyak sawit?
Ester / Ester

(ii) Substance Y is another product in the reaction. State the name of substance Y.
Bahan Y ialah hasil lain dalam tindak balas ini. Nyatakan nama bagi bahan Y.

Glycerol / Gliserol

(iii) A student wants to prepare a potassium palmitate soap. What alkali should be used?
Seorang pelajar mahu membuat sabun kalium palmitat. Apakah alkali yang patut digunakan?

Potassium hydroxide / Kalium hidroksida

(d) The diagram below shows the structural formula of a soap anion.
Rajah di bawah menunjukkan struktur formula bagi suatu anion sabun.

CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2

CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 C – O_
O

Hydrophobic Hydrophilic
Hidrofobik Hidrofilik

(i) State the property of hydrophobic and hydrophilic parts of soap anion.

Nyatakan sifat bagi bahagian hidrofobik dan hidrofilik dalam anion sabun.

Hydrophobic dissolves in organic solvent. Hydrophilic dissolves in water.

Hidrofobik larut dalam pelarut organik. Hidrofilik larut dalam air.

(ii) Soap is used to wash oily stains on cloth. Explain the cleansing action of soap on the oily stain.
Sabun digunakan untuk membasuh kotoran berminyak pada kain. Terangkan tindakan pembersihan sabun ke atas

kotoran berminyak.

• Soap reduces the surface tension of water and increases the wetting ability of water on the surface of the oily cloth.

• The hydrophobic part of the soap anion dissolves in the oily stains. The hydrophilic part of the soap anion remains in the water.

• Scrubbing helps to lift the oily stains from the cloth and break the oily stains into small droplets.

• The droplets do not redeposit on the surface of the cloth due to repulsion between the negative charges on their surface. U
• The droplets are suspended in water forming an emulsion. N
• Rinsing washes away these droplets and leaves the surface clean. I
• Sabun mengurangkan ketegangan permukaan air dan meningkatkan kebolehan membasah air ke atas permukaan kain T
berminyak.
5

• Bahagian hidrofobik anion sabun larut dalam kotoran berminyak. Bahagian hidrofilik anion sabun larut dalam air.

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MODULE • Chemistry Form 5

• Gosokan dapat membantu menarik kotoran berminyak dan memecahkan kotoran berminyak kepada titisan kecil.
• Titisan-titisan tersebut tidak termendak pada permukaan kain kerana berlaku penolakan di antara cas negatif pada
bahagian hidrofilik anion sabun.
• Titisan-titisan itu tersebar dalam air dan membentuk emulsi.
• Pembilasan dapat membersihkan titisan-titisan ini dan menjadikan permukaan kain bersih.

2 The diagrams below show the structural formula for the anion part of cleansing agent X and cleansing agent Y particles.
Rajah di bawah menunjukkan formula struktur bahagian anion agen pembersih X dan zarah-zarah agen pembersih Y.

OO

R C O–Na+ R O S O–Na+

Cleansing agent X / Agen pembersih X O

Cleansing agent Y / Agen pembersih Y

(a) Identify cleansing agents X and Y as soap and detergent.
Kenal pasti agen pembersih X dan Y sebagai sabun dan detergen.

Soap: / Sabun: Cleansing agent X / Agen pembersih X

Detergent: / Detergen: Cleansing agent Y / Agen pembersih Y

(b) Draw the hydrophilic part of cleaning agents X and Y. / Lukiskan bahagian hidrofilik agen pembersih X dan Y.

O O
C O– O S O–

O

Cleansing agent X / Agen pembersih X Cleansing agent Y / Agen pembersih Y

(c) State the name of part of the cleansing agents X and Y that is soluble in grease.
Nyatakan nama bahagian bagi agen pembersih X dan Y yang larut dalam gris.

Hydrophobic / Hidrofobik

(d) Soap is ineffective as a cleansing agent in hard water. Explain.
Sabun tidak berkesan sebagai agen pembersih dalam air liat. Terangkan.

The soap anions react with the magnesium ions and calcium ions in hard water to form white precipitate of magnesium and

calcium salt/ scum. Formation of scum reduces the amount of soap for cleaning. / Anion sabun bertindak balas dengan ion
magnesium dan ion kalsium dalam air liat membentuk mendakan putih magnesium dan garam kalsium/kekat. Pembentukan

kekat mengurangkan jumlah sabun yang boleh digunakan untuk membersih.

