The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Discover the best professional documents and content resources in AnyFlip Document Base.
Search
Published by Far Abd, 2020-04-28 03:04:00

Softcopy for Teachers FORM 5

MODUL NILAM KIMIA

Keywords: CHEMISTRY

CONTENTS

KANDUNGAN

Exercise for Common Chemical 1 UNIT 102
Equations in SPM Chemistry Syllabus
Latihan untuk Persamaan Kimia Biasa dalam 3 REDOX REACTIONS
Sukatan Pelajaran Kimia SPM TINDAK BALAS REDOKS

UNIT 7 ❖ The Definition of Redox Reaction
Definisi Tindak Balas Redoks
1 RATE OF REACTION ❖ Redox Reaction in Terms of Loss or Gain of Oxygen
KADAR TINDAK BALAS
and Loss or Gain of Hydrogen
❖ Meaning of Rate of Reaction Tindak Balas Redoks dari Segi Kehilangan atau
Maksud Kadar Tindak Balas Penerimaan Oksigen dan Kehilangan atau
❖ How to Determine the Rate of Reaction From the Penerimaan Hidrogen
❖ Redox Reaction in Terms of Electron Gain / Loss
Graph? Tindak Balas Redoks dari Segi Penerimaan /
Bagaimanakah Cara Menentukan Kadar Tindak Balas Kehilangan Elektron
daripada Graf? ❖ Redox Reaction in Terms of Change of Oxidation
❖ Factors that Affect the Rate of Reaction Number / Tindak Balas Redoks dari segi Perubahan
Faktor yang Mempengaruhi Kadar Tindak Balas Nombor Pengoksidaan
❖ Application of Factors That Affect the Rate of ❖ Nomenclature of Ionic Compounds Using
Reaction / Aplikasi Faktor-Faktor yang Mempengaruhi IUPAC Nomenclature / Penamaan Sebatian Ion
Kadar Tindak Balas Menggunakan Sistem Penamaan IUPAC
❖ The Collision Theory / Teori Perlanggaran ❖ Redox Reaction in Terms of Change in Oxidation
❖ Relationship Between Frequency of Effective Number / Tindak Balas Redoks dari Segi Perubahan
Collision with Factors Affecting Rate of Reaction Nombor Pengoksidaan
Hubung Kait antara Frekuensi Perlanggaran Berkesan ❖ Writing Equations for Redox Reactions
dengan Faktor Mempengaruhi Kadar Tindak Balas Menulis Persamaan untuk Tindak Balas Redoks
❖ Redox Reaction In Electrochemistry
UNIT 45 Tindak Balas Redoks dalam Elektrokimia
❖ Redox Reaction in the Displacement of Metals
2 CARBON COMPOUNDS from its Salt Solution / Persamaan Redoks dalam
SEBATIAN KARBON Penyesaran Logam daripada Larutan Garamnya
❖ Carbon Compounds / Sebatian Karbon ❖ Redox Reaction in Electrolytic and Chemical
❖ Hydrocarbon / Hidrokarbon Cell / Voltaic Cell / Tindak Balas Redoks dalam Sel
❖ Alkane / Alkana Elektrolisis dan Sel Kimia
❖ Physical Properties of Alkane ❖ Redox in Electrochemistry
Sifat-sifat Fizik Alkana Redoks dalam Elektrokimia
❖ Chemical Properties of Alkanes ❖ Redox Reaction in Corrosion of Metal / Rusting of
Sifat-sifat Kimia Alkana Iron / Tindak Balas Redoks dalam Kakisan Logam /
❖ Alkene / Alkena Pengaratan Besi
❖ Physical Properties of Alkenes ❖ Prevention of Rusting / Pencegahan Pengaratan
Sifat-sifat Fizik Alkena ❖ Redox Reaction in Displacement of Halogen from
❖ Chemical Properties of Alkenes its Halide Solution / Tindak Balas Redoks dalam
Sifat-sifat Kimia Alkena Penyesaran Halogen daripada Larutan Halidanya
❖ Homologous Series / Siri Homolog
❖ Alcohol / Alkohol
❖ Preparation of Alcohol / Penyediaan Alkohol
❖ Carboxylic Acid / Asid Karboksilik
❖ Esters / Ester
❖ Fats / Lemak
❖ Natural Rubber / Getah Asli

❖ Redox Reaction in the Change of Fe2+ → Fe3+ and ❖ Numerical Problems Involving Heat of
Fe3+ → Fe2+ Neutralisation

Tindak Balas Redoks dalam Perubahan Fe2+ → Fe3+ dan Pengiraan Melibatkan Haba Peneutralan
Fe3+ → Fe2+ ❖ Comparing the Heat of Neutralisation
Membandingkan Haba Peneutralan
❖ Redox Reaction in the Transfer of Electron at a ❖ Numerical Problems Involving Heat of Combustion
Distance Pengiraan Melibatkan Haba Pembakaran
❖ Comparing the Heat of Combustion of Various Fuels
Tindak Balas Redoks dalam Pemindahan Elektron Membandingkan Haba Pembakaran Pelbagai Bahan
pada Satu Jarak
Api
❖ Redox Reaction in the Reactivity Series of Metals
and Its Applications

Tindak Balas Redoks dalam Siri Kereaktifan Logam
dan Aplikasinya

UNIT 148 UNIT 179

4 THERMOCHEMISTRY 5 CHEMICAL FOR CONSUMERS
TERMOKIMIA BAHAN KIMIA UNTUK PENGGUNA

❖ Energy Changes in Chemical Reactions ❖ Soap / Sabun
Perubahan Tenaga dalam Tindak Balas Kimia ❖ Preparing a Sample of Soap in the Laboratory
❖ The Heat of Reaction (∆H) Menyediakan Sabun dalam Makmal
Haba Tindak Balas (∆H) ❖ Detergent / Detergen
❖ Activity / Experiment to Determine Heat of Reaction ❖ The Cleansing Action of Soap and Detergent
Aktiviti / Eksperimen untuk Menentukan Haba Tindak Tindakan Pembersihan Sabun dan Detergen
❖ The Effectiveness of the Cleansing of Soap and
Balas
❖ Numerical Problems Involving Heat of Displacement Detergent
Pengiraan Melibatkan Haba Penyesaran Keberkesanan Pembersihan Sabun dan Detergen
❖ Numerical Problems Involving Heat of Precipitation ❖ Food Additives / Bahan Tambah Makanan
Pengiraan Melibatkan Haba Pemendakan ❖ Medicine / Ubat

Guideline to Scan AR for 3D Model
Garis Panduan untuk Mengimbas AR bagi Model 3D

1 2 Download the free ‘AR Chemistry Nilam

Download the free QR reader Publication’ application by scanning the QR
application from Play Store. code below.
Muat turun aplikasi ‘QR reader’ Muat turun aplikasi ‘AR Chemistry Nilam
dari ‘Play Store’. Publication’ dengan mengimbas kod QR di bawah.

4 3 Find a page that have the

Scan the icon on that page with following icon.
your smartphone and enjoy the 3D Cari muka surat yang mempunyai
model that appear. ikon berikut.
Imbas ikon pada muka surat
tersebut menggunakan telefon
pintar anda dan nikmati model 3D
yang terpapar.

Exercises for Common Chemical Equations in SPM Chemistry Syllabus
Latihan untuk Persamaan Kimia Biasa dalam Sukatan Pelajaran Kimia SPM

1 Chemical Properties of Acid / Sifat Kimia Asid
(a) Metal + Acid → Salt + Hydrogen: / Logam + Asid → Garam + Hidrogen:

Application: / Digunakan dalam:

Acid and base Preparation of salt Rate of reaction
Asid dan bes Penyediaan garam Kadar tindak balas

Wooden splinter Metal
Kayu uji Logam

Acid Acid Water
Asid Asid Air

Metal Acid Metal
Logam Asid Logam

Acid Metal Balanced chemical equation Ionic equation
Asid Logam Persamaan kimia seimbang Persamaan ion
HNO3 Mg + 2H+ → Mg2+ + H2
Mg Mg + 2HNO3 → Mg(NO3)2 + H2 2Al + 6H+ → 2Al3+ + 3H2
MODULE • Chemistry Form 5H2SO4AlZn + 2H+ → Zn2+ + H2
HCl Zn 2Al + 6HNO3 → 2Al(NO3)3 + 3H2 Pb + 2H+ → Pb2+ + H2
1 © Nilam Publication Sdn BhdPb Mg + 2H+ → Mg2+ + H2
CH3COOH Mg Zn + 2HNO3 → Zn(NO3)2 + H2 2Al + 6H+ → 2Al3+ + 3H2
Al Zn + 2H+ → Zn2+ + H2
Zn Pb + 2HNO3 → Pb(NO3)2 + H2 Mg + 2H+ → Mg2+ + H2
Mg 2Al + 6H+ → 2Al3+ + 3H2
Al Mg + H2SO4 → MgSO4 + H2 Zn + 2H+ → Zn2+ + H2
Zn Mg + 2H+ → Mg2+ + H2
Mg 2Al + 3H2SO4 → Al2(SO4)3 + 3H2 2Al + 6H+ → 2Al3+ + 3H2
Al Zn + 2H+ → Zn2+ + H2
Zn Zn + H2SO4 → ZnSO4 + H2 Pb + 2H+ → Pb2+ + H2
Pb
Mg + 2HCl → MgCl2 + H2

2Al + 6HCl → 2AlCl3 + 3H2

Zn + 2HCl → ZnCl2 + H2

Mg + 2CH3COOH → (CH3COO)2Mg + H2

2Al + 6CH3COOH → 2(CH3COO)3Al + 3H2

Zn + 2CH3COOH → (CH3COO)2Zn + H2

Pb + 2CH3COOH → (CH3COO)2Pb + H2

© Nilam Publication Sdn Bhd (b) Metal Carbonate + Acid → Salt + Carbon dioxide + Water / Logam karbonat + Asid → Garam + Karbon dioksida + Air MODULE • Chemistry Form 5
Application: / Digunakan dalam:

Chemical properties of acid Preparation and qualitative analysis of salt Rate of reaction
Sifat kimia asid Penyediaan dan analisis kualitatif garam Kadar tindak balas

Metal carbonate
Logam karbonat

Metal carbonate + acid Acid Water
Logam karbonat + asid Acid Asid Air
Asid Metal
Lime water
Air kapur Logam

2 Acid Metal carbonate Balanced chemical equation Ionic equation
Asid Logam karbonat Persamaan kimia seimbang Persamaan ion
HNO3 CaCO3 + 2HNO3 → Ca(NO3)2 + H2O + CO2 CaCO3 + 2H+ → Ca2+ + H2O + CO2
CaCO3 MgCO3 + 2HNO3 → Mg(NO3)2 + H2O + CO2 MgCO3 + 2H+ → Mg2+ + H2O + CO2
H2SO4 MgCO3 Al2(CO3)3 + 6HNO3 → 2Al(NO3)3 + 3H2O + 3CO2 Al2(CO3)3 + 6H+ → 2Al3+ + 3H2O + 3CO2
Al2(CO3)3 ZnCO3 + 2HNO3 → Zn(NO3)2 + H2O + CO2 ZnCO3 + 2H+ → Zn2+ + H2O + CO2
HCl ZnCO3 FeCO3 + 2HNO3 → Fe(NO3)2 + H2O + CO2 FeCO3 + 2H+ → Fe2+ + H2O + CO2
FeCO3 PbCO3 + 2HNO3 → Pb(NO3)2 + H2O + CO2 PbCO3 + 2H+ → Pb2+ + H2O + CO2
PbCO3 CuCO3 + 2HNO3 → Cu(NO3)2 + H2O + CO2 CuCO3 + 2H+ → Cu2+ + H2O + CO2
CuCO3 K2CO3 + H2SO4 → K2SO4 + H2O + CO2 CO32– + 2H+ → H2O + CO2
K2CO3 Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2 CO32– + 2H+ → H2O + CO2
Na2CO3 MgCO3 + 2H2SO4 → MgSO4 + H2O + CO2 MgCO3 + 2H+ → Mg2+ + H2O + CO2
MgCO3 Al2(CO3)3 + 3H2SO4 → Al2(SO4)3 + 3H2O + 3CO2 Al2(CO3)3 + 6H+ → 2Al3+ + 3H2O + 3CO2
Al2(CO3)3 ZnCO3 + H2SO4 → ZnSO4 + H2O + CO2 ZnCO3 + 2H+ → Zn2+ + H2O + CO2
ZnCO3 CuCO3 + H2SO4 → CuSO4 + H2O + CO2 CuCO3 + 2H+ → Cu2+ + H2O + CO2
CuCO3 CaCO3 + 2HCl → CaCl2 + H2O + CO2 CaCO3 + 2H+ → Ca2+ + H2O + CO2
CaCO3 MgCO3 + 2HCl → MgCl2 + H2O + CO2 MgCO3 + 2H+ → Mg2+ + H2O + CO2
MgCO3 Al2(CO3)3 + 6HCl → 2AlCl3 + 3H2O + 3CO2 Al2(CO3)3 + 6H+ → 2Al3+ + 3H2O + 3CO2
Al2(CO3)3 ZnCO3 + 2HCl → ZnCl2 + H2O + CO2 ZnCO3 + 2H+ → Zn2+ + H2O + CO2
ZnCO3 CuCO3 + 2HCl → CuCl2 + H2O + CO2 CuCO3 + 2H+ → Cu2+ + H2O + CO2
CuCO3

(c) Acid + Base → Salt + Water: / Asid + Bes → Garam + Air: MODULE • Chemistry Form 5
Application: / Digunakan dalam:
Thermochemistry / Termokimia
Acid and base / Asid dan bes Preparation of salt / Penyediaan garam

Acid Metal oxide 34 35 36 37 38 39 40 41 42 43
Asid Logam oksida

Acid Polystyrene cup
Asid Cawan polisterina

Alkali Acid + alkali
Alkali Asid + alkali

Acid Base Balanced chemical equation Ionic equation
Asid Bes Persamaan kimia seimbang Persamaan ion
H+ + OH– → H2O
KOH KOH + HNO3 → KNO3 + H2O H+ + OH– → H2O
CaO + 2H+ → Ca2+ + H2O
NaOH NaOH + HNO3 → NaNO3 + H2O MgO + 2H+ → Mg2+ + H2O
Al2O3 + 6H+ → 2Al3+ + 3H2O
CaO CaO + 2HNO3 → Ca(NO3)2 + H2O ZnO + 2H+ → Zn2+ + H2O
CuO + 2H+ → Cu2+ + H2O
HNO3 MgO MgO + 2HNO3 → Mg(NO3)2 + H2O H+ + OH– → H2O
H+ + OH– → H2O
Al2O3 Al2O3 + 6HNO3 → 2Al2(NO3)3 + 3H2O MgO + 2H+ → Mg2+ + H2O
Al2O3 + 6H+ → 2Al3+ + 3H2O
ZnO ZnO + 2HNO3 → Zn(NO3)2 + H2O ZnO + 2H+ → Zn2+ + H2O
CuO + 2H+ → Cu2+ + H2O
CuO CuO + 2HNO3 → Cu(NO3)2 + H2O H+ + OH– → H2O
H+ + OH– → H2O
KOH 2KOH + H2SO4 → K2SO4 + 2H2O CaO + 2H+ → Ca2+ + H2O
MgO + 2H+ → Mg2+ + H2O
NaOH 2NaOH + H2SO4 → Na2SO4 + 2H2O Al2O3 + 6H+ → 2Al3+ + 3H2O
ZnO + 2H+ → Zn2+ + H2O
H2SO4 MgO MgO + H2SO4 → MgSO4 + H2O CuO + 2H+ → Cu2+ + H2O
Al2O3 Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O

ZnO ZnO + 2H2SO4 → ZnSO4 + H2O

CuO CuO + H2SO4 → CuSO4 + H2O

KOH KOH + HCl → KCl + H2O

NaOH NaOH + HCl → NaCl + H2O

CaO CaO + 2HCl → CaCl2 + H2O

HCl MgO MgO + 2HCl → MgCl2 + H2O

Al2O3 Al2O3 + 6HCl → 2AlCl3 + 3H2O

ZnO ZnO + 2HCl → ZnCl2 + H2O

CuO CuO + 2HNO3 → Cu(NO3)2 + H2O

3 © Nilam Publication Sdn Bhd

© Nilam Publication Sdn Bhd 2 Displacement Reactions / Tindak Balas Penyesaran (iii) Thermochemistry MODULE • Chemistry Form 5
Application. / Digunakan dalam: Termokimia

(i) Electrochemistry / Elektrokimia
(ii) Redox / Redoks

More electropositive metal
Logam lebih elektropositif

Salt solution of less 34 35 36 37 38 39 40 41 42 43 More electropositive
electropositive metal metal / Logam lebih
Larutan garam bagi elektropositif
logam kurang
Salt solution of elektropositif
less electropositive
metal
Larutan garam
bagi logam kurang
elektropositif

4 Salt solution Metal Balanced chemical equation Ionic equation
Larutan garam Logam Persamaan kimia seimbang Persamaan ion
Mg + 2AgNO3 → Mg(NO3)2 + 2Ag Mg + 2Ag+ → Mg2+ + 2Ag
AgNO3 Mg Al + 3AgNO3 → Al(NO3)3 + 3Ag Al + 3Ag+ → Al3+ + 3Ag
Al Zn + 2AgNO3 → Zn(NO3)2 + 2Ag Zn + 2Ag+ → Zn2+ + 2Ag
CuSO4/ CuCl2 / Cu(NO3)2 Zn Fe + 2AgNO3 → Fe(NO3)2 + 2Ag Fe + 2Ag+ → Fe2+ + 2Ag
* Choose one solution Fe Pb + 2AgNO3 → Pb(NO3)2 + 2Ag Pb + 2Ag+ → Pb2+ + 2Ag
Pb Cu + 2AgNO3 → Cu(NO3)2 + 2Ag Cu + 2Ag+ → Cu2+ + 2Ag
Pilih satu larutan Cu Mg + CuSO4 → MgSO4 + Cu Mg + Cu2+ → Mg2+ + Cu
Mg 2Al + 3CuSO4 → Al2(SO4)3 + 3Cu 2Al + 3Cu2+ → 2Al3+ + 3Cu
Pb(NO3)2 Al Zn + CuSO4 → ZnSO4 + Cu Zn + Cu2+ → Zn2+ + Cu
Zn Fe + CuSO4 → FeSO4 + Cu Fe + Cu2+ → Fe2+ + Cu
Zn(NO3)2 / ZnCl2 / ZnSO4 Fe Pb + CuSO4 → PbSO4 + Cu Pb + Cu2+ → Pb2+ + Cu
* Choose one solution Pb Mg + Pb(NO3)2 → Mg(NO3)2 + Pb Mg + Pb2+ → Mg2+ + Pb
Mg 2Al + 3Pb(NO3)2 → 2Al(NO3)3 + 3Pb 2Al + 3Pb2+ → 2Al3+ + 3Pb
Pilih satu larutan Al Zn + Pb(NO3)2 → Zn(NO3)2 + Pb Zn + Pb2+ → Zn2+ + Pb
Zn Fe + Pb(NO3)2 → Fe(NO3)2 + Pb Fe + Pb2+ → Fe2+ + Pb
Fe Mg + Zn(NO3)2 → Mg(NO3)2 + Zn Mg + Zn2+ → Mg2+ + Zn
Mg 2Al + 3Zn(NO3)2 → 2Al(NO3)3 + 3Zn 2Al + 3Zn2+ → 2Al3+ + 3Zn
Al

3 Double Decomposition Reaction / Tindak Balas Penguraian Ganda Dua
Application / Digunakan dalam:

(i) P reparation of insoluble of salt (ii) Q ualitative analysis of salt (iii) Thermochemistry
Penyediaan garam tak terlarutkan Analisis kualitatif garam Termokimia

NaCl(aq/ak)/Na2SO4/KI(aq/ak) Mixture of salt solution
contains cation and anion of
Salt solution Salt solution Salt solution 34 35 36 37 38 39 40 41 42 43 insoluble salt
contains anion contains cation contains Campuran larutan garam
of insoluble salt of insoluble salt Larutan garam yang mengandungi kation
Larutan garam Larutan garam mengandungi dan anion garam tak
mengandungi mengandungi Al3+/ Pb2+ terlarutkan
garam tak garam tak
terlarutkan terlarutkan Ionic equation
Persamaan ion
Insoluble salt Pb2+ + 2Cl– → PbCl2
Garam tak Pb2+ + SO42– → PbSO4
terlarutkan Pb2+ + CO32– → PbCO3
Pb2+ + 2I– → PbI2
Salt solution (i) Salt solution (ii) Balanced chemical equation Ba2+ + SO42– → BaSO4
Larutan garam (i) Larutan garam (ii) Persamaan kimia seimbang Ba2+ + CO32– → BaCO3
Ag+ + Cl– → AgCl
MODULE • Chemistry Form 5NaClPb(NO3)2 + 2NaCl → PbCl2 + 2NaNO32Ag+ + CO32– → Ag2CO3

Pb(NO3)25 © Nilam Publication Sdn BhdNa2SO4Pb(NO3)2 + Na2SO4 → PbSO4 + 2NaNO3
Na2CO3 Pb(NO3)2 + Na2CO3 → PbCO3 + 2NaNO3

