Module & More Kimia Tg 5

  Kimia Tingkatan 5  Bab 3 Termokimia

(a) Tulis persamaan kimia seimbang. TP 2 (e) Tulis persamaan termokimia. TP 2
Write a balanced chemical equation. Write a thermochemical equation.
CuSO4(ak) + Fe(p) → FeSO4(ak) + Cu(p) CuSO4(ak) + Fe(p) → FeSO4(ak) + Cu(p)
CuSO4(aq) + Fe(s) → FeSO4(aq) + Cu(s) ΔH = –252 kJ mol–1
CuSO4(aq) + Fe(s) → FeSO4(aq) + Cu(s)
(b) Tulis persamaan ion. TP 2 ΔH = –252 kJ mol–1
Write an ionic equation.
Cu2+(ak) + Fe(p) → Fe2+(ak) + Cu(p) Cu2+(ak) + Fe(p) → Fe2+(ak) + Cu(p)
Cu2+(aq) + Fe(s) → Fe2+(aq) + Cu(s) ΔH = –252 kJ mol–1
Cu2+(aq) + Fe(s) → Fe2+(aq) + Cu(s)
ΔH = –252 kJ mol–1

(c) Tulis tiga pemerhatian yang diperoleh daripada (f ) Lukis gambar rajah aras tenaga untuk tindak balas di
tindak balas di atas. TP 3 atas. TP 3

Write three observations obtained from the above reaction. Draw the energy level diagram for the above reaction.
– Pepejal perang terbentuk. Tenaga / Energy
Brown solid is formed. Fe(p) + Cu2+(ak)
– Paku besi menjadi nipis.
Iron nail becomes thinner. Fe(s) + Cu2+(aq)
– Larutan biru bertukar warna menjadi hijau.
Blue solution turns green. ∆H = –252 kJ mol–1

Fe2+(ak) + Cu(p)
Fe2+(aq) + Cu(s)

BAB 3 (d) Hitung haba penyesaran kuprum oleh paku besi. (g) Nyatakan definisi secara operasi untuk haba
Calculate the heat of displacement of copper by the iron nail. penyesaran di atas. TP 3

TP 3 State the operational definition for the above heat of
displacement.
Perubahan suhu / Temperature change, θ
= 40.5 – 28.5 Suhu meningkat apabila 1 mol kuprum disesarkan
= 12.0 oC daripada larutan kuprum(II) sulfat oleh paku besi.

Perubahan haba / Heat change, Q The temperature rises when 1 mole of copper is displaced from
= (50) (4.2) (12.0) copper(II) sulphate solution by the iron nail.
= 2520 J

Bilangan mol CuSO4 / Number of moles of CuSO4,
(0.2)(50)
n = 1000

n = 0.01 mol

Bilangan mol Cu disesarkan

Number of moles of Cu displaced
= Bilangan mol Cu2+ / Number of moles of Cu2+
= 0.01 mol

Haba penyesaran Cu / Heat of displacement of Cu
2520
= 0.01

= –252000 J mol–1
= –252 kJ mol–1

© Penerbitan Pelangi Sdn. Bhd. 94

Kimia Tingkatan 5  Bab 3 Termokimia 

Haba peneutralan

Heat of neutralisation

1. Peneutralan ialah tindak balas antara asid dengan alkali atau bes untuk membentuk garam dan air . TP 1
an alkali or a base to form a salt and water
Neutralisation is a reaction when an acid reacts with .

2. Asid kuat ialah bahan yang mengion dengan lengkap dalam air untuk membentuk kepekatan ion
hidrogen yang tinggi
. TP 2
high
A strong acid is a substance that ionises completely in water to form concentrations of

hydrogen ions.

3. Asid lemah ialah bahan yang mengion separa dalam air untuk membentuk kepekatan ion
hidrogen yang rendah . TP 2
in water to form low concentrations of hydrogen ions.
A weak acid is a substance that ionises partially

4. Alkali kuat ialah bahan yang mengion lengkap dalam air untuk membentuk kepekatan ion hidroksida
yang tinggi
. TP 2

A strong alkali is a substance that ionises completely in water to form high concentrations of hydoxide ions .

5. Alkali lemah ialah bahan yang mengion separa dalam air untuk membentuk kepekatan ion

hidroksida yang rendah . TP 2

A weak alkali is a substance that ionises partially in water to form low concentrations of hidroxide ions .

6. Haba peneutralan ialah perubahan haba apabila 1 mol air terbentuk daripada tindak balas antara asid
dengan alkali
.
The heat of neutralisation is the heat change when 1 mole of water is formed from the neutralisation reaction between
an acid and an alkali
. TP 1 BAB 3

7. Tulis persamaan kimia seimbang dan persamaan ion untuk tindak balas berikut.
Write balance chemical equations and ionic equations for the following reactions. TP 3
(a) Asid hidroklorik bertindak balas dengan larutan natrium hidroksida.

Hydrochloric acid reacts with sodium hydroxide solution.

Persamaan kimia

Chemical equation:

HCl(ak) + NaOH(ak) → NaCl(ak) + H2O(ce) / HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Persamaan ion / Ionic equation:
H+(ak) + OH–(ak) → H2O(ce) / H+(aq) + OH–(aq) → H2O(l)

(b) Asid etanoik bertindak balas dengan larutan kalium hidroksida.

Ethanoic acid reacts with potassium hydroxide solution.

Persamaan kimia

Chemical equation:

CH3COOH(ak) + KOH(ak) → CH3COOK(ak) + H2O(ce)

CH3COOH(aq) + KOH(aq) → CH3COOK(aq) + H2O(l)

Persamaan ion / Ionic equation:
H+(ak) + OH–(ak) → H2O(ce) / H+(aq) + OH–(aq) → H2O(l)

(c) Asid sufurik bertindak balas dengan larutan natrium hidroksida.

Sulphuric acid reacts with sodium hydroxide solution.

Persamaan kimia

Chemical equation:

H2SO4(ak) + 2NaOH(ak) → Na2SO4(ak) + 2H2O(ce) / H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

Persamaan ion / Ionic equation:
H+(ak) + OH–(ak) → H2O(ce) / H+(aq) + OH–(aq) → H2O(l)

95 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 3 Termokimia Membandingkan Haba Peneutralan antara Asid dengan Alkali
yang Berlainan Kekuatan
Eksperimen 3.2
Comparing the Heat of Neutralisation between Acids and Alkalis of Different
Strengths

Tujuan / Aim: Haba
Untuk membandingkan haba peneutralan antara asid dengan alkali yang berlainan kekuatan. Peneutralan
Heat of
To compare the heat of neutralisation between acids and alkalis of different strengths. VIDEO 4 Neutralisation

Pernyataan masalah / Problem statement:
Bagaimanakah asid dan alkali yang berlainan kekuatan mempengaruhi haba peneutralan?

How do acids and alkalis of different strengths affect the heat of neutralisation?

Hipotesis / Hypothesis: asid kuat dengan alkali kuat adalah paling tinggi diikuti dengan haba

Haba peneutralan antara
peneutralan antara asid lemah dengan alkali kuat atau asid kuat dengan alkali lemah manakala
haba peneutralan antara asid lemah dengan alkali lemah adalah paling rendah.
The heat of neutralisation between a strong acid and a strong alkali is the highest followed by the heat of neutralisation
weak acid strong alkali
between a and a or a strong acid and a weak alkali while the heat of neutralisation
weak acid weak alkali
between a and a is the lowest.

BAB 3 Pemboleh ubah / Variables:
Dimanipulasikan / Manipulated
Asid dan alkali yang berlainan kekuatan /Acids and alkalis of different strengths

Bergerak balas / Responding
Haba peneutralan /Heat of neutralisation
Dimalarkan / Constant
Kepekatan dan isi padu asid dan alkali / Concentration and volume of the acid and alkali used

Bahan / Materials:

Asid hidroklorik 1.0 mol dm–3, asid etanoik 1.0 mol dm–3, larutan natrium hidroksida 1.0 mol dm–3 dan larutan

ammonia 1.0 mol dm–3.

1.0 mol dm–3 hydrochloric acid, 1.0 mol dm–3 ethanoic acid, 1.0 mol dm–3 sodium hydroxide solution and 1.0 mol dm–3 ammonia
solution.

Radas / Apparatus:

Cawan polistirena, penutup cawan plastik, termometer dan silinder penyukat 50 cm3.

Polystyrene cups, plastic cup covers, thermometers and 50 cm3 measuring cylinder.

Termometer Kacau menggunakan
Thermometer termometer
Stir using a thermometer

50 cm3 asid hidroklorik, Cawan 50 cm3 larutan natrium Larutan campuran
Mixture solution
HCl, 1.0 mol dm–3 A polistirena B hidroksida, NaOH
50 cm3 of 1.0 mol dm–3 Polystyrene 1.0 mol dm–3
hydrochloric acid, HCl cup 50 cm3 of 1.0 mol dm–3

sodium hydroxide, NaOH
solution

Prosedur / Procedures:
1. 50 cm3 asid hidroklorik, HCl 1.0 mol dm–3 disukat dan dituangkan ke dalam sebuah cawan polistirena A dan

ditutup dengan penutup plastik dan dibiarkan selama 5 minit.

50 cm3 of 1.0 mol dm–3 hydrochloric acid, HCl is measured and poured into the polystyrene cup A and covered with a plastic
cover and left for 5 minutes.

2. Termometer dimasukkan ke dalam cawan polistirena untuk mengukur suhu awal.

A thermometer is inserted into the polystyrene cup to measure the initial temperature.

© Penerbitan Pelangi Sdn. Bhd. 96

Kimia Tingkatan 5  Bab 3 Termokimia 

3. 50 cm3 larutan natrium hidroksida, NaOH 1.0 mol dm–3 disukat dan dituangkan ke dalam sebuah cawan
polistirena B dan ditutup dengan penutup plastik dan dibiarkan selama 5 minit.

50 cm3 of 1.0 mol dm–3 sodium hydroxide, NaOH solution is measured and poured into the polystyrene cup B and covered
with a plastic cover and left for 5 minutes.

4. Termometer dimasukkan ke dalam cawan polistirena untuk mengukur suhu awal.

A thermometer is inserted into the polystyrene cup to measure the initial temperature.

5. Suhu awal kedua-dua larutan dicatatkan dalam jadual.

The initial temperatures of both solutions are recorded in the table.

6. Penutup plastik ditanggalkan dan asid hidroklorik dituangkan dengan cermat ke dalam cawan polistirena B.

The plastic cover is removed and the hydrochloric acid is poured carefully into the polystyrene cup B.

7. Larutan campuran dikacau serta-merta dengan termometer.

The mixture solution is stirred immediately with a thermometer.

8. Suhu tertinggi yang diperoleh dicatatkan dalam jadual.

The highest temperature obtained is recorded in the table.

9. Langkah 1 – 8 diulangi dengan pasangan larutan asid dan alkali seperti berikut:

Steps 1 – 8 are repeated with the following pairs of acid and alkaline solutions:

Set II 50 cm3 asid etanoik, CH3COOH 1.0 mol dm–3 + 50 cm3 larutan natrium hidroksida, NaOH 1.0 mol dm–3
Set III
Set IV 50 cm3 of 1.0 mol dm–3 ethanoic acid, CH3COOH + 50 cm3 of 1.0 mol dm–3 sodium hydroxide, NaOH solution

50 cm3 asid hidroklorik, HCl 1.0 mol dm–3 + 50 cm3 larutan ammonia, NH3 1.0 mol dm–3

50 cm3 of 1.0 mol dm–3 hydrochloric acid, HCl + 50 cm3 of 1.0 mol dm–3 ammonia, NH3 solution

50 cm3 asid etanoik, CH3COOH 1.0 mol dm–3 + 50 cm3 larutan ammonia, NH3 1.0 mol dm–3

50 cm3 of 1.0 mol dm–3 ethanoic acid, CH3COOH + 50 cm3 of 1.0 mol dm–3 ammonia, NH3 solution

Keputusan / Results: BAB 3

Set I II III IV

Campuran bahan tindak balas HCl CH3COOH HCl CH3COOH
+ + + +
Reactant mixture
NaOH NaOH NH3 NH3
Suhu awal asid (oC)
28.0 28.5 28.5 28.0
Initial temperature of acid (oC)
29.0 28.5 28.5 29.0
Suhu awal alkali (oC)
(28.0 + 29.0) (28.5 + 28.5) (28.5 + 28.5) (28.0 + 29.0)
Initial temperature of alkali (oC) 2 2 2 2

Suhu awal purata asid dan alkali = 28.5 = 28.5 = 28.5 = 28.5
(oC)
35.2 35.0 34.5 34.0
Average initial temperatures of acid and
alkali (oC) 35.2 – 28.5 35.0 – 28.5 34.5 – 28.5 34.0 – 28.5
= 6.7 = 6.5 = 6.0 = 5.5
Suhu tertinggi campuran (oC)

Highest temperature of mixture (oC)

Kenaikan suhu (°C)

Increase in temperature (°C)

Penghitungan / Calculation:

[Anggapan: Muatan haba tentu larutan = Muatan haba tentu air = 4.2 J g–1 oC–1;

Ketumpatan larutan = Ketumpatan air = 1 g cm–3

Assumption: Specific heat capacity of the solutions = Specific heat capacity of water = 4.2 J g–1 oC–1; Density of the solutions =
Density of water = 1 g cm–3]

Set I II III IV

Campuran bahan tindak balas HCl CH3COOH HCl CH3COOH
+ + + +
Reactant mixture
NaOH NaOH NH3 NH3
Perubahan haba, Q = mcθ (J)
(50+50)(4.2)(6.7) (50+50)(4.2)(6.5) (50+50)(4.2)(6.0) (50+50)(4.2)(5.5)
Heat change, Q = mc θ (J) = 2814 = 2730 = 2520 = 2310

97 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 3 Termokimia

Bilangan mol asid atau alkali (1.0)(50) (1.0)(50) (1.0)(50) (1.0)(50)
1000 1000 1000 1000
Number of moles of acid or alkali
= 0.05 mol = 0.05 mol = 0.05 mol = 0.05 mol
Bilangan mol H2O yang terbentuk
= Bilangan mol asid atau alkali, n 0.05 mol 0.05 mol 0.05 mol 0.05 mol

Number of moles of H2O formed 2814 2730 2520 2310
= Number of moles of acid or alkali, n 0.05 0.05 0.05 0.05
= –56280 J mol–1 = –54600 J mol–1 = –50400 J mol–1 = –46200 J mol–1
Haba peneutralan, ∆H = Q
n = –56.28 kJ mol–1 = –54.6 kJ mol–1 = –50.4 kJ mol–1 = –46.2 kJ mol–1
Q
Heat of neutralisation, ∆H = n

[Catatan: Terdapat peningkatan suhu semasa tindak balas peneutralan; haba dibebaskan ke persekitaran. Oleh itu,
eksotermik negatif(–)
tindak balas peneutralan ialah tindak balas dan ∆H bernilai .

Remarks: Since there is an increase in temperature during the neutralisation reaction, heat is released to the surroundings. Thus,
exothermic negative(–)
the neutralisation reaction is an reaction and ∆H is .]

Set Campuran bahan tindak balas Persamaan kimia seimbang untuk tindak balas peneutralan

Reactant mixture Balanced chemical equation for neutralisation reaction

I HCl + NaOH HCl(ak) + NaOH(ak) → NaCl(ak) + H2O(ce)

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

II CH3COOH + NaOH CH3COOH(ak) + NaOH(ak) → CH3COONa(ak) + H2O(ce)

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

BAB 3 III HCl + NH3 HCl(ak) + NH4+(ak) + OH–(ak) → NH4Cl(ak) + H2O(ce)

HCl(aq) + NH4+(aq) + OH–(aq) → NH4Cl(aq) + H2O(l)

IV CH3COOH + NH3 CH3COOH(ak) + NH4+(ak) + OH–(ak) → CH3COONH4(ak) + H2O(ce)

CH3COOH(aq) + NH4+(aq) + OH–(aq) → CH3COONH4(aq) + H2O(l)

Set Campuran bahan tindak balas Persamaan ion untuk tindak balas peneutralan

Reactant mixture Ionic equation for neutralisation reaction

I HCl + NaOH H+(ak) + OH–(ak) → H2O(ce) / H+(aq) + OH–(aq) → H2O(l)

II CH3COOH + NaOH H+(ak) + OH–(ak) → H2O(ce) / H+(aq) + OH–(aq) → H2O(l)

III HCl + NH3 H+(ak) + OH–(ak) → H2O(ce) / H+(aq) + OH–(aq) → H2O(l)

IV CH3COOH + NH3 H+(ak) + OH–(ak) → H2O(ce) / H+(aq) + OH–(aq) → H2O(l)

Set Campuran bahan tindak balas Persamaan termokimia untuk tindak balas peneutralan

Reactant mixture Thermochemical equation for neutralisation reaction

I HCl + NaOH HCl(ak) + NaOH(ak) → NaCl(ak) + H2O(ce) ∆H = –56.28 kJ mol–1

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56.28 kJ mol–1

II CH3COOH + NaOH CH3COOH(ak) + NaOH(ak) → CH3COONa(ak) + H2O(ce) ∆H = –54.6 kJ mol–1

CH3COOH (aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) ∆H = –54.6 kJ mol–1

III HCl + NH3 HCl(ak) + NH4+(ak) + OH–(ak) → NH4Cl(ak) + H2O(ce) ∆H = –50.4 kJ mol–1

HCl(aq) + NH4+(aq) + OH–(aq) → NH4Cl(aq) + H2O(l) ∆H = –50.4 kJ mol–1

IV CH3COOH + NH3 CH3COOH(ak) + NH4+(ak) + OH–(ak) → CH3COONH4(ak) + H2O(ce)

∆H = –46.2 kJ mol–1
CH3COOH(aq) + NH4+(aq) + OH–(aq) → CH3COONH4(aq) + H2O(l) ∆H = –46.2 kJ mol–1

© Penerbitan Pelangi Sdn. Bhd. 98

Kimia Tingkatan 5  Bab 3 Termokimia 

Discussion / Perbincangan:

1. Keempat-empat tindak balas peneutralan di atas ialah tindak balas eksotermik kerana haba

dibebaskan ke persekitaran. exothermic reactions because heat is released to the

All of the four neutralisation reactions above are

surroundings.