U (e) State one advantage of detergent over soap. / Nyatakan satu kelebihan detergen berbanding sabun.
N Detergent anions react with the magnesium ions and calcium ions in hard water to form soluble salt/no scum. The cleansing

I action of detergent is more effective than soap in hard water. / Detergen tidak membentuk kekat dengan ion magnesium dan

5T ion kalsium dalam air liat. Tindakan pembersihan detergen lebih berkesan berbanding sabun dalam air liat.
(f) One of the properties of soap and detergent is that they can form lather with water. What is the function of lather?

Salah satu sifat sabun dan detergen ialah ia membentuk buih dengan air. Apakah fungsi buih?

It helps to suspend the grease particles. / Membantu mengapungkan titisan gris.

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MODULE • Chemistry Form 5

3 A student carried out two experiments to investigate the cleansing effect of soap and detergent on oily stained cloth in hard
water.

Seorang pelajar menjalankan dua eksperimen untuk mengkaji kesan pembersihan sabun dan detergen ke atas kain kotor yang
berminyak di dalam air liat.

Experiment / Eksperimen Experiment I / Eksperimen I Experiment II / Eksperimen II

Set up of apparatus Soap + hard water Detergent + hard water
Susunan alat radas Sabun + air liat Detergen + air liat

Cloth with oily stains Cloth with oily stains
Kain dengan kotoran berminyak Kain dengan kotoran berminyak

Observation / Pemerhatian Oily stain remained / Kotoran berminyak kekal Oily stain disappeared / Kotoran berminyak hilang

(a) Compare the cleansing effect between experiment I and experiment II.
Bandingkan kesan pembersihan antara eksperimen I dengan eksperimen II.

Soap in experiment I is not effective as cleansing agent in hard water whereas detergent in experiment II is effective as

cleansing agent in hard water. / Sabun dalam eksperimen I tidak berkesan sebagai agen pembersih dalam air liat manakala

detergen dalam eksperimen II berkesan sebagai agen pembersih dalam air liat.

(b) Explain why there are differences in the observations.
Terangkan mengapa terdapat perbezaan dalam pemerhatian.

•  Hard water contains calcium ions, Ca2+ and magnesium ions, Mg2+.
•  Soap anions in hard water react with magnesium ions or calcium ions to form scum, insoluble precipitate, no foam is formed.
•  Detergent anions react with Ca2+ ions and Mg2+ ions to form soluble salt, no precipitate, no scum, foam is formed.
•  Air liat mengandungi ion kalsium, Ca2+ dan ion magnesium, Mg2+.
•  Anion sabun dalam air liat bertindak balas dengan ion magnesium dan ion kalsium membentuk kekat, mendakan tak larut,
  tiada buih yang terbentuk.
•  Anion detergen bertindak balas dengan ion Ca2+ dan ion Mg2+ untuk membentuk garam terlarut, tiada mendakan, tiada kekat.
  Buih terbentuk.

(c) State the substance which is more suitable as a cleansing agent to remove stains in hard water.
Nyatakan bahan yang lebih sesuai sebagai agen pembersih untuk membuang kotoran dalam air liat.

Detergent is more effective than soap as a cleansing agent in hard water.

Detergen lebih berkesan berbanding sabun sebagai agen pembersih dalam air liat.

Food Additives / Bahan Tambah Makanan

Why are food additives added to (a) Improve its appearance, taste or texture U
food? Memperbaiki rupa, rasa dan teksturnya N
Mengapa bahan tambah makanan (b) Preserve the food I
ditambah kepada makanan? Mengawet makanan T

How are food additives classified? Food additives are classified according to their functions. 5
Bagaimana bahan tambah Bahan tambah makanan dikelaskan mengikut fungsi-fungsinya.
makanan dikelaskan? © Nilam Publication Sdn Bhd

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What are the types of food additives, their functions and examples? / Apakah jenis bahan tambah makanan, fungsi-fungsi dan contoh-contohnya? MODULE • Chemistry Form 5

Food additive / Bahan tambah makanan
Type of food additives / Jenis bahan tambah makanan