KI Pb(NO3)2 + 2KI → PbI2 + 2KNO3

Ba(NO3)2 Na2SO4 Ba(NO3)2 + Na2SO4 → BaSO4 + 2NaNO3
or / atau Na2CO3 BaCl2 + Na2CO3 → BaCO3 + 2NaCl
BaCl2

AgNO3 NaCl AgNO3 + NaCl → AgCl + NaNO3
Na2CO3 2AgNO3 + Na2CO3 → Ag2CO3 + 2NaNO3

© Nilam Publication Sdn Bhd 4 Combustion of Carbon Compound / Pembakaran Sebatian Karbon MODULE • Chemistry Form 5
Carbon compound + Oxygen → Carbon dioxide + Water

Sebatian karbon Oksigen Karbon dioksida Air

Application / Digunakan dalam: (ii) Thermochemistry / Termokimia
(i) Carbon Compound / Sebatian Karbon

Soot Copper can 34 35 36 37 38 39 40 41 42 43 Thermometer
Jelaga Tin kuprum Termometer
Water
Air

Alkane / alkene / alcohol Alcohol ------------------------------------
Alkana / alkena / alkohol Alkohol

Carbon Compound / Sebatian Karbon Balanced chemical equation / Persamaan kimia seimbang
CH4
C2H6 CH4 + 2O2 → CO2 + 2H2O
C3H8
6 C4H10 C2H6 + 7 O2 → 2CO2 + 3H2O
C5H12 2
C2H4
C3H6 C3H8 + 5O2 → 3CO2 + 4H2O
C4H8
C5H10 C4H10 + 13 O2 → 4CO2 + 5H2O
2
CH3OH
C2H5OH C5H12 + 8O2 → 5CO2 + 6H2O
C3H7OH
C4H9OH C2H4 + 3O2 → 2CO2 + 2H2O
C5H11OH
C3H6 + 9 O2 → 3CO2 + 3H2O
2

C4H8 + 6O2 → 4CO2 + 4H2O

C5H10 + 15 O2 → 5CO2 + 5H2O
2

CH3OH + 3 O2 → CO2 + 2H2O
2

C2H5OH + 3O2 → 2CO2 + 3H2O

C3H7OH + 9 O2 → 3CO2 + 4H2O
2

C4H9OH + 6O2 → 4CO2 + 5H2O

C5H11OH + 15 O2 → 5CO2 + 6H2O
2

MODULE • Chemistry Form 5

UNIT RATE OF REACTION

1 KADAR TINDAK BALAS

Concept Map / Peta Konsep

Daily activities Industry processes Average rate of reaction Rate of reaction at any given time U
Aktiviti harian Proses industri Kadar tindak balas purata Kadar tindak balas pada masa tertentu N
I
Examples in Measuring rate of reaction T
Contoh dalam Mengukur kadar tindak balas
1
Application Measurement of the change in quantity of reactant or product per unit time
Aplikasi Ukuran perubahan dalam kuantiti bahan atau hasil tindak balas per unit masa

Meaning / Maksud

Factors affect rate of reaction RATE OF REACTION
Faktor mempengaruhi kadar tindak balas KADAR TINDAK BALAS

Experiments on the effect of Can be explained using
Eksperimen tentang kesan Boleh diterangkan menggunakan

Size Temperature Concentration Catalyst Collision Particles possess
Saiz Suhu Kepekatan Mangkin Theory activation energy
Teori perlanggaran
The larger the size The higher the The higher the The presence of Happen if Zarah-zarah
of solid reactant, temperature, the concentration of catalyst lower the Effective collision Berlaku jika mencapai
the less total higher the kinetic solution, the higher activation energy Perlanggaran
surface area energy of particles the number of of the reaction berkesan tenaga pengaktifan
expose to collision Semakin tinggi particles per unit Kehadiran
Semakin besar suhu, semakin volume mangkin Related to Particles collide at
saiz pepejal bahan tinggi tenaga Semakin tinggi merendahkan Berkait dengan the correct
tindak balas, kinetik zarah kepekatan tenaga orientation
semakin larutan, semakin pengaktifan Frequency of Zarah-zarah
berkurangan tinggi bilangan tindak balas effective collision
jumlah luas zarah per unit isi berlanggar pada
permukaan padu Frekuensi orientasi yang betul
yang terdedah perlanggaran
kepada
perlanggaran berkesan

Related to / Berkaitan Cause
Menyebabkan

Rate of reaction increases
Kadar tindak balas meningkat

Learning objective / Objektif pembelajaran

• Analysing rate of reaction / Menganalisis kadar tindak balas
• Synthesising factors affect rate of reaction
Mensintesis faktor-faktor yang mempengaruhi kadar tindak balas
• Synthesising idea on collision theory / Mensintesis idea tentang teori perlanggaran
• Practising scientific knowledge to improve quality of life
Mempraktikkan pengetahuan saintifik untuk meningkatkan kualiti kehidupan

7 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Meaning of Rate of Reaction / Maksud Kadar Tindak Balas

State the meaning of rate of The rate of reaction is a measurement of the change in quantity of reactant or product per unit time.
reaction. Kadar tindak balas ialah pengukuran perubahan kuantiti bahan atau hasil tindak balas per unit masa.
Nyatakan maksud kadar tindak
balas.

U State the relationship between 1 The rate of reaction is high if the reaction occurs fast within a short period
N rate of reaction and time. of time. low if the reaction occurs slowly within a long period
Nyatakan hubungan antara kadar
tindak balas dan masa. 2 The rate of reaction is
of time.
1I
3 The rate of reaction is inversely proportional to time:
T
Rate of reaction ∝ 1

Time taken

1 Kadar tindak balas tinggi jika tindak balas berlaku dengan cepat dalam

jangka masa yang pendek.

2 Kadar tindak balas rendah jika tindak balas berlaku dengan perlahan dalam

jangka masa yang panjang.

3 Kadar tindak balas berkadar songsang dengan masa:

Kadar tindak balas ∝ 1

Masa yang diambil

Give examples of fast reactions. – Reaction of marble chip with hydrochloric acid.
Berikan contoh tindak balas – Reaction of magnesium with sulphuric acid.
cepat. – Reaction of potassium with water.
– Burning of fuel.
– Tindak balas ketulan marmar dengan asid hidroklorik.
– Tindak balas magnesium dengan asid sulfurik.
– Tindak balas kalium dengan air.
– Pembakaran bahan api.

Give examples of slow reactions. – Rusting of iron in the air.
Berikan contoh tindak balas – Photosynthesis.
perlahan. – Fermentation of fruit juice to form alcohol.
– Pengaratan besi dalam udara.
– Fotosintesis
– Penapaian jus buah untuk dijadikan alkohol.

How to determine rate of reaction? Rate of reaction can be determined by calculating the rate of chemical change or measured quantity in a
Bagaimanakah menentukan kadar chemical change per unit time.
tindak balas? Kadar tindak balas boleh ditentukan dengan mengira kadar perubahan kimia atau kuantiti yang diukur
dalam perubahan kimia per unit masa.

Rate of reaction = Change in quantity of reactant/product
Time taken for the change to occur

Kadar tindak balas = Perubahan kuantiti bahan/hasil

Masa yang diambil untuk perubahan berlaku

How to identify the change in The change in amount of reactant or product in any reaction which is chosen for the purpose of measuring
quantity of reactant / product for rate of reaction must be observable and measurable.
measuring rate of reaction? Perubahan jumlah bahan atau hasil dalam tindak balas yang dipilih untuk mengukur kadar tindak balas
Give example. mestilah boleh diperhatikan dan diukur.
Bagaimanakah cara mengenal
pasti perubahan dalam kuantiti Example / Contoh:
bahan / hasil tindak balas untuk (a) Decrease in the mass of reactant.
mengukur kadar tindak balas? Pengurangan dalam jisim bahan tindak balas.
Berikan contoh. (b) Increase in the mass of product.
Peningkatan dalam jisim hasil tindak balas.
(c) Increase in volume of gas released.
Peningkatan dalam isi padu gas yang dibebaskan.
(d) Formation of precipitate as a product.
Pembentukan mendakan sebagai hasil.

© Nilam Publication Sdn Bhd 8

MODULE • Chemistry Form 5

How to measure the observable Chemical reaction Observable changes Method of measuring the
changes when a reaction produce Tindak balas kimia Perubahan yang boleh observable changes
gas?
Bagaimanakah cara mengukur Reaction between diperhatikan Kaedah pengukuran perubahan
perubahan yang dapat magnesium and yang boleh diperhatikan
diperhatikan apabila tindak balas hydrochloric acid: Decrease in mass of
menghasilkan gas? Tindak balas antara magnesium
magnesium dan asid Pengurangan jisim
hidroklorik: magnesium Hydrochloric acid Magnesium
Asid hidroklorik Magnesium

100 g U
N
Mg(s/p) + 2HCl(aq/ak) → Reading from the balance is recorded in I
MgCl2(aq/ak) + H2(g/g) every 30 seconds. T
Bacaan daripada penimbang direkodkan
Increase in volume of setiap 30 saat. 1
hydrogen
Peningkatan isi padu
hidrogen

Hydrochloric acid Water
Asid hidroklorik Air

Magnesium
Magnesium

Hydrogen gas is collected by water
displacement in a burette. The volume of
hydrogen gas collected is recorded every
30 seconds.
Gas hidrogen diperoleh dengan cara
penyesaran air di dalam buret. Isi padu gas
hidrogen yang diperoleh direkod setiap
30 saat.
* This apparatus set-up can also be used to

measure the increase in volume of other
gases that are insoluble for example
oxygen, hydrogen and carbon dioxide.
Susunan alat radas ini juga boleh
digunakan untuk mengukur peningkatan isi
padu gas lain yang tak larut seperti oksigen,
hidrogen dan karbon dioksida.

How to measure the observable Chemical reaction Observable changes Method of measuring the observable
change when a reaction produces Tindak balas kimia Perubahan yang boleh changes
precipitate?
Bagaimanakah cara mengukur diperhatikan Kaedah pengukuran perubahan yang
perubahan yang dapat boleh diperhatikan
diperhatikan apabila tindak balas
menghasilkan mendakan? Reaction between sodium Formation of sulphur as a
thiousulphate and precipitate
hydrochloric acid: Pembentukan sulfur Sodium thiosulphate
Tindak balas antara natrium sebagai mendakan solution + hydrochloric acid
tiosulfat dan asid hidroklorik: * Volume of sulphur Larutan natrium tiosulfat
+ asid hidroklorik
Na2S2O3(aq/ak) + 2HCl(aq/ak) dioxide gas, SO2 cannot
→ 2NaCl(aq/ak) + H2O(l/ce ) + be measured by water Amount of solid sulphur formed is
SO2(g/g) + S(s/p) displacement because measured by the time taken for the mark
sulphur dioxide is ‘X’ placed under the conical flask can no
soluble in water longer be seen.
Isi padu gas sulfur Jumlah pepejal sulfur yang terbentuk
dioksida, SO2 tidak diukur dengan masa yang diambil untuk
boleh diukur dengan tanda ‘X’ yang diletak di bawah kelalang
cara penyesaran air kon hilang dari penglihatan.
kerana sulfur dioksida
larut dalam air.

9 © Nilam Publication Sdn Bhd

© Nilam Publication Sdn Bhd How to determine the rate of reaction from the graph? MODULE • Chemistry Form 5
Bagaimanakah cara menentukan kadar tindak balas daripada graf?
U
Measurement of Rate of Reaction / Pengukuran Kadar Tindak Balas N
I
T

1
Rate of reaction at any given time/temperature/concentration Average rate of reaction
Kadar tindak balas pada masa /suhu/kepekatan tertentu Kadar tindak balas purata

Rate of reaction is measured by volume of gas Rate of reaction is measured by time taken for Average rate of reaction from 0 second: Average rate of reaction within certain
released in every 30 seconds by water the formation of precipitate. Kadar tindak balas purata dari 0 saat: period of time:
displacement in a burette. Kadar tindak balas diukur dengan masa yang Rate of reaction is measured by volume of Kadar tindak balas purata dalam suatu jangka
Kadar tindak balas diukur dengan isi padu gas diambil bagi pembentukan mendakan. gas released in every 30 seconds by water masa:
terbebas dalam setiap 30 saat melalui displacement in a burette. Rate of reaction is measured by volume of gas
penyesaran air dalam buret. Example / Contoh Kadar tindak balas diukur dengan isi padu gas released in every 30 seconds by water
terbebas dalam setiap 30 saat melalui displacement in a burette.
Example / Contoh penyesaran air dalam buret. Kadar tindak balas diukur dengan isi padu gas
terbebas dalam setiap 30 saat melalui
Sodium thiosulphate solution + hydrochloric acid Example / Contoh penyesaran air dalam buret.
Larutan natrium tiosulfat + asid hidroklorik
Example / Contoh
Sketch of graph / Lakaran graf:
Hydrochloric acid Water (a) Concentration of sodium thiosulphate solution/ mol dm–3
Asid hidroklorik Air
Calcium carbonate Kepekatan natrium tiosulfat / mol dm–3
10 Kalsium karbonat Hydrochloric acid Water
KCeopnecekanMttar1antinoantroiuf msotdioiusmulftahtio/smuolplhdamte–3solution/ mol dm–3 Asid hidroklorik Air
Sketch of graph: Calcium carbonate
Lakaran graf: Kalsium karbonat Hydrochloric acid Water
Asid hidroklorik Air
Volume of carbon dioxide gas/cm3 Sketch of graph: Calcium carbonate
Isi padu gas karbon dioksida/cm3 M1 t1 Time/s Lakaran graf: Kalsium karbonat
Masa/s
Volume of carbon dioxide gas/cm3 Sketch of graph:
Rate of reaction for sodium thiosulphate Isi padu gas karbon dioksida /cm3 Lakaran graf:
sKtiooalsduuatilTSforeaunmthtiunpdwendreaaiatttrnuhikurtgem1cbaolofaanrsnuloaktcdasieneumptnbioettsahrukiagolfasatiuittoll/apa°nhnCrautMeMtas1o1nlmumtniooonall/tdr°dMTCimuimmams–e–a3/3/ss
∆y V Volume of carbon dioxide gas/cm3
Isi padu gas karbon dioksida /cm3
∆x Time /s = 1 = x s–1
Masa /s t1 s V1
t1 V2

(b) TempeTr1ature of sodium thiosulphate solution / °C
Suhu natrium larutan tiosulfat / °C
The rate of reaction at t1 second
Kadar tindak balas pada masa t1 saat T1 t1 Time/s 1 234 5 6 7 8 Time/s
= T he gradient of tangent to the curve at t1 s Masa/s Masa/s

Kecerunan tangen pada lengkung pada t1 s Average rate of reaction in the first 4 minutes 1 234 5 6 7 8 Time/s
Masa/s
= D y cm3 Kadar tindak balas purata dalam 4 minit
Dxs Average rate of reaction in the fourth minute
Time/s pertama
t1 Masa/s (V – 0) cm3 Kadar tindak balas purata dalam minit keempat
= (4 – 0) s = x cm3 s–1 (V2 – V1) cm3
Rate of reaction for sodium thiosulphate = (4 – 3) s = y cm3 s–1

solution at temperature T1 °C

Kadar tindak balas bagi larutan natrium

tiosulfat pada suhu T1 °C
1
= t1 s = y s–1

MODULE • Chemistry Form 5

Exercise / Latihan

1 An experiment is carried out to determine the rate of reaction of 20 cm3 of 0.5 mol dm–3 hydrochloric acid with excess calcium
carbonate. The results are shown below.

Suatu eksperimen dijalankan untuk menentukan kadar tindak balas 20 cm3 asid hidroklorik 0.5 mol dm–3 dengan kalsium
karbonat berlebihan. Keputusannya ditunjukkan di bawah.

Time/s 0 15 30 45 60 75 90 105 120 135 150 165
Masa/s

Volume of CO2/cm3 0.00 10.00 16.00 22.00 27.00 31.50 36.00 39.50 42.00 44.00 44.00 44.00 U
Isi padu CO2 /cm3 N
I
(a) (i) Write a chemical equation for the above reaction. T
Tuliskan persamaan kimia bagi tindak balas di atas.
1
CaCO3 + 2HCl → CaCl2 + H2O + CO2

(ii) State the observable and measurable changes in the experiment.
Nyatakan perubahan yang boleh dilihat dan diukur dalam eksperimen ini.

Increase in volume of carbon dioxide/decrease in mass of calcium carbonate

Peningkatan isi padu gas karbon dioksida/penurunan jisim kalsium karbonat

(iii) State the meaning of the rate of reaction for the above reaction.
Nyatakan maksud kadar tindak balas bagi tindak balas di atas.

Change in volume of carbon dioxide gas in one second/change in mass of calcium carbonate in one second.

Perubahan isi padu gas karbon dioksida dalam satu saat/ perubahan jisim kalsium karbonat dalam satu saat.

(iv) Draw an apparatus set-up to measure rate of reaction in the given reaction.
Lukiskan susunan alat radas yang digunakan untuk mengukur kadar tindak balas dalam tindak balas tersebut.

Hydrochloric acid / Asid hidroklorik Water
Air
Calcium carbonate
Kalsium karbonat

11 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

(b) Draw the graph of the volume of carbon dioxide gas collected against time.
Lukiskan graf isi padu gas karbon dioksida yang terkumpul melawan masa.

Volume of CO2 /cm3
Isi padu CO2 /cm3

50

U
N
I
T

1 40

30

20

10

0 Time/s
30 60 90 120 150 180 Masa/s

© Nilam Publication Sdn Bhd 12

MODULE • Chemistry Form 5

(c) From the graph, determine: / Daripada graf, tentukan:
(i) the average rate of reaction in the first minute.
kadar tindak balas purata dalam minit pertama

Total volume of carbon dioxide gas collected in the first minute
Jumlah isi padu gas karbon dioksida yang terkumpul dalam minit pertama
=

Time taken for the change to occur
Masa yang diambil bagi perubahan berlaku

= 27 U
60 N
I
= 0.45 cm3 s–1 T

(ii) the average rate of reaction in the second minute / kadar tindak balas purata dalam minit kedua 1

Total volume of carbon dioxide gas collected between first minute and the second minute
Jumlah isi padu gas karbon dioksida yang terkumpul antara minit pertama dan minit kedua
=

Time taken for the change to occur
Masa yang diambil bagi perubahan berlaku

= 42 – 27
60

= 0.25 cm3 s–1

(iii) the time when the reaction has completed / masa apabila tindak balas selesai
135 s

(iv) the average rate of reaction for overall reaction / kadar tindak balas purata bagi tindak balas keseluruhannya
Total volume of carbon dioxide collected / Jumlah isi padu gas karbon dioksida yang terkumpul

=
Time taken for the change to occur / Masa yang diambil bagi perubahan berlaku

= 44
135

= 0.326 cm3 s–1

(v) the rate of reaction at 30 seconds / kadar tindak balas pada masa 30 saat
= the gradient of the graph at 30 seconds / kecerunan graf pada masa 30 saat

= 0.405 ± 0.1 cm3 s–1

(vi) the rate of reaction at 105 seconds / kadar tindak balas pada masa 105 saat
= the gradient of the graph at 105 seconds / kecerunan graf pada masa 105 saat

= 0.217 ± 0.1 cm3 s–1

(d) Compare the rate of reaction at 30 seconds and 105 seconds. Explain your answer.
Bandingkan kadar tindak balas pada masa 30 saat dan 105 saat. Terangkan jawapan anda.

Rate of reaction at 30 seconds is higher than at 105 seconds because the concentration of hydrochloric acid decreases as time
increases. / Kadar tindak balas pada masa 30 saat lebih tinggi daripada pada masa 105 saat kerana kepekatan asid hidroklorik
berkurang dengan masa.

13 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

2 Excess of zinc powder is added to 50 cm3 of 1 mol dm–3 hydrochloric acid. The volume of hydrogen gas collected and time taken
are recorded. Complete the following table.

Serbuk zink berlebihan ditambah kepada 50 cm3 asid hidroklorik 1 mol dm–3. Isi padu gas yang dikumpul dan masa yang
diambil direkodkan. Lengkapkan jadual berikut.

Sketch a curve for volume of hydrogen gas Volume of hydrogen/cm3
collected against time for the reaction between Isi padu hidrogen/cm3
excess of zinc powder with 50 cm3 of 1 mol dm–3

hydrochloric acid. The tangents on the curve at V

t1, t2 and t3 are shown. Tangent on the curve at t2 t3 Time/min
Lakarkan graf bagi isi padu gas hidrogen yang t1, t2 and t3 respectively Masa/min
Tangen bagi graf pada
U terkumpul melawan masa untuk tindak balas t1, t2 dan t3
N antara serbuk zink berlebihan dengan 50 cm3
I asid hidroklorik 1 mol dm–3. Tangen bagi graf 0 t1

1T pada t1, t2 dan t3 ditunjukkan.

Write the balanced equation for the reaction. Zn + 2HCl → ZnCl2 + H2
Tuliskan persamaan seimbang bagi tindak balas.

Calculate the volume of hydrogen gas collected From the equation / Dari persamaan,
in the experiment at room conditions. 2 mol of HCl : 1 mol H2
Hitung isi padu gas hidrogen yang terkumpul 0.05 mol HCl : 0.025 mol H2
dalam eksperimen pada keadaan bilik.
Volume of H2 / Isi padu H2
Compare the gradient of the curve at t1 and t2. = 0.025 mol × 24 dm3 mol–1
Explain your answer. = 0.6 dm3
Bandingkan kecerunan graf pada t1 dan t2. = 600 cm3
Terangkan jawapan anda.
The gradient of tangent on the curve at t2 is lower than t1. The rate of reaction at
What is the gradient at t3? Explain your answer.
Apakah kecerunan pada t3? Terangkan jawapan t2 is lower than at t1. The rate of reaction decreases as the time increases
anda.
because mass of zinc and concentration of hydrochloric acid decreases .

