PROGRAM SMART SPM 2021 PERINGKAT DKBB

Bab 03 - Termokimia
Chap. 03 Thermochemistry

4.1 Menilai perubahan tenaga dalam tindak balas kimia

Evaluating energy changes in chemical reactions
4.2 Memahami haba pemendakan
Understanding heat of precipitation
4.3 Memahami haba penyesaran
Understanding heat of displacement
4.4 Memahami haba peneutralan
Understanding heat of neutralisation
4.5 Memahami haba pembakaran
Understanding Heat of Combustion

A. Perubahan Tenaga Dalam Tindak Balas Kimia

Energy Changes in Chemical Reactions

1. Tenaga berubah apabila tindak balas kimia berlaku.

The energy changes when a chemical reaction takes place.
2. Tenaga tersebut adalah dalam bentuk haba yang dibebaskan atau diserap semasa
tindak balas kimia.
The energy is in the form of heat, could be released or absorbed during the chemical
reaction.

B. Tindak Balas Eksotermik/ Exothermic Reaction

Haba/ Heat • Ialah tindak balas kimia yang membebaskan
haba ke persekitaran.
Is the chemical reaction that releases heat to
surroundings.

Haba Haba
• Apabila tenaga dibebaskan semasa tindak
Heat Heat balas kimia, persekitaran akan menjadi
panas dan suhu campuran akan naik.
When energy is given off during a chemical
Tindak balas eksotermik membebaskan haba
reaction, the surrounding will become hot
Exothermic reaction giving out heat and the temperature of the reaction mixture
will rise.

Contoh tindak balas eksotermik / Example of exothermic reaction:

a. Respirasi a Respiration
b. Tindak balas peneutralan b. Neutralization reactions
c. Tindak balas antara asid dengan logam c. Reaction between acid and metals
d. Tindak balas antara asid dengan d. Reaction between acid and metal
karbonat logam carbonates
e. Pembakaran bahan api seperti minyak e. Burning offuels such as petrol, wood,
petrol, kayu, arang batu dan coal, etc
sebagainya f Dissolving sodium hydroxide, NaOH in
f. Melarutkan natrium hidroksida , NaOH water
dalam air. g. Adding concentrated acids into water.
g. Penambahan asid pekat pada air h. Reaction between reactive metals such
h. Tindak balas antara logam reaktif as potassium, K and sodium, Na with
seperti kalium, K dan natrium, Na water
dengan air.

1
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

C. Tindak Balas Endotermik/ Endothermic Reaction

• Ialah tindak balas kimia yang menyerap haba
Haba / Heat
daripada persekitaran.
Is the chemical reaction that absorbs heat from
surroundings.

Haba Haba • Apabila tenaga diserap semasa tindak balas

Heat Heat kimia, persekitaran akan menjadi sejuk dan
suhu campuran akan menurun.
When energy is taken in during a chemical
reaction, the surrounding will become cold
Tindak balas endotermik menyerap haba
and the temperature of the reaction mixture
Endoothermic reaction absorbing heat
will fall.

Contoh tindak balas endotermik / Example of endothermic reaction:

a. Fotosintesis a. Photosynthesis
b. Melarutkan garam ammonium dalam air b. Di,ssolving ammonium salts in water
c. Menggoreng telur c. Frying an egg
d. Penguraian karbonat logam d. Decomposition of metal carbonates
e. Penguraian nitrat logam e. Decomposition of metal nitrates

Aktiviti 1: Kelaskan tindak balas kimia berikut kepada tindak balas eksotermik
atau endotermik
Activity 1: Classify the following chemical reactions into exothermic or endothermic

Bil Tindak balas kimia Jenis tindak balas

No Chemical Reaction Type of reaction
Tindak balas peneutralan
1.
Neutralization reactions
Tindak balas antara asid dengan logam
2. Reaction between acid and metals
Tindak balas antara asid dengan karbonat logam
3.
Reaction between acid and metal carbonates
Melarutkan garam ammonium dalam air
4.
Dissolving ammonium salts in water
Melarutkan natrium hidroksida, NaOH dalam air
5.
Dissolving sodium hydroxide, NaOH in water
Menambahkan asid pekat ke dalam air
6.
Adding concentrated acids into water
Penguraian logam karbonat
7.
Decomposition of metal carbonates
Penguraian logam nitrat
8.
Decomposition of metal nitrates
Pembakaran bahan api
9.
Burning offuels

2
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

D. Gambar Rajah Aras Tenaga/ Energy Level Diagram

1. Apabila tindak balas berlaku, kuantiti tertentu haba dibebaskan atau diserap.
When a chemical reaction takes places, a certain quantity of heat is given off or absorb.

2. Kuantiti haba yang diberikan simbol, LiH.

The quantity of heat is given the symbol, Lili.

3. LiH ialah perbezaan antara tenaga bahan tindak balas dengan tenaga hasil tindak
balas.
LJH is the difference between the energy of the reactants and the energy of products.