1.  Flavouring agent / Agen perasa 2.  Antioxidants / Antioksidan 3.  Colouring / Pewarna
Function / Fungsi Function / Fungsi
Function / Fungsi
(a) Enhance the natural (b) Produce artificial (c) Sweeten the (a) Add or restore the colour in (b) Make the food appearance
• Prevent oxidation that causes rancid fats and food look more attractive
flavour of the food flavour food brown fruits
Menambah atau mengekalkan Untuk menjadikannya lebih
Meningkatkan rasa Menghasilkan Memaniskan Menghalang pengoksidaan yang menyebabkan warna dalam makanan menarik
lemak tengik dan buah menjadi perang
asli makanan rasa tiruan makanan
Example / Contoh
Example / Contoh Example / Contoh

• Monosodium glutamate (MSG) • Ester • Aspartame • Ascorbic acid (vitamin C) • Citric acid • Azo compound such as tartrazine and • Triphenyl compounds such
Mononatrium glutamat (MSG) Ester Aspartam Asid askorbik (vitamin C) Asid sitrik ‘sunset yellow’ / Sebatian azo seperti as brilliant blue / Sebatian
tartrazine dan ‘sunset yellow’ trifenil seperti warna biru
Food / Makanan

– Soup / Sup – Cake / Kek – Drink Food / Makanan
– Frozen food – Jam / Jem Minuman
Makanan beku – Ice cream – Juice – Cake / Kek – Margarine / Marjerin Food / Makanan
– Meat / Daging Aiskrim Jus
– Biscuit / Biskut – Fruit juice / Jus buah – Orange juice / Jus oren
– Jam / Jem
192

4.  Preservatives / Pengawet 5.  Thickening agent / Pemekat 6.  Stabilisers / Penstabil
Function / Fungsi Function / Fungsi
Function / Fungsi
(a) Prevent food from being spoilt (b) Prevent food from being spoilt by: (c) Prevent food from (a) Improves the texture of (b) Improve the
being spoilt by (a) To thicken food food by preventing an consistency of food
by preventing or slow down (i) Remove water providing acidic Memekatkan makanan emulsion* from separating by giving a firmer,
condition out into a layer of water more uniform and
the growth of microorganisms (ii) Retard the growth of Example / Contoh and oil. (To emulsify food) smoother texture
Menghalang
Mengelakkan makanan microorganism makanan daripada Memperbaikkan tekstur Memperbaiki
rosak dengan makanan dengan struktur makanan
daripada rosak dengan Mengelakkan makanan menyediakan • Acacia gum • Pectin / Gelatin menghalang emulsi* dengan menjadikan
keadaan berasid Gam Acacia Pektin / Gelatin daripada terpisah kepada teksturnya lebih
menghalang atau daripada rosak dengan: lapisan air dan minyak sekata, halus dan
Example (Mengemulsikan makanan) licin
memperlahankan (i) Mengeluarkan air Contoh Food / Makanan

pertumbuhan mikroorganisma (ii) Membantutkan pertumbuhan • Vinegar – Chewing gum – Jam / Jem
Cuka Gula-gula getah – Jelly /Jeli
Example and food mikroorganisma – Cheese cream
Food Krim keju
Contoh dan makanan Example Makanan – Low fat milk
Dadih rendah
• Sodium nitrate / Natrium nitrat’ Contoh – Pickled
Jeruk lemak
– Sausage, burger, processed meat • Salt • Sugar Example and food
Sosej, burger, daging yang diproses Garam Gula Contoh dan makanan
• Sodium benzoate / Natrium benzoat’
– Chilli and tomato sauce Food • Lecithin, Gelatin / Lesitin, Gelatin
Sos cili dan tomato Makanan – Chocolate / Coklat – Butter /Mentega
• Sulphur dioxide / Sulfur dioksida’ – Ice cream /Aiskrim – Salad dressing
– Fruit juice, jam / Jus buah, jem – Salted fish – Jam Sos salad
Ikan masin Jem

MODULE • Chemistry Form 5

Exercise / Latihan

1 The diagram below shows a label of ingredients present on a packaging of a food.
Rajah di bawah menunjukkan kandungan bahan-bahan yang terdapat dalam suatu bungkusan makanan.

Solid milk, tartazine, sugar, …...
Pepejal susu, tartazin, gula, ……

(a) What is the function of tartazine?
Apakah fungsi tartazin?