Kecerunan tangen pada graf di t2 lebih rendah berbanding di t1. Kadar

tindak balas di t2 lebih rendah berbanding di t1. Kadar tindak balas

berkurang apabila masa meningkat kerana jisim zink dan

kepekatan asid hidroklorik berkurang

The gradient of tangent on the curve at t3 is zero , the rate of reaction at t3 is

zero . The reaction is completed at t3. All hydrochloric acid has

reacted because zinc powder used is in excess . At t3,

maximum volume of hydrogen gas is collected. The maximum volume of hydrogen

gas collected is 600 cm3 .

Kecerunan tangen bagi graf di t3 adalah sifar , kadar tindak balas di t3

adalah sifar . Tindak balas ini lengkap di t3. Semua asid

hidroklorik telah bertindak balas kerana serbuk zink yang digunakan adalah

berlebihan . Pada masa t3, isi padu gas hidrogen yang maksimum dikumpulkan.

Isi padu maksimum gas hidrogen yang dikumpulkan adalah 600 cm3 .

© Nilam Publication Sdn Bhd 14

MODULE • Chemistry Form 5

Sketch a curve for mass of zinc against time. Mass of zinc / g
Lakarkan lengkung bagi jisim zink melawan Jisim zink / g
masa.

Time / s U
Masa / s N
I
Sketch a curve for concentration of hydrochloric Concentration of hydrochloric acid / mol dm–3 T
acid against time. Kepekatan asid hidroklorik / mol dm–3
Lakarkan lengkung bagi kepekatan asid 1
hidroklorik melawan masa.

Time / s
Masa / s

Factors that Affect the Rate of Reaction
Faktor yang Mempengaruhi Kadar Tindak Balas

1 The rate of reaction is affected by:
Kadar tindak balas dipengaruhi oleh:

(a) size of solid reactant
saiz bahan tindak balas pepejal
(b) concentration of solution (for the reactant used in the form of solution)
kepekatan larutan (bagi bahan tidak balas yang digunakan dalam bentuk larutan)
(c) temperature of solution at which the reaction occurs
suhu larutan ketika tindak balas berlaku
(d) presence of catalyst (for a particular reaction)
kehadiran mangkin (untuk tindak balas tertentu)
(e) pressure of gas reactant
tekanan gas bahan tindak balas

15 © Nilam Publication Sdn Bhd

U
N
I
T

1

© Nilam Publication Sdn Bhd Planning of Experiments to Study the Factor that Affect the Rate of Reaction MODULE • Chemistry Form 5

Size of solid reactant Catalyst Concentration of solution Temperature

Sulphuric acid Sodium 34 35 36 37 38 39 40 41 42 43 Sodium thiosulphate Conical
+ thiosulphate solution flask
+
Copper(II) sulphate solution
sulphuric acid

Hydrochloric Water Sodium thiosulphate solution Paper X
acid + Heat

Calcium carbonate Zinc Sulphuric acid ‘X’ mark

Water Problem statement:
How does the concentration of sodium
Problem statement: Problem statement: thiousulphate solution affect the rate of its reaction Problem statement:
with sulphuric acid?
16 How does the size of calcium carbonate chips affect Does the presence of copper(II) sulphate solution Hypothesis: How does the temperature of sodium thiousulphate
the rate of its reaction with dilute hydrochloric acid? affect the rate of reaction between zinc and When the concentration of sodium thiosulphate solution affect the rate of its reaction with sulphuric
sulphuric acid? solution increases, the rate of its reaction with acid?
Hypothesis: Hypothesis: sulphuric acid increases. Hypothesis:
Manipulated variable:
The smaller the size of calcium carbonate, the The presence of copper(II) sulphate solution in the Concentration of sodium thiosulphate solution When the temperature of sodium thiosulphate
higher the rate of reaction. reaction between zinc and sulphuric acid increases solution increases, the rate of its reaction with
the rate of reaction. sulphuric acid increases.
Manipulated variable: Manipulated variable: Manipulated variable:

The size of calcium carbonate The presence of copper(II) sulphate Temperature of sodium thiosulphate solution

Responding variable: Responding variable: Responding variable: Responding variable:
The rate of reaction The rate of reaction
Fixed variables: Fixed variables: The rate of reaction The rate of reaction
Mass of zinc, the volume and concentration of Fixed variables: Fixed variables:
Mass of calcium carbonate chips, volume and sulphuric acid, H2SO4, the temperature of the
concentration of hydrochloric acid, the temperature reaction mixture. Volume of sodium thiosulphate solution, volume and Volume and concentration of sodium thiosulphate
of the reaction mixture.
concentration of dilute sulphuric acid, temperature solution, volume and concentration of dilute

of mixture, size of conical flask. sulphuric acid, size of conical flask.

Apparatus: Apparatus: Apparatus: Apparatus:

Conical flask, basin, delivery tube and rubber 100 cm3 of conical flask, basin, burette, delivery 100 cm3 conical flask, 50 cm3 and 5 cm3 100 cm3 conical flask, 50 cm3 and 5 cm3
stopper, retort stand and clamp, measuring tube and rubber stopper, retort stand and measuring cylinder, stop watch, white paper measuring cylinder, stop watch, thermometer.
cylinder, burette, stop watch and weighing clamp, measuring cylinder and weighing with mark ‘X’.
balance. balance.

Materials: Materials: Materials: Materials:
Large and small calcium carbonate chips, Granulated zinc, 0.1 mol dm–3 dilute sulphuric 0.2 mol dm–3 sodium thiosulphate solution, 0.2 mol dm–3 sodium thiosulphate solution,
0.2 mol dm–3 of hydrochloric acid. acid, 0.5 mol dm–3 copper(II) sulphate solution. 1 mol dm–3 sulphuric acid, distilled water. 1 mol dm–3 sulphuric acid, distilled water.

PROCEDURE: PROCEDURE: PROCEDURE: PROCEDURE:
1 A basin is filled with water until half full. 1 A basin is filled with water until half full.
2 A burette full with water is inverted into the 2 A burette full with water is inverted into the 1 50 cm3 of sodium thiosulphate solution is 1 50 cm3 of sodium thio­sul­phate solution is

basin. basin. measured and poured into a conical flask. measured and poured into a conical flask.
3 It is then clamped vertically using retort 3 It is then clamped vertically using retort stand.
4 5 g of granulated zinc is weight and put into 2 The conical flask is placed on top of a piece 2 The temperature of this solution is measured
stand.
4 5 g of large calcium carbonate chips is the conical flask. of paper with a mark ‘X’ at the centre. using thermometer and recorded.
5 50 cm3 of 0.2 mol dm–3 sulphuric acid is
weighed and put inside the conical flask as 3 5 cm3 of sulphuric acid 1 mol dm–3 is 3 The conical flask is placed on top of a piece of
shown in the diagram. measured and poured into the flask as
5 50 cm 3 of hydrochloric acid 0.2 mol dm –3 is shown in the diagram. measured and poured quickly and carefully paper with a mark ‘X’ at the centre.
measured and poured into the conical flask. 6 1 cm3 of 1.0 mol dm–3 copper(II) sulphate
6 The conical flask is closed immediately with solution is measured and poured in the into the conical flask. Swirl the conical flask 4 5 cm3 of sulphuric acid is measured and
a stopper fitted with delivery tube as shown conical flask.
in the diagram. 7 The conical flask is closed immediately with at the same time start the stop watch. poured quickly and carefully into the conical
7 A stop watch is started immediately. a stopper fitted with delivery tube as shown
8 The volume of gas released is recorded for in the diagram. 4 The mark ‘X’ as shown in the above diagram flask. Swirl the conical flask at the same time
every 30 seconds. 8 A stop watch is started immediately.
9 Steps 1 to 8 are repeated using 5 g small 9 The volume of gas released is recorded for is observed. start the stop watch.
calcium carbonate chips to replace 5 g of every 30 seconds.
large calcium carbonate chips. 10 Steps 1 to 9 are repeated without adding 5 The stop watch is stopped immediately when 5 The mark ‘X’ as shown in the above diagram
10 The graphs of volume of carbon dioxide copper(II) sulphate solution.
against time for the two experiments are 11 The graphs of volume of hydrogen gas the mark ‘X’ is no longer visible. is observed.
plotted on the same axes. against time for the two experiments are
plotted on the same axes. 6 The time taken for the mark ‘X’ is no longer 6 The stop watch is stopped immediately when

visible is recorded. the mark ‘X’ is no longer visible.

7 Steps 1 to 6 are repeated using different 7 The time taken for the mark ‘X’ is no longer

volumes of sodium thiosulphate solution with visible is recorded.

different volumes of distilled water as shown 8 Steps 1 to 7 are repeated using same volume and

in the table. conc­ entration of sodium thiosulphate solution but
MODULE • Chemistry Form 5
8 Graphs of concentration of sodium heate­d gently to a higher temperature of 35°C,
U
Nthiosulphate against time and concentration40°C, 45°C, 50°C and 55°C. All other conditions
I1
T ofsodiumthiosulphateagainsttimeareremain unchanged.

1 9 A graph of temperature against time is plotted
plotted. 1
17 © Nilam Publication Sdn Bhd and temperature against time are plotted.

TABULATION OF DATA: TABULATION OF DATA: TABULATION OF DATA: TABULATION OF DATA:
A Large calcium carbonate chips A Granulated zinc with copper(II) sulphate
Volume of Na2S2O3 / 50 40 30 20 10 Temperature of 55 50 45 40 35 30
Time/s Time/s cm3 Na2S2O3, solution /
Burette reading/ °C
Burette reading/ cm 3
cm 3 Volume of gas/ Volume of water/ 0 10 20 30 40
Volume of gas/ cm3 cm3
cm3 Time taken for ‘X’
B Granulated zinc without copper(II) sulphate Concentration of to disappear/ s
B Small calcium carbonate chips Na2S2O3 solution /
Time/s mol dm –3
Time/s Burette reading/
Burette reading/ cm 3 Time taken for ‘X’
cm 3 Volume of gas/ to disappear/s
Volume of gas/ cm3
cm3 1 1 /s–1
time time
/s–1

U
N
I
T

1

© Nilam Publication Sdn Bhd Perancangan Eksperimen untuk Mengkaji Faktor yang Mempengaruhi Kadar Tindak Balas MODULE • Chemistry Form 5

Saiz bahan tindak balas pepejal Mangkin Kepekatan larutan Suhu

Asid sulfurik Larutan natrium 34 35 36 37 38 39 40 41 42 43 Larutan natrium Kelalang
+ tiosulfat tiosulfat kon
+
Kuprum(II) sulfat
asid sulfurik

Asid hidroklorik Air Larutan natrium tiosulfat Kertas
+

Asid sulfurik

X

Kalsium karbonat Zink Panaskan Tanda ‘X’

Air

Pernyataan masalah: Pernyataan masalah: Pernyataan masalah: Pernyataan masalah:

18 Bagaimanakah saiz ketulan kalsium karbonat Adakah kehadiran larutan kuprum(II) sulfat Bagaimanakah kepekatan larutan natrium tiosulfat Bagaimanakah suhu larutan natrium tiosulfat
mempengaruhi kadar tindak balas dengan asid mempengaruhi kadar tindak balas antara zink mempengaruhi kadar tindak balas dengan mempengaruhi kadar tindak balas dengan asid
hidroklorik? dengan asid sulfurik? asid sulfurik? sulfurik?
Hipotesis: Hipotesis: Hipotesis: Hipotesis:

Semakin kecil saiz kalsium karbonat, semakin tinggi Kehadiran larutan kuprum(II) sulfat dalam tindak Apabila kepekatan larutan natrium tiosulfat Apabila suhu larutan natrium tiosulfat meningkat,
kadar tindak balas. balas antara zink dengan asid sulfurik meningkat, kadar tindak balasnya dengan asid kadar tindak balasnya dengan asid sulfurik juga
meningkatkan kadar tindak balas. sulfurik juga meningkat. meningkat.
Pemboleh ubah dimanipulasi: Pemboleh ubah dimanipulasi: Pemboleh ubah dimanipulasi: Pemboleh ubah dimanipulasi:

Saiz kalsium karbonat Kehadiran kuprum(II) sulfat Kepekatan larutan natrium tiosulfat. Suhu larutan natrium tiosulfat
Pemboleh ubah bergerak balas: Pemboleh ubah bergerak balas: Pemboleh ubah bergerak balas: Pemboleh ubah bergerak balas:

Kadar tindak balas Kadar tindak balas Kadar tindak balas Kadar tindak balas
Pemboleh ubah dimalarkan: Pemboleh ubah dimalarkan: Pemboleh ubah dimalarkan: Pemboleh ubah dimalarkan:

Jisim kalsium karbonat, isi padu dan kepekatan asid Jisim zink, isi padu dan kepekatan asid sulfurik, Isi padu larutan natrium tiosulfat, isi padu dan Isi padu dan kepekatan larutan natrium tiosulfat, isi
kepekatan asid sulfurik cair, suhu campuran, saiz padu, kepekatan asid sulfurik cair, saiz kelalang kon
hidroklorik, suhu campuran tindak balas. suhu campuran tindak balas. kelalang kon

Radas: Radas: Radas: Radas:
Kelalang kon 100 cm3, silinder penyukat 50 cm3 Kelalang kon 100 cm3, silinder penyukat 50 cm3
Kelalang kon, besen, salur penghantar dengan Kelalang kon, besen, buret, salur penghantar dan 5 cm3, jam randik, kertas putih bertanda ‘X’ dan 5 cm3, jam randik, termometer
penyumbat, kaki retort, silinder penyukat, buret, dengan penyumbat, kaki retort, silinder
jam randik dan penimbang. penyukat dan penimbang.

Bahan: Bahan: Bahan: Bahan:
Ketulan kalsium karbonat besar dan kecil, Butiran zink, asid sulfurik 0.1 mol dm–3, 0.2 mol dm–3 larutan natrium tiosulfat, 0.2 mol dm–3 larutan natrium tiosulfat,
asid hidroklorik 0.2 mol dm–3. larutan kuprum(II) sulfat 0.5 mol dm–3 1 mol dm–3 asid sulfurik, air suling air suling, 1 mol dm–3 asid sulfurik.

PROSEDUR: PROSEDUR: PROSEDUR: PROSEDUR:
1 Sebuah besen diisi dengan air sehingga
1 Sebuah besen diisi dengan air sehingga 1 50 cm3 larutan natrium tiosulfat disukat dan 1 50 cm3 larutan natrium tiosulfat disukat dan
separuh penuh.
2 Sebuah buret yang telah diisi penuh dengan separuh penuh. dituang ke dalam kelalang kon. dituang ke dalam kelalang kon.

air diterbalikkan di dalam besen tersebut. 2 Sebuah buret yang telah diisi penuh dengan 2 Kelalang kon tersebut diletakkan di atas 2 Suhu larutan tersebut disukat menggunakan
3 Buret tersebut kemudiannya diapitkan air diterbalikkan di dalam besen tersebut.
sekeping kertas yang telah ditanda dengan termometer dan direkodkan.
menegak menggunakan kaki retort. 3 Buret tersebut kemudiannya diapitkan
4 5 g ketulan kalsium karbonat besar ditimbang 4 menegak menggunakan kaki retort. tanda ‘X’ di tengah. 3 Kelalang kon tersebut diletakkan di atas
5 g butiran zink ditimbang dan dimasukkan 3 5 cm3 asid sulfurik 1 mol dm–3 disukat dan
dan dimasukkan ke dalam kelalang kon 5 ke dalam kelalang kon seperti yang sekeping kertas yang telah ditanda dengan
seperti yang ditunjukkan dalam rajah. ditunjukkan dalam rajah. dituang dengan cepat dan cermat ke dalam
5 50 cm 3 asid hidroklorik 0.2 mol dm–3 disukat 50 cm3 asid sulfurik 0.2 mol dm–3 disukat dan kelalang kon. Goncangkan kelalang kon. tanda ‘X’ di tengah.
dan dituang ke dalam kelalang kon. Pada masa yang sama jam randik dimulakan.
6 Kelalang kon ditutup dengan serta merta 4 5 cm3 asid sulfurik disukat dan dituang
menggunakan penyumbat bersama salur
penghantar. 4 Tanda ‘X’ seperti dalam rajah di atas dengan cepat dan cermat ke dalam kelalang
7 Jam randik dimulakan dengan serta merta.
8 Isi padu gas yang dibebaskan direkod setiap dituang ke dalam kelalang kon. diperhatikan. kon. Goncangkan kelalang kon dan pada
30 saat.
9 Langkah 1 hingga 8 diulang menggunakan 6 1 cm3 larutan kuprum(II) sulfat 1.0 mol dm–3 masa yang sama jam randik dimulakan.
5 g ketulan kalsium karbonat kecil disukat dan dituang ke dalam kelalang kon. 5 Jam randik diberhentikan sebaik sahaja
menggantikan 5 g ketulan kalsium karbonat tanda ‘X’ tidak kelihatan. 5 Tanda ‘X’ seperti yang ditunjukkan dalam
besar. 7 Kelalang kon ditutup dengan serta merta
10 Graf isi padu karbon dioksida melawan menggunakan penyumbat bersama salur 6 Masa yang diambil untuk tanda ‘X’ untuk rajah diperhatikan.
masa diplot pada paksi yang sama.
penghantar. tidak kelihatan direkod. 6 Jam randik diberhentikan sebaik sahaja
PENJADUALAN DATA:
A Ketulan besar kalsium karbonat 8 Jam randik dimulakan dengan serta merta. 7 Langkah 1 hingga 6 diulang menggunakan isi tanda ‘X’ tidak kelihatan.

Masa/s 9 Isi padu gas yang dibebaskan direkod setiap padu larutan natrium tiosulfat yang berbeza 7 Masa yang diambil untuk tanda ‘X’ untuk

Bacaan buret/cm 3 30 saat. dan isi padu air suling yang berbeza seperti tidak kelihatan direkod.

Isi padu gas/cm 3 10 Langkah 1 hingga 9 diulang tanpa 8 Lang­kah 1 hingga 7 diulang menggunakan isi
MODULE • Chemistry Form 5yang ditunjukkan di dalam jadual di bawah.
B Ketulan kecil kalsium karbonat menambah larutan kuprum(II) sulfat. padu dan kepekatan larutan natrium tiosulfat
U8 Graf kepekatan larutan natrium tiosulfat
Masa/s N11 Graf isi padu gas hidrogen melawan masayang sama tetapi dipanaskan kepada suhu
Ibagi kedua-dua eksperimen diplotkan dalammelawan masa dan kepekatan larutan
Bacaan buret/cm 3 Tpaksi yang sama.1yang lebih tinggi iaitu 35°C, 40°C, 45°C, 50°C
natrium tiosulfat melawan masa diplot.
Isi padu gas/cm 3 1 dan 55°C. Semua pemboleh ubah lain tidak

19 © Nilam Publication Sdn Bhd berubah.

PENJADUALAN DATA: 9 Graf suhu larutan natrium tiosulfat melawan
A Butiran zink dengan larutan kuprum(II) sulfat
masa dan suhu larutan natrium tiosulfat
1
PENJADUALAN DATA: melawan masa diplot.