2. Asid etanoik ialah asid lemah dan larutan ammonia ialah alkali lemah . Asid

etanoik dan larutan ammonia mengion separa dalam air menyebabkan masih banyak asid etanoik dan
molekul
ammonia wujud dalam bentuk dalam air.
weak acid weak alkali
Ethanoic acid is a and ammonia solution is a . Ethanoic acid and
ammonia ionise partially in water causing most of the ethanoic acid and ammonia exist as molecules

in water.

3. Sebahagian haba yang dibebaskan semasa peneutralan diserap semula untuk mengion semua

molekul asid lemah dan molekul alkali lemah dengan lengkap dalam air untuk menghasilkan ion hidrogen
menurun
dan ion hidroksida masing-masing menyebabkan magnitud haba peneutralan .
reabsorbed ionise
Part of the heat released during neutralisation is to all weak acid and weak alkali
completely
molecules in water to produce hydrogen ions and hydroxide ions respectively causing the magnitude
drop
of heat of neutralisation to .

4. Lebih banyak haba dibebaskan apabila asid hidroklorik digunakan untuk meneutralkan larutan natrium
hidroksida kerana asid hidroklorik ialah asid monoprotik kuat yang mengion lengkap dalam air untuk
ion hidrogen
membentuk kepekatan yang lebih tinggi.

More heat is released when hydrochloric acid is used to neutralise sodium hydroxide solution because hydrochloric acid is a
strong monoprotic acid that ionises completely in water to form higher concentrations of hydrogen ions
.

5. Gambar rajah aras tenaga untuk keempat-empat tindak balas peneutralan: BAB 3

The energy level diagrams for the four neutralisation reactions:

Tenaga / Energy Tenaga / Energy
HCI(ak) + NaOH(ak) CH3COOH(ak) + NaOH(ak)
HCI(aq) + NaOH(aq) CH3COOH(aq) + NaOH(aq)

∆H = –56.28 kJ mol–1 ∆H = –54.6 kJ mol–1
NNaaCCI(I(aaqk) )++HH2O2(Ol)(ce) CH3COONa(ak) + H2O(ce)
CH3COONa(aq) + H2O(l)
Set 1
Set II
Tenaga / Energy
HCI(ak) + NH4+(ak) + OH–(ak) Tenaga / Energy
HCI(aq) + NH4+(aq) + OH–(aq) CH3COOH(ak) + NH4+(ak) + OH–(ak)
CH3COOH(aq) + NH4+(aq) + OH–(aq)

∆H = –50.4 kJ mol–1 ∆H = 46.2 kJ mol–1
NH4CI(ak) + H2O(ce) CH3COONH4(ak) + H2O(ak)
NH4CI(aq) + H2O(l) CH3COONH4(aq) + H2O(aq)

Set III Set IV

99 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 3 Termokimia

6. Haba peneutralan yang diperoleh daripada eksperimen lebih rendah daripada nilai teori kerana sebahagian
haba:

The heat of neutralisation obtained from the experiment is lower than the theoritical value because some of the heat:

•  hilang ke persekitaran / is lost to the surroundings

•  diserap oleh radas / is absorbed by the apparatus.

7. Langkah berjaga-jaga:

Precautionary steps:

(a) Asid dicampurkan dengan segera dan berhati-hati kepada alkali untuk mengelakkan tumpahan.

Acid is added immediately and carefully to alkali to prevent spillage.

(b) Suhu awal larutan diambil setelah 5 minit masa menunggu untuk memastikan larutan telah mencapai
suhu tetap
.
make sure the solution has
The initial temperature of the solution is taken after the 5-minute waiting period to
reached a constant temperature
.

(c) Kacau campuran dengan termometer sepanjang eksperimen untuk penyebaran haba yang sekata agar
suhu campuran yang sekata diperoleh .
Stir the mixture with a thermometer throughout the experiment for even heat distribution to obtain a uniform
temperature of the mixture
.

Kesimpulan / Conclusion:
Haba peneutralan bagi tindak balas antara asid kuat dengan alkali kuat adalah paling tinggi , diikuti dengan

haba peneutralan bagi tindak balas antara asid kuat dengan alkali lemah atau asid lemah dengan alkali kuat. Haba
peneutralan bagi tindak balas antara asid lemah dengan alkali lemah adalah paling rendah .
highest
The heat of neutralisation for the reaction between a strong acid and a strong alkali is the , followed by the heat of

BAB 3 neutralisation for the reaction between a strong acid and a weak alkali or a weak acid and a strong alkali. The heat of neutralisation
lowest
for the reaction between a weak acid and a weak alkali is the .

4Tugasan

Farah menjalankan eksperimen Set I dengan menambahkan 100 cm3 larutan kalium hidroksida 1.0 mol dm–3 pada
suhu 28.0oC kepada 100 cm3 asid hidroklorik 1.0 mol dm–3 pada suhu 29.0oC. Haba peneutralan antara larutan kalium
hidroksida dengan asid hidroklorik ialah –57 kJ mol–1.
Sementara dalam eksperimen Set II, Suzannah menambahkan 100 cm3 larutan kalium hidroksida 1.0 mol dm–3 pada
suhu 28.5oC kepada 100 cm3 asid sulfurik 0.5 mol dm–3 pada suhu 29.5oC. Bacaan termometer tertinggi yang dicapai ialah
42.5oC.
[Anggapan bahawa muatan haba tentu larutan = muatan haba tentu air = 4.2 J g–1 oC–1; ketumpatan larutan = ketumpatan
air = 1 g cm–3]
Farah carried out Set I experiment by adding 100 cm3 of 1.0 mol dm–3 potassium hydroxide solution at 28.0oC to 100 cm3 of 1.0 mol dm–3
hydrochloric acid at 29.0OC. The heat of neutralisation between potassium hydroxide solution and hydrochloric acid is – 57 kJ mol–1.
In Set II experiment, Suzannah adds 100 cm3 of 1.0 mol dm–3 potassium hydroxide solution at 28.5oC to 100 cm3 of 0.5 mol dm–3 sulphuric acid
at 29.5oC. The highest thermometer reading achieved is 42.5oC.
[Assume that specific heat capacity of solutions = specific heat capacity of water = 4.2 J g–1 oC–1; density of the solutions = density of water
= 1 g cm–3].

(a)  Eksperimen Set I / Set I experiment (b)  Eksperimen Set II / Set II experiment

(i) Tulis persamaan kimia seimbang. TP 2 (i) Tulis persamaan kimia seimbang. TP 2
Write a balanced chemical equation. Write a balanced chemical equation.
HCl(ak) + KOH(ak) → KCl(ak) + H2O(ce) H2SO4(ak) + 2KOH(ak) → K2SO4(ak) + 2H2O(ce)
HCl(aq) + KOH(aq) → KCl(aq) + H2O(l) H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)

© Penerbitan Pelangi Sdn. Bhd. 100

Kimia Tingkatan 5  Bab 3 Termokimia 

(ii) Hitung bilangan mol untuk kedua-dua larutan asid (ii) Hitung bilangan mol untuk kedua-dua larutan asid
dan alkali. / Calculate the number of moles for both acid and dan alkali. / Calculate the number of moles for both acid and

alkaline solution. TP 3 alkaline solution. TP 3

Bilangan mol HCl / Number of moles of HCl Bilangan mol H2SO4 / Number of moles of H2SO4

n= (1.0)(100) n= (0.5)(100)
1000 0.05
= 0.1 mol = 0.05 mol

Bilangan mol KOH / Number of moles of KOH Bilangan mol KOH / Number of moles of KOH

n= (1.0)(100) n= (1.0)(100)
1000 1000
= 0.1 mol = 0.1 mol

(iii) Hitung perubahan suhu untuk tindak balas di atas.

TP 3

Calculate the temperature change for the above reaction.

Purata bacaan awal termometer
Average initial thermometer reading

(iii) Hitung perubahan haba bagi tindak balas peneutralan = 28.5 + 29.5
2
di atas. TP 3 = 29.0 oC
Calculate the heat change for the neutralisation reaction
above. Perubahan suhu /Temperature change, θ
Q = 42.5 – 29.0
Gunakan formula ΔH = n = 13.5 oC BAB 3
Apply the formula

Perubahan haba / Heat change, Q (iv) Hitung perubahan haba bagi tindak balas peneutralan
= 57 × 1000 × 0.1 di atas. TP 3
= 5700 J
Calculate the heat change for the neutralisation reaction
above.

Perubahan haba/ Heat change,
Q = mcθ
= (100 + 100)(4.2)(13.5)
(iv) Hitung bacaan termometer tertinggi dalam tindak = 11340 J

balas peneutralan di atas. TP 3
Calculate the highest thermometer reading for the
neutralisation reaction above.
(v) Hitung haba peneutralan antara asid sulfurik dengan
Perubahan haba/ Heat change, larutan kalium hidroksida. TP 3
Q = mcθ
5700 = (100 + 100) (4.2) θ Calculate the heat of neutralisation between sulphuric acid
θ = 6.8 oC and potassium hydroxide solution.

Purata bacaan termometer awal Bilangan mol H2O terbentuk
Average initial thermometer reading Number of moles of H2O formed
= bilangan mol KOH / number of moles of KOH
= 28.0 + 29.0 = 28.5 oC = 0.1 mol
2
Bacaan termometer tertinggi Gunakan formula / Apply the formula ΔH = Q
Highest thermometer reading Haba peneutralan / Heat of neutralisation n

= 28.5 + 6.8 = 35.3 oC = 11340
0.1
= –113400 J mol–1
= –113.4 kJ mol–1

101 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 3 Termokimia

(v) Lukis gambar rajah aras tenaga untuk tindak balas di (vi) Lukis gambar rajah aras tenaga untuk tindak balas di
atas. TP 3 atas. TP 3

Draw the energy level diagram for the above reaction. Draw the energy level diagram for the above reaction.
Tenaga / Energy Tenaga / Energy

KOH(ak) + HCI(ak) 2KOH(ak) + H2SO4(ak)
KOH(aq) + HCI(aq) 2KOH(aq) + H2SO4(aq)

∆H = –57 kJ mol–1 ∆H = –113.4 kJ mol–1
KCI(ak) + H2O(ce) K2SO4(ak) + 2H2O(ce)
KCI(aq) + H2O(l) K2SO4(aq) + 2H2O(l)

(c) Bandingkan dan terangkan perbezaan dalam haba peneutralan antara Set I dengan Set II. TP 4
Compare and explain the difference in the heat of neutralisation between Set I and Set II.

(i) Haba peneutralan bagi Set II adalah dua kali ganda haba peneutralan bagi Set I.
The heat of neutralisation for Set II is double of the heat of neutralisation for Set I.

(ii) Dalam Set I, asid hidroklorik ialah asid monoprotik yang mengion lengkap dalam air untuk
membentuk 1 mol ion hidrogen.
In Set I, hydrochloric acid is a monoprotic acid that ionises completely in water to form 1 mole of hydrogen ions.

1 mol molekul air terbentuk pada akhir tindak balas peneutralan dengan pembebasan 57000 J
tenaga haba.

BAB 3 1 mole of water molecules is formed at the end of the neutralisation reaction with a release of 57000 J of heat energy.

(iii) Dalam Set II, asid sulfurik ialah asid diprotik yang mengion lengkap dalam air untuk membentuk 2
mol ion hidrogen, iaitu 2 kali ganda kepekatan ion hidrogen berbanding dengan Set I.
In Set II, sulphuric acid is a diprotic acid that ionises completely in water to form 2 moles of hydrogen ions, or double

concentration of hydrogen ions of Set I.

2 mol molekul air terbentuk pada akhir tindak balas peneutralan dengan pembebasan

dua kali ganda tenaga haba.
2 moles of water molecules are formed at the end of the neutralisation reaction with a release of double
amount of heat energy.

(d) Nyatakan definisi secara operasi untuk haba peneutralan dalam Set II. TP 3
State the operational definition for the heat of neutralisation in Set II.
Suhu meningkat apabila 2 mol air terbentuk dalam tindak balas peneutralan antara asid sulfurik dengan
larutan kalium hidroksida.
The temperature rises when 2 moles of water are formed in the neutralisation reaction between sulphuric acid and
potassium hydroxide solution.

Haba pembakaran

Heat of combustion

1. Gas oksigen diperlukan untuk tindak balas pembakaran berlaku. TP 1

Oxygen gas is needed for combustion reaction to occur.

2. Dalam pembakaran lengkap, gas oksigen berlebihan diperlukan untuk menghasilkan gas karbon dioksida
dan air . TP 1
water .
In a complete combustion, excess oxygen gas is needed to produce carbon dioxide gas and

3. Dalam pembakaran tidak lengkap, hasil yang terbentuk ialah gas karbon monoksida atau karbon

dan air . TP 1 carbon and water

In an incomplete combustion, products formed are carbon monoxide gas or .

© Penerbitan Pelangi Sdn. Bhd. 102

Kimia Tingkatan 5  Bab 3 Termokimia 

4. Alkohol , alkana dan alkena ialah tiga contoh bahan api yang

boleh digunakan dalam tindak balas pembakaran. alkenes
Alcohols
, alkanes and are three examples of combustion fuels

that can be used in combustion reactions. TP 1

5. Haba pembakaran ialah haba yang dibebaskan apabila 1 mol bahan api dibakar lengkap dalam gas oksigen
yang berlebihan . TP 1

The heat of combustion is the heat released when 1 mole of combustion fuel is burnt completely in excess oxygen gas .

6. Tulis persamaan kimia seimbang untuk 1 mol bahan api berikut terbakar dalam gas oksigen berlebihan. TP 2

Write the balanced chemical equations for 1 mole of the following fuels burnt in excess oxygen gas.

(a) Karbon C(p) + O2(g) → CO2(g)

Carbon C(s) + O2(g) → CO2(g)

(b) Etanol C2H5OH(ce) + 3O2(g) → 2CO2(g) + 3H2O(ce)

Ethanol C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

(c) Pentana C6H12(ce) + 8O2(g) → 5CO2(g) + 6H2O(ce)

Pentane C6H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l)

(d) Butena C4H8(ce) + 6O2(g) → 4CO2(g) + 4H2O(ce)

Butene C4H8(l) + 6O2(g) → 4CO2(g) + 4H2O(l)

Eksperimen 3.3 Membandingkan Haba Pembakaran bagi Alkohol dengan
Bilangan Atom Karbon yang Berlainan per Molekul

Comparing the Heat of Combustion for Alcohols with Different Number of Carbon
Atoms per Molecule

Tujuan / Aim: BAB 3

Untuk membandingkan haba pembakaran bagi alkohol dengan bilangan atom karbon yang berlainan per molekul.

To compare the heat of combustion for alcohols with different number of carbon atoms per molecule.

Pernyataan masalah / Problem statement:

Bagaimanakah alkohol dengan bilangan atom karbon yang berlainan per molekul mempengaruhi haba pembakaran?

How do alcohols with different number of carbon atoms per molecule affect the heat of combustion?

Hipotesis / Hypothesis:
Semakin tinggi bilangan atom karbon per molekul alkohol, semakin tinggi haba pembakaran .
the higher the heat of combustion
The higher the number of carbon atoms per alcohol molecule, .

Pemboleh ubah / Variables:
Dimanipulasikan / Manipulated:

Alkohol dengan bilangan atom karbon yang berlainan per molekul

Alcohols with different number of carbon atoms per molecule

Bergerak balas / Responding:
Haba pembakaran / Heat of combustion

Dimalarkan / Constant: Pengadang angin Termometer
Isi padu air, tin kuprum / Volume of water, copper can Windshield Thermometer

Bahan / Materials: Tin kuprum 200 cm3 air
Metanol, etanol, propanol, butanol dan air. Copper can 200 cm3 of water
Tungku kaki tiga
Methanol, ethanol, propanol, butanol and water. Bongkah kayu Tripod stand
Wooden block Lampu spirit + metanol
Radas / Apparatus: Spirit lamp + methanol
Tin kuprum, termometer, lampu spirit, pengadang
angin, blok kayu, tungku kaki tiga, penimbang
elektronik dan silinder penyukat 50 cm3.

Copper can, thermometer, spirit lamp, windshield, wooden
block, tripod stand, electronic balance and 50 cm3 measuring
cylinder.

103 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 3 Termokimia

Prosedur / Procedures:
1. 200 cm3 air disukat dan dituangkan ke dalam tin kuprum dan dibiarkan selama 5 minit.

200 cm3 of water is measured and poured into a copper can and left for 5 minutes.

2. Tin kuprum itu diletakkan di atas tungku kaki tiga.

The copper can is placed on a tripod stand.

3. Suhu awal air diukur dan direkodkan dalam jadual.

The initial temperature of water is measured dan recorded in the table.

4. Lampu spirit diisi dengan metanol dan jisim awalnya diukur sebagai m1.

A spirit lamp is filled with methanol and its initial mass is measured as m1.

5. Lampu spirit berisi metanol diletakkan di atas blok kayu di bawah tin kuprum.

The spirit lamp with methanol is placed on a wooden block under the copper can.

6. Pengadang angin diletakkan di sekeliling tin kuprum dan lampu spirit.

A windshield is placed around the copper can and spirit lamp.

7. Sumbu lampu spirit dinyalakan untuk memanaskan air di dalam tin kuprum.

The wick in the spirit lamp is lighted to heat the water in the copper can.