Add colour to the food. / Menambah warna makanan.

(b) How does sugar make the food last longer?
Bagaimana gula membolehkan makanan tahan lebih lama?

Sugar make food last longer by removing water from the cell of microorganism and retards its growth.

Gula menyebabkan makanan tahan lebih lama dengan mengeluarkan air dari sel mikroorganisma dan menghalang

pertumbuhannya.

(c) The table below shows the functions for two examples of food additives.
Jadual di bawah menunjukkan fungsi bagi dua contoh bahan tambah makanan.

Food additive Function Type of food additive
Bahan tambah makanan Fungsi Jenis bahan tambah makanan

Sodium benzoate Controls and inhibits the growth of microorganisms P
Natrium benzoat Mengawal dan menghalang pertumbuhan mikroorganisma
Q
Lecithine Improve texture of the food
Lecitin Memperbaiki tekstur makanan

What are P and Q?
Apakah P dan Q?

P : Preservatives / Pengawet

Q : Stabiliser / Penstabil

2 The table below shows the types and examples of food additives.
Jadual di bawah menunjukkan jenis-jenis dan contoh-contoh bahan tambah makanan.

Types of food additives / Jenis bahan tambah Examples / Contoh U
V Sodium nitrate / Natrium nitrat N
Sodium chloride / Natrium klorida I
Antioxidants / Antioksidan T
X
Flavouring agents / Agen perasa Y 5
MSG
W Octyl ethanoate / Oktil etanoat
Food colouring / Pewarna makanan Acacia gum / Gam Acacia
Z

193 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

(a) What is V, W, X, Y and Z?
Apakah V, W, X, Y dan Z?

V : Preservatives / Pengawet
W : Stablisers and thickening agent / Penstabil dan pemekat
X : Ascorbic acid / Asid askorbik
Y : Saccharin/ aspartame / Sakarin/ aspartam
Z : Azo compound/ triphenyl compound / Sebatian azo/ sebatian trifenil

(b) (i) Give an example of food that uses sodium nitrate as food additive.
Berikan satu contoh makanan yang menggunakan natrium nitrat sebagai bahan tambah makanan.

Sausage/ burger/ meat / Sosej/ burger/ daging

(ii) Explain how sodium nitrate works as food additive.
Terangkan bagaimana natrium nitrat bertindak sebagai bahan tambah makanan.

Sodium nitrate prevents sausage/ burger/ meat from being spoilt by slowing down the growth of microorganism / Natrium

nitrat menghalang sosej/ burger/ daging daripada rosak dengan memperlahankan pertumbuhan mikroorganisma

(iii) Suggest another example of the same type of food additive as sodium nitrate that is used in chilli and tomato sauce.
Cadangkan contoh lain bahan tambah makanan dari jenis yang sama seperti natrium nitrat yang digunakan dalam

sos cili dan sos tomato.
Sodium benzoate / Natrium benzoat

(c) (i) Give one example of food that uses sodium chloride as food additive.
Berikan satu contoh makanan yang menggunakan natrium klorida sebagai bahan tambah makanan.

Salted fish / Ikan masin

(ii) Explain how sodium chloride works as food additive.
Terangkan bagaimana natrium klorida bertindak sebagai bahan tambah makanan.

Sodium chloride prevents fish from being spoilt by removing water from the cell of microorganism and retards its growth.

Natrium klorida menghalang ikan daripada rosak dengan mengeluarkan air daripada sel mikroorganisma dan

membantutkan pertumbuhannya.

(d) (i) What is the function of Y as a flavouring agent?
Apakah fungsi Y sebagai agen perasa?

Sweeten the food with less calories. / Memaniskan makanan dengan kalori yang lebih rendah.

(ii) What is the side effect of MSG on our health?
Apakah kesan sampingan MSG ke atas kesihatan kita?

Cause headache/ falling hair. / Menyebabkan sakit kepala/ rambut gugur.

(iii) What is the function of octyl ethanoate as a flavouring agent?
Apakah fungsi oktil etanoat sebagai agen perasa?