Masa/s Isi padu larutan 50 40 30 20 10 PENJADUALAN DATA:
Bacaan buret/cm 3 natrium tiosulfat /
cm3
Suhu larutan 55 50 45 40 35 30
Isi padu gas/cm 3 Isi padu air/cm 3 0 10 20 30 40 Na2S2O2 /°C
B Butiran zink tanpa larutan kuprum(II) sulfat
Kepekatan larutan Masa yang
Masa/s natrium tiosulfat / diambil untuk ‘X’
Bacaan buret/cm 3 mol dm–3 hilang/ s

Masa yang 1 /s–1
diambil untuk ‘X’ masa
hilang/s

Isi padu gas/cm 3 1 /s–1
masa

© Nilam Publication Sdn Bhd 2 Intepretation of data and conclusion for the experiments to investigate factors that affect the rate of reaction. MODULE • Chemistry Form 5
Pentafsiran data dan kesimpulan bagi eksperimen untuk mengkaji faktor-faktor yang mempengaruhi kadar tindak balas.
U
(a) Factor: Size of reactant / Faktor: Saiz bahan tindak balas N
I
T

1
Sketch of the graph / Lakaran graf Interpretation and conclusion / Tafsiran dan kesimpulan

Experiment I: Volume of carbon dioxide /cm3 1 From the sketch of graph in Diagram 1:
Eksperimen I: Isi padu karbon dioksida /cm3
Dari lakaran graf dalam Rajah 1:
50 cm3 of 0.2 mol dm–3 hydrochloric acid + excess of small calcium Experiment I
carbonate chips. Eksperimen I (a) The average rate of reaction for the first t1 seconds in experiment I
50 cm3 asid hidroklorik 0.2 mol dm–3 + ketulan kecil kalsium karbonat V1
berlebihan. Experiment II Kadar tindak balas purata dalam masa t1 saat pertama dalam
V2 Eksperimen II
Experiment II: eksperimen I
Eksperimen II: V1 cm3
= t1 s = X cm3 s–1

(b) The average rate of reaction for the first t1 seconds in experiment II

Kadar tindak balas purata dalam masa t1 saat pertama dalam

50 cm3 of 0.2 mol dm–3 hydrochloric acid + excess of large calcium eksperimen II
carbonate chips. V2 cm3
50 cm3 asid hidroklorik 0.2 mol dm–3 + ketulan besar kalsium karbonat t1 Time/s = t1 s = Y cm3 s–1
berlebihan. Diagram 1 / Rajah 1 Masa/s

The value of V1 is larger than V2 / Nilai V1 lebih besar dari V2

Balanced equation: • Total volume of carbon dioxide gas released ⇒ The average rate of reaction for the first t1 seconds in experiment
Persamaan seimbang:
20 CaCO3(s/p) + 2HCl(aq/ak) → CaCl2(aq/ak) + H2O(l/ce) + CO2(g/g) for the first t1 seconds in experiment I = V1 I is higher than experiment II. / Kadar tindak balas purata dalam
Jumlah isi padu karbon dioksida yang t1 saat pertama dalam eksperimen I adalah lebih tinggi dari
Observable changes to measure rate of reaction: dibebaskan pada t1 saat yang pertama dalam eksperimen II.
Perubahan yang diperhatikan untuk mengukur kadar tindak balas: eksperimen I = V1
Volume of carbon dioxide collected in every 30 seconds by water displacement • Total volume of carbon dioxide gas released 2 From the sketch of graph in Diagram 2: / Dari lakaran graf dalam Rajah 2:
in the burette / Isi padu gas karbon dioksida terkumpul setiap 30 saat melalui for the first t1 seconds in experiment II = V2
Jumlah isi padu karbon dioksida yang (a) The tangent at t1 in experiment I is steeper than experiment II.
dibebaskan pada t1 saat yang pertama dalam Tangen pada t1 dalam eksperimen I adalah lebih curam dari
eksperimen II = V2
eksperimen II.
(b) The gradient of tangent at t1 for experiment I is greater than

experiment II. / Kecerunan tangen pada t1 untuk eksperimen I
adalah lebih tinggi dari eksperimen II.
(c) The rate of reaction at t1 for experiment I is higher than experiment
penyesaran air dalam buret Volume of CO2 /cm3
Isi padu CO2 /cm3 II. / Kadar tindak balas pada t1 untuk eksperimen I adalah
* The size of CaCO3 is changed in both experiments. The volume and lebih tinggi dari eksperimen II.
concentration of HCl are kept constant. / Saiz CaCO3 diubah dalam Tangent at t1 for experiments I and II
kedua-dua eksperimen. Isi padu dan kepekatan HCl ditetapkan. Tangen pada t1 untuk eksperimen I dan II 3 Since calcium carbonate used is in excess, all hydrochloric acid has
reacted . / Oleh sebab kalsium karbonat yang digunakan adalah
Experiment I
Eksperimen I berlebihan, semua asid hidroklorik telah bertindak balas .

Experiment II The number of mole of hydrochloric acid in both experiments
Eksperimen II
Bilangan mol asid hidroklorik dalam kedua-dua eksperimen

= 50 u 0.2 = 0.01 mol
1 000

t1 Time/s 4 The volume of hydrogen gas collected for both experiments is
Diagram 2 / Rajah 2 Masa/s equal because number of mol of hydrochloric acid in experiments

I and II is equal . / Isi padu gas hidrogen yang dikumpulkan dalam
kedua-dua eksperimen adalah sama kerana bilangan mol asid
hidroklorik dalam eksperimen I dan eksperimen II adalah sama .
5 Conclusion: / Kesimpulan:
The rate of reaction of the small calcium carbonate chips is higher

than the larger calcium carbonate chips. / Kadar tindak balas ketulan
kecil kalsium karbonat lebih tinggi dari ketulan besar kalsium

karbonat.

(b) Factor: Concentration of solution / Faktor: Kepekatan larutan Sketch of the graph / Lakaran graf Interpretation and conclusion / Tafsiran dan kesimpulan

Experiment: / Eksperimen: Concentration of sodium thiosulphate solution / mol dm–3 1 From the sketch of graph in Diagram 3: / Dari lakaran graf dalam Rajah 3:
45 cm3 of 0.2 mol dm–3 sodium thiosulphate solution + 5 cm3 of Kepekatan larutan natrium tiosulfat / mol dm–3 (a) As the concentration of sodium thiosulphate solution decreases, a
1.0 mol dm–3 sulphuric acid. Experiment is repeated four more times using
0.2 mol dm–3 sodium thiosulphate solution diluted with different volume of longer time is needed for the mark ‘X’ to disappear from view.
distilled water. / 45 cm3 larutan natrium tiosulfat 0.2 mol dm–3 + 5 cm3 asid
sulfurik 1.0 mol dm–3. Eksperimen diulang empat kali menggunakan Apabila kepekatan larutan natrium tiosulfat berkurang, masa yang
0.2 mol dm–3 larutan natrium tiosulfat yang dicairkan dengan isi padu air
suling yang berbeza. lebih lama diperlukan untuk tanda ‘X’ hilang.

Balanced equation: / Persamaan seimbang: (b) The higher the concentration, the shorter is the time taken

Na2S2O3(aq/ak) + H2SO4(aq/ak) → Na2SO4(aq/ak) + H2O(l/ce) + SO2(g/g) + S(s/p) for the yellow sulphur precipitate to appear and the faster for

Observable changes to measure rate of reaction: the “X” sign to disappear. / Semakin tinggi kepekatan, semakin
Perubahan yang diperhatikan untuk mengukur kadar tindak balas:
Time/s pendek masa yang diambil untuk mendakan kuning sulfur
Time taken for the mark ‘X’ placed under the conical flask to disappear Masa/s kelihatan dan lebih cepat untuk tanda “X” untuk hilang.
from view. Fixed quantity of solid sulphur is formed in every experiment.
Masa yang diambil untuk tanda ‘X’ di bawah kelalang kon untuk hilang Diagram 3 / Rajah 3 2 From the sketch of graph in Diagram 4: / Dari lakaran graf dalam Rajah 4:
dari penglihatan. Kuantiti pepejal sulfur yang terbentuk dalam setiap
eksperimen adalah tetap. Concentration of sodium thiosulphate solution / mol dm–3 tAmAimsp1eaentbhiniienlgackcrkaeoetan.pscmeeekasn1a.sttraaatnitmm1iolaeenrwuroateafkpnisrlieonsdaeiturnimkutsamdttahhtrieiootissnuudrllaapfatkhetabomtafelearensinainccgtrieok.anast,e.nsi,latihme a1vsaaluteurouft
* The concentration of Na2S2O3(aq) is changed in all experiments. The Kepekatan larutan natrium tiosulfat / mol dm–3

volume and temperature of sulphuric acid are kept constant. / Kepekatan 1 /s–1 3 Conclusion: / Kesimpulan:
Na2S2O3(ak) ditukar dalam semua eksperimen. Isi padu dan suhu asid Time The higher the concentration of sodium thiosulphate solution, the
sulfurik dikekalkan. 1 higher is the rate of reaction . / Semakin tinggi kepekatan larutan
Masa /s–1

Diagram 4 / Rajah 4 natrium tiosulfat, semakin tinggi kadar tindak balas .

MODULE • Chemistry Form 5(c) F actor: Temperature of reaction mixtureSketch of the graph / Lakaran grafInterpretation and conclusion / Tafsiran dan kesimpulan
Faktor: Suhu campuran bahan tindak balas
UTemperature of sodium thiosulphate solution /°C1 From the sketch of graph in Diagram 5: / Dari lakaran graf dalam Rajah 5:
NExperiment: / Eksperimen:Suhu larutan natrium tiosulfat /°C (a) As the temperature of sodium thiosulphate solution decreases, a
I50 cm3 of 0.2 mol dm–3 sodium thiosulphate solution at 30°C + 5 cm3 of
T1.0 mol dm–3 sulphuric acid. Experiment is repeated using 50 cm3 oflonger time is needed for the mark ‘X’ to disappear from view. /
0.2 mol dm–3 sodium thiosulphate solution at 35°C, 40°C, 45°C and 50°C Apabila suhu larutan natrium tiosulfat berkurang, masa yang lebih
1respectively + 5 cm3 of 1.0 mol dm–3 sulphuric acid. / 50 cm3 larutan panjang diperlukan untuk tanda ‘X’ hilang dari pandangan.
natrium tiosulfat 0.2 mol dm–3 pada suhu 30°C + 5 cm3 asid sulfurik
21 © Nilam Publication Sdn Bhd1.0 mol dm–3. Eksperimen diulang menggunakan 50 cm3 larutan natrium
tiosulfat 0.2 mol dm–3 pada suhu 35°C, 40°C, 45°C dan 50°C + 5 cm3 asid (b) The higher the temperature, the shorter is the time taken for the
sulfurik 1.0 mol dm–3.
yellow sulphur precipitate to appear and the faster for the mark ‘X’
Balanced equation: / Persamaan seimbang:
to disappear. / Semakin tinggi suhu, semakin pendek masa yang
Na2S2O3(aq/ak) + H2SO4(aq/ak) → Na2SO4(aq/ak) + H2O(l/ce) + SO2(g/g)
Time/s diperlukan untuk mendakan kuning sulfur untuk kelihatan dan
+ S(s/p) Masa/s semakin cepat tanda ‘X’ hilang dari pandangan.

Observable changes to measure rate of reaction: Diagram 5 / Rajah 5 2 From the sketch of graph in Diagram 6: / Dari lakaran graf dalam Rajah 6:
Perubahan yang diperhatikan untuk mengukur kadar tindak balas:
Time taken for the mark ‘X’ placed under the conical flask to disappear Temperature of sodium thiosulphate solution /°C As the temperature of sodium thiosulphate solution increases, the
from view. Fixed quantity of solid sulphur is formed in every experiment. Suhu larutan natrium tiosulfat / °C 1 1
Masa yang diambil untuk tanda ‘X’ di bawah kelalang kon untuk hilang value of time increases . time represents the rate of reaction.
dari penglihatan. Kuantiti pepejal sulfur yang terbentuk dalam setiap nilai 1
eksperimen adalah tetap. Apabila suhu larutan natrium tiosulfat meningkat, masa turut
1
meningkat . masa mewakili kadar tindak balas.

1 /s–1 3 Conclusion: / Kesimpulan:
Time The higher the temperature of sodium thiosulphate solution, the higher
1
Masa /s–1 is the rate of reaction . / Semakin tinggi suhu larutan natrium tiosulfat,

Diagram 6 / Rajah 6 semakin tinggi kadar tindak balas .

© Nilam Publication Sdn Bhd MODULE • Chemistry Form 5

U
N
I
T

1
3 The curve for graph of amount of product formed against time in a chemical reaction consists of two parts:
Graf bagi kuantiti hasil tindak balas yang terbentuk melawan masa bagi suatu tindak balas kimia terbahagi kepada dua bahagian:

(a) The maximum quantity of product: It depends on the number of mol of reactants react in the chemical reaction.
Kuantiti hasil maksimum: Ia bergantung pada bilangan mol bahan tindak balas yang bertindak balas dalam tindak balas kimia tersebut.

(b) The gradient of the curve: It depends on the factors that affect the rate of reaction.
Kecerunan graf: Ia bergantung pada faktor-faktor yang mempengaruhi kadar tindak balas.

Factor Effect on the gradient of the curve
Faktor Kesan ke atas kecerunan graf

The size of solid reactant The smaller the size of a solid reactant, the higher the rate of reaction, the greater is the
Saiz bahan tindak balas gradient of the curve. tinggi
pepejal

Semakin kecil saiz bahan tindak balas pepejal, semakin tinggi kadar tindak balas, semakin
kecerunan graf.

22 Quantity of product (g/mol/cm3) The concentration of The higher the concentration of a solution, the higher the rate of reaction, the greater is the gradient of the curve.
Kuantiti hasil (g/mol/cm3) solution Semakin tinggi kepekatan larutan, semakin tinggi kadar tindak balas, semakin tinggi kecerunan graf.
Kepekatan larutan
(a) The maximum quantity of product
Kuantiti hasil maksimum The temperature of The higher the temperature of a solution, the higher the rate of reaction, the greater is the
reaction mixture
V Suhu campuran tindak gradient of the curve. tinggi kadar tindak balas, semakin tinggi kecerunan graf.
balas Semakin tinggi suhu larutan, semakin
(b) The gradient of the curve
Kecerunan lengkung The presence of catalyst The presence of catalyst in certain chemical reaction increases the rate of reaction, the gradient of the curve
Kehadiran mangkin
Time (minute/second) becomes greater .
Masa (minit/saat)
Kehadiran mangkin dalam suatu tindak balas kimia meningkatkan kadar tindak balas, kecerunan graf menjadi lebih

tinggi .

An increase in the quantity of catalyst chemical reaction increases the rate of reaction, the gradient of the curve

becomes greater .

Peningkatan dalam kuantiti mangkin yang digunakan akan meningkatkan kadar tindak balas, kecerunan graf menjadi

lebih tinggi .

Exercise: Sketch the graph of volume of gas produced against time for following experiments.
Latihan: Lakar graf isi padu gas yang dihasilkan melawan masa bagi eksperimen-eksperimen berikut.

(a) Experiment / Eksperimen Number of mol of reactant / Quantity of product / Factor Sketch of the graph
Bilangan mol bahan tindak balas / Kuantiti hasil / Faktor Lakaran graf

Experiment I: 1 Zinc is in excess in experiments I and II, the volume of hydrogen gas collected is not affected by the quantity of Volume of hydrogen/cm3
Eksperimen I: Isi padu hidrogen/cm3
Excess of zinc powder + 100 cm3 of zinc.
1.0 mol dm–3 sulphuric acid at 40°C Experiment I
Serbuk zink berlebihan + 100 cm3 asid Zink adalah berlebihan dalam eksperimen I dan II, isi padu gas hidrogen yang terkumpul tidak dipengaruhi oleh Eksperimen I
sulfurik 1.0 mol dm–3 pada 40°C Experiment II
kuantiti zink. Eksperimen II
Experiment II:
Eksperimen II: 2 Volume of H2 collected depends on number of mol of H2SO4 Time/s
Excess of zinc powder + 50 cm3 of Masa/s
1.0 mol dm–3 sulphuric acid at 30°C Isi padu H2 yang terkumpul bergantung pada bilangan mol H2SO4
Serbuk zink berlebihan + 50 cm3 asid
sulfurik 1.0 mol dm–3 pada 30°C (a) Number of mol of H2SO4 in experiment I = 100 u 1 = 0.1 mol
1 000
Balanced equation: Bilangan mol H2SO4 dalam eksperimen I
Persamaan seimbang:
Zn(s/p) + H2SO4(aq/ak) → ZnSO4(aq/ak) (b) Number of mol of H2SO4 in experiment II = 50 u 1 = 0.05 mol
+ H2(g/g) Bilangan mol H2SO4 dalam eksperimen II 1 000

Observable changes to measure rate ⇒ The maximum volume of hydrogen gas collected in experiment I is double of experiment II .
of reaction:
Perubahan yang dapat dilihat untuk Isi padu maksimum gas hidrogen yang terkumpul dalam eksperimen I adalah dua kali ganda dari
mengukur kadar tindak balas: eksperimen II .
Volume of hydrogen gas collected in every
MODULE • Chemistry Form 5Compare rate of reaction:
Bandingkan kadar tindak balas:
U
NExperimentType of zincConcentration of H2SO4Temperature
IEksperimenJenis zinkKepekatan H2SO4Suhu
T
I Powder / Serbuk 1.0 mol dm–3 40°C
1
II Powder / Serbuk 1.0 mol dm–3 30°C
23 © Nilam Publication Sdn Bhd
30 seconds by water displacement in the ⇒ The rate of reaction in experiments I and II is not affected by size of zinc and concentration of sulphuric
burette. / Isi padu gas hidrogen terkumpul
setiap 30 saat melalui kaedah penyesaran air acid.
dalam buret. Kadar tindak balas dalam eksperimen I dan II tidak dipengaruhi oleh saiz zink dan kepekatan asid

sulfurik.
⇒ Initial rate of reaction in experiment I is higher than experiment II because the temperature of reaction

mixture in experiment I higher than experiment II, the gradient of the curve for experiment I is greater

than experiment II.
Kadar tindak balas awal dalam eksperimen I lebih tinggi berbanding eksperimen II kerana suhu
campuran tindak balas dalam eksperimen I lebih tinggi berbanding eksperimen II , kecerunan graf bagi
eksperimen I adalah lebih tinggi berbanding eksperimen II.

© Nilam Publication Sdn Bhd Number of mol of reactant / Quantity of product / Factor MODULE • Chemistry Form 5
Bilangan mol bahan tindak balas / Kuantiti hasil / Faktor
U
N
I
T

1
(b) Experiment / Eksperimen 1 Calcium carbonate is in excess in experiment I, II and III, the volume of carbon dioxide gas collected is not affected Sketch of the graph
Lakaran graf
Experiment I: by the quantity of calcium carbonate.
Eksperimen I: Volume of carbon dioxide /cm3
Excess calcium carbonate chips and Kalsium karbonat adalah berlebihan dalam eksperimen I, II dan III, isi padu gas karbon dioksida yang terkumpul Isi padu karbon dioksida/cm3
25 cm3 of 1.0 mol dm–3 hydrochloric acid
Ketulan kalsium karbonat berlebihan dan tidak dipengaruhi oleh kuantiti kalsium karbonat. Experiment III
25 cm3 asid hidroklorik 1.0 mol dm–3 Eksperimen III
2 Volume of CO2 collected depends on the number of mol of HCl Experiment I
Experiment II: Eksperimen I
Eksperimen II: Isi padu CO2 yang terkumpul bergantung pada bilangan mol HCl Experiment II
Excess calcium carbonate chips and 25 u 1.0 Eksperimen II
25 cm3 of 0.5 mol dm–3 hydrochloric acid (a) Number of mol of HCl in experiment I = 1 000 = 0.025 mol
Ketulan kalsium karbonat berlebihan dan Bilangan mol HCl dalam eksperimen I Time/s
25 cm3 asid hidroklorik 0.5 mol dm–3 Masa/s
(b) Number of mol of HCl in experiment II = 25 u 0.5 = 0.0125 mol
Experiment III: Bilangan mol HCl dalam eksperimen II 1 000
Eksperimen III:
Excess calcium carbonate chips and (c) Number of mol of HCl in experiment III = 100 u 0.5 = 0.05 mol
100 cm3 of 0.5 mol dm–3 hydrochloric acid Bilangan mol HCl dalam eksperimen III 1 000
Ketulan kalsium karbonat berlebihan dan
100 cm3 asid hidroklorik 0.5 mol dm–3 ⇒ The maximum volume of carbon dioxide gas in experiment III is double of experiment I .
Isi padu maksimum karbon dioksida yang terkumpul dalam eksperimen III adalah dua kali ganda eksperimen I.
Balanced equation:
Persamaan seimbang: ⇒ The maximum volume of carbon dioxide gas in experiment I is double of experiment II .
Isi padu maksimum karbon dioksida yang terkumpul dalam eksperimen I adalah dua kali ganda eksperimen II.
CaCO3(s/p) + 2HCl(aq/ak) →
24 3 Compare rate of reaction: / Bandingkan kadar tindak balas:
CaCl2(aq/ak) + H2O(l/ce) + CO2(g/g)
Experiment Type of CaCO3 Concentration of HCl
Observable changes to measure rate Eksperimen Jenis CaCO3 Kepekatan HCl
of reaction:
Perubahan yang diperhatikan untuk I Chips / Ketulan 1.0 mol dm–3
mengukur kadar tindak balas:
II Chips / Ketulan 0.5 mol dm–3
Volume of carbon dioxide collected in every
III Chips / Ketulan 0.5 mol dm–3
30 seconds by water displacement in the
⇒ The rate of reaction in experiments I, II and III is not affected by size of calcium carbonate.
burette. / Isi padu gas hidrogen terkumpul Kadar tindak balas dalam eksperimen I, II dan II tidak dipengaruhi oleh saiz kalsium karbonat.

setiap 30 saat melalui kaedah penyesaran ⇒ Initial rate of reaction in experiment I is higher than experiment II because the concentration of HCl in
experiment I is higher than experiment II, the gradient of the curve for experiment I is greater than
air dalam buret.
experiment II.
Kadar tindak balas awal bagi eksperimen I lebih tinggi berbanding eksperimen II kerana kepekatan HCl
dalam eksperimen I lebih tinggi berbanding eksperimen II, kecerunan graf bagi eksperimen I lebih tinggi

berbanding eksperimen II.
⇒ Initial rate of reaction in experiment II is equal to experiment III because the concentration of HCl in

experiments II and III is the same , the gradient of the curve in experiments II and III is the same .
Kadar tindak balas awal dalam eksperimen II adalah sama dengan eksperimen III kerana kepekatan HCl
dalam eksperimen II dan III adalah sama , kecerunan graf dalam eksperimen II dan III adalah sama .