LiH = H hasil tindak balas - H bahan tindak balas

LJH = H products - H reactants

4. Perubahan tenaga dalam tindak balas kimia boleh diwakili dengan menggunakan
gambar rajah aras tenaga.
The energy change in chemical reaction can be presented using an energy level diagram.

5. Satu gambar rajah aras tenaga adalah graf yang menunjukkan perubahan tenaga
dalam tindak balas kimia.
An energy level diagram it is a graph that shows the energy change in chemical
reaction.

Eksotermik/ Exothermic Endotermik/ Endothermic

Tenaga / Energy Tenaga / Energy
a
�Bahan tindak balas Hasil tindak balas
reactant product
Rajah '
Aras Tenaga 6H negatif 6H positif
Energy Level negative positive
Diagram
Hasil tindak balas Bahan tindak balas
,,product reactant

Hhasil tindak balas< Hbahan tindak balas Hhasil tindak balas > Hbahan tindak balas
LiH
Hproducts < Hreactants Hproducts > Hreactants
Tenaga dibebaskan
Tenaga diserap
Tenaga hasil tindak balas,
Tenaga hasil tindak balas,
Hhasil tindak balas kurang daripada
Hhasil tindak balas lebih daripada tenaga
tenaga bahan tindak balas,
Penerangan bahan tindak balas, Hbahan tindak balas
Hbahan tindak balas
Explanation Energy is absorbed.
Energy is given off
The energy of the products, Hproducts
The energy of the products, Hproducts
is more than the energy if reactants,
is less than the energy of
Hreactants
reactants, Hreactants

3
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

Aktiviti 2: Bina gambar rajah aras tenaga bagi setiap tindak balas yang berikut:
Activity 2: Construct an energy level diagram for each of the following reaction.

a. Zn + 2HCI ➔ ZnCb + H2;�H = - 126 kJ

(d) C + 2S ➔ CS2; �H = + 88 kJ

E. Perubahan tenaga dan ikatan kimia/ Energy change and chemical bond

Jenis Tindak Balas Perubahan tenaga Tanda �H

Type of Reaction Energy change Sign of AH
Eksotermik Tenaga pemecahan ikatan < Tenaga pembentukan ikatan Negatif
Exothermic Energy for bond breaking < Energy from bond formation Negative
Endotermik Tenaga pemecahan ikatan > Tenaga pembentukan ikatan Positif
Endothermic Energy for bond breaking > Energy from bond formation Positive

4
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

F. Aplikasi Ekso termik dan Endo termik/ Application of Exothermic and Endothermic

Contoh dalam kehidupan harian ialah pek panas dan pek sejuk yang digunakan
untuk melegakan kesakitan
The example in our daily life is hot packs and cold packs which are used to relieve pain.

Contoh Eksotermik/ Example Exothermic Contoh Endotermik/ Example Endothermic

Pek Panas/ Hot Packs Pek Sejuk/ Cold Packs
Mengandungi kalsium klorida, CaCb atau Mengandungi ammonium nitrat, NH4NQ3
Magnesium sulfat, MgS04 dengan air dan air dalam ruangan berasingan.
dalam ruang berasingan Contains ammonium nitrate, NH4NOs and
Contains calcium chloride, CaCb or water with separate compartments.
magnesium sulphate, MgS04 with water
Kedua-dua ruang berasingan itu dipecahkan
with separate compartments.
dengan memicit beg luar
Kedua-dua ruang berasingan itu The barrier between two compartments is
dipecahkan dengan memicit beg luar broken by squeezing the outer bag
The barrier between two compartments is
Haba diserap daripada persekitaran dan
broken by squeezing the outer bag
memberikan kesan sejuk
Haba dibebaskan ke persekitaran iaitu Heat is absorbed from the surroundings and
kawasan yang cedera gives a cold effect.
Heat is released to the surroundings such
Boleh digunakan oleh ahli sukan yang
as the injured area.
mengalami kecederaan ringan seperti
Boleh digunakan untuk meredakan sakit terseliuh atau lebam.
otot dan bengkak Used by athletes for light injuries like sprains
Used to relieve muscle aches and swelling or bruises

G. Pengiraan Melibatkan Termo kimia/ Calculation Involved Thermochemistry

1. Haba yang dibebaskan atau diserap dalam eksperimen atau perubahan haba boleh
dihitung menggunakan formula.
Heat released or absorb by experiment or heat change can be calculate by using the
formula.