Produce artificial orange flavour. / Menghasilkan rasa oren tiruan

U (e) Name the example of food additives that cause the children to become hyperactive.
N
I Namakan contoh bahan tambah makanan yang menyebabkan kanak-kanak menjadi hiperaktif.
T Azo compound/ triphenyl compound / Sebatian azo/ sebatian trifenil
5

© Nilam Publication Sdn Bhd 194

MODULE • Chemistry Form 5

Medicine / Ubat

What is the function of medicine? A medicine is used to prevent or cure disease or to relieve pain .
Apakah fungsi ubat? atau mengurangkan
Ubat digunakan untuk menghalang atau menyembuhkan penyakit
What are the sources of traditional
medicines? kesakitan .
Apakah sumber ubat-ubatan
tradisional? Traditional medicines are obtained from natural sources ( plants or animals), without chemical reactions.
Ubat tradisional diperoleh daripada sumber semula jadi (tumbuhan atau binatang), tanpa tindak balas
What are the examples of kimia.
traditional medicines and their
uses? Traditional medicine Function
Apakah contoh ubat tradisional Ubat tradisional Fungsi
dan kegunaanya?
Aloe vera Its juice is used to treat skin wounds and burns.
Lidah buaya Jusnya digunakan untuk merawat luka kulit dan kesan bakar.

Bitter gourd Fruits are used to treat diabetes.
Peria Buahnya digunakan untuk merawat diabetes.

Ginger Rhizomes are used to treat stomach wind, improves blood circulation and
Halia digestion.
Rizomnya digunakan untuk merawat angin dalam perut, memperbaikkan
Garlic pengaliran darah dan pencernaan.
Bawang putih
Used to lower blood pressure and also has antibiotic properties.
Hibiscus Digunakan untuk mengurangkan tekanan darah dan mempunyai sifat-sifat
Bunga raya antibiotik.

Turmeric Leaves relieve headaches and hair loss.
Kunyit Daunnya digunakan untuk meredakan demam dan merawat keguguran
rambut.

Treats pimples.
Merawat jerawat.

Tamarind Juice reduces coughing.
Asam jawa Jus asam jawa boleh mengurangkan batuk.

Centella asiatica Its leaves are used as herbal tea to treat pain and inflammation.
Pegaga Daun pegaga digunakan sebagai teh herba untuk merawat sakit dan
pembengkakan.

What is modern medicines? Modern medicines are chemicals that are extracted from plants and animals or synthetic chemicals. U
Apakah ubat-ubatan moden? Ubat moden bahan kimia yang diekstrak daripada tumbuhan dan binatang atau bahan kimia buatan. N
I
What are the types of modern Remark: / Catatan: T
medicines? Modern medicines are manufactured in the form of liquid, capsules, powders and tablet.
Apakah jenis ubat-ubatan moden? Ubat moden dihasilkan dalam bentuk cecair, kapsul, serbuk atau pil. 5

(i) Analgesics
Analgesik
(ii) Antibiotics
Antibiotik
(iii) Psychotherapeutic
Psikoteraputik

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What are the types of modern medicines, their functions and examples? / Apakah jenis ubatan moden, fungsi-fungsi dan contoh-contohnya? MODULE • Chemistry Form 5