(c) Experiment / Eksperimen Number of mol of reactant / Quantity of product / Factor Sketch of the graph
Bilangan mol bahan tindak balas / Kuantiti hasil / Faktor Lakaran graf
Experiment I:
Eksperimen I: 1 Magnesium is in excess in experiments I and II, the volume of hydrogen gas collected is not affected by the Volume of hydrogen /cm3
Excess of magnesium powder + 100 cm3 quantity of magnesium. Isi padu hidrogen /cm3
of 1.0 mol dm–3 hydrochloric acid Magnesium adalah berlebihan dalam eksperimen I dan II, isi padu gas hidrogen yang terkumpul tidak
Serbuk magnesium berlebihan + dipengaruhi oleh kuantiti magnesium. Experiment II
100 cm3 asid hidroklorik 1.0 mol dm–3 Eksperimen II
2 Volume of H2 collected depends on the number of mol of HCl. Experiment I
Experiment II: Eksperimen I
Eksperimen II: Isi padu H2 yang terkumpul bergantung pada bilangan mol HCl.
Excess of magnesium ribbon + 200 cm3 100 u 1 Time/s
of 1.0 mol dm–3 hydrochloric acid (a) Number of mol of HCl in experiment I = 1 000 = 0.1 mol Masa/s
Pita magnesium berlebihan + 200 cm3 Bilangan mol HCl dalam eksperimen I
asid hidroklorik 1.0 mol dm–3
(b) Number of mol of HCl in experiment II = 200 u 1 = 0.2 mol
Balanced equation: Bilangan mol HCl dalam eksperimen II 1 000
Persamaan seimbang:
Mg(s/p) + 2HCl(aq/ak) → MgCl2(aq/ak) ⇒ The maximum volume of hydrogen gas collected in experiment II is double of experiment I .
Isi padu maksimum gas hidrogen yang terkumpul dalam eksperimen II adalah dua kali ganda eksperimen I .
Observable changes to measure rate
of reaction: 3 Compare rate of reaction: / Bandingkan kadar tindak balas:
Perubahan yang dapat dilihat untuk
mengukur kadar tindak balas: Experiment Type of Mg Concentration of HCl
Volume of hydrogen gas collected in every Eksperimen Jenis Mg Kepekatan HCl

30 seconds by water displacement in the

burette. / Isi padu gas hidrogen terkumpul

setiap 30 saat melalui kaedah penyesaran

air dalam buret.
MODULE • Chemistry Form 5I Powder / Serbuk1.0 mol dm–3

UII Ribbon / Pita1.0 mol dm–3
N
I
T

1

25 © Nilam Publication Sdn Bhd
4 Initial rate of reaction in experiment I is higher than experiment II because the total surface area of magnesium
powder in experiment I is higher than magnesium ribbon in experiment II, the gradient of the curve for experiment
I is greater than experiment II.
Kadar tindak balas awal dalam eksperimen I lebih tinggi daripada eksperimen II kerana jumlah luas permukaan
serbuk magnesium dalam eksperimen I lebih tinggi daripada pita magnesium dalam eksperimen II, kecerunan
graf bagi eksperimen I lebih tinggi daripada eksperimen II.

© Nilam Publication Sdn Bhd (d) Experiment / Eksperimen Number of mol of reactant / Quantity of product / Factor MODULE • Chemistry Form 5Sketch of the graph
Bilangan mol bahan tindak balas / Kuantiti hasil / Faktor Lakaran graf
U
Experiment I: 1 Zinc is in excess in experiments I, II and III, the volume of hydrogen gas collected is not affected by the quantityN
Eksperimen I: I
Excess of zinc granules and 100 cm3 of of zinc. T
1.0 mol dm-3 sulphuric acid + 5 cm3 of Zink adalah berlebihan dalam eksperimen I, II dan III, Isi padu gas hidrogen yang terkumpul tidak dipengaruhi
copper(II) sulphate solution oleh kuantiti zink. 1
Butiran zink berlebihan dan 100 cm3 asid
sulfurik 1.0 mol dm–3 + 5 cm3 larutan 2 Volume of H2 depends on the number of mol of H2SO4.
kuprum(II) sulfat
Isi padu H2 bergantung pada bilangan mol H2SO4.
Experiment II:
Eksperimen II: (a) Number of mol of H2SO4 in experiment I = 100 u 1 = 0.1 mol
Excess of zinc granules and 100 cm3 of Bilangan mol H2SO4 dalam eksperimen I 1 000
0.5 mol dm–3 sulphuric acid + 5 cm3 of
copper(II) sulphate solution (b) Number of mol of H2SO4 in experiment II = 100 u 0.5 = 0.05 mol
Butiran zink berlebihan dan 100 cm3 asid Bilangan mol H2SO4 dalam eksperimen II 1 000
sulfurik 0.5 mol dm–3 + 5 cm3 larutan
kuprum(II) sulfat (c) Number of mol of H2SO4 in experiment III = 100 u 0.5 = 0.05 mol
Bilangan mol H2SO4 dalam eksperimen III 1 000
Experiment III:
26 Eksperimen III: ⇒ The maximum volume of hydrogen gas collected in experiment I is double of experiment II . Volume of hydrogen /cm3
Excess of zinc granules and 100 cm3 of Isi padu maksimum gas hidrogen yang terkumpul dalam eksperimen I adalah dua kali ganda eksperimen II . Isi padu hidrogen /cm3
0.5 mol dm–3 sulphuric acid
Butiran zink berlebihan dan 100 cm3 asid ⇒ The maximum volume of hydrogen gas collected in experiment II is equal to experiment III. Experiment I
sulfurik 0.5 mol dm–3 Isi padu maksimum gas hidrogen yang terkumpul dalam eksperimen II adalah sama dengan yang Eksperimen I
terkumpul dalam eksperimen III.
Balanced equation: Experiment II
Persamaan seimbang: 3 Compare rate of reaction: / Bandingkan kadar tindak balas: Eksperimen II

Experiments I and II: Experiment Type of zinc Concentration of H2SO4 Presence of catalyst Experiment III
Eksperimen I dan II: Eksperimen Jenis zink Kepekatan H2SO4 Kehadiran mangkin Eksperimen III

Zn + H2SO4 → ZnSO4 + H2 I Granules / Butiran 1.0 mol dm–3 Copper(II) sulphate Time/s
Kuprum(II) sulfat Masa/s
Experiment III:
Eksperimen III: II Granules / Butiran 0.5 mol dm–3 Copper(II) sulphate
Kuprum(II) sulfat
Zn + H2SO4 → ZnSO4 + H2

Observable changes to measure rate III Granules / Butiran 0.5 mol dm–3 –
of reaction:
Perubahan yang dapat dilihat untuk ⇒ Initial rate of reaction in experiment I is higher than experiment II because the concentration of H2SO4 in
mengukur kadar tindak balas: experiment I higher than experiment II, the gradient of the curve for experiment I is greater than experiment II.
Volume of hydrogen gas collected in every Kadar tindak balas awal dalam eksperimen I adalah lebih tinggi berbanding dalam eksperimen II kerana
kepekatan H2SO4 dalam eksperimen I lebih tinggi berbanding dalam eksperimen II, kecerunan graf bagi
30 seconds by water displacement in the eksperimen I lebih tinggi daripada eksperimen II.

burette. / Isi padu gas hidrogen terkumpul ⇒ Initial rate of reaction in experiment II is higher than experiment III because the catalyst copper(II) sulphate
is present in experiment II. The gradient of the curve in experiment II is greater than experiment III.
setiap 30 saat melalui kaedah penyesaran Kadar tindak balas awal dalam eksperimen II lebih tinggi daripada eksperimen III kerana kehadiran
mangkin kuprum(II) sulfat dalam eksperimen II. Kecerunan graf bagi eksperimen II lebih tinggi daripada
air dalam buret.
eksperimen III.

MODULE • Chemistry Form 5

Application of factors that affect the rate of reaction
Aplikasi faktor-faktor yang mempengaruhi kadar tindak balas

Give examples of the application (a) Storage of food in a refrigerator – When the food is kept in refrigerator, the food lasts longer.
of factors that affect the rate of The lower temperature in the refrigerator slows down the activity of the bacteria . The
reaction in daily activities.
Berikan contoh aplikasi faktor- bacteria produces less toxin , hence the rate of decomposition of food is lower .
faktor yang mempengaruhi kadar
tindak balas dalam aktiviti harian. (b) Burning of charcoal – When food is cooked with smaller pieces of charcoal, the food cooked faster. U
The smaller pieces of charcoal have a larger total exposed surface area. Hence, smaller pieces N
Give examples of the application charcoal burns faster to produce more heat. I
in industrial processes. T
Berikan contoh aplikasi dalam (c) Cooking food in a pressure cooker – The high pressure in pressure cooker increases the
proses industri. boiling point of water to a temperature above 100ºC. The kinetic energy of the particles in 1

the food is increase/higher . Hence time taken for the food to be cooked is shorter . Thus the
food cooked faster at a higher temperature in a pressure cooker.
(d) Cooking of smaller pieces of food – The total surface area on a smaller cut pieces of food is

larger . The food can absorb more heat. Hence, the time taken for the food to be cooked is
shorter .

(a) Penyimpanan makanan dalam peti sejuk – Apabila makanan disimpan dalam peti sejuk, makanan

tahan lebih lama. Suhu yang lebih rendah dalam peti sejuk memperlahankan aktiviti bakteria .

Bakteria menghasilkan kurang toksin , kadar penguraian makanan lebih rendah .
(b) Pembakaran arang – Apabila makanan dimasak dengan ketulan arang yang kecil, makanan masak

dengan lebih cepat. Ketulan kecil arang mempunyai jumlah luas permukaan terdedah yang lebih
besar. Oleh itu, ketulan kecil arang terbakar lebih cepat untuk menghasilkan lebih haba.

(c) Memasak makanan dalam periuk tekanan – Tekanan tinggi dalam periuk tekanan meningkatkan

takat didih air kepada suhu yang melebihi 100°C. Tenaga kinetik zarah-zarah dalam

makanan meningkat/lebih tinggi . Maka masa untuk makanan masak berkurang . Oleh itu,

makanan masak dengan lebih cepat pada suhu yang lebih tinggi dalam periuk tekanan.

(d) Memasak kepingan makanan yang lebih kecil – Jumlah luas permukaan pada kepingan

makanan yang lebih kecil adalah lebih besar . Makanan dapat menyerap lebih haba.

Maka, masa yang diambil untuk makanan dimasak berkurang .

(a) Haber process is an industrial process to manufacture ammonia gas . The optimum conditions

to run the process at a higher rate are:

Proses Haber adalah proses dalam industri untuk menghasilkan gas ammonia . Keadaan

optimum bagi proses ini supaya berkadar tinggi adalah:

(i) The temperature is 400ºC – 500ºC. / Suhu 400°C – 500°C.

(ii) The pressure is 200 – 300 atm. / Tekanan 200 – 300 atm.

(iii) The catalyst is iron , Fe. / Mangkin adalah besi , Fe.

Chemical equation: / Persamaan kimia:

N2(g) + 3H2(g) 400ºC – 500ºC 2NH3(g)
Fe, 200 – 300 atm



(b) Contact process is an industrial process to manufacture sulphuric acid. The optimum conditions to run
the process at a higher rate are: / Proses Sentuh adalah proses dalam industri untuk menghasilkan
asid sulfurik. Keadaan optimum bagi proses ini supaya berkadar tinggi adalah:
(i) The temperature is 450ºC. / Suhu 450°C.
(ii) The pressure is 1 atm. / Tekanan 1 atm.

(iii) The catalyst is vanadium(V) oxide , V2O5. / Mangkin adalah vanadium(V) oksida , V2O5.

Chemical equation: / Persamaan kimia:

2SO2 + O2 V2O5, 450ºC 2SO3
1 atm



27 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

The Collision Theory / Teori Perlanggaran

The energy changes are always During a chemical reaction,

caused by two processes that Semasa tindak balas kimia,

occur during chemical reaction 1 Chemical bonds in the reactants are broken.

when reactants change to Ikatan kimia dalam bahan tindak balas terputus.

products. What are these two 2 New bonds in the products are formed.

processes? / Perubahan tenaga Ikatan baru dalam hasil tindak balas terbentuk.

adalah disebabkan oleh dua

proses yang berlaku semasa

U tindak balas kimia apabila bahan
N tindak balas bertukar menjadi
I hasil tindak balas. Apakah dua
T proses ini?

1 How can a reaction take place According to the collision theory, a chemical reaction can only occur when a reacting particles collide with
according to the collision theory? one another with certain amount of kinetic energy.
Bagaimanakah suatu tindak balas Berdasarkan teori perlanggaran, suatu tindak balas kimia boleh berlaku apabila satu zarah tindak balas

berlaku berdasarkan teori berlanggar dengan satu zarah lain dengan jumlah tenaga kinetik tertentu.

perlanggaran?

What is activation energy, Ea? It is the minimum amount of energy needed for the reacting particles to react.
Apakah tenaga pengaktifan, Ea? Tenaga minima yang diperlukan oleh zarah bahan tindak balas untuk bertindak balas.
Are all chemical reactions have
the same activation energy, Ea? No, different chemical reactions have different activation energy.
Adakah semua tindak balas kimia Tidak, tindak balas kimia yang berbeza mempunyai tenaga pengaktifan yang berbeza.
mempunyai tenaga pengaktifan, Ea
yang sama? The collisions that lead to a chemical reaction and result in the formation of products.
Define effective collision. Perlanggaran yang membawa kepada tindak balas kimia dan membentuk hasil tindak balas.
Nyatakan maksud perlanggaran
berkesan. The reactants particles just bounce off each other and no reaction occur.
What happens when the energy of
collision between particles less Zarah bahan tindak balas hanya akan melantun antara satu sama lain dan tiada tindak balas berlaku.
than activation energy, Ea? Bounce / Melantun
Apakah yang berlaku apabila Collision Below Ea
tenaga perlanggaran antara zarah Perlanggaran Bawah Ea
kurang daripada tenaga
pengaktifan, Ea? Reactant A Reactant B No new products formed
Bahan A Bahan B Tiada hasil baru terbentuk
What happens when the collision
between particles does not occur ⇒ Collision energy of particles < activation energy. / Tenaga perlanggaran zarah < tenaga pengaktifan.
at the correct orientation? ⇒ The chemical bonds in the reactants are not broken.
Apakah yang berlaku apabila Ikatan kimia dalam bahan tindak balas tidak terputus.
perlanggaran antara zarah tidak ⇒ No reaction. / Tiada tindak balas.
berlaku pada orientasi yang betul? ⇒ The collision is an ineffective collision. / Perlanggaran tidak berkesan.

What happens when the energy of The reactants particles just bounce off each other and no reaction occur.
collision between the reacting
particles have equal or more than Zarah bahan tindak balas hanya akan melantun antara satu sama lain dan tiada tindak balas berlaku.
the activation energy, Ea and Collision
collide at correct orientation? Perlanggaran Wrong orientation Bounce / Melantun
Apakah yang berlaku apabila Orientasi salah
tenaga perlanggaran antara zarah
tindak balas adalah sama atau Reactant A Reactant B No new products formed
lebih daripada tenaga pengaktifan, Bahan A Bahan B Tiada hasil baru terbentuk
Ea dan berlanggar pada orientasi
yang betul? ⇒ The chemical bonds in the reactants are not broken. No reaction.
Ikatan kimia dalam bahan tindak balas tidak terputus. Tiada tindak balas.
⇒ The collision is an ineffective collision. / Perlanggaran tidak berkesan.

Collision Achieves Ea / Mencapai Ea
Perlanggaran Correct orientation / Orientasi betul

Reactant A Reactant B New products are formed
Bahan A Bahan B Hasil baru terbentuk

⇒ Collision energy of particles ≥ activation energy. / Tenaga perlanggaran zarah ≥ tenaga pengaktifan.
⇒ The chemical bonds in the reactants are broken. / Ikatan kimia dalam bahan tindak balas terputus.
⇒ Reaction occurs. / Tindak balas berlaku.
⇒ The collision is an effective collision. / Perlanggaran berkesan.

© Nilam Publication Sdn Bhd 28

MODULE • Chemistry Form 5

What are the conditions for a For a reaction to occur, the collision must satisfy two conditions:
reaction to occur? Untuk suatu tindak balas berlaku, perlanggaran tersebut mesti mematuhi dua syarat:
Apakah syarat untuk tindak balas (a) The colliding particle must have enough energy i.e equal or more than a minimum amount of energy
berlaku?
known as *activation energy, Ea.
Zarah-zarah yang berlanggar mesti mempunyai tenaga yang cukup, iaitu sama atau lebih daripada U
N
tenaga minimum yang dikenali sebagai *tenaga pengaktifan, Ea. I
* The activation energy differs in different chemical reaction. The lower the activation energy, the higher T

the rate of reaction. 1
Tenaga pengaktifan berbeza bagi tindak balas kimia yang berbeza. Semakin rendah tenaga

pengaktifan, semakin tinggi kadar tindak balas.
(b) The colliding particles must also have the right orientation of collisions.
Zarah-zarah yang berlanggar juga mestilah mempunyai orientasi perlanggaran yang betul.

How to show the energy changes The energy changes of the reacting particles and the activation energy of a reaction is shown in an
of reacting particles and the energy profile diagram.
activation energy, Ea in a reaction? Perubahan tenaga zarah-zarah yang bertindak balas dan tenaga pengaktifan tindak balas ditunjukkan
Bagaimana cara menunjukkan dalam gambar rajah profil tenaga.
perubahan tenaga zarah bahan (a) * Exothermic reaction / Tindak balas eksotermik
bertindak balas dan tenaga
pengaktifan, Ea dalam suatu tindak Energy / Tenaga
balas?
Ea

Reactants Products Reaction path
Bahan tindak balas Hasil tindak balas Lintasan tindak balas

(b) * Endothermic reaction / Tindak balas endotermik
Energy / Tenaga

Products

Ea Hasil tindak balas

Reactants Reaction path
Bahan tindak balas Lintasan tindak balas


Ea – The minimum energy the reactant particles must possess before the collision between them can
result in a chemical reaction.

*Exothermic and endothermic reactions will studied in topic 4, Thermochemistry.
Ea – Tenaga minimum yang mesti dimiliki oleh zarah-zarah bahan tindak balas sebelum perlanggaran di

antara mereka dapat menghasilkan tindak balas kimia.
* Tindak balas eksotermik dan endotermik akan dipelajari dalam tajuk 4, Termokimia.

How to relate the frequency of The effective collisions will result in chemical reaction. When frequency of effective collision increases, the
effective collision with the rate of rate of reaction will also increase.
reaction? Perlanggaran berkesan menyebabkan tindak balas kimia berlaku. Apabila frekuensi perlanggaran
Bagaimanakah cara mengaitkan berkesan meningkat, kadar tindak balas turut meningkat.
frekuensi perlanggaran berkesan
dengan kadar tindak balas? Remark / Catatan:
Frequency of collision is the number of collisions in one second. When the frequency of collision between particles of
reactants increases, the frequency of effective collisions between particles will also increase.
Frekuensi bagi perlanggaran ialah bilangan perlanggaran dalam satu saat. Apabila frekuensi bagi perlanggaran di antara
zarah-zarah bahan tindak balas meningkat, frekuensi bagi perlanggaran berkesan di antara zarah-zarah turut meningkat.

How to change the frequency of Frequency of effective collision can be changed by changing the following:
effective collision? Frekuensi perlanggaran berkesan boleh diubah dengan mengubah perkara berikut:
Bagaimana cara mengubah 1 The frequency of collision between particles (number of collisions per unit time).
frekuensi perlanggaran berkesan? Frekuensi perlanggaran antara zarah (bilangan perlanggaran per unit masa).

Or / Atau

2 The activation energy of the chemical reaction.
Tenaga pengaktifan tindak balas kimia.

29 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

State factors that can change the 1 Size of solid reactant.
frequency of effective collision. Saiz pepejal bahan tindak balas.
Nyatakan faktor yang boleh 2 Concentration of reacting solution.
mengubah frekuensi perlanggaran Kepekatan larutan yang bertindak balas.
berkesan. 3 Temperature of reaction mixture.
Suhu campuran tindak balas.
4 The presence of catalyst.
Kehadiran mangkin.

Remark / Catatan:

Factors Affect Frequency of collision or Affect Frequency of effective Affect Rate of
activation energy collision / Frekuensi reaction
NU Faktor
mempengaruhi Frekuensi perlanggaran mempengaruhi perlanggaran mempengaruhi Kadar
atau tenaga pengaktifan berkesan tindak balas

IT

1 Relationship Between Frequency of Effective Collision with Factors
Affecting Rate of Reaction / Hubung Kait antara Frekuensi Perlanggaran
Berkesan dengan Faktor Mempengaruhi Kadar Tindak Balas

1 Size of Solid Reactant
Saiz Pepejal Bahan Tindak Balas

How does the change in size of The smaller the size of solid reactant, the larger total surface area exposed to collision that allows
solid reactant affect rate of more collision between reacting particles.
reaction? Semakin kecil saiz pepejal bahan tindak balas, semakin besar jumlah luas permukaan yang
Bagaimana perubahan saiz terdedah kepada perlanggaran yang membenarkan lebih banyak perlanggaran antara zarah tindak
pepejal bahan tindak balas balas.
mempengaruhi kadar tindak
balas? – A chemical reaction takes place on the surface of a solid reactant as it is where collision between

Explain how this factor affects the reacting particles take place.
rate of reaction.
Terangkan bagaimana faktor ini Tindak balas kimia berlaku pada permukaan pepejal bahan tindak balas kerana ia adalah tempat
mempengaruhi kadar tindak balas.
berlakunya perlanggaran antara zarah-zarah tindak balas.

– When the size of solid reactant is larger, the area exposed to collisions is smaller because collisions

are limited on the surface of solid. Thus, the total surface area at which reaction occur is smaller.