8. Air di dalam tin kuprum dikacau sepanjang eksperimen sehingga suhu akhir air mencapai 70oC.

The water in the copper can is stirred throughout the experiment until the final temperature of water achieves 70oC.

9. Nyalaan lampu spirit dipadamkan dan jisim akhir lampu spirit bersama metanol diukur sebagai m2.

The flame of the spirit lamp is put out and the final mass of the spirit lamp with the methanol is measured again as m2.

10. Langkah 1 – 9 diulangi dengan etanol, propanol dan butanol masing-masing.

Steps 1 – 9 are repeated with ethanol, propanol and butanol respectively.

Penjadualan data / Tabulation of data:

Set I II III IV

BAB 3 Alkohol Metanol Etanol Propanol Butanol

Alcohol Methanol Ethanol Propanol Butanol

Suhu awal air (oC) / Initial temperature of water (oC) 28.0 29.0 28.0 28.5

Suhu akhir air (oC) / Final temperature of water (oC) 70.0 70.0 70.0 70.0

Perubahan suhu (oC) / Temperature change (oC) 42.0 41.0 42.0 41.5

Jisim awal lampu spirit + alkohol, m1 (g) 165.16 160.50 161.30 155.83
163.88 159.49 160.40 155.09
Initial mass of spirit lamp + alcohol, m1 (g)
1.28 1.01 0.90 0.74
Jisim akhir lampu spirit + alkohol, m2 (g)

Final mass of spirit lamp + alcohol, m2 (g)

Jisim alkohol yang digunakan = m1 – m2 (g)

Mass of alcohol used = m1 – m2 (g)

Penghitungan / Calculation:
[Muatan haba tentu air = 4.2 J g–1 oC–1; Ketumpatan air = 1 g cm–3; Jisim atom relatif: H = 1, C = 12, O = 16

Specific heat capacity of water = 4.2 J g–1 oC–1; Density of water = 1 g cm–3; Relative atomic mass: H = 1, C = 12, O = 16]

Set I II III IV

Alkohol Metanol Etanol Propanol Butanol

Alcohol Methanol Ethanol Propanol Butanol

Perubahan haba (200)(4.2)(42.0) (200)(4.2)(41.0) (200)(4.2)(42.0) (200)(4.2)(41.5)
= 35280 = 34440 = 35280 = 34860
Heat change
Q = mc θ(J) 1.28 1.01 0.90 0.74
32 46 60 74
Bilangan mol alkohol, n = 0.040 mol = 0.022 mol = 0.015 mol = 0.010 mol

Number of moles of alcohol, n 34440 35280 34860
0.022 0.015 0.010
Haba pembakaran 35280 = –1565454.5 J mol–1 = –2352000 J mol–1 = –3486000 J mol–1
Heat of combustion 0.040 = –1565.45 kJ mol–1 = –2352.00 kJ mol–1 = –3486.00 kJ mol–1
Q = –882000 J mol–1
∆H = n = –882.00 kJ mol–1

© Penerbitan Pelangi Sdn. Bhd. 104

Kimia Tingkatan 5  Bab 3 Termokimia 

[Catatan: Terdapat peningkatan suhu air semasa tindak balas pembakaran; haba dibebaskan ke persekitaran. Oleh
itu, tindak balas pembakaran ialah tindak balas eksotermik dan ∆H bernilai negatif(–) .

Remarks: Since there is an increase in temperature during the combustion reaction, heat is released to the surroundings. Thus, the
exothermic negative(-)
combustion reaction is an reaction and ∆H is .]

Set Alkohol Persamaan termokimia untuk tindak balas pembakaran

Alcohol Thermochemical equation for combustion reactions

I Metanol CH3OH(ce) + 32O2(g) → CO2(g) + 2H2O(ce) ∆H = –882.00 kJ mol–1

Methanol CH3OH(l) + 3 O2(g) → CO2(g) + 2H2O(l) ∆H = –882.00 kJ mol–1
2
II Etanol
C2H5OH(ce) + 3O2(g) → 2CO2(g) + 3H2O(ce) ∆H = –1565.45 kJ mol–1
Ethanol
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ∆H = –1565.45 kJ mol–1
III Propanol
C3H7OH(ce) + 92O2(g) → 3CO2(g) + 4H2O(ce) ∆H = –2352.00 kJ mol–1
Propanol
C3H7OH(l) + 9 O2(g) → 3CO2(g) + 4H2O(l) ∆H = –2352.00 kJ mol–1
IV Butanol 2

Butanol C4H9OH(ce) + 6O2(g) → 4CO2(g) + 5H2O(ce) ∆H = –3486.00 kJ mol–1

C4H9OH(l) + 6O2(g) → 4CO2(g) + 5H2O(l) ∆H = –3486.00 kJ mol–1

Gambar rajah aras tenaga untuk keempat-empat tindak balas pembakaran:

Energy level diagrams for the four combustion reactions:

Tenaga / Energy Tenaga / Energy

CCHH33OOHH(l)(c+e)23 +O2(23gO) 2(g) C2H5OH(ce) + 3O2(g)
C2H5OH(l) + 3O2(g)

∆H = –882 kJ mol–1 ∆H = –1565.45 kJ mol–1 BAB 3
CO2(g) + 2H2O(ce) 2CO2(g) + 3H2O(ce)
CO2(g) + 2H2O(l) 2CO2(g) + 3H2O(l)

Gambar rajah aras tenaga untuk pembakaran Gambar rajah aras tenaga untuk pembakaran
metanol etanol

Energy level diagram for combustion of methanol Energy level diagram for combustion of ethanol
Tenaga / Energy
Tenaga / Energy
CC33HH77OOHH(l()c+e92) +O292(gO) 2(g) C4H9OH(ce) + 6O2(g)
C4H9OH(l) + 6O2(g)

∆H = –2352 kJ mol–1 ∆H = –3486.00 kJ mol–1
3CO2(g) + 4H2O(ce) 4CO2(g) + 5H2O(ce)
3CO2(g) + 4H2O(l) 4CO2(g) + 5H2O(l)

Gambar rajah aras tenaga untuk pembakaran Gambar rajah aras tenaga untuk pembakaran
propanol butanol

Energy level diagram for combustion of propanol Energy level diagram for combustion of butanol

Perbincangan / Discussion:

1. Tindak balas pembakaran methanol, etanol, propanol dan butanol dalam gas oksigen berlebihan ialah tindak
eksotermik
balas kerana haba dibebaskan ke persekitaran.
exothermic
The combustion reactions of methanol, ethanol, propanol and butanol in excess oxygen gas are all

reactions because heat is released to the surroundings.

105 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 3 Termokimia

2. Lebih banyak haba dibebaskan apabila butanol dibakar dalam gas oksigen berlebihan kerana butanol
bilangan atom karbon per molekul yang lebih tinggi
mempunyai .
a higher number of
More heat is released when butanol is burned in excess oxygen gas because butanol has
carbon atoms per molecule
.

3. Semakin tinggi bilangan atom karbon dan atom hidrogen per molekul alkohol, semakin tinggi nilai

haba pembakaran alkohol itu kerana lebih banyak molekul karbon dioksida dan air dihasilkan dan
lebih banyak haba terbebas .

The higher the number of carbon and hydrogen atoms per alcohol molecule, the higher the value of heat of combustion of
more heat is released
alcohol because more carbon dioxide and water molecules are produced and .

4. Graf haba pembakaran alkohol melawan bilangan atom karbon per molekul alkohol ditunjukkan di bawah.

Graph of heat of combustion of alcohols against the number of carbon atoms per alcohol molecule is shown below.

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

3500

x 3486.00

3000

2500 x 2352.00

2000 x 1565.45
1500

BAB 3 1000 x 882
500

O 1234
Bilangan atom karbon per molekul
Number of carbon atoms per molecule

5. Haba pembakaran yang diperoleh melalui eksperimen lebih rendah daripada nilai teori kerana:

The heat of combustion obtained in the experiment is lower than the theoretical value because:
(a) Sebahagian haba yang terbebas telah hilang ke sekeliling .
is lost to the surroundings .
Some of the heat released

(b) Sebahagian haba yang terbebas diserap oleh tin kuprum .

Some of the heat released is absorbed by the copper can .

(c) Pembakaran alkohol yang tidak lengkap .
Incomplete
combustion of the alcohol.

6. Tin kuprum digunakan kerana kuprum ialah konduktor haba yang baik yang dapat mengalirkan haba
yang terbebas daripada pembakaran alkohol kepada air.

A copper can is used because copper is a good heat conductor which helps to transfer the heat released from the
combustion of the alcohol to water.

7. Langkah berjaga-jaga:

Precautionary steps:
Pengadang api
(a) digunakan untuk melindungi nyalaan pembakaran alkohol daripada

tiupan angin yang mempercepat kehilangan haba ke sekeliling.
Windshield
is used to protect the flame of alcohol combustion from wind blow which will

fasten the heat loss to the surroundings.

© Penerbitan Pelangi Sdn. Bhd. 106

Kimia Tingkatan 5  Bab 3 Termokimia 

(b) Naikkan lampu spirit menggunakan bongkah kayu sehingga nyalaan pembakaran alkohol bersentuhan
langsung dengan dasar tin kuprum untuk pemindahan haba pembakaran alkohol yang lebih berkesan
kepada air
.

Raise up the spirit lamp using the wooden block so that the combustion flame of alcohol can contact directly with the
more efficient transfer of heat of alcohol combustion to water
bottom of the copper can for .

(c) Kasa dawai tidak digunakan dalam eksperimen ini agar tidak menyerap sebahagian haba yang terbebas
daripada pembakaran alkohol
.
Wire gauze is not used in this experiment so that it will not absorb part of the heat released from alcohol combustion .

(d) Lampu spirit ditimbang serta-merta selepas nyalaan pembakaran alkohol dipadamkan untuk mendapatkan
jisim alkohol yang dibakar yang lebih tepat kerana alkohol adalah sangat meruap .

The spirit lamp is weighed immediately after the combustion flame of the alcohol is put out in order to get a more accurate
the alcohol is very volatile
mass of the alcohol burnt because .

(e) Air dalam tin kuprum hendaklah dikacau sepanjang eksperimen supaya suhu air adalah sekata .

The water in the copper can must be stirred throughout the experiment to get a uniform temperature of water .

Kesimpulan / Conclusion:
Semakin tinggi bilangan atom karbon per molekul alkohol, semakin tinggi nilai haba pembakaran alkohol.

The higher the number of carbon atoms per alcohol molecule, the higher the value of heat of combustion of alcohol.

5Tugasan

Danish membakar 1.341 g serbuk glukosa (C6H12O6) dengan lengkap untuk menghasilkan gas karbon dioksida dan air. BAB 3
Haba yang terbebas menaikkan suhu 500 cm3 air sebanyak 10 oC.
[Muatan haba tentu air = 4.2 J g–1 oC–1; Ketumpatan air = 1 g cm–3; Jisim atom relatif: H = 1, C = 12, O = 16]
Danish burned 1.341 g of glucose powder (C6H12O6) completely to form carbon dioxide gas and water. The heat released raised up the
temperature of 500 cm3 of water by 10 oC.
[Specific heat capacity of water = 4.2 J g–1 oC–1; Density of water = 1 g cm–3; Relative atomic mass: H = 1, C = 12, O = 16]

(a) Tulis persamaan kimia seimbang. TP 2 (c) Hitung bilangan mol glukosa yang terbakar.
Write a balanced chemical equation. Calculate the number of moles of glucose burned.
C6H12O6(p) + 6O2(g) → 6CO2(g) + 6H2O(ce)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) n= 1.341
[6(12) + 12(1) + 6(16)]

= 0.00745 mol

(b) Hitung haba yang dibebaskan. TP 3 (d) Hitung haba pembakaran glukosa
Calculate the heat released. Calculate the heat of combustion of glucose.
Q = mcθ
= (500)(4.2)(10) Haba pembakaran C6H12O6
Heat of combustion of C6H12O6
= 21000 J 21
= 21 kJ = 0.00745

= –2818.79 kJ mol–1

107 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 3 Termokimia (f ) Lukis gambar rajah aras tenaga untuk tindak balas di
atas. TP 3
(e) Tulis persamaan termokimia. TP 2
Write the thermochemical equation. Draw the energy level diagram for the above reaction.
C6H12O6(p) + 6O2(g) → 6CO2(g) + 6H2O(ce) Tenaga / Energy
ΔH = –2818.79 kJ mol–1
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) C6H12O6(p) + 6O2(g)
ΔH = –2818.79 kJ mol–1 C6H12O6(s) + 6O2(g)

∆H = –2818.79 kJ mol–1
6CO2(g) + 6H2O(ce)
6CO2(g) + 6H2O(l)

3.3 Aplikasi Tindak Balas Endotermik dan Eksotermik dalam Kehidupan Harian
Application of Endothermic and Exothermic Reactions in Daily Life

1. Kelaskan aplikasi harian berikut kepada tindak balas eksotermik dan tindak balas endotermik. TP 2

Classify the following everyday applications into exothermic reaction and endothermic reaction.

BAB 3 Pek sejuk Menggoreng telur Fotosintesis Pengaratan

Cold pack Frying eggs Photosynthesis Rusting

Tindak balas termit Pek panas Membakar kertas Mencairkan ketulan ais

Thermite reaction Hot pack Burning paper Melting ice cubes

Tindak balas eksotermik Tindak balas endotermik

Exothermic reaction Endothermic reaction

• Pengaratan • Pek sejuk

Rusting Cold pack

• Tindak balas termit • Fotosintesis

Thermite reaction Photosynthesis

• Pek panas • Mencairkan ketulan ais

Hot pack Melting ice cubes

• Membakar kertas • Menggoreng telur

Burning paper Frying eggs

2. Bahan api ialah bahan yang diperoleh daripada Bumi yang mudah dibakar untuk membebaskan tenaga

haba dan digunakan oleh manusia untuk melakukan banyak pekerjaan berguna. TP 1

A fuel is a substance obtained from Earth that is easily burnt to release heat energy and used by humans to do
many useful works.

3. Bahan api mengandungi tenaga kimia yang dapat diukur sebagai nilai bahan api . TP 1
fuel value
Fuels contain the chemical energy that can be measured as .

4. Nilai bahan api ditakrifkan sebagai jumlah tenaga haba yang dibebaskan apabila 1 g bahan api dibakar
dengan lengkap dalam gas oksigen yang berlebihan
. TP 1

Fuel value is defined as the amount of heat energy released when 1 g of fuel is completely burned in excess oxygen gas .

5. Unit untuk nilai bahan api ialah kJ g–1 . TP 1

The unit for fuel value is kJ g–1 .

6. Tandakan (✓) pada ciri-ciri bahan api yang baik. TP 2
Tick (✓) the characteristics of a good fuel.

© Penerbitan Pelangi Sdn. Bhd. 108

Kimia Tingkatan 5  Bab 3 Termokimia 

Ciri-ciri bahan api ✓

Characteristics of fuels

(a) Menghasilkan sejumlah besar tenaga haba apabila terbakar dengan lengkap.

Produces a large amount of heat energy when burnt completely.

(b) Mengakibatkan pencemaran udara.

Causes air pollution.

(c) Murah dan mudah didapati. ✓

Cost-friendly and easily available.

(d) Boleh dibakar dengan mudah. ✓

Can be burnt easily.

(e) Dapat diangkut dengan mudah. ✓

Can be transported easily.

(f) Dapat disimpan dengan mudah tetapi dengan kos yang tinggi.

Can be stored easily but with high costs.

6Tugasan

Haba pembakaran gas hidrogen ialah –286 kJ mol–1 manakala haba pembakaran metanol ialah –728 kJ mol–1. BAB 3
The heat of combustion of hydrogen gas is –286 kJ mol–1 while the heat of combustion of metanol is –728 kJ mol–1.

(a) Hitungkan nilai bahan api untuk gas hidrogen dan metanol. TP 3
Calculate fuel value of hydrogen gas and methanol.

Gunakan formula / Use the formula
Nilai bahan api / Fuel value

= Haba pembakaran / Heat of combustion
Jisim molar / molas mass

Nilai bahan api gas hidrogen
Fuel value of hydrogen gas

= 286 = 143 kJ g–1
2(1)

Nilai bahan api metanol
Fuel value of methanol

= 728 = 22.75 kJ g–1
[12 + 3(1) + 16 + 1]

(b) Daripada jawapan anda di (a), bahan api yang manakah adalah lebih baik? Wajarkan jawapan anda. TP 5
From your answers in (a), which is a better fuel? Justify your answer.
Gas hidrogen ialah bahan api yang lebih baik kerana mempunyai nilai bahan api yang lebih tinggi.
Pembakaran lengkap 1 g gas hidrogen menghasilkan lebih banyak haba berbanding dengan pembakaran lengkap
1 g metanol.
Hydrogen gas is a better fuel because it has a higher fuel value. The complete combustion of 1 g of hydrogen gas produces
more heat compared to the complete combustion of 1 g of methanol.