Modern medicine / Ubat moden
Type of modern medicine / Jenis ubat moden

1  Analgesic / Analgesik 2  Antibiotic / Antibiotik 3  Psychotherapeutic / Psikoteraputik
Function / Fungsi
Function / Fungsi Function / Fungsi
Used to treat mental illness.
To relieve pain without causing (a) Used to treat infections caused Digunakan untuk merawat
unconsciousness by bacteria or fungi / Digunakan sakit mental
Untuk melegakan kesakitan tanpa untuk merawat jangkitan yang
menyebabkan pesakit tidak sedar disebabkan bakteria dan kulat Type of Psychotherapeutic
Jenis Psikoteraputik
Example / Contoh (b) Antibiotics kill or slow down
the growth of bacteria or fungi
Paracetamol Aspirin / Aspirin Codeine Stimulants Antidepressant Antipsychotics
Parasetamol (acetylsalicylic acid) Kodeina Antibiotik membunuh atau Stimulan Antidepresen Antipsikotik
(asid asetilsalisilik) memperlahankan pertumbuhan
bakteria atau kulat Example
196 • Neutral • Acidic / Berasid • Cough Contoh Example Example
Neutral • Relieves pain caused by medicine Example / Contoh Contoh Contoh
• Relieves pain and Amphetamine
headache, toothache and Ubat batuk Penicilline and Streptomycin Amfetamin Barbiturates and Chloropromazine
reduces fever arthritis • Causes Penisilin dan Streptomisin tranquilizer Kloropromazine
Meredakan sakit Meredakan sakit yang
disebabkan sakit kepala, sleepiness • Side effects of antibiotics are Barbiturat dan
dan mengurangkan sakit gigi dan artritis Menyebabkan headache, allergic reaction and trankuilizer
demam • Reduces fever and dizziness.
• Does not reduce inflammation caused by mengantuk • To reduce fatigue • To reduce tension • To treat
inflammation infection • Misuse of Kesan sampingan antibiotik and elevate mood and anxiety psychiatric
Tidak Mengurangkan demam adalah sakit kepala, alergi dan illness with
mengurangkan dan keradangan yang codeine pening. Mengurangkan Mengurangkan severe mental
keradangan disebabkan jangkitan causes keletihan dan tekanan dan disorder
• Does not irritate the • Causes internal addiction • Patient must take full course merangsang kegelisahan
stomach bleeding ulceration Salah guna of the antibiotic prescribed by perasaan Merawat sakit
Tidak menyakitkan (not suitable for gastric kodeina boleh the doctor to make sure all the • Can cause jiwa dan sakit
perut patients) menyebabkan bacteria are killed, otherwise • Can cause addiction mental yang
Menyebabkan ketagihan they may become resistant to addiction teruk
pendarahan dalaman the antibiotic. Boleh
(tidak sesuai untuk Boleh menyebabkan
pesakit gastrik) Pesakit mesti mengambil menyebabkan ketagihan
semua antibiotik yang ketagihan
dipreskripsi oleh doktor supaya
semua bakteria dibunuh, jika
tidak, bakteria akan menjadi
imun terhadap antibiotik.

MODULE • Chemistry Form 5

Exercise / Latihan

1 The diagram below shows the molecular structure of aspirin. / Rajah di bawah menunjukkan struktur molekul bagi aspirin.

OH
H

HC CC H
O

CC H

CC O

HC C

HO

H
(a) (i) What is the molecular formula for aspirin? / Apakah formula molekul bagi aspirin?

C9H8O4

(ii) What is the scientific name for aspirin? / Apakah nama saintifik bagi aspirin?
Acetylsalicylic acid / Asid asetilsalisilik

(iii) State which type of medicine does aspirin belong to. / Nyatakan jenis ubat bagi aspirin.
Analgesic / Analgesik

(iv) What is the side effect of aspirin on children below 12 years old?
Apakah kesan sampingan aspirin kepada kanak-kanak di bawah 12 tahun?

Irritates lining of the stomach and causes bleeding / Radang pada perut dan menyebabkan pendarahan

(v) Suggest another example of medicine that can be used in replacing aspirin to reduce fever.
Cadangkan satu contoh ubat lain yang boleh digunakan untuk menggantikan aspirin bagi meredakan demam.

Paracetamol / Parasetamol

(b) (i) Antibiotic is one of the medicines that is always given by a doctor to a patient. What is the function of antibiotic?
Antibiotik adalah salah satu ubat yang selalu diberi oleh doktor kepada pesakit. Apakah fungsi antibiotik?

To kill or inhibit growth of infectious bacteria.

Untuk membunuh atau menghalang pertumbuhan bakteria yang mudah berjangkit.

(ii) Explain why the patient must take full course of the antibiotic prescribed by the doctor.
Terangkan mengapa pesakit mesti menghabiskan semua antibiotik yang telah dipreskripsi oleh doktor.

To make sure that all the bacteria are killed. If not, the bacteria will become immune to the medicine or it will cause

further infection. / Untuk memastikan semua bakteria telah dibunuh. Jika tidak, bakteria akan menjadi imun kepada ubat

dan menyebabkan jangkitan yang teruk.

(c) (i) State the name of one type of medicine that changes the emotions and behaviour of the patient.
Namakan satu jenis ubat lain yang mengubah emosi dan kelakuan pesakit.