Apabila saiz pepejal bahan tindak balas besar, luas kawasan yang terdedah kepada perlanggaran

adalah lebih kecil kerana perlanggaran adalah terhad pada permukaan pepejal. Oleh itu, jumlah luas

permukaan di mana tindak balas berlaku adalah lebih kecil.

– When the solid reactant is broken into smaller size, the area exposed to collision is larger when surface

area in the inner part of the solid becomes available for collisions. Hence, the total surface area at

which reaction occur becomes larger.

Apabila pepejal bahan tindak balas dipecahkan kepada saiz yang lebih kecil, kawasan yang terdedah

kepada perlanggaran adalah lebih besar apabila luas permukaan bahagian dalam pepejal menjadi

tersedia untuk perlanggaran. Oleh itu, jumlah luas permukaan berlakunya tindak balas menjadi lebih

besar.
Particle of substance A Particle of substance B

Zarah bahan A Zarah bahan B

When the size
substance B decreases

Apabila saiz bahan B
berkurangan

When the size of solid substance B is large, only When size of solid substance B is broken into smaller
outer layer of reacting particles B can react with size, more reacting particles B can react with reacting
reacting particles A particles A
Apabila saiz pepejal bahan B besar, hanya Apabila saiz pepejal bahan B dipecahkan kepada
permukaan luar bahan tindak balas zarah B boleh saiz yang lebih kecil, lebih banyak bahan tindak balas
bertindak balas dengan bahan tindak balas zarah B dapat bertindak dengan bahan tindak balas
zarah A zarah A

– This will increase the frequency of collision between particles, thus, frequency of effective
collisions between particles increases which leads to an increase of rate of reaction.

Ini akan menambahkan frekuensi perlanggaran antara zarah, oleh itu, frekuensi perlanggaran
berkesan antara zarah meningkat yang membawa kepada peningkatan kadar tindak balas.

© Nilam Publication Sdn Bhd 30

MODULE • Chemistry Form 5

2 Concentration / Kepekatan

How does the concentration of – Concentration is the number of solutes dissolved in a given volume of solution.
solution affect rate of reaction? Kepekatan ialah bilangan zat terlarut yang melarut dalam suatu isi padu larutan yang diberi.
Bagaimanakah kepekatan larutan – A higher concentration means that the number of particles of solute per unit volume is also higher, and
mempengaruhi kadar tindak
balas? vice versa.
Kepekatan tinggi bermaksud bilangan zarah zat terlarut per unit isi padu juga tinggi, dan sebaliknya.
Explain how this factor affects the
rate of reaction. – When the concentration of solution is higher, the number of particles per unit volume increases.
Terangkan bagaimana faktor ini Apabila kepekatan larutan tinggi, bilangan zarah per unit isi padu meningkat.
mempengaruhi kadar tindak balas. – This will increase the frequency of collision between particles, thus, frequency of effective

collisions between particles increases. U
Ini akan meningkatkan frekuensi perlanggaran antara zarah-zarah, oleh itu, frekuensi perlanggaran N
I
berkesan antara zarah meningkat. T

Particle of substance A Particle of substance B 1

Zarah bahan A Zarah bahan B

When the concentration
increases

Apabila kepekatan
meningkat

At a lower concentration, the number of particles At a higher concentration, the number of particles of
of A and B in a unit volume is lower, thus the A and B in a unit volume is higher, thus the
frequency of collision between particles is low. frequency of collision between particles is high.
Pada kepekatan yang rendah, bilangan zarah A Pada kepekatan yang tinggi, bilangan zarah A dan
dan B dalam satu unit isi padu adalah rendah, B dalam satu unit isi padu adalah tinggi, oleh itu
oleh itu frekuensi perlanggaran antara zarah frekuensi perlanggaran antara zarah adalah tinggi.
juga rendah.

– This leads to an increase of rate of reaction.
Ini membawa kepada peningkatan kadar tindak balas.

Explain why the monoprotic acid – For diprotic acid, 1 mol dm–3 of the acid ionises to 2 mol dm–3 of hidrogen ions, H+
and diprotic acid have different Bagi asid diprotik, 1 mol dm–3 asid mengion kepada 2 mol dm–3 ion hidrogen, H+
rate of reaction when the
concentration of the acids is the Example / Contoh: H2SO4 2H+ +   SO42–
same. 1 mol dm–3 2 mol dm–3
Terangkan mengapa asid
monoprotik dan asid diprotik
mempunyai kadar tindak balas – For monoprotic acid, 1 mol dm–3 of the acid ionises to 1 mol dm–3 of hidrogen ions, H+
yang berbeza apabila kepekatan Bagi asid monoprotik, 1 mol dm–3 asid mengion kepada 1 mol dm–3 ion hidrogen, H+
asid adalah sama.
Example / Contoh: HCl H+ +   Cl–
1 mol dm–3 1 mol dm–3


– Hence, when the same concentration of sulphuric acid and hydrochloric acid are used, the concentration
of hydrogen ions in sulphuric acid is double.

Oleh itu, apabila kepekatan asid sulfurik dan asid hidroklorik yang sama digunakan, kepekatan ion
hidrogen dalam asid sulfurik adalah dua kali ganda.

– The rate of reaction using sulphuric acid is higher than hydrochloric acid.
Kadar tindak balas menggunakan asid sulfurik adalah lebih tinggi daripada asid hidroklorik.

3 Temperature / Suhu

How does the temperature of The higher the temperature of a substance, the higher the kinetic energy of the reacting particles and vice
solution affect rate of reaction? versa.
Bagaimanakah suhu larutan Semakin tinggi suhu bahan, semakin tinggi tenaga kinetik zarah bahan tindak balas dan sebaliknya.
mempengaruhi kadar tindak
balas? – When the temperature increases, the kinetic energy of reacting particles increases and the particles
move faster.
Explain how this factor affects the
rate of reaction. Apabila suhu meningkat, tenaga kinetik zarah bahan tindak balas meningkat dan zarah bergerak lebih
Terangkan bagaimana faktor ini pantas.
mempengaruhi kadar tindak balas.
– This will increase the frequency of collision between particles, thus, frequency of effective collisions
between particles increases.

Ini akan meningkatkan frekuensi perlanggaran antara zarah, oleh itu, frekuensi perlanggaran
berkesan antara zarah meningkat.

– This leads to an increase of rate of reaction.
Ini akan membawa kepada peningkatan kadar tindak balas.

31 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

4 Catalyst / Mangkin

Define catalyst. It is a substance that change the rate of reaction without itself being chemically changed in the reaction.
Nyatakan maksud mangkin. Ialah sejenis bahan yang mengubah kadar tindak balas tanpa ianya berubah secara kimia dalam tindak
balas.

How does catalyst affect the rate Catalyst provides alternative pathway of a reaction at lower activation energy for the reacting particles to
of reaction? react.
Bagaimanakah mangkin Mangkin menyediakan lintasan alternatif suatu tindak balas pada tenaga pengaktifan yang lebih rendah
mempengaruhi kadar tindak bagi zarah tindak balas untuk bertindak balas.
balas?

U Explain how this factor affects the 1 In a reaction, if the reacting particles collide with energy lower than the activation energy, there will be
1N rate of reaction. no reaction.
Terangkan bagaimana faktor ini
I mempengaruhi kadar tindak balas. 2 When catalyst is added to the same reaction, its provides alternative pathway with a lower activation
T energy for the reacting particles to react.

3 With this lower activation energy, more reacting particles will have enough energy to react when they

collide.

4 This will increase the frequency of effective collisions between particles which leads to an increase of

rate of reaction.

1 Dalam suatu tindak balas, jika zarah-zarah bahan tindak balas berlanggar dengan tenaga yang lebih

rendah daripada tenaga pengaktifan, tiada tindak balas berlaku.

2 Apabila mangkin ditambah kepada tindak balas yang sama, ia memberikan laluan alternatif dengan

tenaga pengaktifan yang lebih rendah untuk zarah-zarah bahan tindak balas bertindak balas.

3 Dengan tenaga pengaktifan yang lebih rendah ini, lebih banyak zarah-zarah bahan tindak balas

mempunyai tenaga yang mencukupi untuk bertindak balas semasa perlanggaran.

4 Ini akan meningkatkan frekuensi perlanggaran berkesan antara zarah-zarah yang menyebabkan

peningkatan kadar tindak balas.

Energy / Tenaga Uncatalysed reaction pathway
Laluan tindak balas tanpa mangkin

Catalysed Ea
reaction pathway Ea'
Lintasan tindak
balas bermangkin

Reactants
Bahan tindak balas

Products / Hasil

Reaction path
Lintasan tindak balas

– Ea: The minimum energy the reactant particle must possess before collision between them can result in a
chemical reaction.

Tenaga minimum yang mesti dimiliki oleh zarah-zarah bahan tidak balas sebelum perlanggaran di antara
mereka menyebabkan tindak balas kimia.

– Ea’: The lower activation energy in the presence of a catalyst. Catalyst is a substance that increases the rate of a
chemical reaction without itself undergoing any chemical change.

Tenaga pengaktifan yang lebih rendah disebabkan kehadiran mangkin. Mangkin adalah bahan yang
meningkatkan kadar tindak balas tanpa ianya mengalami perubahan kimia.

Remark / Catatan:
Catalyst provide lower activation energy for the reacting particles to collide effectively.
Mangkin menyediakan tenaga pengaktifan yang lebih rendah bagi zarah tindak balas untuk berlanggar secara berkesan.

State the characteristics of 1 A catalyst does not change the amount of products of a reaction.
catalyst. 2 Catalyst is specific to a particular reaction, different chemical reactions need different catalyst.
Nyatakan ciri-ciri mangkin. 3 Catalyst remains unchanged chemically during a reaction.
4 Catalyst may undergo physical change during a reaction.
5 Only a small amount of catalyst needed to achieve big increase in rate of reaction.
6 More amount of catalyst used can further increase the rate of reaction.
7 Catalyst in powder form can further increase the rate of reaction.
1 Mangkin tidak mengubah jumlah hasil tindak balas.
2 Mangkin adalah khusus untuk tindak balas tertentu, tindak balas kimia yang berbeza memerlukan

mangkin yang berbeza.
3 Mangkin tidak berubah secara kimia semasa tindak balas.
4 Mangkin boleh mengalami perubahan fizikal semasa tindak balas.
5 Hanya sedikit mangkin yang diperlukan untuk mencapai peningkatan besar dalam kadar tindak balas.
6 Lebih banyak mangkin yang digunakan dapat meningkatkan kadar tindak balas.
7 Mangkin dalam bentuk serbuk dapat meningkatkan lagi kadar tindak balas.

© Nilam Publication Sdn Bhd 32

MODULE • Chemistry Form 5

Conclusion / Kesimpulan

Steps to explain factors that affect the rate of reaction based on collision theory:
Kaedah untuk menerangkan faktor-faktor yang mempengaruhi kadar tindak balas berdasarkan teori perlanggaran:

Size of Reactant Concentration of Reactant Temperature of Reaction Mixture Catalyst U
Saiz Bahan Tindak Balas Kepekatan Bahan Tindak Balas Suhu Campuran Tindak Balas Mangkin N
The smaller the size of The higher the concentration of Catalyst provides an alternative I
reactant, reactants, The higher the temperature, path of reaction which needs T
the total surface area exposed the number of particles in a unit the kinetic energy of reacting particles is lower activation energy (Ea’).
higher. 1
to collision is larger. volume is higher. Mangkin menyediakan jalan
Semakin tinggi suhu, alternatif bagi tindak balas
Semakin kecil saiz bahan Semakin tinggi kepekatan bahan semakin tinggi tenaga kinetik zarah-zarah yang memerlukan
tindak balas, tindak balas, yang bertindak balas. tenaga pengaktifan yang
semakin besar jumlah luas semakin banyak bilangan zarah
The reacting particles move faster. lebih rendah (Ea’)
permukaan yang terdedah dalam satu unit isi padu. Zarah yang bertindak balas bergerak
dengan lebih laju. More colliding particles achive
kepada perlanggaran. activation energy .

Lebih banyak zarah yang
berlanggar mencapai

tenaga pengaktifan .

The frequency of collision between particles increases.
meningkat.
Frekuensi perlanggaran di antara zarah-zarah

The frequency of effective collision between*particles increases.
Frekuensi perlanggaran berkesan di antara*zarah-zarah meningkat.

The rate of reaction increases . / Kadar tindak balas meningkat .

*State the particles that collide based on ionic equation.
*Nyatakan zarah yang berlanggar berdasarkan persamaan ion.

Rate of Reaction

Kadar Tindak Balas

https://goo.gl/xQWlIn

33 © Nilam Publication Sdn Bhd

© Nilam Publication Sdn Bhd Steps to Compare And Explain Factor that Affects Rate of Reaction Between Any Two Experiments MODULE • Chemistry Form 5

U
N
I
T

1
1 Compare factor that affects the rate of reaction in both experiments. 2 Compare how the factor affects the rate of reaction in both
34 Example: experiments.
(a) Factor size solid reactant
(a) Factor size of solid reactant (for reaction between different size of calcium The total surface area of calcium carbonate chips in experiment I is
carbonate and hydrochloric acid) larger than experiment II
(b) Factor concentration
The size of calcium carbonate chips in experiment I is smaller than experiment II The number of hydrogen ions per unit volume in experiment I is
(b) Factor concentration (for the reaction between different concentration of higher than experiment II
(c) Factor temperature
hydrochloric acid with zinc) The kinetic energy of hydrogen ion in experiment I is higher than
The concentration of hydrochloric acid in experiment I is higher than experiment experiment II
(d) Factor catalyst
II
(c) Factor temperature (for the reaction between different temperature of • Copper(II) sulphate lower the activation energy for the reaction
between zinc and sulphuric acid in experiment I
hydrochloric acid and zinc)
The temperature of reaction mixture in experiment I is higher than experiment II • More colliding particles achieve the activation energy
(d) Factor catalyst (for the reaction between zinc and sulphuric acid)
Copper(II) sulphate present as a catalyst for the reaction between zinc and 3 Compare frequency of collision between number of *particles in
both experiments.
sulphuric acid in experiment I but not in experiment II (a) Factor size solid reactant
The frequency of collision between calcium carbonate and
• Analyse the condition for both experiments given in the form of description/balanced hydrogen ion in experiment I is higher than experiment II
equation/diagram to identify factor that affects the rate of reaction. (b) Factor concentration
The frequency of collision between hydrogen ions and zinc
4 Compare frequency of effective collision between*particles in both experiments. atoms in experiment I is higher than experiment II
(a) Factor size solid reactant (c) Factor temperature
The frequency of effective collision between calcium carbonate and hydrogen The frequency of collision between hydrogen ions and zinc atoms
ion in experiment I is higher than experiment II in experiment I is higher than experiment II
(b) Factor concentration (d) **Factor catalyst
The frequency of effective collision between hydrogen ions and zinc atoms in
experiment I is higher than experiment II * State the particles that collide in the reaction based on the ionic equation
(c) Factor temperature **Catalyst does not increase the frequency of collision between particles
The frequency of effective collision between hydrogen ions and zinc atoms in
experiment I is higher than experiment II
(d) Factor catalyst
The frequency of effective collision between hydrogen ions and zinc atoms in
experiment I is higher than experiment II

5 Compare rate of reaction in both experiments
Rate of reaction in experiment I is higher than experiment II.

Langkah untuk Membanding dan Menerangkan Faktor yang Mempengaruhi
Kadar Tindak Balas antara Sebarang Dua Eksperimen

1 Banding faktor yang mempengaruhi kadar tindak balas dalam kedua-dua eksperimen. 2 Banding bagaimana faktor itu mempengaruhi kadar tindak balas
Contoh: dalam kedua-dua eksperimen.
(a) Faktor saiz bahan tindak balas pepejal
(a) Faktor saiz bahan tindak balas pepejal (untuk tindak balas antara saiz ketulan Jumlah luas permukaan ketulan kalsium karbonat dalam
kalsium karbonat yang berbeza dengan asid hidroklorik) eksperimen I lebih besar dari eksperimen II
(b) Faktor kepekatan
Saiz ketulan kalsium karbonat dalam eksperimen I lebih kecil dari eksprimen II Bilangan ion hidrogen dalam satu unit isi padu dalam eksperimen
(b) Faktor kepekatan (untuk tindak balas antara asid hidroklorik berkepekatan I lebih tinggi dari eksperimen II
(c) Faktor suhu
berbeza dengan zink) Tenaga kinetik ion hidrogen dalam eksperimen I lebih tinggi
Kepekatan asid hidroklorik dalam eksperimen I lebih tinggi dari eksperimen II dari eksperimen II
(c) Faktor suhu (untuk tindak balas antara asid hidroklorik bersuhu berbeza (d) Faktor mangkin

dengan zink) • Kuprum(II) sulfat merendahkan tenaga pengaktifan untuk
Suhu campuran tindak balas dalam eksperimen I lebih tinggi dari eksperimen II tindak balas antara zink dan asid sulfurik dalam experiment I
(d) Faktor mangkin (untuk tindak balas antara zink dan asid sulfurik)
Kuprum(II) sulfat hadir sebagai mangkin untuk tindak balas antara zink dan • Lebih banyak zarah yang berlanggar mencapai tenaga
pengaktifan
asid sulfurik dalam eksperimen I tetapi tiada dalam eksperimen II
3 Banding frekuensi perlanggaran antara*zarah dalam kedua-dua
• Analisis keadaan bagi kedua-dua eksperimen yang diberi dalam bentuk huraian/ eksperimen.
persamaan kima seimbang/susunan radas untuk mengenal pasti faktor yang (a) Faktor saiz bahan tindak balas pepejal
mempengaruhi kadar tindak balas Frekuensi perlanggaran antara kalsium karbonat dan ion
hidrogen dalam eksperimen I lebih tinggi dari eksperimen II
4 Banding frekuensi perlanggaran berkesan antara bilangan*zarah dalam kedua- (b) Faktor kepekatan
dua eksperimen Frekuensi perlanggaran antara ion hidrogen dan atom zink
(a) Faktor saiz bahan tindak balas pepejal dalam eksperimen I lebih tinggi dari eksperimen II
Frekuensi perlanggaran berkesan antara kalsium karbonat dan ion hidrogen (c) Faktor suhu
dalam eksperimen I lebih tinggi dari eksperimen II Frekuensi perlanggaran antara ion hidrogen dan atom zink
(b) Faktor kepekatan dalam eksperimen I lebih tinggi dari eksperimen II
Frekuensi perlanggaran berkesan antara ion hidrogen dan atom zink dalam (d) **Faktor mangkin
eksperimen I lebih tinggi dari eksperimen II
(c) Faktor suhu * Nyatakan zarah yang berlanggar dalam tindak balas berdasarkan
Frekuensi perlanggaran berkesan antara ion hidrogen dan atom zink dalam persamaan ion
eksperimen I lebih tinggi dari eksperimen II
(d) Faktor mangkin **Mangkin tidak meningkatkan frekuensi perlanggaran antara zarah
Frekuensi perlanggaran berkesan antara ion hidrogen dan atom zink dalam
eksperimen I lebih tinggi dari eksperimen II

5 Banding kadar tindak balas dalam kedua-dua eksperimen.
Kadar tindak balas dalam eksperimen I lebih tinggi dari ekspeimen II
MODULE • Chemistry Form 5

U
N
I
T

1

35 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Exercise / Latihan

1 Complete the following table. / Lengkapkan jadual berikut.

Chemical equation Ionic equation *Particles that collide in the reaction
Persamaan kimia Persamaan ion *Zarah-zarah yang berlanggar dalam tindak balas

(i) Mg + 2HCl → MgCl2 + H2 Mg + 2H+ → Mg2+ + H2 Magnesium atom and hydrogen ion
Atom magnesium dan ion hidrogen
(ii) Mg + 2HNO3 → Mg(NO3)2 + H2 Mg + 2H+ → Mg2+ + H2
Zn + 2H+ → Zn2+ + H2 Magnesium atom and hydrogen ion
U (iii) Zn + H2SO4 → ZnSO4 + H2 Zn + 2H+ → Zn2+ + H2 Atom magnesium dan ion hidrogen
N CaCO3 + 2H+ → H2O + CO2 + Ca2+
TI (iv) Zn + 2CH3COOH → (CH3COO)2Zn + H2 Zinc atom and hydrogen ion
– Atom zink dan ion hidrogen
1 (v) CaCO3 + 2HCl → CaCl2 + H2O + CO2
(vi) 2H2O2 → 2H2O + O2 Zinc atom and hydrogen ion
Atom zink dan ion hidrogen
(vii) Na2S2O3 + H2SO4 → Na2SO4 + H2O + SO2 + S S2O32– + 2H+ → H2O + SO2 + S
Calcium carbonate and hydrogen ion
Kalsium karbonat dan ion hidrogen

Hydrogen peroxide molecules
Molekul hidrogen peroksida

Thiosulphate ion and hydrogen ion
Ion tiosulfat dan ion hidrogen

2 Compare rate of reaction in Experiment I and Experiment II. Explain based on collision theory.
Bandingkan kadar tindak balas dalam Eksperimen I dan Eksperimen II. Terangkan berdasarkan teori perlanggaran.