109 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 3 Termokimia SPM 3

PRAKTIS

Soalan Objektif A Haba diserap dalam tindak balas.
Heat is absorbed in the reaction.
1. Brandon menambahkan satu spatula serbuk natrium
hidroksida ke dalam bikar berisi 100 cm3 air suling. B Merupakan tindak balas endotermik.
Apabila serbuk natrium hidroksida larut dalam air, It is an endothermic reaction.
bikar menjadi panas. Apakah sebabnya?
C Jumlah kandungan tenaga bahan tindak balas
Brandon adds a spatula of sodium hydroxide powder into adalah lebih rendah daripada jumlah kandungan
a beaker with 100 cm3 of distilled water. When the sodium tenaga hasil tindak balas.
hydroxide powder dissolves in water, the beaker turns hot.
What is the reason? Total energy content of the reactants is lower than total
A Haba diserap daripada air. energy content of the products.
Heat is absorbed from the water.
B Haba dibebaskan ke dalam air. D Campuran di dalam kelalang kon menyebabkan
Heat is released into the water. kelalang kon menjadi panas.
C Bikar bukan penebat haba yang baik.
Beaker is not a good heat insulator. Mixture in the conical flask makes the conical flask
hot.
2. Antara yang berikut, yang manakah bukan tindak
BAB 3 balas eksotermik? 4. Antara bahan berikut, yang manakah boleh digunakan
dalam pek sejuk?
Which of the following is not an exothermic reaction?
A C2H5OH + 3O2 → 2CO2 + 3H2O 2011 Which of the following substances can be used in a cold pack?
B NaOH + HCl → NaCl + H2O
C 2Cu(NO3)2 → 2CuO + 4NO2 + O2 A Natrium hidroksida / Sodium hydroxide
D Mg + CuSO4 → MgSO4 + Cu B Magnesium klorida / Magnesium chloride
C Kalium nitrat / Potassium nitrate
3. Rajah 1 menunjukkan gambar rajah aras tenaga bagi D Natrium klorida / Sodium chloride
tindak balas penyesaran antara zink dengan larutan
kuprum(II) sulfat. 5. Berikut menunjukkan persamaan termokimia untuk
tindak balas pemendakan antara ion plumbum(II)
Diagram 1 shows the energy level diagram for the displacement dengan ion iodida.
reaction between zinc and copper(II) sulphate solution.
The following shows a thermochemical equation for the
Tenaga / Energy precipitation reaction between lead(II) ion and iodide ion.
Zn(p) + Cu2+(ak)
Zn(s) + Cu2+(aq) Pb2+(ak) + 2I–(ak) → PbI2(p) ΔH = –y kJ mol–1
Pb2+(aq) + 2I–(aq) → PbI2(s) ΔH = –y kJ mol–1
∆H = –210 kJ mol–1
Pernyataan yang manakah benar mengenai
Zn2+(ak)+ Cu(p) persamaan di atas?
Zn2+(aq)+ Cu(s)
Which of the following statements is true about the above
Rajah 1 / Diagram 1 equation?
A Heat is absorbed from the surroundings.
Antara yang berikut, yang manakah dapat dirumuskan Haba diserap dari persekitaran.
daripada Rajah 1? B Mendakan kuning terbentuk.
Yellow precipitate is formed.
Which of the following can be deduced from Diagram 1? C Bacaan termometer akhir lebih rendah daripada
bacaan termometer awal.
The final thermometer reading is lower than the initial
thermometer reading.
D Tindak balas endotermik berlaku.
An endothermic reaction occurs.

© Penerbitan Pelangi Sdn. Bhd. 110

Kimia Tingkatan 5  Bab 3 Termokimia 

6. Tindak balas penyesaran logam berlaku apabila III Bacaan termometer akhir lebih tinggi daripada
satu spatula serbuk magnesium ditambahkan ke bacaan termometer awal.
dalam 100 cm3 larutan ferum(II) klorida 0.1 mol dm–3.
Diketahui bahawa haba penyesaran ialah –200 kJ Final thermometer reading is higher than initial
mol–1, hitung perubahan suhu. thermometer reading.

[Muatan haba tentu larutan = 4.2 J g–1 °C–1; Ketumpatan IV Melarutkan serbuk ammonium nitrat dalam air
larutan = 1 g cm–3] ialah tindak balas eksotermik.

A metal displacement reaction occurs when a spatula of Dissolving ammonium nitrate powder in water is an
magnesium powder is added to 100 cm3 of 0.1 mol dm–3 iron(II) exothermic reaction.
chloride solution. Knowing that the heat of displacement is
–200 kJ mol–1, calculate the temperature change. A I dan II C II dan IV

[Specific heat capacity of solution = 4.2 J g–1°C–1; Density of I and II II and IV
solution = 1 g cm–3]
A 4.0 oC B I dan III D III dan IV
B 4.8 oC
C 8.0 oC I and III III and IV
D 9.6 oC
8. Apakah nilai bahan api? BAB 3
7. Antara pernyataan berikut, yang manakah benar What is a fuel value?
tentang tindak balas eksotermik dan endotermik?
A Perubahan haba apabila 1 mol bahan api dibakar
Which of the following statements are true about exothermic dengan lengkap.
and endothermic reactions?
I Tindak balas eksotermik ialah tindak balas kimia Heat change when 1 mole of fuel is burnt completely.
yang membebaskan haba ke persekitaran. B Jumlah tenaga haba yang dihasilkan apabila 1
An exothermic reaction is a chemical reaction that
releases heat to the surroundings. g bahan api dibakar dengan lengkap dalam gas
II Jumlah kandungan tenaga bahan tindak balas oksigen berlebihan.
lebih tinggi daripada jumlah kandungan tenaga The amount of heat energy produced when 1 g of a fuel is
hasil tindak balas. completely burnt in excess of oxygen.
Total energy content of the reactants is higher than total C Proses membakar benda hidup dan benda bukan
energy content of the products. hidup dalam gas oksigen yang berlebihan.
Process of burning both living and non-living things in
excessive oxygen gas.
D Tenaga haba yang dihasilkan apabila sebatian
karbon dibakar.
The heat produced when a carbon compound is burnt.

Soalan Struktur Tenaga / Energy

Bahagian A 2Hg(ce) + O2(g)
1. Rajah 1.1 dan Rajah 1.2 ialah dua gambar rajah aras tenaga. 2Hg(l) + O2(g)
Diagram 1.1 and Diagram 1.2 are two energy level diagrams.

Tenaga / Energy
Mg(p) + H2SO4(ak)
Mg(s) + H2SO4(aq)

∆H = –467 kJ mol–1 ∆H = +182 kJ mol–1
2HgO(p)
MgSO4(ak) + H2(g) 2HgO(s)
MgSO4(aq) + H2(g)
Rajah 1.2 / Diagram 1.2
Rajah 1.1 / Diagram 1.1
[1 markah / mark]
(a) Berdasarkan Rajah 1.1, / Based on Diagram 1.1,
(i) Nyatakan jenis tindak balas. / State the type of reaction.
Tindak balas eksotermik / Exothermic reaction

111 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 3 Termokimia

(ii) Ramalkan dua pemerhatian. / Predict two observations.
• Bacaan termometer akhir lebih tinggi daripada bacaan termometer awal.

Final thermometer reading is higher than initial thermometer reading.

• Gelembung gas tidak berwarna terbentuk. / Colourless gas bubbles are formed.

• Pepejal kelabu berkilat terlarut dalam larutan. / Shiny grey solid dissolves in solution.

(Mana-mana dua jawapan. / Any two answers.)
[2 markah / marks]

(iii) Nyatakan dua maklumat yang diperoleh daripada gambar rajah aras tenaga.
State two information obtained from the energy level diagram.

• Haba yang dibebaskan adalah lebih daripada haba yang diserap.

Heat released is more than heat absorbed.

• Jumlah kandungan tenaga Mg dan H2SO4 lebih tinggi daripada jumlah kandungan tenaga MgSO4 dan H2.

Total energy content of Mg and H2SO4 is higher than total energy content of MgSO4 and H2.

• 467 kJ haba dibebaskan apabila 1 mol Mg bertindak balas dengan 1 mol H2SO4 untuk membentuk 1 mol

MgSO4 dan 1 mol H2.

467 kJ heat is released when 1 mole of Mg reacts with 1 mole of H2SO4 to form 1 mole of MgSO4 and 1 mole of H2.

(Mana-mana dua jawapan. / Any two answers.)
[2 markah / marks]
BAB 3
(b) Berdasarkan Rajah 1.2, / Based on Diagram 1.2, [1 markah / mark]
(i) Nyatakan jenis tindak balas. / State the type of reaction.
Tindak balas endotermik / Endothermic reaction

(ii) Ramalkan dua pemerhatian. / Predict two observations.
• Bacaan termometer akhir lebih rendah daripada bacaan termometer awal.

Final thermometer reading is lower than initial thermometer reading.

• Gelembung gas tidak berwarna terbentuk. / Colourless gas bubbles are formed. [2 markah / marks]

(iii) Nyatakan dua maklumat yang diperoleh daripada gambar rajah aras tenaga.
State two information obtained from the energy level diagram.
• Haba yang diserap adalah lebih daripada haba yang dibebaskan.

Heat absorbed is more than heat released.

• Jumlah kandungan tenaga HgO lebih rendah daripada jumlah kandungan tenaga Hg dan O2.
Total energy content of HgO is lower than total energy content of Hg and O2.
• 182 kJ haba diserap apabila 2 mol HgO terurai untuk membentuk 2 mol Hg dan 1 mol O2.
182 kJ heat is absorbed when 2 moles of HgO decomposes to form 2 moles of Hg and 1 mole of O2.

(Mana-mana dua jawapan. / Any two answers.)
[2 markah / marks]

© Penerbitan Pelangi Sdn. Bhd. 112

Kimia Tingkatan 5  Bab 3 Termokimia 

Bahagian B

2. (a) Tindak balas kimia melibatkan pemecahan ikatan bahan tindak balas dan pembentukan ikatan hasil tindak balas.
Diketahui bahawa hidrogen klorida dapat dihasilkan daripada tindak balas antara gas hidrogen dengan gas
klorin,

[Diberi tenaga ikatan: H – H = 436 kJ mol–1; Cl – Cl = 242 kJ mol–1; H – Cl = 431 kJ mol–1]
Chemical reactions involve the breaking of bonds of the reactants and the formation of bonds of the products. Knowing that

hydrogen chloride can be produced from the reaction between hydrogen gas and chlorine gas,
[Given the bond energy: H – H = 436 kJ mol–1; Cl – Cl = 242 kJ mol–1; H – Cl = 431 kJ mol–1]

(i) Hitung perubahan tenaga keseluruhan.
Calculate the overall energy change.

[3 markah / marks]

(ii) Tuliskan persamaan termokimia seimbang.
Write a balanced thermochemical equation.

[2 markah / marks]

(iii) Lukis gambar rajah aras tenaga.
Draw an energy level diagram.

[2 markah / marks]

(iv) Kenal pasti sama ada pembentukan hidrogen klorida adalah tindak balas eksotermik atau endotermik.
Terangkan jawapan anda dari segi pemecahan dan pembentukan ikatan.

Identify whether formation of hydrogen chloride is an exothermic or endothermic reaction. Explain your answer in terms of
breaking and formation of bonds.
[3 markah / marks]

(b) Dalam suatu eksperimen, 25 cm3 larutan barium nitrat 1.0 mol dm–3 ditambahkan kepada 25 cm3 larutan natrium BAB 3
sulfat 1.0 mol dm–3 dalam cawan polistirena. Campuran dikacau dan bacaan termometer tertinggi dicatatkan.
Jadual 2 menunjukkan semua data yang diperoleh sepanjang eksperimen.

In an experiment, 25 cm3 of 1.0 mol dm–3 barium nitrate solution is added to 25 cm3 of 1.0 mol dm–3 sodium sulphate solution in a
polystyrene cup. The mixture solution is stirred and the highest thermometer reading is recorded. Table 2 shows all the data obtained
throughout the experiment.

Suhu awal larutan barium nitrat 28.4 oC
Initial temperature of barium nitrate solution

Suhu awal larutan natrium sulfat 28.2 oC
Initial temperature of sodium sulphate solution

Suhu tertinggi larutan campuran 33.3 oC
Highest temperature of the mixture solution

Jadual 2 / Table 2

(i) Apakah fungsi cawan polistirena? [1 markah / mark]
What is the function of the polystyrene cup?

(ii) Nyatakan satu pemerhatian untuk eksperimen di atas.
State an observation for the above experiment.

[1 markah / mark]

(iii) Tuliskan persamaan ion seimbang untuk tindak balas di atas. Kemudian, hitung haba pemendakan bagi
tindak balas ini.

Write a balanced ionic equation for the above reaction. Then, calculate the heat of precipitation for this reaction.
[6 markah / marks]

(iv) Nyatakan definisi secara operasi bagi haba pemendakan.
State the operational definition for the heat of precipitation.

[2 markah / marks]

113 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 3 Termokimia

Bahagian C
3. Rachel diberi serbuk ammonium nitrat dalam cawan polistirena A manakala serbuk natrium hidroksida dalam cawan

polistirena B. Dia menyukat dan menuangkan 100 cm3 air suling ke dalam setiap cawan polistirena.
Rachel is given ammonium nitrate powder in polystyrene cup A while sodium hydroxide powder in polystyrene cup B. She measured and

poured 100 cm3 of distilled water into each polystyrene cup.

Cik Chong : Rachel, masukkan sebatang termometer ke dalam setiap cawan polistirena dan kacaukan campuran
tersebut.

Miss Chong Rachel, insert a thermometer into each polystyrene cup and stir the mixture.
Rachel : Baiklah, cikgu.
Rachel Okay, teacher.
Cik Chong : Apakah pemerhatian kamu?
Miss Chong What is your observation?
Rachel : Bacaan termometer dalam cawan polistirena A menurun manakala bacaan termometer dalam

cawan polistirena B meningkat.
Rachel The thermometer reading in polystyrene cup A drops while the thermometer reading in polystyrene cup B rises.

BAB 3 Berdasarkan perbualan antara Cik Chong dengan Rachel, jawab soalan-soalan berikut.
Based on the conversation between Miss Chong and Rachel, answer the following questions.

(a) Nyatakan definisi secara operasi untuk kedua-dua tindak balas endotermik dan eksotermik.
State the operational definition for both endothermic and exothermic reactions.

[4 markah / marks]

(b) Tindak balas kimia melibatkan pemecahan ikatan kimia dalam bahan tindak balas dan pembentukan ikatan kimia
dalam hasil tindak balas. Dengan menggunakan bahan kimia yang sama yang digunakan oleh Rachel, terangkan
konsep tindak balas eksotermik dan endotermik dari segi pemecahan dan pembentukan ikatan kimia.

Chemical reactions involve breaking of chemical bonds in the reactants and formation of chemical bonds in the products. By using
the same chemical substances used by Rachel, explain the concept of exothermic and endothermic reactions in terms of breaking
and formation of chemical bonds.
[6 markah / marks]

(c) Asha dan Priyanka melakukan eksperimen untuk menentukan haba pembakaran heksena. Jadual 3 menunjukkan
semua maklumat yang direkodkan oleh Asha semasa eksperimen.

Asha and Priyanka carried out an experiment to determine the heat of combustion of hexene. Table 3 shows all the information
recorded by Asha during the experiment.

Jisim awal lampu pelita dengan heksena (g) / Initial mass of spirit lamp with hexene (g) 148.822
Jisim akhir lampu pelita dengan heksena (g) / Final mass of spirit lamp with hexene (g) 146.078
Isi padu air dalam tin kuprum (cm3) / Volume of water in the copper can (cm3)
Suhu awal air (oC) / Initial temperature of water (oC) 500
Suhu akhir air (oC) / Final temperature of water (oC) 30.0
80.0
Jadual 3 / Table 3

Hitungkan haba pembakaran bagi heksena. Sertakan persamaan kimia seimbang. Terangkan mengapa nilai
eksperimen yang dikira berbeza denga nilai teori.

[Muatan haba tentu larutan = 4.2 J g–1 oC–1; Ketumpatan larutan = 1.0 g cm–3]
Calculate the heat of combustion of hexene. Include a balanced chemical equation. Explain why the calculated experimental value

is different from the theoretical value.
[Specific heat capacity of water = 4.2 J g–1 oC–1; Density of water = 1.0 g cm–3]

[10 markah / marks]

Kuiz 3

© Penerbitan Pelangi Sdn. Bhd. 114

BAB Kimia Polimer

4 Polymer Chemistry

PETA Konsep Sumber polimer Semula jadi

Proses pempolimeran Sources of polymer Natural

Polymerisation process Sintetik

• Pempolimeran penambahan Synthetic

Addition polymerisation • Kegunaan

• Pempolimeran kondensasi Uses

Condensation polymerisation • Kesan-kesan negatif

POLIMER Negative effects
POLYMER

Plastik Gentian sintetik Elastomer

Plastic Synthetic fibre Elastomer

Termoplastik Nilon Getah sintetik

Thermoplastic Nylon Synthetic rubber

Termoset Getah asli

Thermoset Natural rubber

Pemvulkanan getah asli Penggumpalan lateks Penamaan monomer

Vulcanisation of natural rubber Coagulation of latex Naming of monomer

• Cara • Kesan asid • Formula struktur

Method Effect of acid Structural formula

• Ciri-ciri • Kesan alkali • Ciri-ciri

Properties Effect of alkali Properties

115 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 4 Kimia Polimer

4.1 Polimer
Polymer

1. Polimer ialah molekul besar yang terdiri daripada rangkaian unit molekul kecil berulang yang

dinamakan monomer . TP 1
molecule made from a chain of repeating units of small molecules known as monomers .
A polymer is a large

2. Monomer merupakan molekul-molekul kecil serupa yang terikat secara kimia untuk menghasilkan
polimer
yang besar dan berantai panjang.
small identical molecules
Monomers are that are bonded chemically to form a large and long-chain
polymer
.

3. Unit-unit monomer yang berulang terikat oleh ikatan kovalen .

The repeating units of monomers are joined together by covalent bonds .

4. Pempolimeran ialah tindak balas kimia apabila monomer-monomer bergabung menjadi polimer.

Polymerisation is a chemical reaction in which monomers join together to form a polymer. TP 2

5. Penyahpolimeran merujuk proses pemisahan molekul polimer berantai panjang kepada monomer-

monomer kecil.
Depolymersation refers to the process of
separation of a long-chain polymer into small monomers.

6. Polimer sintetik dihasilkan daripada petroleum , manakala polimer semula jadi diperoleh daripada
tumbuhan atau haiwan
.
Synthetic polymers are synthesised from , while natural polymers are derived from plants
petroleum
animals
or .

7. Selulosa dalam dinding sel tumbuhan terdiri daripada rangkaian unit glukosa yang berulang.

Cellulose in plant cell walls consists of repeating units of glucose .