Psychotherapeutic / Psikoteraputik

(ii) A patient is suffering from hallucination, delusion or other symptoms of mental illness. Suggest other example of U
medicine that is suitable to treat the patient. / Seorang pesakit mengalami masalah halusinasi, khayalan dan gejala- N
gejala lain sakit mental. Cadangkan contoh ubat lain yang sesuai untuk merawat pesakit. I
T
Antipsychotic / Antipsikotik
5
(d) Medicines that are obtained from plants and animals are known as traditional medicines. State the Additional Questions
name of a traditional medicine that can be used to cure diabetes.
Soalan Tambahan
Ubat yang diperoleh daripada tumbuhan dan binatang dikenali sebagai ubat tradisional. Namakan
satu ubat tradisional yang digunakan untuk merawat diabetes.

Bitter gourd / Peria

197 © Nilam Publication Sdn Bhd

The Periodic Table of Elements / Jadual Berkala Unsur

1 Symbol of the element Proton number / Nombor proton 18
1 Simbol unsur
2
1 H Name of the element / Nama unsur
He
Hydrogen
Helium
Hidrogen 2 13 14 15 16 17 Helium

1 4 Relative atomic mass 5 6 7 8 9 4
Jisim atom relatif
3 Be B C N O F 10

2 Li Beryllium Boron Carbon Nitrogen Oxygen Fluorine Ne
Fluorin
Lithium Neon
19 Neon
Litium Berilium Transition elements / Unsur peralihan Boron Karbon Nitrogen Oksigen
7 9 11 12 14 16 17 20

11 12 13 14 15 16 Cl 18

3 Na Mg Al Si P S Chorine Ar
Klorin
Sodium Magnesium Aluminium Silicone Phosphorus Sulphur 35.5 Argon
Argon
Natrium Magnesium 3 4 5 6 7 8 9 10 11 12 Aluminium Silikon Fosforus Sulfur 35
23 24 27 28 31 32 40
Br
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 36
Bromine
4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Bromin Kr

Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium 80 Krypton
53 Kripton
Kalium Kalsium Skandium Titanium Vanadium Kromium Mangan Ferum Kobalt Nikel Kuprum Zink Galium Germanium Arsenik Selenium
I 84
39 40 45 48 51 52 55 56 59 59 64 65 70 73 75 79 54
Iodine
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 Iodin Xe
127
5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te Xenon
85 Xenon
Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium
Argentum Kadmium Indium Stanum Antimoni Telurium At 131
Rubidium Strontium Yttrium Zirkonium Niobium Molibdenum Teknetium Rutenium Rodium Paladium
88 89 91 93 96 98 101 103 106 108 112 115 119 122 128 Astatine 86
85.5 Astatin
Rn
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 210
Radon
Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po Radon
6 Cesium
Barium Lanthanum Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth Polonium 222
Sesium Aurum Merkuri Talium Plumbum Bismut Polonium
Barium Lantanum Hafnium Tantalum Tungsten Renium Osmium Iridium Platinum
133 137 139 178.5 181 184 186 190 192 195 197 201 204 207 209 210

87 88 89 104 105 106 107 108 109 110 111 112

Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Uub
7 Francium
Radium Actinium Rutherfordium Dubnium Scaborgium Bohrium Hassium Meitnerium Darmstadtium Roentgenium Ununbium

Fransium Radium Aktinida Rutherfordium Dubnium Siborgium Bohrium Hassium Meitnerium Darmstadtium Roentgenium Ununbium

223 226 227 257 260 262 262 265 266 271 272 285

58 59 60 61 62 63 64 65 66 67 68 69 70 71

Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Lantanida
Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium
Actinides Serium Praseodimium Neodimium Prometium Samarium Europium Gadolinium Terbium Disprosium Holmium Erbium
Aktinida Tulium Itterbium Lutetium
140 141 144 147 150 152 157 159 162.5 165 167
169 173 175
90 91 92 93 94 95 96 97 98 99 100
101 102 103
Th Pa U Np Pu Am Cm Bk Cf Es Fm
Md No Lr
Thorium Proactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium
Torium Proaktinium Uranium Neptunium Plutonium Amerisium Kurium Berkelium Kalifornium Einsteinium Fermium
Mendelevium Nobelium Lawrensium
232 231 238 237 244 243 247 247 249 254 253
256 254 257

Key: Metal Semi-metal Non-metal
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