Reactant Experiment I / Eksperimen I Experiment II / Eksperimen II
Bahan tindak balas
20 cm3 of 0.5 mol dm–3 hydrochloric acid + 20 cm3 of 0.5 mol dm–3 hydrochloric acid +
excess of calcium carbonate powder at 30°C excess of calcium carbonate chips at 30°C
20 cm3 asid hidroklorik 0.5 mol dm–3 + serbuk 20 cm3 asid hidroklorik 0.5 mol dm–3 + ketulan
kalsium karbonat berlebihan pada 30°C kalsium karbonat berlebihan pada 30°C

Balanced equation CaCO3 + 2HCl → CaCl2 + H2O + CO2
Persamaan seimbang
CaCO3 + 2H+ → H2O + CO2 + Ca2+
Ionic equation
Persamaan ion Number of moles of HCl / Bilangan mol HCl = 20 × 0.5 = 0.01 mol
1 000
Calculate volume of carbon dioxide gas
released in both experiments at room From the equation / Daripada persamaan
conditions. 1 mol of gas occupies 24 dm3 at 2 mol HCl : 1 mol CO2
room conditions 0.01 mol HCl : 0.005 mol CO2
Hitung isi padu gas karbon dioksida yang V olume of CO2 / Isi padu CO2 = 0.005 mol × 24 dm3 mol–1
dibebaskan dalam kedua-dua eksperimen = 0.12 dm3
pada suhu bilik. 1 mol gas memenuhi 24 dm3 = 120 cm3
pada suhu bilik

Factor that affects rate of reaction Size of calcium carbonate in experiment I is smaller than experiment II.
Faktor yang mempengaruhi tindak balas Saiz kalsium karbonat dalam eksperimen I lebih kecil berbanding dalam eksperimen II.

Compare how the factor affects rate of The total surface area of calcium carbonate exposed to collision in experiment I is higher than
reaction experiment II.
Bandingkan bagaimana faktor mempengaruhi Jumlah luas permukaan kalsium karbonat yang terdedah kepada perlanggaran dalam
kadar tindak balas eksperimen I lebih besar berbanding dalam eksperimen II.

Compare the frequency of collision between The frequency of collisions between calcium carbonate and hydrogen ions in experiment I is
* particles higher than experiment II.
Bandingkan frekuensi perlanggaran di antara Frekuensi perlanggaran di antara kalsium karbonat dan ion hidrogen dalam eksperimen I lebih
* zarah-zarah tinggi daripada eksperimen II.

Compare the frequency of effective collisions The frequency of effective collisions between calcium carbonate and hydrogen ions in
between *particles experiment I is higher than experiment II.
Bandingkan frekuensi perlanggaran berkesan Frekuensi perlanggaran berkesan di antara kalsium karbonat dan ion hidrogen dalam
di antara *zarah-zarah eksperimen I lebih tinggi daripada eksperimen II.

© Nilam Publication Sdn Bhd 36

MODULE • Chemistry Form 5

Experiment I / Eksperimen I Experiment II / Eksperimen II

Compare rate of reaction Rate of reaction in experiment I is higher than experiment II.
Bandingkan kadar tindak balas Kadar tindak balas dalam eksperimen I lebih tinggi berbanding dalam eksperimen II

Sketch of the graph Volume of carbon dioxide gas/cm3
Lakaran graf Isi padu gas carbon dioksida/cm3

Experiment I
Eksperimen I

Experiment II U
Eksperimen II N
I
Time/s T
Masa/s
1
* particles / zarah-zarah – State the type of particles (atom/ion/molecule) that collide based on the ionic equation for the reaction.

Nyatakan jenis zarah (atom/ion/molekul) yang berlanggar berdasarkan persamaan ion untuk tindak balas tersebut.

3 Complete the following table to compare and explain the rate of reaction in Experiment I and Experiment II based on collision
theory. / Lengkapkan jadual berikut bagi membanding dan menerangkan berdasarkan teori perlanggaran kadar tindak balas
dalam Eksperimen I dan Eksperimen II.

Reactant Experiment I / Eksperimen I Experiment II / Eksperimen II
Bahan tindak balas
20 cm3 of 0.5 mol dm–3 hydrochloric acid + excess 20 cm3 of 0.5 mol dm–3 sulphuric acid + excess
of magnesium powder at 30°C of magnesium powder at 30°C
20 cm3 asid hidroklorik 0.5 mol dm–3 + serbuk 20 cm3 asid sulfurik 0.5 mol dm–3 + serbuk
magnesium berlebihan pada 30°C magnesium berlebihan pada 30°C

Balanced equation Mg + 2HCl → MgCl2 + H2 Mg + H2SO4 → MgSO4 + H2
Persamaan seimbang

Ionic equation Mg + 2H+ → Mg2+ + H2 Mg + 2H+ → Mg2+ + H2
Persamaan ion
Number of moles of HCl / Bilangan HCl Number of moles of H2SO4 / Bilangan mol H2SO4
Calculate volume of gas released in
each experiment at room condition = 20 × 0.5 = 0.01 mol = 20 × 0.5 = 0.01 mol
Hitung isi padu gas yang dibebaskan 1 000 1 000
dalam setiap eksperimen pada suhu
bilik From the equation / Daripada persamaan From the equation / Daripada persamaan
2 mol HCl : 1 mol H2 1 mol H2SO4 : 1 mol H2
0.01 mol HCl : 0.005 mol H2 0.01 mol H2SO4 : 0.001 mol H2
Volume of H2 / Isi padu H2 = 0.005 mol × 24 dm3 mol–1 Volume of H2 / Isi padu H2
= 0.12 dm3 = 0.01 mol × 24 dm3 mol–1 = 0.24 dm3
= 120 cm3 = 240 cm3

Ionisation equation of acid and HCl → H+ + Cl– H2SO4 → 2H+ + SO42–
concentration of H+ ion 1 mol of hydrochloric acid ionise to 1 mol H+ 1 mol of sulphuric acid ionise to 2 mol H+
Persamaan pengionan asid dan 1 mol dm–3 hydrochloric acid ionise to 1 mol dm–3 H+ 1 mol dm–3 sulphuric acid ionise to 2 mol dm–3 H+
kepekatan ion H+ 1 mol asid hidroklorik mengion kepada 1 mol ion H+ 1 mol asid sulfurik mengion kepada 2 mol ion H+

1 mol dm–3 asid hidroklorik mengion kepada 1 mol dm–3 asid sulfurik mengion kepada
1 mol dm–3 ion H+ 2 mol dm–3 ion H+

Compare the concentration of Concentration of hydrogen ion, H+ in experiment II is double of experiment I.
reactant / Bandingkan kepekatan Kepekatan ion hidrogen, H+ dalam eksperimen II adalah dua kali ganda eksperimen I
bahan tindak balas
The number of hydrogen ion per unit volume in experiment II is double of experiment I.
Compare how the factor affects rate Bilangan ion hidrogen dalam satu unit isi padu dalam eksperimen II adalah dua kali ganda eksperimen
of reaction / Bandingkan faktor yang I.
mempengaruhi kadar tindak balas
Frequency of collisions between hydrogen ions and magnesium atoms in experiment II is higher than
Compare the frequency of collision experiment I.
between *particles / Bandingkan Frekuensi perlanggaran di antara ion hidrogen dan atom magnesium dalam eksperimen II lebih tinggi
frekuensi perlanggaran di antara daripada eksperimen I.
*zarah-zarah
37 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Experiment I / Eksperimen I Experiment II / Eksperimen II

Compare the frequency of effective Frequency of effective collisions between hydrogen ions and magnesium atoms in experiment II
collisions between *particles is higher than experiment I. / Frekuensi perlanggaran berkesan di antara ion hidrogen dan atom
Bandingkan frekuensi perlanggaran magnesium dalam eksperimen II lebih tinggi daripada eksperimen I.
berkesan di antara *zarah-zarah
Rate of reaction in experiment II is higher than experiment I.
Compare rate of reaction Kadar tindak balas dalam eksperimen II lebih tinggi daripada eksperimen I.
Bandingkan kadar tindak balas

Sketch of the graph Volume of hydrogen gas /cm3
Lakaran graf Isi padu gas hidrogen /cm3

NU 240 Experiment II
Eksperimen II

I1T 120

Experiment I
Eksperimen I

Time/s
Masa/s

Compare the gradient and amount of (i) The gradient of the curve for experiment II is greater than experiment I because the rate of
product for the curve in both
experiments. Explain. / Bandingkan reaction in experiment II is higher than experiment I.
kecerunan dan kuantiti hasil bagi graf Kecerunan graf bagi eksperimen II lebih tinggi daripada eksperimen I kerana kadar tindak balas
dalam kedua-dua eksperimen.
Terangkan. dalam eksperimen II lebih tinggi daripada eksperimen I.

(ii) The volume of hydrogen gas collected in experiment II is double of experiment I.

Sulphuric acid is diprotic acid while hydrochloric acid is monoprotic acid. One mol of

sulphuric acid ionises to two mol of H+ ions, one mol of hydrochloric acid ionises to

one mol of H+ ions. / Isi padu gas hidrogen yang terkumpul dalam eksperimen II adalah

dua kali ganda lebih banyak daripada eksperimen I. Asid sulfurik ialah asid diprotik

manakala asid hidroklorik ialah asid monoprotik . Satu mol asid sulfurik mengion kepada

dua mol ion H+, satu mol asid hidroklorik mengion kepada satu mol ion H+.

(iii) The number of H+ ions in the same volume and concentration of both acids is double in
sulphuric acid. / Bilangan mol ion H+ dalam isi padu dan kepekatan asid yang sama adalah

dua kali ganda lebih banyak dalam asid sulfurik.

* particles / zarah-zarah – State the type of particles (atom/ion/molecule) that collide based on the ionic equation for the reaction.

Nyatakan jenis zarah (atom/ion/molekul) yang berlanggar berdasarkan persamaan ion untuk tindak balas tersebut.

4 Complete the following table to compare and explain how does copper(II) sulphate solution affect the rate of reaction.
Lengkapkan jadual berikut untuk membanding dan menerangkan berdasarkan teori perlanggaran bagaimana larutan

kuprum(II) sulfat mempengaruhi kadar tindak balas.

Experiment I / Eksperimen I Experiment II / Eksperimen II

Excess of Water Excess of Water
zinc granules Air zinc granules Air
Butiran zink Butiran zink
berlebihan berlebihan

50 cm3 of 1.0 mol dm–3 hydrochloric acid 50 cm3 of 1.0 mol dm–3 hydrochloric acid + Copper(II)
50 cm3 asid hidroklorik 1.0 mol dm–3 sulphate solution / 50 cm3 asid hidroklorik 1.0 mol dm–3
+ kuprum(II) sulfat

Balanced equation Zn(s/p) + 2HCl(aq/ak) → ZnCl2(aq/ak) + H2(g/g) Zn(s/p) + 2HCl(aq/ak) CuSO4 ZnCl2(aq/ak) + H2(g/g)
Persamaan seimbang
Zn + 2H+ → Zn2+ + H2
Ionic equation
Persamaan ion 38

© Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

Experiment I / Eksperimen I Experiment II / Eksperimen II

Calculate volume of hydrogen gas Number of moles of HCl / Bilangan mol HCl = 50 × 1 = 0.05 mol
released in experiments I and II at 1 000
room conditions. 1 mol of gas
occupies 24 dm3 at room conditions From the equation / Daripada persamaan
Hitung isi padu gas hidrogen yang 2 mol HCl : 1 mol H2
dibebaskan dalam eksperimen I 0.05 mol HCl : 0.025 mol H2
dan II pada suhu bilik. 1 mol gas
memenuhi 24 dm3 pada suhu bilik Volume of H2 / Isi padu H2 = 0.025 mol H2 × 24 dm3 mol–1
= 0.6 dm3
= 600 cm3

Compare the factor that affects Reaction between zinc and hydrochloric acid in experiment I is without catalyst while copper(II) sulphate U
rate of reaction added in experiment II acts as a catalyst. N
Bandingkan faktor yang Tindak balas antara zink dan asid hidroklorik dalam eksperimen I adalah tanpa mangkin manakala I
mempengaruhi kadar tindak balas kuprum(II) sulfat hadir dalam eksperimen II sebagai mangkin. T

Compare how the factor affects Copper(II) sulphate lowers the activation energy for the reaction between zinc and hydrochloric acid in 1
rate of reaction experiment II.
Bandingkan bagaimana faktor Kuprum(II) sulfat merendahkan tenaga pengaktifan bagi tindak balas antara zink dan asid hidroklorik
tersebut mempengaruhi kadar dalam eksperimen II.
tindak balas

Compare the frequency of effective Frequency of effective collisions between zinc atom and hydrogen ion in experiment II is higher than
collisions between *particles experiment I.
Bandingkan frekuensi Frekuensi perlanggaran berkesan di antara atom zink dan ion hidrogen dalam eksperimen II adalah lebih
perlanggaran berkesan di antara tinggi daripada eksperimen I.
*zarah-zarah

Compare rate of reaction Rate of reaction in experiment II is higher than experiment I.
Bandingkan kadar tindak balas Kadar tindak balas dalam eksperimen II lebih tinggi daripada eksperimen I.

Sketch of the graph for experiment Volume of hydrogen gas /cm3
I and experiment II Isi padu gas hidrogen /cm3
Lakaran graf bagi eksperimen I
dan eksperimen II Experiment II
Eksperimen II

Experiment I
Eksperimen I

Time/s
Masa/s

Compare the gradient and amount 1 The gradient of the curve for experiment II is greater than experiment I because the rate of reaction
of product for the curve in both in reaction experiment II is higher than experiment I.
experiments. Explain.
Bandingkan kecerunan dan 2 The volume of hydrogen gas released in experiment I is equal to experiment II. The volume and
kuantiti hasil untuk graf dalam
kedua-dua eksperimen. concentration of hydrochloric acid in experiments I and II are similar. Copper(II) sulphate in experiment
Terangkan. II that is used as a catalyst does not affect the total volume of hydrogen gas produced.
1 Kecerunan graf bagi eksperimen II lebih tinggi daripada eksperimen I kerana kadar tindak balas
dalam eksperimen II lebih tinggi daripada eksperimen I.
2 Isi padu gas hidrogen yang dibebaskan dalam eksperimen I adalah sama dengan eksperimen

II. Isi padu dan kepekatan asid hidroklorik dalam eksperimen I dan II adalah sama. Kuprum(II) sulfat

dalam eksperimen II yang digunakan sebagai mangkin tidak mempengaruhi jumlah isi padu gas
hidrogen yang dihasilkan.

39 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

5 The table below shows the data for three experiments that have been carried out to determine the effect of catalyst on the
decomposition of hydrogen peroxide to water and oxygen. / Jadual di bawah menunjukkan data bagi tiga eksperimen yang
dijalankan untuk menentukan kesan mangkin ke atas penguraian hidrogen peroksida kepada air dan oksigen.

Volume of H2O2 / cm3 Concentration of H2O2 / mol dm–3 Mass of MnO2 /g
Isi padu H2O2 / cm3 Kepekatan H2O2 / mol dm–3 Jisim MnO2 /g

I 20 1.0 –

II 20 1.0 1.0

III 20 1.0 2.0

U (a) Write the chemical equation for the decomposition of hydrogen peroxide.
N
I Tuliskan persamaan kimia bagi penguraian hidrogen peroksida.
1T 2H2O2 → 2H2O + O2
(b) Draw the set-up of apparatus for the experiment. / Lukiskan susunan radas bagi eksperimen ini.

Manganese(IV) oxide Water
Mangan(IV) oksida Air

H2O2

(c) Between experiments I, II and III, which has the highest rate of reaction? Explain your answer. / Antara eksperimen I, II
dan III, yang manakah mempunyai kadar tindak balas yang paling tinggi? Terangkan jawapan anda.
Experiment III. The amount of catalyst used in experiment III is more than experiment II.

Eksperimen III. Kuantiti mangkin yang digunakan dalam eksperimen III adalah lebih banyak dari eksperimen II.

(d) Explain how does manganese(IV) oxide affect the rate of decomposition of hydrogen peroxide by using collision theory.
Terangkan bagaimana mangan(IV) oksida mempengaruhi kadar penguraian hidrogen peroksida menggunakan teori

perlanggaran.
• Manganese(IV) oxide provides an alternative path with lower activation energy for the decomposition of hydrogen peroxide

• The frequency of effective collisions between hydrogen peroxide molecules increases.
• The rate of decomposition of hydrogen peroxide increases.
• Mangan(IV) oksida menyediakan laluan alternatif dengan tenaga pengaktifan yang lebih rendah bagi penguraian hidrogen

peroksida.

• Frekuensi perlanggaran berkesan antara molekul hidrogen peroksida meningkat.

• Kadar penguraian hidrogen peroksida meningkat.

(e) Calculate the volume of oxygen gas released in experiment II at room temperature and pressure.

Hitung isi padu gas oksigen yang dibebaskan dalam eksperimen II pada suhu dan tekanan bilik.

Number of moles of H2O2 / Bilangan mol H2O2 = 20 × 1 = 0.02 mol
1 000

From the equation, / Daripada persamaan,

2 mol H2O2 : 1 mol O2
0.02 mol H2O2 : 0.01 mol O2

Volume of O2 / Isi padu O2 = 0.01 mol × 24 dm3 mol–1
= 0.24 dm3
= 240 cm3

© Nilam Publication Sdn Bhd 40

MODULE • Chemistry Form 5

(f) Sketch the graph of volume of oxygen gas against time for experiments I, II and III.
Lakarkan graf isi padu gas oksigen melawan masa bagi eksperimen I, II dan III.

Volume of oxygen gas /cm3
Isi padu gas oksigen /cm3

III U
II I N
I
Time/s T
Masa/s
1
(g) State one factor other than concentration and catalyst that can affect the rate of decomposition of hydrogen peroxide.
Nyatakan satu faktor lain selain kepekatan dan mangkin yang boleh mempengaruhi kadar penguraian hidrogen peroksida.

Temperature of hydrogen peroxide. / Suhu hidrogen peroksida

6 A student carried out three sets of experiment to investigate the factors affecting the rate of reaction. The table below shows the
information and the result of the experiment.

Seorang pelajar menjalankan tiga set eksperimen untuk mengkaji faktor-faktor yang mempengaruhi kadar tindak balas. Jadual
di bawah menunjukkan maklumat dan keputusan eksperimen itu.

Time taken to collect

Set Reactants maximum volume of gas / s
Set Bahan tindak balas Masa diambil untuk

mengumpulkan isi padu

gas maksimum / s

I 3 g magnesium ribbon and 50 cm3 of 1 mol dm–3 hydrochloric acid 100
3 g pita magnesium dan 50 cm3 asid hidroklorik 1 mol dm–3

II 3 g magnesium powder and 50 cm3 of 1 mol dm–3 hydrochloric acid 60
3 g serbuk magnesium dan 50 cm3 asid hidroklorik 1 mol dm–3

III 3 g magnesium powder and 50 cm3 of 1 mol dm–3 hydrochloric acid and copper(II) sulphate solution 30
3 g serbuk magnesium dan 50 cm3 asid hidroklorik 1 mol dm–3 dan larutan kuprum(II) sulfat

(a) Write a chemical equation to show the reaction between magnesium and hydrochloric acid.
Tuliskan persamaan kimia untuk menunjukkan tindak balas antara magnesium dan asid hidroklorik.

Mg + 2HCl → MgCl2 + H2

(b) Calculate the number of mole of / Hitung bilangan mol bagi
(i) Magnesium / Magnesium

[Relative atomic mass of Mg = 24] / [Jisim atom relatif Mg = 24]

Number of moles Mg / Bilangan Mol Mg = 3 = 0.125 mol
24

(ii) Hydrochloric acid / Asid hidroklorik

Number of moles HCl / Bilangan Mol HCl = 1 × 50 = 0.05 mol
1 000

(c) Calculate the maximum volume of hydrogen gas produced at room condition.

Hitung isi padu maksimum gas hidrogen yang dihasilkan pada keadaan bilik.

[1 mole of gas occupies the volume of 24 dm3 at room condition] / [1 mol gas menempati isi padu 24 dm3 pada keadaan bilik]

From the equation / Daripada persamaan:

2 mol HCl : 1 mol H2

0.05 mol HCl : 0.025 mol H2

Volume of hydrogen gas / Isi padu gas hidrogen = 0.025 × 24 dm3 = 0.6 dm3 = 600 cm3

41 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

(d) What is the average rate of reaction for / Apakah kadar tindak balas purata bagi

(i) Set I? 0.6 600
100 100
Rate of reaction / Kadar tindak balas = = 0.006 dm3 s–1 // = 6 cm3 s–1

(ii) Set II? 0.6 600
60 60
Rate of reaction / Kadar tindak balas = = 0.01 dm3 s–1 // = 10 cm3 s–1

U (iii) Set III? 0.6 600
N 30 30
I Rate of reaction / Kadar tindak balas = = 0.02 dm3 s–1 // = 20 cm3 s–1
T
1 (e) (i) The diagram below shows the curve obtained for set I when the volume of hydrogen gas liberated against time is
plotted. On the same axes, sketch the curve that you would expect to obtain if the experiment is repeated using 3 g

magnesium ribbon and 50 cm3 of 2 mol dm–3 hydrochloric acid.

Rajah di bawah menunjukkan lengkung yang diperoleh bagi Set I apabila isi padu gas hidrogen yang dibebaskan

melawan masa diplotkan. Pada paksi yang sama, lakarkan lengkung yang anda jangka diperoleh sekiranya eksperimen

diulang menggunakan 3 g pita magnesium dan 50 cm3 asid hidroklorik 2 mol dm–3.