8. Nilon dan terilena ialah contoh polimer sintetik.

Nylon and terylene are examples of synthetic fibre.

9. Contoh polimer semula jadi dan polimer sintetik: TP 2

Examples of natural polymers and synthetic polymers:

Polimer semula jadi / Natural polymers Polimer sintetik / Synthetic polymers

BAB 4 (a) Protein / Protein Termoplastik / Thermoplastic

(b) Karbohidrat / Carbohydrate Termoset / Thermoset

(c) Getah asli / Natural rubber Gentian sintetik / Synthetic fibres

(d) Sutera / Silk Elastomer / Elastomer

10. Ciri-ciri polimer:

Properties of polymers:

Ciri-ciri Jenis polimer
Properties Types of polymers

(a) Boleh dipanaskan semula beberapa kali sehingga mendapat acuan yang sempurna. Plastik

Can be reheated several times until a perfect mould is created. Plastic

(b) Polimer berantai panjang yang tahan regangan. Gentian sintetik

Long-chain polymers that withstand stretching. Synthetic fibre

(c) Kembali semula kepada bentuk asal setelah diregang atau ditekan. Elastomer

Regain its original shape after being stretched or pressed. Elastomer

© Penerbitan Pelangi Sdn. Bhd. 116

11. Pelbagai jenis plastik: / Different types of plastics: Kimia Tingkatan 5  Bab 4 Kimia Polimer 

Plastik / Plastic Monomer / Monomer
Etena
(a) Polietena (PE)
Ethene
Polyethene (PE)
Propena
(b) Polipropena (PP)
Propene
Polypropene (PP)
Kloroetena
(c) Polivinil klorida (PVC) / Polikloroetena
Chloroethene
Polyvinylchloride (PVC) / Polychloroethene
Feniletena
(d) Polistirena (PS)
Phenylethene
Polystyrene (PS)
Tetrafluoroetena
(e) Teflon / Politetrafluoroetilena (PTFE)
Tetrafluoroethene
Teflon / Polytetrafluoroethylene (PTFE)
Metil 2-metilpropenoat
(f) Polimetil metakrilat (PMMA) / Perspeks (PP)
Methyl 2-methylpropenoate
Polymethyl methacrylate (PMMA) / Perspex (PP)
(c)
12. Label rajah berikut sebagai termoplastik, termoset atau elastomer.

Label the following diagram as a thermoplastic, a thermoset or an elastomer.

(a) (b)

Termoplastik Elastomer Termoset

Thermoplastic Elastomer Thermoset

13. Perbandingan antara ciri termoplastik dengan termoset. TP 2 TP 3

Comparison between the properties of thermoplastics and thermosets.

Termoplastik Ciri-ciri Termoset

Thermoplastic Properties Thermoset

Pempolimeran penambahan Cara polimer dihasilkan Pempolimeran kondensasi BAB 4

Addition polymerisation Way of producing the polymer Condensation polymerisation

Tidak Kehadiran rangkai silang Ya

No Presence of cross-link Yes

Boleh diacu berulang kali Hanya boleh diacu sekali dan
dan boleh dikitar semula . tidak boleh dikitar semula.
Kebolehan diacu Can be moulded only one time and
Can be moulded several times and
recycled repeatedly Mouldability / Ability to be moulded

cannot be recycled.

Tidak berwarna dan lut sinar. Warna Legap

Does not have colour and transparent. Colour Opaque

Melebur apabila dipanaskan
tetapi menjadi keras apabila
Kesan haba Tidak melebur apabila dipanaskan.
Melts sejuk.
Effect of heat Do not melt when heated.
when heated, and
hardens when cooled.

Polietena, polipropena, polivinil Contoh-contoh Bakelit, epoksi, melamina

klorida, polistirena, perspeks Examples Bakelit, epoxy, melamine

Polyethene, polypropene, polyvinyl © Penerbitan Pelangi Sdn. Bhd.
chloride, polystyrene, perspex

117

  Kimia Tingkatan 5  Bab 4 Kimia Polimer

Tindak balas pempolimeran

Polymerisation reactions

14. Terdapat dua jenis tindak balas pempolimeran iaitu pempolimeran penambahan dan pempolimeran
kondensasi .
addition polymerisation and condensation polymerisation.
There are two types polymerisation reactions;

15. Jadual menunjukkan perbezaan antara dua jenis pempolimeran tersebut.

The table shows the differences between these two types of polymerisation.

Pempolimeran penambahan Pempolimeran kondensasi

Addition polymerisation Condensation polymerisation

Melibatkan sejenis monomer. Melibatkan dua jenis monomer.

Involves one monomer. Involves two kinds of monomers.

Monomer mesti mempunyai ikatan ganda dua atau Monomer mesti mempunyai dua kumpulan berfungsi .
ikatan ganda tiga .
Monomers must have two functional groups .
Monomers must have either double bond or triple bond .

Tidak menyebabkan kehilangan molekul kecil. Menyebabkan kehilangan molekul kecil seperti
H2O dan HCl .
Does not result in the loss of small molecules. H2O

Results in the loss of small molecules such as and
HCI .

Formula empirik polimer adalah sama dengan Formula empirik polimer adalah berbeza dengan

monomer. identical with its monomer. different from its

Empirical formula of the polymer is Empirical formula of the polymer is
monomers. monomers.

16. Termoplastik dihasilkan daripada pempolimeran penambahan , manakala termoset dihasilkan daripada
kondensasi . / Thermoplastic is synthesised by addition
pempolimeran polymerisation, while
thermoset is synthesised by condensation polymerisation.

17. Rajah menunjukkan pelbagai struktur polimer yang dihasilkan daripada proses pempolimeran penambahan.

Lukis formula struktur monomer masing-masing. / The diagram shows the structures of different polymers produced

by addition polymerisation. Draw the structural formulae of their monomers.

Polimer / Polymer Monomer / Monomer
(a)
HH
HH

BAB 4 CC CC
HH
H Hn
H CI
(b)

H CI

CC CC
HH
H Hn
H CO2CH3
(c) CO2CH3 CC
HH
CH2 C
HH
Hn CC
H
(d)

CH2 CH

n

© Penerbitan Pelangi Sdn. Bhd. 118

Kimia Tingkatan 5  Bab 4 Kimia Polimer 

18. Tandakan (✓) pada polimer kondensasi. / Put a tick (✓) next to the condensation polymer.

Polimer / Polymers Polimer kondensasi / Condensation polymers
Lateks / Latex
Nilon / Nylon ✓
Perspeks / Perspex
Terilena / Terylene ✓
Kanji / Starch ✓

Tugasan 1

Rajah di bawah menunjukkan sejenis pempolimeran. Isi tempat kosong berdasarkan rajah. TP 4 TP 5
The diagram below shows a type of polymerisation. Fill in the blanks based on the diagram.

OO OO

C C + HO OH + C C + HO OH

HO OH HO OH

OO OO

C CO OC CO O

(a) Proses ini dikenali sebagai pempolimeran kondensasi kerana melibatkan penyingkiran
water molecules
molekul air apabila polimer yang besar dihasilkan.

This process is known as condensation polymerisation as it involves the removal of
when a large polymer is formed.

(b) Polimer ini dihasilkan daripada monomer dengan dua kumpulan berfungsi yang berlainan, iaitu
asid karboksilik dan alkohol
.
functional groups , which are carboxylic acid
The polymer is formed from monomers of two different and
alcohol
.

(c) Polimer ini dinamakan poliester kerana pembentukan ikatan ester di antara dua monomer. BAB 4
This polymer is known as polyester as ester linkage is formed between the monomers.

19. Protein ialah polimer yang penting untuk pertumbuhan dan pembaikan tisu manusia. Asid amino bertindak
balas bersama untuk menghasilkan polimer tersebut. Molekul air turut dihasilkan semasa pempolimeran.
Namakan jenis pempolimeran ini.

Protein is a polymer which is essential for growth and repair of tissue in the human body. Amino acids react together to form
this polymer. Water molecules are produced during the polymerisation. Name this type of polymerisation.

Pempolimeran kondensasi / Condensation polymerisation

20. Rajah menunjukkan struktur bagi suatu asid amino. H HO

The diagram shows the structure of an amino acid. HN C C OH

Lukis sebahagian protein yang terbentuk daripada tiga unit asid amino Kumpulan amino Kumpulan karboksil
berulang ini. TP 5 Amino group Carboxyl group

Draw a section of protein composed of three repeating units of this amino acid. R

Rantai sisi
Side chain

119 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 4 Kimia Polimer

H H OHH OH HO

N C CNC CN CC

RR Rn

Aktiviti 4.1 Menghasilkan Nilon dan Mengkaji Sifat Nilon
Synthesise Nylon and Investigate Its Properties

Tujuan / Aim:

Menghasilkan nilon dan mengkaji sifatnya melalui tindak balas antara 1,6-heksanadiamina dengan dekanadioil

diklorida.

To synthesise nylon and investigate its properties through the reaction between 1,6-diaminohexane and decanedioyl dichloride.

Bahan /Materials:

Air suling, 1,6-heksanadiamina, dekanadioil diklorida, dan sikloheksana.

Distilled water, 1,6-diaminohexane, decanedioyl dichloride, and cyclohexane.

Radas / Apparatus:

Bikar, penimbang elektronik, silinder penyukat, rod kaca, dan forseps.

Beaker, electronic balance, measuring cylinder, glass rod, and forceps.

Prosedur / Procedures:
1. 2.2 g 1,6-heksanadiamina ditimbang di dalam bikar dengan penimbang elektronik .
electronic balance
2.2 g of 1,6-diaminohexane is weighed in a beaker using the .

2. 50 cm3 air suling disukat dengan silinder penyukat dan digunakan untuk melarutkan

1,6-heksanadiamina. Bikar ini dilabel sebagai larutan A.
distilled water
50 cm3 of is measured using a measuring cylinder and is used to dissolve 1,6-diaminohexane.

This beaker is labelled as solution A.

3. 1.5 g dekanadioil diklorida ditimbang di dalam bikar dengan penimbang elektronik.

1.5 g of decanedioyl dichloride is weighed in a beaker using the electronic balance.

BAB 4 4. 50 cm3 sikloheksana diukur dengan silinder penyukat dan digunakan untuk melarutkan
dekanadioil diklorida . Bikar ini dilabel sebagai larutan B.

50 cm3 of cyclohexane is measured using a measuring cylinder and is used to dissolve the decanedioyl dichloride . This
beaker is labelled as solution B.

5. 10 cm3 larutan A dituangkan ke dalam bikar yang lain. 10 cm3 larutan B kemudiannya dituangkan dengan
perlahan-lahan melalui dinding bikar tersebut supaya larutan B berada di atas larutan A.
gently
10 cm3 of solution A is transferred into another beaker. 10 cm3 of solution B is poured down the wall

of the beaker so that solution B is on top of solution A.

6. Perubahan tekstur dan warna hasil yang terbentuk di dalam bikar diperhatikan dan
dicatatkan.
texture and colour of the product formed in the beaker are observed and
The changes in
recorded.

7. Hasil yang terbentuk dikutip dengan forseps dan ditarik perlahan-lahan mengelilingi rod kaca.

The product formed is picked up with a pair of forceps and pulled gently over a glass rod.

8. Hasil akhir dibilas dengan air suling dan dikeringkan di dalam ketuhar sebelum diuji kebolehan
untuk diregangkan
. is tested.
stretchability
The end product is rinsed with the distilled water and dried in an oven before its

© Penerbitan Pelangi Sdn. Bhd. 120

Kimia Tingkatan 5  Bab 4 Kimia Polimer 

Keputusan / Results:
1. Satu lapisan putih terbentuk di antara dua larutan. / A whitish layer is formed at the interface of both solutions.
nilon
2. Benang dapat ditarik beberapa meter panjang secara berterusan.
nylon
Several metres of thread can be pulled continuously.
3. Benang nilon dapat diregangkan tanpa putus.
stretched
The nylon thread can be without breaking easily.

Perbincangan / Discussion : TP 3 TP 4

1. Nilon merupakan gentian sintetik yang dihasilkan daripada polimer berantai panjang yang tahan regangan .
Nylon is a synthetic fibre made from long-chain polymers that withstands
stretching .

2. Nilon-6,10 dihasilkan daripada tindak balas antara 1,6-heksanadiamina dan dekanadioil diklorida.

Nylon-6,10 is synthesised from the chemical reaction between 1,6-diaminohexane and decanedioyl dichloride.

3. Formula molekul 1,6-heksanadiamina ialah C6H16N2 dan boleh ditulis sebagai (CH2)6(NH2)2 .
(CH2)6(NH2)2
The molecular formula of 1,6-diaminohexane is C6H16N2 and also can be written as .

4. Formula molekul dekanadioil diklorida ialah C10H16Cl2O2 dan boleh ditulis sebagai (CH2)10(COCl)2 .

The molecular formula of decanedioyl dichloride is C10H16Cl2O2 and also can be written as (CH2)10(COCl)2 .

5. Penghasilan nilon-6,10 boleh dilukis sebagai: / The synthesis of nylon-6,10 can be drawn as:

HH OO

n H N (CH2)6 N H + n CI C (CH2)8 C CI

1, 6-heksanadiamina Dekanadioil diklorida
1,6-diaminohexane Decanedioyl dichloride

H HO O

N (CH2)6 N C (CH2)8 C + n HCI

Nilon-6, 10 n
Nylon-6, 10

Kesimpulan / Conclusion:
Nilon boleh dihasilkan daripada pempolimeran kondensasi antara 1,6-heksanadiamina dan dekanadioil diklorida .

Nylon can be synthesised through the condensation polymerisation between 1,6-diaminohexane and decanedioyl dichloride .

Kegunaan polimer dalam kehidupan harian

Uses of polymers in daily life

21. Senaraikan kegunaan polimer. / List the uses of polymers. TP 1 TP 2 BAB 4

Polimer / Polymers Kegunaan / Uses

Polietena (PE ) / Polyethene (PE) beg plastik, mainan plastik, bekas plastik / plastic bags, plastic toy, plastic container

Polipropena (PP)/ Polypropene (PP) botol plastik, perabot plastik, picagari / plastic bottle, plastic furniture, syringes

Polivinil klorida (PVC) paip air, kot hujan, kulit tiruan, penebat wayar

Polyvinyl chloride (PVC) water pipe, rain coat, artificial leather, wire casing

Polistirena (PS) / Polystyrene (PS) bekas makanan, pembungkus pelindung / food container, protective packaging

Teflon / Teflon peralatan memasak kalis lekat / non-stick cookware

Perspeks (PP) / Perspex (PP) tingkap kapal terbang, lampu utama kereta / airplane window, headlamp

Bakelit / Bakelite soket elektrik, pemegang periuk / power socket, cooking pot handle

Epoksi / Epoxy gam epoksi, cat epoksi / epoxy glue, epoxy paint

Melamina / Melamine lamina, perkakas makanan / laminate, dinnerware

Nilon / Nylon tali, tali joran, stoking, berus gigi, tali gitar, payung terjun

rope, fishing line, stocking, toothbrush, guitar string, parachute

121 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 4 Kimia Polimer

22. Apakah polimer tidak terbiodegradasikan? TP 3

What are non-biodegradable polymers?

Molekul besar yang tidak dapat diuraikan oleh bakteria.

Large molecules which cannot be decomposed/broken down by bacteria.

23. Nyatakan tiga masalah yang timbul disebabkan oleh pembuangan polimer sintetik. TP 4 TP 5

State three problems that arise as a result of the disposal of synthetic polymers.

• Kekurangan tapak pelupusan.

Shortage of landfill sites.

• Membahayakan hidupan marin / akuatik.

Endanger marine / aquatic life.

• Menghasilkan gas toksik / rumah hijau apabila dibakar.

Release toxic / greenhouse gases when burnt.

2Tugasan

1. Tangki air sejuk biasanya diperbuat daripada polimer kerana lebih murah
berbanding dengan tangki logam. Cadangkan dua kelebihan yang lain bagi
penggunaan polimer untuk membuat tangki air sejuk. TP 5

Cold water storage tanks are commonly made from polymers because they are cheaper
compared to metal tanks. Suggest two other advantages of using polymers for manufacturing
cold water tank.
• tahan kakisan / lengai / resistant to corrosion / chemically inert
• ringan / lightweight
• boleh diacu kepada pelbagai bentuk / can be moulded into different shapes
• penebat haba yang baik / good heat insulator

2. Tandakan (✓) pada dua ciri polimer sintetik yang menyebabkan pencemaran.
Place a tick (✓) against two properties of synthetic polymers that cause pollution.

BAB 4 Ciri-ciri Ciri-ciri yang menyebabkan pencemaran
Properties Properties which cause pollution

Kuat dan ringan
Strong and lightweight

Lengai terhadap tindak balas kimia ✓
Inert to chemical reactions

Mudah diacu
Easily moulded

Membebabskan gas beracun apabila terbakar ✓
Release poisonous gas when burnt

© Penerbitan Pelangi Sdn. Bhd. 122

Kimia Tingkatan 5  Bab 4 Kimia Polimer 

3. Isi tempat kosong dalam perenggan berikut yang menerangkan kesan beg plastik terhadap penyu.
Fill in the blanks of the paragraphs below which describe the effect of plastic bags on the sea turtles.

(a) Beg plastik dibuat daripada polimer sintetik yang dikenali sebagai polietena .
kecil
Polimer ialah molekul besar yang terdiri daripada rangkaian unit molekul

berulang yang dinamakan monomer. large molecule
Plastic bags are made from a synthetic polymer known as polyethene . A polymer is a
made from a chain of repeating units of small molecules known as monomers.

(b) Monomer polietena dikenali sebagai etena dengan formula molekul C2H4 . Rantaian

karbon-karbon yang panjang menyebabkan beg plastik tidak terbiodegradasikan . C2H4 . The
ethene
The monomer of polyethene is known as , with the chemical formula of
long carbon-carbon chains cause the plastic bags to be non-biodegradable .

(c) Apabila beg plastik dibuang ke dalam laut, penyu sering tersalah anggap beg plastik sebagai

makanan . Beg plastik akan menyumbat usus dan menembusi dinding usus penyu sehingga

menyebabkan pendarahan dalaman. food . The plastic bags can
When the plastic bags are being disposed at sea, sea turtles often mistake them for
cause blockages in their intestines and even pierce the intestinal wall causing internal bleeding.

(d) Oleh itu, kita perlu mengurangkan penggunaan beg plastik demi meminimumkan kesan negatif ke atas
haiwan akuatik .
reduce the use of plastic bags to minimise the negative impacts on the aquatic animals .
Hence, we should

4.2 Getah Asli BAB 4
Natural Rubber

1. Getah asli merupakan polimer berantai panjang yang terdiri daripada rangkaian unit monomer
berulang yang dinamakan isoprena
. TP 2 isoprene .
Natural rubber is a long-chain polymer
made from repeating units of monomer known as

2. Nama IUPAC isoperena ialah 2-metilbuta-1, 3-diena .

The IUPAC name of isoprene is 2-methylbuta-1, 3-diene .

3. Monomer getah asli dinamakan –diena kerana mempunyai dua ikatan ganda dua . TP 4
two double bonds
The monomer of natural rubber is known as –diene as it has .

4. Formula empirik monomer getah asli ialah C5H8 .

The empirical formula of the monomer of natural rubber is C5H8 .

5. Getah sintetik dihasilkan daripada petroleum, manakala getah asli dihasilkan daripada susu getah / lateks .
latex
Synthetic rubber is synthesised from petroleum, while natural rubber is made from the of rubber trees.

6. Isi tempat kosong dengan formula struktur apabila monomer getah asli menjalani proses pempolimeran

penambahan. TP 5

Fill in the empty space with the structural formula of natural rubber after its monomers undergo the addition polymerisation
process.

123 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 4 Kimia Polimer

H CH3H H Pempolimeran penambahan H CH3H H
Addition polymerisation
n(H – C = C – C = C – H) ( C – C = C – C )n
HH
Isoprena (monomer)
Isoprene (monomer) Poli(isoprena) – getah asli
Poly(isoprene) – natural rubber

Penggumpalan lateks

Coagulation of latex

7. Lateks merupakan larutan koloid yang mengandungi molekul getah , air dan bahan bukan getah.

Latex is a colloidal solution containing rubber molecules , water and non-rubber materials.

8. Lateks berada dalam keadaan cecair kerana zarah-zarah getah tidak bergumpal apabila

membran bercas negatif menolak antara satu sama lain. negatively -charged
liquid
Latex remains in form as the rubber particles do not coagulate when their
repel
membranes each other.

9. Getah asli ialah satu contoh elastomer kerana dapat berbalik kepada bentuk asalnya apabila diregangkan .

Natural rubber is an example of elastomer as it can return to its original shape after being stretched .

10. Elastomer mempunyai ciri-ciri termoplastik dan termoset . Elastomer bersifat seperti termoplastik
kerana terdiri daripada rantaian polimer yang berasingan . Selepas pemvulkanan, elastomer
membentuk rangkai silang yang serupa dengan termoset.
Elastomers share the properties of both thermoplastic and thermosets . Elastomers behave like the thermoplastic as
separate chains of polymers . After vulcanisation, elastomers form a  cross-link
they are composed of

network similar to the thermoset.

11.

(a)  Kekenyalan / Elasticity (b)  Pengoksidaan / Oxidation
Boleh kembali kepada bentuk asal selepas diregangkan .
• Menjadi rapuh apabila ikatan ganda dua
Able to return to its original shape after being stretched .
BAB 4 dioksidakan oleh oksigen dan ozon.
brittle when its double bonds are
Becomes

oxidised by oxygen and ozone.

Ciri-ciri getah asli TP 4

Properties of natural rubber

(c)  Kesan Haba / Effect of heat (d)  Kesan larutan / Effect of solvent

• Tidak tahan haba. • Tidak larut dalam air.
Insoluble
Not resistant to heat. in water.

• Menjadi keras dan rapuh pada suhu rendah . • Larut dalam larutan organik .
pada
Becomes hard and brittle at low temperatures . Soluble in organic solvents .

• Menjadi lembut dan melekit

suhu tinggi. soft and sticky

Becomes

at high temperatures.

© Penerbitan Pelangi Sdn. Bhd. 124

Kimia Tingkatan 5  Bab 4 Kimia Polimer 

12. Penggumpalan lateks / Coagulation of latex

Zarah-zarah getah Membran protein
Rubber particles Protein membrane

– – (a) Membran protein zarah-zarah getah adalah bercas negatif .
– – The protein membranes of the rubber particles is negatively-charged .

– – (b) Ini menghalang penggumpalan zarah-zarah getah yang
– menolak antara satu sama lain.
– –– coagulation
–– – –– This prevents the of rubber particles when they

– – repel each other.
–
––
––

Ion hidrogen
Hydrogen ions

+ +

+ + (c) Apabila asid ditambahkan, ion hidrogen akan meneutralkan
+– –
+– –+ + membran protein yang bercas negatif.
–
+ + – –+
–
+ – + – –+ When acid is added, the hydrogen ions will neutralise the
– –+ – – negative charges on the protein membrane .
–+ –+
+ – + + –
+ –

+–– + +

+ ++ +
+
+
+

Membran protein pecah
Protein membrane ruptures

(d) Zarah-zarah getah yang neutral berlanggar antara satu sama

lain dan menyebabkan membran protein pecah .

The neutral rubber particles then collide with each other causing

the protein membrane to break .

(e) Penggumpalan lateks berlaku apabila polimer getah bergabung

bersama-sama.
Coagulation
of latex takes place when the rubber polymers combine

together.

13. Mengapakah lateks bergumpal selepas dibiarkan beberapa hari? BAB 4

Why does latex coagulate after being left aside for a few days?

(a) Bakteria di dalam lateks menghasilkan asid yang mengion kepada ion hidrogen.
acid
The bacteria in latex produces which ionises to form hydrogen ions.

(b) Cas positif pada ion hidrogen meneutralkan cas negatif pada membran protein getah.

The positive charges of hydrogen ions neutralise the negative charges on the protein membrane of rubber.

3Tugasan

1. (a) Namakan satu bahan kimia yang ditambahkan untuk mencegah penggumpalan lateks sebelum dihantar ke
kilang. TP 3

Name a chemical added to prevent the coagulation of latex before the latex is sent to the factory.
Larutan ammonia / Ammonia solution

125 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 4 Kimia Polimer

(b) Huraikan tindakan bahan kimia tersebut. TP 5
Describe the action of the chemical.
Ion hidroksida daripada alkali lemah meneutralkan ion hidrogen daripada asid yang dihasilkan oleh bakteria.
The hydroxide ions from the weak alkali will neutralise the hydrogen ions from the acid produced by the bacteria.
2. Apabila lateks sampai di kilang, lateks dituangkan ke dalam acuan dan suatu bahan kimia ditambah untuk

menggumpalkan lateks. Namakan bahan kimia tersebut dan tulis formula kimianya.
Once the latex arrives at the factory, it is poured into a mould and a substance is added to coagulate the latex. Name the chemical formula

of the substance and write its chemical formula.
Asid asetik or asid etanoik, CH3COOH / Acetic acid atau ethanoic acid, CH3COOH
Asid formik or asid metanoik, HCOOH / Formic acid atau ethanoic acid, HCOOH

Penggumpalan Lateks dan Kaedah untuk Mengelakkan
Penggumpalan
Eksperimen 4.1
Coagulation of Latex and the Methods to Prevent Coagulation

BAB 4 Tujuan / Aim:
Mengkaji kesan asid dan alkali terhadap penggumpalan lateks.

To investigate the effect of acid and alkali on the coagulation of latex.

Pernyataan masalah / Problem statement:
Apakah kesan asid dan alkali terhadap penggumpalan lateks?

What is the effect of acid and alkali on the coagulation of latex?

Hipotesis / Hypothesis:
Asid akan menggumpalkan lateks sementara alkali akan mencegah penggumpalan lateks.

Acid will coagulate latex while alkali will prevent the coagulation of latex.

Pemboleh ubah / Variables:
Dimanipulasikan / Manipulated:
Kehadiran asid atau akali/ Presence of acid or alkali

Bergerak balas / Responding:
Penggumpalan lateks / Coagulation of latex

Dimalarkan / Constant:
Isi padu lateks / Volume of Latex

Bahan / Materials:
Lateks, asid etanoik, ammonia, kertas litmus merah dan biru. / Latex, ethanoic acid, ammonia, red and blue litmus papers.

Radas / Apparatus:
Bikar, penitis, dan rod kaca / Beaker, dropper, and glass rod.

Prosedur / Procedures: Rod kaca Lateks + asid etanoik Lateks + ammonia
Glass rod Latex + ethanoic acid Latex + ammonia
1. Tiga bikar dilabel sebagai A, B dan C masing-
Lateks sahaja B C
masing. Latex only

Three beakers are labelled as A, B and C A

respectively. lateks

2. 50 cm3 dituangkan ke dalam

setiap bikar. latex

50 cm3 of is poured into each

beaker.

© Penerbitan Pelangi Sdn. Bhd. 126

Kimia Tingkatan 5  Bab 4 Kimia Polimer 

3. 5 cm3 asid etanoik cair ditambah ke dalam lateks di dalam bikar B dan dikacau dengan rod kaca.
stirred
5 cm3 of dilute ethanoic acid are added to the latex in beaker B and the solution is with a glass rod.

4. 5 cm3 ammonia ditambah ke dalam lateks di dalam bikar C dan dikacau dengan rod kaca.

5 cm3 of ammonia solution are added into beaker C and the solution is stirred with a glass rod.

5. Keadaan lateks di dalam setiap bikar diperhatikan dan direkod selepas 30 minit.

The condition of latex in each beaker is observed and recorded after 30 minutes.

6. Keadaan lateks juga diperhatikan dan direkod sehari kemudian.

The condition of latex is observed and recorded again after the next day.

Keputusan / Results:

Bikar Penambahan asid atau alkali Keadaan lateks selepas 30 minit Penggumpalan lateks

Beaker Addition of acid or alkali Condition of latex after 30 minutes Coagulation of latex

A Tiada Cecair Bergumpal dengan perlahan

None Liquid Coagulate slowly

B Asid etanoik Pepejal putih Bergumpal dengan cepat

Ethanoic acid White solid Coagulate rapidly

C Ammonia Cecair Tidak bergumpal

Ammonia Liquid Does not coagulate

Perbincangan / Discussion:

1. Asid etanoik mengion separa dalam air untuk menghasilkan ion hidrogen yang bercas positif.

Ethanoic acid ionises partially in water to produce positively-charged hydrogen ions.

2. Ion hidrogen daripada asid etanoik meneutralkan cas negatif pada membran protein.

Hydrogen ions from ethanoic acid neutralise the negative charges on the protein membrane.

3. Zarah-zarah getah berlanggar antara satu sama lain dan memecahkan membran protein .
collide with each other and break the protein membrane .
The rubber particles then
separa
4. Ammonia mengion dalam air untuk menghasilkan ion hidroksida yang bercas negatif.
Ammonia ionises partially in water to produce negatively-charged hydroxide ions .

5. Ion hidroksida meneutralkan ion hidrogen dalam asid yang dihasilkan oleh bakteria . Oleh itu,
penggumpalan lateks dapat dicegah dengan penambahan larutan ammonia.
bacteria
Hydroxide ions neutralise the hydrogen ions of the acid produced by . Hence, coagulation of latex can be
prevented by adding ammonia solution.

Kesimpulan / Conclusion: alkali
Asid menggumpalkan lateks sementara
acid alkali mencegah penggumpalan lateks.

An coagulates latex while an prevents its coagulation. BAB 4

Pemvulkanan getah

Vulcanisation of rubber

14. (a)  Pemvulkanan getah asli dengan Sejarah

rangkai silang sulfur Pemvulkanan Getah
The History of
Vulcanisation of natural rubber by Vulcanisation of
sulphur cross-links
VIDEO 5 Rubber

(b)  Panaskan getah asli dengan sulfur, pada (c) Rendam getah asli dalam larutan
1400C dengan kehadiran zink oksida sebagai disulfur diklorida (S2Cl2) dalam metilbenzena
mangkin .
Heat (C7H8). disulphur dichloride
natural rubber with sulphur at 1400C in the
Dip natural rubber into a solution of
presence of zinc oxide as catalyst . (S2Cl2) in methylbenzene (C7H8).

127 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 4 Kimia Polimer

15. Pemvulkanan sinaran getah asli merupakan proses pemvulkanan yang menggunakan sinaran pengion .

Radiation vulcanisation of natural rubber is a vulcanisation process which uses Ionising radiation .

16. Pemvulkanan sinaran digunakan untuk menghasilkan sarung tangan getah dan puting botol susu yang bebas
sulfur dan logam oksida yang menyebabkan alahan.
daripada
sulphur
Radiation vulcanisation is used to manufacture rubber gloves and baby teats which are free from and
metal oxide allergies
that cause .

17. Pemvulkanan peroksida melibatkan penghasilan radikal bebas yang membentuk rangkaian

silang karbon di antara rantaian polimer.

Peroxide vulcanisation involves the generation of free radicals which form carbon cross-

links between adjacent polymer chains.

18. Logam oksida digunakan sebagai mangkin dan agen pengukuh yang meningkatkan ciri-ciri

mekanikal dan terma getah tervulkan.
Metal oxides catalyst
are used as and reinforcing agents which enhance the mechanical and thermal

properties of vulcanised rubber.

19. Ciri-ciri getah tervulkan dan getah tak tervulkan:

Properties of vulcanised rubber and unvulcanised rubber:

Getah tervulkan Ciri-ciri Getah tak tervulkan

Vulcanised rubber Properties Unvulcanised rubber

Kurang ikatan ganda dua kerana pembentukan Ikatan ganda dua Lebih
rangkai silang sulfur .
double bond More
Less double bonds due to the formation of
sulphur cross-links
.

Takat lebur yang lebih tinggi kerana kehadiran Takat lebur Lebih rendah
rangkai silang sulfur .
Melting point Lower
Higher melting point due the presence of
sulphur cross-links
.

Ketahanan terhadap haba yang lebih tinggi Ketahanan terhadap haba yang lebih
rendah ; getah menjadi lembut
kerana kehadiran rangkai silang sulfur . Ketahanan terhadap dan melekit apabila dipanaskan.
More
heat resistant due to the presence of haba Less heat resistant; rubber becomes
sulphur cross-links
. Resistance to heat

soft and sticky when heated.

BAB 4 Lebih kenyal kerana rangkaian silang sulfur Kekenyalan Kurang kenyal kerana rantaian getah
menarik rantaian balik kepada bentuk asal . boleh bergelongsor antara satu sama
Elasticity
More elastic as sulphur cross-links pull the chains lain dengan senang.
back to their original arrangement . Less elastic
as rubber chains can
slide over
each other easily.

Lebih keras kerana kehadiran rangkai
silang sulfur .
sulphur Kekerasan Lembut
Harder due to the presence of
Hardness Softer

cross-links .

Tidak mudah teroksida kerana ikatan ganda Ketahanan terhadap Mudah teroksida kerana kehadiran ikatan
dua yang berkurang . ganda dua
pengoksidaan .
Easily oxidised
Not easily oxidised due to reduced Resistance to oxidation because of the presence of

number of double bonds. many double bonds .

© Penerbitan Pelangi Sdn. Bhd. 128

Kimia Tingkatan 5  Bab 4 Kimia Polimer 

Aktiviti 4.2 Penyediaan Getah Tervulkan

Preparation of the Vulcanised Rubber

Tujuan / Aim:
Untuk menyediakan getah tervulkan.

To prepare the vulcanised rubber.

Bahan / Materials:
Lateks, larutan disulfur diklorida dalam metilbenzena.

Latex, disulphur dichloride solution in methylbenzene.

Radas/ Apparatus:
Kepingan kaca, rod kaca, bikar, penyepit, kertas turas, klip kertas, kaki retort, pembaris, dan pemberat.

Glass plate, glass rod, beaker, tongs, filter papers, paper clips, retort stand, ruler, and weight.

Prosedur / Procedures:

(a) (b) (c)

Lateks Rod kaca Penyepit
Latex Glass rod Tongs

Kepingan kaca Lapisan lateks
Glass plate Latex layer

Larutan disulfur klorida
Disulphur chloride solution
Jalur getah asli
Natural rubber strip

1. 10 cm3 lateks dituangkan ke atas kepingan kaca.

10 cm3 of latex is poured onto a glass plate.

2. Rod kaca digolekkan di atas kepingan kaca untuk menghasilkan satu lapisan lateks yang
rata dengan ketebalan lebih kurang 1 mm.
glass rod flat
A is rolled on the glass plate to produce a latex layer with a thickness of

approximately 1 mm.

3. Lapisan lateks dibiarkan selama dua hari untuk membenarkan berlakunya proses penggumpalan .
coagulation
The latex layer is then left aside for two days to allow for the process to take place.

4. Kepingan getah yang terbentuk digunting kepada dua jalur getah yang sama saiz . BAB 4
size .
The rubber sheet formed is cut into two strips of equal

5. Salah satu jalur getah dijepit oleh penyepit dan direndam dalam larutan disulfur diklorida dalam
metilbenzena selama 2 minit untuk menghasilkan jalur getah tervulkan.

One of the rubber strips is held by a pair of tongs and dipped into a solution of disulphur dichloride in methylbenzene
for 2 minutes to produce a vulcanised rubber strip.