Volume of hydrogen gas / dm3
Isi padu gas hidrogen / dm3

1.2

0.6

Time / s
Masa / s

(ii) Explain how you obtain the curve in (e)(i).
Terangkan bagaimana anda memperoleh lengkung di (e)(i).

1 Initial rate of reaction of the experiment is higher because the concentration of hydrochloric acid is higher, therefore
the curve is steeper.
2 Hydrochloric is the limiting factor // Magnesium is in excess.
3 Maximum volume of hydrogen gas collected is double because the number of mole of hydrochloric acid is double.
1 Kadar tindak balas awal eksperimen adalah lebih tinggi kerana kepekatan asid hidroklorik lebih tinggi, maka
lengkung lebih curam.
2 Hidroklorik adalah faktor pengehad // Magnesium berlebihan.
3 Isi padu maksimum gas hidrogen yang dikumpul adalah dua kali ganda kerana bilangan mol asid hidroklorik adalah

dua kali ganda.

(f) (i) Compare the rate of reaction in Set II and Set III.
Bandingkan kadar tindak balas dalam Set II dan Set III.

Rate of reaction in Set III is higher than Set II.
Kadar tindak balas dalam Set III lebih tinggi daripada Set II

© Nilam Publication Sdn Bhd 42

MODULE • Chemistry Form 5

(ii) By using the collision theory, explain the difference in the rate of reaction between Set II and Set III. U
Dengan menggunakan teori perlanggaran, terangkan perbezaan kadar tindak balas antara Set II dan Set III. N
I
– Reaction between magnesium and hydrochloric acid in Set II is without catalyst while copper(II) sulphate presence in T
Set III as a catalyst.
– Copper(II) sulphate lower the activation energy for the reaction between magnesium and hydrochloric acid in Set III. 1
– Frequency of effective collisions between magnesium atom and hydrogen ion in Set III is higher than Set II.
– Tindak balas antara magnesium dan asid hidroklorik dalam Set II tanpa mangkin manakala kuprum(II) sulfat hadir
sebagai mangkin dalam Set III.
– Kuprum(II) sulfat merendahkan tenaga pengaktifan untuk tindak balas antara magnesium dan asid hidroklorik dalam
Set III.
– Frekuensi perlanggaran berkesan antara atom magnesium dan ion hidrogen dalam Set III lebih tinggi daripada Set II.

(iii) Sketch energy profile diagram which shows an activation energy for the reactions in Set II and Set III.
Lakarkan gambar rajah profil tenaga yang menunjukkan tenaga pengaktifan untuk Set II dan Set III.

Energy – Ea : The activation energy in Set II.
Tenaga Tenaga pengaktifan Set II.

Ea – Ea;: The activation energy in Set III.
Tenaga pengaktifan Set III.

Mg + 2HCl Ea;
MgCl2 + H2
Reaction path
Lintasan tindak balas

7 The diagram below shows a sketch of curve I and curve II for volume of carbon dioxide gas collected against time. Curve I is for
the reaction in Experiment I between 2.0 g marble chips added to 50 cm3 of 1 mol dm–3 hydrochloric acid at room temperature.

Rajah di bawah menunjukkan lakaran lengkung I dan lengkung II untuk isi padu gas karbon dioksida yang dikumpulkan
melawan masa. Lengkung I adalah tindak balas dalam Eksperimen I di antara 2.0 g ketulan marmar ditambah kepada 50 cm3
asid hidroklorik 1 mol dm–3 pada suhu bilik.
Volume of carbon dioxide gas / cm3
Isi padu gas karbon dioksida / cm3

II

I

Time / s
Masa / s
(a) Write a chemical equation to show the reaction between marble chips and hydrochloric acid.
Tuliskan persamaan kimia untuk menunjukkan tindak balas antara ketulan marmar dan asid hidroklorik.
CaCO3 + 2HCl → CaCl2 + H2O + CO2

43 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

(b) State two ways to obtain curve II for the reaction of 50 cm3 of 1 mol dm–3 hydrochloric acid at room temperature.
Nyatakan dua cara untuk mendapatkan lengkung II bagi tindak balas 50 cm3 asid hidroklorik 1 mol dm–3 pada suhu bilik.

(i) 2.0 g marble powder is used to replace 2.0 g marble chips
(ii) The experiment is carried out a 40 °C (higher than room temperature)
(i) 2.0 g serbuk marmar digunakan untuk menggantikan 2.0 g ketulan marmar
(ii) Eksperimen dijalankan pada suhu 40 °C (lebih tinggi daripada suhu bilik)

(c) Explain each of your answer in (b).
U Terangkan setiap jawapan anda dalam (b).
N
I (i) 2.0 g marble powder in Experiment II has a larger total surface area than 2.0 g marble chips in Experiment I, hence

T initial rate of reaction of Experiment II is higher than Experiment I. Maximum volume of carbon dioxide gas collected in
1 Experiment I and II are the same because the quantity of marble and hydrochloric acid are the same in both
experiments.

(ii) Experiment II is conducted at higher temperature compared to Experiment I, hence intial rate of reaction

Experiment II is higher than Experiment I. Maximum volume of carbon dioxide gas collected in Experiment I and II are

the same because the quantity of marble and hydrochloric acid are the same in both experiments.

(i) 2.0 g serbuk marmar dalam Eksperimen II mempunyai jumlah luas permukaan yang lebih besar daripada 2.0 g

ketulan marmar dalam Eksperimen I, oleh itu kadar awal tindak balas Eksperimen II adalah lebih tinggi daripada

Eksperimen I. Isi padu maksimum gas karbon dioksida yang dikumpulkan dalam Eksperimen I dan II adalah sama kerana

kuantiti marmar dan asid hidroklorik adalah sama dalam kedua-dua eksperimen.

(ii) Eksperimen II dijalankan pada suhu yang lebih tinggi berbanding Eksperimen I, maka kadar tindak balas awal

Eksperimen II adalah lebih tinggi daripada Eksperimen I. Isi padu maksimum gas karbon dioksida yang dikumpulkan

dalam Eksperimen I dan II adalah sama kerana kuantiti marmar dan asid hidroklorik adalah sama dalam kedua-dua

eksperimen.

(d) Which of the two reactants in Experiment I is in excess? Explain your answer.

Antara dua bahan tindak balas dalam Eksperimen I, yang manakah berlebihan? Terangkan jawapan anda.

[Relative atomic mass: C = 12, O = 16, Ca = 40] / [Jisim atom relatif: C = 12, O = 16, Ca = 40]

Number of moles CaCO3 / Bilangan mol CaCO3 = 2g = 0.02 mol
100 g mol–1

Number of moles HCl / Bilangan mol HCl = 50 × 1 = 0.05 mol
1 000

From the equation / Daripada persamaan:
1 mol of CaCO3 react with 2 mol of HCl / 1 mol CaCO3 bertindak balas dengan 2 mol HCl
0.02 mol of CaCO3 react with 0.04 mol of HCl / 0.02 mol CaCO3 bertindak balas dengan 0.04 mol HCl
→ Hydrochloric acid is in excess / Asid hidroklorik berlebihan

Additional Questions

Soalan Tambahan

© Nilam Publication Sdn Bhd 44

CARBON COMPOUND / SEBATIAN KARBON Amino acid Glucose Concept Map / Peta Konsep UNIT
Asid amino Glukosa
ORGANIC CARBON COMPOUND INORGANIC CARBON COMPOUND 2
SEBATIAN KARBON ORGANIK SEBATIAN KARBON TAK ORGANIK Monomer Monomer

Type / Jenis Protein Carbohydrate
Protein Karbohidrat

Hydrocarbon / Hidrokarbon Non hydrocarbon / Bukan hidrokarbon Natural polymer Examples / Contoh
Type / Jenis Polimer semula jadi

Saturated Hydrocarbon Unsaturated hydrocarbon Homologous series CnH2n + 1 COOH Natural MODULE • Chemistry Form 5
Hidrokarbon Tepu Hidrokarbon tak tepu Siri homolog n = 1, 2, 3, 4… rubber
Getah asli CARBON COMPOUNDS
General formula
Homologous series Homologous series SEBATIAN KARBON
Hydration / Penghidratan Formula am Monomer
Siri homolog Siri homolog Dehydration / Pendehidratan Alcohol Oxidation Isoprene
Alkohol Pengoksidaan Carboxylic acid
Alkane / Alkana Hydrogenation Alkene / Alkena Asid karboksilik
Penghidrogenan
Isoprena
General formula General formula CnH2n + 1OH General formula Esterification Functional group
Formula am Formula am n = 1, 2, 3… Formula am Pengesteran Kumpulan berfungsi Vulcanisation
Pemvulkanan
CnH2n + 2 CnH2n Functional Carboxyl, / Karboksil,
n = 1, 2, 3… n = 2, 3, 4… group Ester –COOH

45 © Nilam Publication Sdn Bhd Functional group Functional group Hydroxyl, / Hidroksil, Kumpulan Ester Homologous series Vulcanised
Kumpulan Kumpulan –OH berfungsi rubber
berfungsi berfungsi Siri homolog Getah
CnH2nCOOCn’H2n’ + 1 Esterification of fatty acid
Single bond Double bond n = 0, 1, 2.… tervulkan
between carbon between carbon n’ = 1,2 … with alcohol glycerol
atoms: atoms:
Ikatan tunggal Ikatan ganda dua General formula Pengesteran asid lemak
Formula am dengan alkohol gliserol

Saturated fat / Lemak tepu

antara atom antara atom Carboxylate, Fat / Lemak Hydrogenation / Penghidrogenan
karbon: karbon: Karbosilat, Unsaturated fat / Lemak tak tepu
–C – C– –C = C– –COOC– Functional group
Kumpulan berfungsi

Focus of study for all the homologous Learning objective / Objektif pembelajaran
series are:
Fokus pembelajaran bagi semua siri 1 Understand carbon compound 6 Analysing carboxylic acid
homolog ialah: Memahami sebatian karbon Menganalisis asid karboksilik
1 General formula, molecular formula, 2 Analysing alkane 7 Analysing ester / Menganalisis ester
Menganalisis alkana 8 Assessing fat / Menilai lemak
structural formula, IUPAC 3 Analysing alkene 9 Analysing natural rubber / Menganalisis getah asli
nomenclature and Isomerism Menganalisis alkena 10 Recognising the order in homologous series
Persamaan am, formula molekul, 4 Synthesising idea of isomerism Menyedari ketertiban dalam siri homolog
formula struktur, penamaan IUPAC Mensintesis idea keisomeran 11 Appreciating existence of various organic materials
dan keisomeran 5 Analysing alcohol
2 Physical properties and chemical Menganalisis alkohol in nature / Menghargai kewujudan pelbagai bahan
properties (included preparation of organik dalam alam
each of homologous series)
Ciri fizik dan kimia (termasuk
penyediaan setiap siri homolog)

U
N
I
T

2

MODULE • Chemistry Form 5

CARBON COMPOUNDS (HYDROCARBON) / SEBATIAN KARBON (HIDROKARBON)
Carbon Compounds / Sebatian Karbon

What are carbon compounds? Carbon compounds are compounds that contain the carbon element.
Apakah sebatian karbon? Sebatian karbon adalah sebatian yang mengandungi unsur karbon.

How many groups can carbon These compounds can be classified into two groups:
compounds be classified to? Sebatian ini boleh dikelaskan kepada dua kumpulan:
Berapa kumpulankah sebatian (a) Organic compounds / Sebatian organik
karbon dapat diklasifikasikan? (b) Inorganic compounds / Sebatian tak organik

What are organic compounds? Organic compounds are carbon compounds that are obtained from living things such as sugar, starch,
Give examples. protein, vitamin, enzym etc. They are obtained from plants and animals.
Apakah sebatian organik? Berikan Sebatian organik adalah sebatian karbon yang diperoleh daripada benda hidup seperti gula, kanji, protein,
contoh. vitamin, enzim dan lain-lain. Bahan-bahan ini diperoleh daripada tumbuhan dan binatang.

What are inorganic compounds? Inorganic compounds are carbon compounds that usually do not contain carbon to carbon bonds such as
Give examples. carbon dioxide (CO2), carbon monoxide (CO), calcium carbonate (CaCO3) etc.
Apakah sebatian tak organik? Sebatian tak organik adalah sebatian karbon yang biasanya tidak mengandungi ikatan karbon-karbon
Berikan contoh. seperti karbon dioksida (CO2), karbon monoksida (CO), kalsium karbonat (CaCO3) dan lain-lain.

What are the combustion products Most organic compounds contain the carbon element and hydrogen element. Complete combustion of

U of organic compounds? Give organic compounds produces carbon dioxide and water.
N examples. Kebanyakan sebatian organik mengandungi unsur karbon dan hidrogen. Pembakaran lengkap sebatian
I Apakah hasil pembakaran organik menghasilkan karbon dioksida dan air.
T sebatian karbon? Berikan contoh.

2 Hydrocarbon / Hidrokarbon

What is hydrocarbon? Hydrocarbons is organic compounds that contain only carbon, C and hydrogen, H.
Apakah hidrokarbon? Hidrokarbon adalah sebatian organik yang hanya mengandungi karbon, C dan hidrogen, H sahaja.

How many groups can Hydrocarbons are classified into two groups:
hydrocarbon be classified to? Hidrokarbon boleh dikelaskan kepada dua kumpulan:
Berapa kumpulankah hidrokarbon (a) Saturated hydrocarbon / Hidrokarbon tepu
dapat diklasifikasikan? (b) Unsaturated hydrocarbon / Hidrokarbon tak tepu

What is saturated hydrocarbon? Hydrocarbon that contains only single covalent bonds between carbon atoms.
Apakah hidrokarbon tepu?
Hidrokarbon yang mengandungi ikatan kovalen tunggal sahaja di antara atom kabon.
What is unsaturated hydrocarbon?
Apakah hidrokarbon tak tepu? Hydrocarbon that contains at least one double or triple covalent bond between carbon
atoms.
Give examples of saturated and
unsaturated hydrocarbons. Hidrokarbon yang mengandungi sekurang-kurangnya satu ikatan kovalen ganda dua atau
Berikan contoh hidrokarbon tepu
dan hidrokarbon tak tepu. ganda tiga di antara atom karbon.

Saturated hydrocarbons Unsaturated hydrocarbons
Hidrokarbon tepu Hidrokarbon tak tepu

HHH HHH H

HC C CH HC C C C

HHH HH H

Single covalent bond between carbon atoms. Double covalent bond between carbon
atoms.
Ikatan kovalen tunggal sahaja di antara Ikatan kovalen ganda dua di antara karbon.
karbon.

What is the main source of The main source of hydrocarbons is petroleum.
hydrocarbon? / Apakah sumber Sumber utama hidrokarbon ialah petroleum.
utama hidrokarbon?
46
© Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

How petroleum is formed? It is formed as a result of decomposition of plants and animals that died millions of years ago.
Bagaimanakah petroleum Ia terbentuk daripada penguraian tumbuhan dan binatang yang telah mati sejak berjuta-juta tahun dahulu.
terbentuk?
Petroleum is a mixture of different molecular size hydrocarbons .
What is petroleum? Petroleum ialah campuran molekul hidrokarbon yang berlainan saiz.
Apakah petroleum?
These hydrocarbons can be separated using fractional distillation of petroleum at different temperature.
How are the mixture of different This process separates hydrocarbons with different molecular size (petroleum gas, petrol, naphta,
molecular size of hydrocarbon in kerosene, diesel and lubricating oil).
petroleum is separated? Hidrokarbon ini boleh diasingkan melalui penyulingan berperingkat petroleum pada suhu yang berlainan.
Bagaimanakah campuran yang Proses ini mengasingkan hidrokarbon yang mempunyai saiz molekul yang berlainan (gas petroleum,
berbeza saiz molekul hidrokarbon petrol, nafta, kerosin, diesel dan minyak pelincir).
dalam petroleum diasingkan?
Different moleculer size of hydrocarbons have different boiling points.
State the physical property on Saiz molekul hidrokarbon yang berbeza mempunyai takat didih berbeza.
which the separation depends.
Nyatakan sifat fizik yang
membolehkan pengasingan
tersebut.

Alkane / Alkana U
N
What is an alkane? Alkanes are hydrocarbons with general formula CnH2n + 2 , where n = 1, 2, 3, … I
Apakah alkana? T
Why alkanes are hydrocarbons? CnH2n + 2
Mengapakah alkana adalah 2
hidrokarbon?
Explain why alkanes are classified Alkana adalah hidrokarbon yang mempunyai formula am , di mana n = 1, 2, 3, ...
as saturated hydrocarbon.
Terangkan mengapa alkana Alkanes are made up of carbon and hydrogen only.
diklasifikasikan sebagai Alkana terdiri daripada karbon dan hidrogen sahaja.
hidrokarbon tepu.
Each carbon atom is bonded to four other atoms by single covalent bonds that is C – C or C – H. It
What is structural formula?
Apakah formula struktur? is classified as saturated hydrocarbons. (A saturated hydrocarbon contains only single covalent
bonds between carbon atoms).

Setiap atom karbon terikat kepada empat atom lain oleh ikatan kovalen tunggal iaitu C – C atau

C – H. Ia dikelaskan sebagai hidrokarbon tepu . (Hidrokarbon tepu mengandungi ikatan
kovalen tunggal sahaja di antara atom karbon).

Example: / Contoh: H Represents ‘•×’ (one pair of
Methane / Metana, CH4 electrons is shared to form

H a single covalent bond)
H C H Mewakili ‘•×’ (Sepasang
HCH
elektron dikongsi untuk
H
H membentuk ikatan kovalen
tunggal)

Electron arrangement in methane molecule, CH4 Structural formula of methane, CH4
Susunan elektron dalam molekul metana, CH4 Formula struktur metana, CH4

Remark / Catatan:
• Each carbon atom in the structural formula must have a total of four pairs of electrons shared with another carbon and

hydrogen atom to achieve octet electron arrangement. (Each carbon atom must have four ‘–’ in the structural formula).
Setiap atom karbon dalam formula struktur mesti mempunyai empat pasang elektron yang dikongsi dengan atom

karbon dan hidrogen yang lain untuk mencapai susunan elektron oktet. (Setiap atom karbon mesti mempunyai empat
‘–’ dalam formula strukturnya).
• Each hydrogen atom in the structural formula must have one pair of electrons shared with carbon to achieve duplet
electron arrangement. (Each hydrogen atom must have one ‘–’ in the structural formula).
Setiap atom hidrogen dalam formula struktur mesti mempunyai satu pasang elektron yang dikongsi dengan karbon
untuk mencapai susunan atom duplet (Setiap hidrogen mesti mempunyai satu ‘–’ dalam formula strukturnya).

A structural formula is a formula that shows how atoms are bonded to each other covalently in a
molecule. / Formula struktur ialah formula yang menunjukkan bagaimana atom terikat antara satu
sama lain secara kovalen dalam satu molekul.

47 © Nilam Publication Sdn Bhd

MODULE • Chemistry Form 5

What is molecular formula? A molecular formula is a chemical formula that shows the actual number of atoms of each element in a
Apakah formula molekul? molecule.
Formula molekul ialah formula kimia yang menunjukkan bilangan sebenar atom bagi setiap unsur dalam
satu molekul.

How to name an alkane? The name of straight chain alkanes (all the carbon atoms are joined in a continuous chain) are made up
Bagaimanakah cara menamakan of two component:
alkana? Nama bagi alkana berantai lurus (semua atom karbon yang bergabung dalam rantai yang berterusan)
terdiri daripada dua komponen:
(i) Stem / root: indicates the number of carbon atoms in the longest continuous carbon chain. The name

of stems for the first ten straight alkanes are:
Awalan / induk: menunjukkan bilangan atom karbon dalam rantai karbon berterusan yang paling

panjang. Nama awalan bagi sepuluh alkana pertama adalah:

Number of carbon, C 1 2 3 4 5 6 7 8 9 10
Bilangan karbon, C

Stem Meth Eth Prop But Pent Hex Hept Oct Non Dec
Awalan Met Et Prop But Pent Heks Hept Okt Non Dek

(ii) Suffix / ending: indicates the group of the compound. For alkane, the suffix is ‘ane’ because it belongs
to the alkane group.

Akhiran: menunjukkan kumpulan sebatian. Bagi alkana, akhiran adalah ‘ana’ kerana ia adalah dalam
kumpulan alkana.

U
N
I What are the molecular formulae, structural formulae and names for the first ten straight chain of alkane?
T Apakah formula molekul, formula struktur dan nama bagi sepuluh rantaian lurus pertama alkana?

2 Number of carbon Molecular formula Structural formula Name of alkane
atoms Formula molekul Formula struktur Nama alkana

Bilangan atom karbon CnH2n + 2

1 CH4 H Methane
( meth / met ) HCH Metana

H

2 C2H6 H H Ethane
( eth / et ) C3H8 HC CH Etana
H
3 H Propane
( prop / prop ) Propana
HHH
HC C CH

HHH

4 C4H10 HHHH Butane
( but / but ) H C C C CH Butana

HHHH

5 C5H12 HHHHH Pentane
( pent / pent ) H C C C C CH Pentana

HHHHH

© Nilam Publication Sdn Bhd 48


Click to View FlipBook Version
Previous Book
Rapid Review of Clinical Medicine for MRCP Part 2 by Sharma, Sanjay Kaushal, Rashmi (z-lib.org)
Next Book
Role of Pre hospital Care Systems and First Responders during the COVID-19 Crisis.