6. Jalur getah tervulkan dikeringkan dengan kertas turas .
filter papers
The vulcanised rubber strip is dried with .

7. Kedua-dua keping jalur getah tervulkan dan jalur getah asli ditekan untuk menguji kekerasannya.

Both vulcanised and natural rubber strips are pressed to test their hardnsess.

Pemerhatian / Observation:

Jenis getah / Type of rubber Pemerhatian / Observation

Jalur getah asli Lebih lembut

Strip of natural rubber Softer

Jalur getah asli yang direndam dalam larutan disulfur diklorida dalam metilbenzena Lebih keras

Strip of natural rubber dipped into disulphur dichloride solution in methylbenzene Harder

129 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 4 Kimia Polimer

Perbincangan / Discussion:
1. Getah tervulkan dihasilkan apabila jalur getah asli direndam dalam larutan disulfur diklorida. Getah tervulkan
keras
adalah lebih .
Vulcanised rubber is formed when the natural rubber strip is dipped into disulphur dichloride solution . Vulcanised rubber
harder
is .

2. Semasa pemvulkanan, atom-atom sulfur membentuk rangkai silang antara rantaian

polimer getah . Hal ini mengurangkan keupayaan polimer getah untuk bergelongsor antara satu

sama lain. sulphur atoms rubber polymers

During vulcanisation, the form cross-links among the chains of .

This reduces the ability of the chains of rubber polymers to slide against one another.

Kesimpulan / Conclusion:
Getah tervulkan
dihasilkan apabila jalur getah asli direndam dalam larutan disulfur diklorida dalam

metilbenzena.

Vulcanised rubber is produced when the natural rubber strip is dipped into disulphur dichloride solution in methylbenzene.

Eksperimen 4.2 Mengkaji Sifat Kekenyalan Getah Tervulkan dan Getah Tak
Tervulkan

Investigating the Elasticity of Vulcanised and Unvulcanised Rubbers

Tujuan / Aim:

Untuk mengkaji kekenyalan getah tervulkan dan getah tak tervulkan.

To investigate the elasticity of vulcanised and unvulcanised rubbers.

Pernyataan masalah / Problem statement :

Bagaimanakah kekenyalan getah tervulkan dan getah tak tervulkan berbeza?

How does the elasticity of unvulcanised rubber differ from that of vulcanised rubber?

Hipotesis / Hypothesis:
Getah tervulkan lebih kenyal daripada getak tak tervulkan. / Vulcanised rubber is more elastic than unvulcanised rubber.

BAB 4 Pemboleh ubah / Variables: Rod logam
Dimanipulasikan / Manipulated: Metal rod
Jenis getah / Types of rubber
Klip kertas
Bergerak balas / Responding: Paper clip
Kekenyalan getah / Elasticity of rubber Jalur getah
Rubber strip
Dimalarkan / Constant:
Jisim pemberat, saiz jalur getah /Mass of weight, size of rubber strip

Bahan / Materials: Tali/ Thread
Jalur getah tervulkan dan jalur getah tak tervulkan. Pemberat
Weight
Vulcanised and unvulcanised rubber strips.

Radas / Apparatus:
Kaki retort, rod logam, klip kertas, tali, pembaris, dan pemberat.

Retort stand, metal rod, paper clip, thread, ruler, and weight.

Prosedur / Procedures: pembaris

1. Panjang asal jalur getak tak tervulkan diukur dan direkod dengan menggunakan ruler .

The initial length of the unvulcanised rubber strip is measured and recorded using a .

2. Kedua-dua hujung jalur getah tak tervulkan diapit dengan klip kertas .
paper clip
Both ends of the unvulcanised rubber strip are attached with a .

3. Pemberat 50 g digantung pada satu hujung jalur getah tak tervulkan, manakala hujung jalur getah yang lain

digantung pada rod logam yang dipasang pada kaki retort.

A weight of 50 g is hung at one end of the unvulcanised rubber strip, while the other end of the rubber strip is hung at the
metal rod which is attached to the retort stand.

© Penerbitan Pelangi Sdn. Bhd. 130

Kimia Tingkatan 5  Bab 4 Kimia Polimer 

4. Panjang akhir getah tak tervulkan setelah pemberat ditanggalkan diukur dan direkod.
removed
The final length of the unvulcanised rubber strip is measured and recorded when the weight is .

5. Langkah 1 ke 4 diulangi dengan menggantikan jalur getah tak tervulkan dengan jalur getah

tervulkan. repeated by using the vulcanised rubber strip instead of the unvulcanised rubber strip.

Steps 1 to 4 are

Keputusan / Results:

Jenis getah Panjang asal Panjang akhir apabila Pemanjangan Panjang ahir setelah
(cm) pemberat digantung (cm) (cm) pemberat ditanggalkan (cm)
Type of rubber
Initial length Final length when the weight is Increase in length Final length when the weight is
Getah tak tervulkan (cm) hung (cm) (cm) removed (cm)

Unvulcanised rubber 5 8 3 6

Getah tervulkan 5 72 5

Vulcanised rubber

Perbincangan / Discussion: kurang

1. Pemanjangan getak tervulkan berbanding dengan getah tak tervulkan kerana getah
keras
tervulkan lebih dengan kehadiran rangkai silang sulfur.
The increase in length of the vulcanised rubber is less than the unvulcanised rubber because the vulcanised
harder
rubber is due to the presence of sulphur cross-links.

2. Setelah pemberat ditanggalkan, panjang getah tervulkan sama seperti panjang asalnya kerana

getah tervulkan adalah lebih kenyal daripada getak tak tervulkan. Rangkai silang sulfur menarik
rantaian polimer getah balik kepada kedudukan asal .
same
After the weigth is removed, the length of the vulcanised rubber is the as its original length because the
more elastic
vulcanised rubber is than the unvulcanised rubber. The sulphur cross-links in the vulcanised rubber
pull the chains of rubber polymers back to their original positions .

Kesimpulan / Conclusion:
Getah tervulkan adalah lebih kenyal daripada getak tak tervulkan.

The vulcanised rubber is more elastic than the unvulcanised rubber.

4.3 Getah Sintetik
Synthetic Rubber

1. Getah sintetik lebih tahan lama berbanding dengan getah asli kerana mempunyai: BAB 4

Synthetic rubbers are more durable compared to natural rubbers as they are:

• ketahanan terhadap haba yang lebih tinggi / more resistant to heat

• ketahanan terhadap bahan kimia / kakisan yang lebih tinggi / more resistant to chemicals / corrosion

• ketahanan terhadap lelasan yang lebih tinggi / more resistant to abrasion

2. Aplikasi pelbagai jenis getah sintetik. / Applications of various types of synthetic rubbers.

Jenis getah sintetik / Types of synthetic rubber Kegunaan / Uses

(a) Neoprena Gasket, pakaian selam, sarung tangan keselamatan

Neoprene Gasket, wetsuit, safety gloves

(b) Getah stirena-butadiena (SBR) Tayar, tapak kasut, gasket

Styrene-butadiene rubber (SBR) Tires, shoe soles, gasket

(c) Getah silikon Perekat silikon, tiub perubatan, produk elektronik, peralatan dapur

Silicon rubber Silicon sealant, medical tubing, electronics products, kitchenware

131 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 4 Kimia Polimer

4Tugasan

1. Rajah menunjukkan sepasang sarung tangan nitril yang diperbuat daripada
getah sintetik dan digunakan secara meluas dalam bidang perubatan. Wajarkan
penggunaan sarung tangan tersebut.

The diagram shows nitrile gloves which are made from a synthetic rubber and widely used in the
medical field. Justify the use of the gloves.
• Lebih tahan lama.
Highly durable.
• Tahan terhadap minyak, pelarut dan bahan kimia.
Resistant to oils, solvents and chemicals.
• Tidak akan menyebabkan tindak balas alahan.
Does not cause allergic reaction.
• Bebas daripada serbuk dan mengurangkan peluh yang akan menyebabkan ketidakselesaan.
Powder-free and reduce sweating which causes discomfort.

2. Tayar terdiri daripada 19% getah asli dan 24% getah sintetik. Namun begitu, hanya sebahagian kecil tayar terpakai
dikitar semula. Huraikan kesan buruk terhadap alam sekitar yang disebabkan oleh tayar.

Tyres are made of about 19% of natural rubber and 24% of synthetic rubber. However, only a very small fraction of used tyres is recycled.
Describe the environmental problems caused by tyres.
• Penghasilan getah sintetik yang digunakan dalam tayar menggunakan petroleum yang merupakan sumber
tenaga yang tidak boleh dibaharui.
The production of synthetic rubber used in tyres requires petroleum which is a non-renewable source of energy.
• Tayar terpakai menakung air yang menyediakan tempat pembiakan yang kondusif kepada nyamuk.
Used tyres collect water which provides good breeding grounds for mosquitoes.
• Permukaan tayar yang haus menghasilkan plastik mikro yang mencemari lautan dan mengancam hidupan akuatik.
Worn-out surfaces of tyres produce microplastics that are washed into the ocean and harm the aquatic lives.

BAB 4 PRAKTIS SPM 4

Soalan Objektif

1. Rajah 1 menunjukkan suatu proses pempolimeran. Antara yang berikut, yang manakah serupa bagi
Diagram 1 shows a polymerisation process. propena and polipropena?

HH HH Which of the following is identical for propene and
polypropene?
nC C CC
A Ketumpatan / Density
H CH3 n B Formula empirik / Empirical formula
H CH3 C Takat lebur / Melting point
Propena D Jism molekul relatif / Relative molecular mass
Propene Polipropena
Polypropene

Rajah 1/ Diagram 1

© Penerbitan Pelangi Sdn. Bhd. 132

2. Pempolimeran kloroetena menghasilkan Kimia Tingkatan 5  Bab 4 Kimia Polimer 
Polymerisation of chloroethene forms
A Membran protein diliputi oleh ion hidroksida
A Polietena / Polyethene Protein membrane is covered with hydroxide ions
B Polipropena / Polypropene B Rangkai silang terbentuk antara rantai
C Polistirena / Polystyrene
D Polivinil klorida / Polyvinyl chloride poliisoprena
Cross-links are formed between chains of polyisoprene
3. Antara yang berikut, yang manakah polimer semula C Rantaian panjang polimer getah bergabung
jadi?
secara rawak
Which of the following is a natural polymer? Long chains of rubber polymers are entangled randomly
D Pempolimeran penambahan berlaku apabila
A Polistirena / Polystyrene
B Polipropena / Polypropene monomer isoprena bergabung antara satu sama
C Poliisoprena / Polyisoprene lain
D Polifeniletena / Polyphenylethene Addition polymerisation takes place when isoprene
monomers combine with each other
4. Seorang tukang paip diminta untuk menggantikan
paip air yang bocor dengan bahan yang tahan lama 8. Lateks boleh disimpan dalam keadaan cecair dengan
dan tidak mudah berkarat. Pilih bahan yang paling menambahkan
sesuai digunakan oleh tukang paip tersebut.
2 015 Latex can be kept in liquid state by adding
A plumber was requested to replace the leaked water pipes A Asid formik / Formic acid
with durable materials which do not rust easily. Choose the B Asik etanoik / Ethanoic acid
most suitable material that can be used by the plumber. C Ammonium sulfat / Ammonium sulphate
D Ammonium hidroksida / Ammonium hydroxide
A Terilena / Terylene
B Perspeks / Perspex 9. Rajah 3 menunjukkan formula struktrur monomer
C Polietena / Polyethene getah asli.
D Polivinil klorida / Polyvinyl chloride
Diagram 3 shows the structural formula of the monomer of
5. Antara yang berikut, yang manakah membentuk natural rubber.
polimer kondensasi?
Which of the following forms a condensation polymer? H CH3 H H

A CH3CH=CH2 C HOCH2CH2COOH CC CC
B CH3CH2NH2 D CH3CH2CH2COOH
HH
6. Rajah 2 menunjukkan struktur satu polimer. BAB 4
Diagram 2 shows the structure of a polymer. Rajah 3 / Diagram 3
Yang manakah nama kimia bagi monomer getah asli
H CH3 H CH3 H CH3 H CH3
mengikut sistem IUPAC?
C C CC CCCC Which of the following is the chemical name for the monomer

H H HH HHHH of natural rubber according to IUPAC nomenclature?
Rajah 2/ Diagram 2 A 2-metilbut-1,4-diena / 2-methylbut-1,4-diene
B 2-metilbut-1,3-diena / 2-methylbut-1,3-diene
Antara yang berikut, yang manakah monomer bagi C 3-metilbut-1,3-diena / 3-methylbut-1,3-diene
polimer tersebut? D 3-metilbut-2,3-diena / 3-methylbut-2,3-diene

Which of the following is the monomer of the polymer? 10. Getah tervulkan lebih keras daripada getak tak
A Etena / Ethene tervulkan kerana polimer getah tervulkan
B Propena / Propene
C Butena / Butene Vulcanised rubbers are harder than unvulcanised rubbers
because the polymers of vulcanised rubbers are
7. Yang manakah antara berikut berlaku semasa A adalah bercas negatif / negatively-charged
pemvulkanan getah? B ditarik oleh rangkai silang sulfur / held by sulphur
cross-links
2 010 Which of the following occurs during vulcanisation of rubber? C bergabung untuk membentuk rantaian yang
lebih panjang / combined to form longer chains
D disusun dengan teratur / arranged in an orderly
manner

133 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 4 Kimia Polimer

Soalan Struktur

Bahagian A

1. Polistirena atau stirofoam biasanya digunakan sebagai bekas makanan CH2 CH CH2 CH CH2 CH
kerana bersifat sebagai penebat haba yang baik. Namun begitu, Rajah 1 / Diagram 1
stirofoam memberikan kesan-kesan negatif kepada alam sekitar. Oleh
itu, kerajaan negeri Selangor melarang penggunaan stirofoam sejak
tahun 2017.

Polystyrene or styrofoam is commonly used for food packaging as it is a good heat
insulator. However, it causes various negative effects on the environment. Hence,
the usage of styrofoam was banned from the year 2017 by the Selangor state
government.

Rajah 1 mewakili formula struktur sebahagian daripada polimer polistirena.
Diagram 1 represents the structural formula of part of a polystyrene polymer.

(a) Apakah maskud pempolimeran?
What is the meaning of polymerisation?

Pempolimeran merupakan tindak balas kimia yang melibatkan unit monomer kecil yang berulang bergabung

menjadi polimer yang besar.

Polymerisation is a chemical reaction of which repeating units of small monomers join together to form a large polymer.
[2 markah / marks]

(b) Lukis formula struktur untuk monomer polistirena.
Draw the structural formula of the monomer of polystyrene.

H2C CH [2 markah / marks]

(c) Namakan dua jenis polimer sintetik yang lain.
Name two other types of synthetic polymers.

BAB 4 Gentian sintetik / elastomer / termoset

Synthetic fibre / elastomer / thermoset

[2 markah / marks]

(d) Selain sifatnya sebagai penebat haba, polistirena juga boleh dipanaskan dan diacu kepada bentuk bekas
makanan yang berbagai-bagai. Oleh itu, polistirena dikelaskan sebagai termoplastik. Huraikan ikatan kimia dalam
termoplastik yang membenarkannya diacu kepada bentuk yang berlainan.

Besides its heat insulating property, polystyrene can also be heated and moulded easily into different shapes of food container.
Hence, polystyrene is classified as thermoplastic. Describe the chemical bonding in thermoplastic which enable it to be moulded into
different shapes.

• Tidak ada rangkai silang antara rantai molekul.

No cross-links between chains of molecules.

• Molekul dapat bergelongsor antara satu sama lain apabila dipanaskan.

The molecules slide over each other easily when heated.

[2 markah / marks]

© Penerbitan Pelangi Sdn. Bhd. 134

Kimia Tingkatan 5  Bab 4 Kimia Polimer 

(e) Apakah ciri polistirena yang menyebabkan masalah semula jadi?
What is the property of polystyrene which results in environmental problem?

Polistirena tidak terbiodegradasikan. / Polystyrene is non-biodegradable.

[1 markah / mark]

(f ) Sebagai seorang pengguna yang bertanggungjawab, cadangkan satu cara untuk mengurangkan penggunaan
bekas makanan plastik pakai buang.

As a responsible consumer, suggest one way to reduce the usage of single-use plastic food containers.

Bawa bekas makanan sendiri. / Bring our own food containers.
Makan di restoran. / Dine-in at restaurants.

[1 markah / mark]

(g) Nyatakan dua kebaikan tindakan kerajaan negeri untuk melarang penggunaan polistirena dalam industri
makanan.

State the two advantages of the action of the state government in banning the use of polystyrene in food industry.

• Memelihara sumber petroleum. / Conserve petroleum.

• Mengurangkan kebergantungan kepada tapak perlupusan. / Reduce the need for landfill sites.
• Mengurangkan pencemaran air. / Reduce water pollution.

• Mengurangkan kesan negatif kepada haiwan akuatik. / Reduce the negative impacts on the aquatic animals. BAB 4
[2 markah / marks]

Bahagian B

2. (a) Gunakan getah asli sebangai contoh, takrifkan
Use natural rubber as an example, define
(i) Pempolimeran /Polymerisation    
(ii)  Pemvulkanan /Vulcanisation
[4 markah / marks]
(b) Pempolimeran boleh dijalankan dalam dua cara yang berbeza bergantung kepada monomer yang terlibat.
Namakan dua cara pempolimeran tersebut dan nyatakan dua perbezaan antara cara tersebut.
Polymerisation can take place in two different methods depending on the monomers involved. Name the two methods of
polymerisation and state two differences between these methods.
[8 markah / marks]

(c) Lengkapkan Rajah 2.1 untuk menggambarkan struktur getah tervulkan selepas proses pemvulkanan.
Complete Diagram 2.1 to illustrate the structure of vulcanised rubber after the vulcanisation process.

H CH3 H H Proses pemvulkanan H CH3 H H
CC CC Vulcanisation process CC C C
HH H
HH Rajah 2.1/ Diagram 2.1 HS S

CCC C HS S H
CC C C
H H CH3 H H H CH3H

Getah asli Getah tervulkan
Natural rubber Vulcanised rubber

[2 markah / marks]

135 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 4 Kimia Polimer [2 markah / marks]

Huraikan proses pemvulkanan getah asli.
Describe the vulcanisation process of natural rubber.
(d) Rajah 2.2 menunjukkan perubahan pada getah asli apabila diregangkan.
Diagram 2.2 shows the change in natural rubber when it is being stretched.

Diregangkan
Stretched

Rajah 2.2/ Diagram 2.2 [2 markah / marks]
(i) Dengan merujuk Rajah 2.2, terangkan sebab getah asli bersifat kenyal. [2 markah / marks]
By referring to Diagram 2.2, explain why natural rubber is elastic.

(ii) Bagaimanakah pemvulkanan mempengaruhi kekenyalan getah asli?
How does vulcanisation affect the elasticity of natural rubber?

2006

Bahagian C

BAB 4 3. Wallace Carothers (1896 – 1937) ialah seorang ahli kimia terkenal yang pertama mencipta getah sintetik yang dikenali
sebagai neoprena dan gentian sintetik yang dikenali sebagai nilon 66. Kedua-dua bahan sintetik memainkan peranan
penting dalam Perang Dunia Kedua.

Wallace Carothers (1896 – 1937) was a renowned polymer chemist who first invented the synthetic rubber known as neoprene as well as
synthetic fibre known as nylon 66. Both synthetic materials played an important role in Second World War.
(a) Monomer bagi neoprena dikenali sebagai kloroprena. Rajah 3.1 menunjukkan formula struktur kloroprena.
The monomer of neoprene is known as chloroprene. Diagram 3.1 shows the structural formula of chloroprene.

H CI H H

nC C C C

HH
Kloroprena
Chloroprene

Rajah 3.1/ Diagram 3.1

(ii) Dengan merujuk Rajah 3.1, lukis formula struktur polimer neoprena.
By referring to Diagram 3.1, draw the structural formula of neoprene polymer.

[2 markah / marks]

(i) Nama IUPAC neoprena ialah 2-klorobuta-1, 3-diena. Terangkan cara untuk memperoleh nama ini.
The IUPAC name of neoprene is 2-chlorobuta-1,3-diene. Explain how this name is being derived.

[4 markah / marks]

(iii) Semasa Perang Dunia Kedua, neoprena digunakan secara meluas untuk menghasilkan tayar dan hos dalam
mesin perang. Pada masa ini, neoprena biasanya digunakan dalam peralatan sokongan perubatan dan sukan.
Dengan menggunakan satu jalur getah asli dan satu jalur neoprena yang sama panjang dan ketebalan,
huraikan prosedur makmal untuk membandingkan kekenyalan getah asli dan neoprena. Cadangkan dua
ciri neoprena yang membolehkannya digunakan dalam pakaian selam yang berkualiti tinggi.

© Penerbitan Pelangi Sdn. Bhd. 136

Kimia Tingkatan 5  Bab 4 Kimia Polimer 

During Second World War, neoprene was used widely in manufacturing tyres and hoses in war machines. Nowadays, neoprene
is commonly used in medical and sport supports. Given a strip of natural rubber and a strip of neoprene of the same length and
thickness, outline the laboratory procedures to compare the elasticity of natural rubber and neoprene. Suggest two properties
of neoprene which enable it to be used as a high-quality wetsuit for deep diving.
[6 markah / marks]

(b) Nilon 66 digunakan untuk menggantikan sutera dalam penghasilan stoking wanita pada mulanya. Namun, nilon
juga digunakan untuk menghasilkan payung terjun tentera semasa Perang Dunia Kedua.

Nylon 66 was initially used to replace silk in manufacturing woman stockings. However, the production of nylon was also diverted to
military parachute during Second World War.
(i) Rajah 3.2 menunjukkan formula struktur polimer nilon 66.
Diagram 3.2 shows the structural formula of nylon 66 polymer.

H HO O

N (CH2)6 N C (CH2)4 C n

Rajah 3.2/ Diagram 3.2
Lukis monomer-monomer bagi nilon 66 dan namakan monomer-monomer tersebut.
Draw the structural formulae of the monomers of nylon 66 and name these monomers.

(ii) Nyatakan kebaikan dan keburukan polimer sintetik. [4 markah / marks]
State the advantages and disadvantages of synthetic polymers. [4 markah / marks]

BAB 4

137 Kuiz 4

© Penerbitan Pelangi Sdn. Bhd.

BAB Kimia Konsumer dan Industri

5 Consumer and Industrial Chemistry

PETA Konsep BAHAN PENCUCI Pengsulfonan dan
CLEANSING AGENTS
MINYAK DAN LEMAK peneutralan
FATS AND OILS
Sulphonation and
neutralisation

Saponifikasi

Saponification

Sabun Detergen

Soap Detergent

Lemak tepu Penghidrogenan Lemak tak tepu Bahan tambah
dalam
Saturated fat Hydrogenation Unsaturated fat
detergen
• Kepentingan lemak dan minyak Pembuatan majerin Tindakan dan
dalam pemakanan manusia keberkesanan pencucian Additives in
Produce margarine detergent
Importance of fats and oils in human Cleansing action and
diet effectiveness

• Kesan buruk kepada kesihatan UBAT-UBATAN BAHAN KOSMETIK
manusia MEDICINES COSMETICS

Negative effects on human health Tradisional Moden • Rias / Make-up
• Perawatan / Treatment
• Bahan api bio Traditional Modern • Pewangi / Fragrances

Biofuel Kesan kepada kesihatan
manusia
BAHAN TAMBAH MAKANAN
FOOD ADDITIVES Effects on human health

• Pengawet / Preservative • Analgesik / Analgesics NANOTEKNOLOGI
• Antioksidan / Antioxidant • Antibiotik / Antibiotics NANOTECHNOLOGY
• Pengemulsi / Emulsifier • Antialergi / Anti-allergy
• Penstabil / Stabiliser • Ubat psikoteraputik / • Definisi / Definition
• Pemekat / Thickening agent • Aplikasi / Application
• Perisa / Flavourings Psychotherapeutic drugs • Sifat fizik dan kimia grafen
• Pewarna / Colourings Physical and chemical
• Kortikosteroid / Corticosteroid
Kesan kepada kesihatan manusia properties of graphene
Kesan sampingan
Effects on human health
Side effects
TEKNOLOGI HIJAU
GREEN TECHNOLOGY Pengurusan sisa Kaedah olahan larutlesapan

Empat tunggak / Four main pillars Waste management Landfill leachate treatment
method

Tenaga Alam sekitar Pengurusan air sisa • 4R
• Bangunan hujau
Energy Environment Wastewater management Green building

Ekonomi Sosial

Economy Social

© Penerbitan Pelangi Sdn. Bhd. 138

Kimia Tingkatan 5  Bab 5 Kimia Konsumer dan Industri 

5.1 Minyak dan Lemak
Fats and Oils

1. Semua lemak dan minyak merupakan ester semula jadi yang terhasil daripada tindak balas
kondensasi (pengesteran) antara gliserol asid lemak
(sebagai alkohol) dengan

berantai panjang yang berbeza (sebagai asid karboksilik). TP 1

All fats and oils are naturally occurring esters , formed from condensation reactions (esterification)

between the glycerol (as an alcohol) and different long chain fatty acids (as carboxylic acids).

2. Berikut menunjukkan kumpulan berfungsi ester. Kenal pasti bahagian yang berasal daripada asid karboksilik

(asid lemak) dan alkohol (gliserol). Kemudian bulatkan dan namakan kumpulan berfungsi ester.

Below is the functional group of esters. Determine the part originated from the carboxylic acid and alcohol. Then circle and
name the functional group of esters.

Asid karboksilik Alkohol

Carboxylic acid Alcohol

Kumpulan berfungsi/ Functional group: COO TP 2

3. Pengesteran ialah tindak balas kondensasi kerana melibatkan penyingkiran 3 mol molekul air . TP 1

Esterification is a condensation reaction as it involves the removal of 3 moles of water molecules .

4. Oleh sebab lemak dan minyak terhasil daripada tiga asid lemak dan satu gliserol,

lemak juga dikenali sebagai trigliserida . TP 1 one glycerol, they are also known as triglycerides .
As fats and oils consist of three fatty acids and

(a) Asid lemak ialah asid karboksilik yang berantai hidrokarbon panjang yang terdiri daripada 4 hingga

28 atom karbon. hydocarbon chain, ranging from 4 to 28 carbon atoms.

A fatty acid is a carboxylic acid with a long

(b) Molekul lemak/ minyak boleh dihasilkan daripada pelbagai jenis asid lemak.

A fat or an oil molecule can be made from different types of fatty acids.

(c) Gliserol merupakan alkohol yang mempunyai tiga kumpulan hidroksil / –OH .

Glycerol is an alcohol with three hydroxyl / –OH groups.

(d) Nama IUPAC gliserol ialah propan-1,2,3 triol .

The IUPAC name of glycerol is propan-1,2,3-triol .

(e) Tiga molekul asid lemak boleh bercantum dengan satu molekul gliserol disebabkan oleh kehadiran tiga
kumpulan hidroksil / –OH dalam gliserol.

Three fatty acid molecules can join with one glycerol molecule due to the presence of three hydroxyl /–OH groups
in the glycerol.

5. Kelaskan yang berikut kepada sumber lemak dan minyak. Kemudian banding dan bezakan sifat fizik dan BAB 5

ketepuan lemak dan minyak. TP 1

Classify the following into sources of fats and oils. Then compare and contrast the physical properties and saturation of fats
and oils.

Mentega Ikan Jagung Bunga matahari Kulit ayam

Butter Fish Corn Sunflower Chicken skin

Lemak lembu Koko Zaitun Kelapa sawit Susu kambing

Beef tallow Cocoa Olive Oil palm Goat milk

139 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 5 Kimia Konsumer dan Industri

Sumber Lemak Minyak

Sources Fats Oils

Mainly animals: Mainly plants:

Haiwan sahaja: Tumbuhan sahaja:

• Mentaga / Butter • Ikan / Fish
• Kulit ayam / Chicken skin • Jagung / Corn
• Lemak lembu / Beef tallow • Bunga matahari / Sunflower
• Susu kambing / Goat milk • Koko / Cocoa
• Zaitun / Olive
• Kelapa sawit / Oil palm

Persamaan

Similarities

• Kedua-duanya mempunyai atom karbon , hidrogen dan oksigen .

Both contain carbon , hydrogen and oxygen atoms.

• Kedua-duanya molekul kovalen yang besar.

Both are large covalent molecules.

Perbezaan

Differences

• Kekal pepejal pada suhu bilik. • Kekal cecair pada suhu bilik.
solids
Remain at room temperature. Remain liquids at room temperature.

Sifat fizik • Takat lebur yang tinggi • Takat lebur yang rendah .

Physical High melting point. Low melting point.
properties
• Asid lemaknya yang tepu • Asid lemaknya yang tidak tepu
( Mempunyai ikatan ganda dua)
( Tiada ikatan gand dua)

Contain saturated fatty acids Contain unsaturated fatty acids
Have
( Have no double bonds) ( double bonds)

6. Semua lemak haiwan berada dalam keadaan pepejal pada suhu bilik, kecuali minyak ikan . TP 3

All animal fats exist as solids at room temperature, except fish oil .

7. Mengapakah minyak tumbuhan wujud dalam keadaan cecair pada suhu bilik manakala lemak haiwan tidak?

Why do plant oils exist as liquids at room temperature while animal fats do not? TP 3

Minyak mempunyai peratusan asid lemak tak tepu yang lebih tinggi secara relatif.

They contain a relative higher percentage of unsaturated fatty acids.

1Tugasan

(a) Lengkapkan Peta Titi berikut: TP 1
Complete the following Bridge Map:

BAB 5 Alkohol Asid karboksilik Ester
Alcohol as Carboxylic acid Ester
Gliserol as

Glycerol Asid lemak Lemak atau minyak
Fatty acid Fat or oil


© Penerbitan Pelangi Sdn. Bhd. 140

Kimia Tingkatan 5  Bab 5 Kimia Konsumer dan Industri 

(b) Lukis formula struktur bagi asid lemak / fatty acid. TP 1
Draw the structural formula of
O
gliserol / glycerol. TP 1 R C OH

H Anggapkan R = Rantai hidrokarbon panjang
Assume R = Long hydrocarbon chain
H C OH

H C OH

H C OH

H

(c) Lukis struktur kimia lemak dalam tindak balas berikut. TP 2 O
Draw the chemical structure of fat in the following chemical reaction.


CH2OH R1COOH CH2 O C R1

CHOH ϩ R2COOH O

CH O C R ϩ 3H2O
2

CH2OH R3COOH O

CH2 O C R
3

Proses penukaran lemak tak tepu kepada lemak tepu

Conversion of unsaturated fats to saturated fats

8. (a) Lengkapkan jadual di bawah untuk mengenal pasti asid lemak tepu dan asid lemak tak tepu. TP 2

Compete the table below to identify the saturated and unsaturated fatty acids.

Nama asid lemak Bilangan atom karbon Formula asid lemak Jenis asid lemak
Name of fatty acid Number of carbon atoms
Formula of fatty acid Type of fatty acid

(i) Asid palmitik 16 CH3(CH2)14COOH Tepu

Palmitic acid Saturated

(ii) Asid oleik 18 CH3(CH2)7CH=CH(CH2)7COOH Tidak tepu

Oleic acid Unsaturated

(iii) Asid stearik 18 CH3(CH2)16COOH Tepu

Stearic acid Saturated BAB 5

(iv) Asid linoleik 18 CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH Tidak tepu

Linoleic acid Unsaturated

141 © Penerbitan Pelangi Sdn. Bhd.

  Kimia Tingkatan 5  Bab 5 Kimia Konsumer dan Industri

(b) Berdasarkan jadual di (a), asid lemak tak tepu berikatan ganda dua manakala asid lemak tepu
tunggal
berikatan . TP 3
double
Based on the table in (a), unsaturated fatty acids have bonds whereas saturated fatty acids have
single
bonds.

9. Asid lemak tak tepu boleh ditukarkan kepada asid lemak tepu melalui tindak balas penghidrogenan atau
dikenal sebagai tindak balas penambahan dengan hidrogen. TP 1
addition
Unsaturated fatty acids can be converted into fatty acids through hydrogenation reactions, namely

reactions with hydrogen.

HH ϩ H2 HH
CC CC
HH

Ikatan ganda dua dalam asid lemak tak tepu
Double bond in unsaturated fatty acids

10. Lengkapkan Peta Berbuih di bawah untuk menghuraikan proses penghasilan marjerin melalui tindak balas

penghidrogenan. TP 1

Complete the Bubble Map below to describe the process of manufacturing margarine through hydrogenation process.

(g) Dijalankan pada 200oC (a) Proses pengerasan (b) Menghasilkan marjerin
dan 4 atm minyak tumbuhan daripada minyak tumbuhan

Carried out at 200oC and Hardening process of Manufacture margarine
4 atm vegetable oils from vegetable oil

(f ) Menggunakan platinum Penghidrogenan (c) Ikatan ganda dua ditukar
atau nikel sebagai Hydrogenation kepada ikatan tunggal
mangkin
Double bond is converted into
Uses platinum or single bond
nickle as catalyst

BAB 5 (e) Proses penambahan (d) Asid lemak tak tepu
ditukarkan kepada asid
Addition process lemak tepu

Unsaturated fatty acid is
converted into saturated
fatty acid

© Penerbitan Pelangi Sdn. Bhd. 142

Kimia Tingkatan 5  Bab 5 Kimia Konsumer dan Industri 

11. Penggunaan lemak dan minyak dalam kehidupan harian: / Uses of fats and oils in our daily life: TP 3

(a) Kepentingan lemak dan minyak dalam (b) Kesan negatif terhadap kesihatan manusia akibat
pemakanan manusia
pengambilan terlalu banyak lemak dan minyak
Importance of fats and oils in the human diet
Negative effects of consuming too much fats and oils on human
health

(i) Membekalkan tenaga untuk aktiviti harian. • Obesiti
energy
Provide for daily activities. Obesity

(ii) Melarutkan vitamin A, D, E, K. • Tekanan darah tinggi

Dissolve vitamin A, D, E, K. High blood pressure

(iii) Melindungi organ dalaman. • Serangan jantung
protect
Heart attack

• Strok

Stroke

To internal organs.

(iv) Sebagai komponen membran plasma.

As a component of plasma membrane .

12. Hampas dan sisa buah kelapa sawit boleh digunakan sebagai bahan api bio yang merupakan
sumber tenaga yang boleh dibaharui . TP 3
biofuel which is a renewable source of energy .
The residues and wastes of palm fruits are used as a

5.2 Bahan Pencuci
Cleansing Agents

Sabun dan detergen

Soaps and detergents

1. Bahan pencuci ialah bahan kimia yang digunakan untuk menanggalkan gris dan kotoran . TP 1
Cleansing agents are chemical substances that are used to remove grease and dirt .

2. Sabun dan detergen sebagai bahan pencuci: / Soaps and detergents as cleansing agents: TP 2

Bahan pencuci / Cleansing agent

(a)  Sabun / Soap (b)  Detergen / Detergent

(i) Garam natrium atau kalium bagi asid lemak (i) Garam natrium atau kalium bagi asid alkil sulfonik
berantai panjang Sodium or potassium salt of alkyl sulphonic acid

Sodium or potassium salt of long-chain fatty acids

(ii) Dihasilkan daripada lemak haiwan atau (ii) Dihasilkan daripada pecahan petroleum BAB 5
minyak tumbuhan Synthesised from petroleum fractions

Synthesised from animal fats or OSO3– Na+
plant oils

COO– Na+



143 © Penerbitan Pelangi Sdn. Bhd.