Jisim larutan/ Mass of solution, m

1. Haba pemendakan:
m = V1 + V2 (2 larutan)
Q = mc0 Heat of precipitation:
m = Vi+Vi (2 solution)
Q = haba dibebas atau diserap dalam eksperimen 2. Haba penyesaran:
heat released or absorbed by experiment m = V (hanya 1 larutan)
m = jisim larutan Heat of displacement:
mass of solution m = V (only 1 solution)

c = muatan haba tentu air 3. Haba peneutralan:

specific heat capacity of water m = Vacid + Valkali
Heat of neutralisation:
8 = perubahan suhu m = Vacid + Valkali
change of temperature
4. Haba pembakaran:
m = isi padu air
Heat of combustion
m = volume of water

5
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

Andaian dalam pengiraan/ The assumptions in this calculation:

1. Ketumpatan, p, campuran akueus bahan bertindak balas ialah 1 g cm-3 , iaitu
ketumpatan air. Ini bermakna 1 cm3 campuran akueus bahan bertindak balas ialah
lg.
The density, p, of the aqueous reaction mixture is 1 g cm-3 , the density of water. That
mean 1 cm3 of aqueous reaction mixture is 1 g.
11. Muatan haba tentu, c, campuran akueus bahan bertindak balas adalah sama dengan
muatan haba tentu air, 4.2 Jg-1 °c-1
The specific heat capacity, c, of the aqueous reaction mixture is same as the specific
heat capacity of water, is 4.2 Jg-1 0
c 1-

iii. Tiada haba dihilangkan ke persekitaran atau diserap oleh radas eksperimen
No heat is lost to the surroundings or absorbed by the apparatus of the experiment

2. Bilangan mol bahan/ The mole of the substance, n

Mol, n = MV @ Mol, n = J1s1m

1000 jisim molar
Mole, n = MV Mole, n = mass
1000 molar mass

M = kemolaran/ molarity HANYA untuk PEMBAKARAN

V = isi padu larutan dalam cm3 ONLY for COMBUSTION
volume of solution in cm3

3. Haba tindak balas, L1H boleh dihitung dengan menggunakan formula,

Heat of reaction, L1H can be calculated by using the formula,
Q haba dibebas atau diserap dalam eksperimen
AH=Q
=
heat released or absorb by experiment

n n = bilangan mol
number of mole

"Haba ..... " haba untuk 1 mol bahan tindak balas yang digunakan atau hasil tindak
balas yang terbentuk
"Heat of ... " heat for 1 mole of reactants used or product produce.

4. Garis paduan untuk penghitungan haba tindak balas

Guidelines for the calculation of the heat of reaction

Langkah-langkah untuk diikuti/ Steps to follow:

Langkah 1 - Hitungkan perubahan haba dengan menggunakan formula,

Step 1 - Calculate the heat change using the formula,
Q = mc0 (p Joule)

Langkah 2 - Tuliskan persamaan kimia atau persamaan ion untuk tindak balas yang
berlaku
Step 2 - Write chemical equation or ionic equation for the reaction that occurs

6
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

Langkah 3 - Hitungkan bilangan mol bahan tindak balas menggunakan formula

berikut:
Step 3 Calculate the number of moles of reactant that reacts using either the
following formulae :

Mol, n = MV Mol, n = J1s1m

1000 @ jisim molar
= q mol = q mol
Mole, n = MV Mole, n = mass
1000 molar mass

Langkah 4 - Kaitkan bilangan mol bahan tindak balas (langkah 3) dengan perubahan
haba (langkah 1)
Step 4 - Link the number of moles of reactants (step 3) with the heat change
(step 1)

q mol bahan tindak balas bertindak balas ➔ haba dibebas / diserap ialah p J
q mol of reactants react ➔ heat lost/ gain is p J

:. 1 mol of bahan tindak balas bertindak balas ➔ haba dibebas/diserap ialah 12. J
q
.: 1 mol of reactants react ➔ heat lost/ gain is Q_ J
q
xlOOO untuk
Haba tindak balas, t.H = +/- p kJ mol-1 tukar J kepada
1000 X q
kJ
= +/- r kJ mol-1

Heat of reaction , t.H = + /- p kJ mol-1

xl 000 for convert
q
1000 X
= +/- r kJ mol-1 Jto kJ

Nota : "+" digunakan untuk tindak balas endotermik dan

is used for endothermic reaction and
"" digunakan untuk tindak balas eksotermik
is used for exothermic reaction

Langkah 5 - Lukiskan gambar rajah aras tenaga

Step 5 - Draw the energy level diagram

Dalam termokimia, melibatkan 4 tindak balas

In thermochemistry, involving 4 reactions

(i) Haba pemendakan [tindak balas penyediaan garam tak terlarutkan]

Heat of precipitation [prepare insoluble salts reaction]

(ii) Haba penyesaran [tindak balas penyesaran dalam siri elektrokimia, bah 6]
Heat of displacement [displacement reaction in electrochemical series chap. 06]

(iii) Haba peneutralan [peneutralan bah 7 asid bes]

Heat of neutralisation [neutralisation in chap. 07 acid bases]

(iv) Haba pembakaran [melibatkan bah 2 Tingkatan 5 sebatian karbon]

Heat of combustion [involving chap 2 Form 5 carbon compound]

7
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

1.Latihan Haba Pemendakan / Exercise Heat of Precipitation

1. 50 cm3 larutan plumbum(II) nitrat), Pb(NO3)2 0.5 mol dm-3 dicampurkan kepada 50 cm3
larutan natrium sulfat, Na2SO4, 0.5 mol dm-3. Suhu awal larutan plumbum(II) nitrat ialah
27.4 0C dan larutan natrium sulfat ialah 27.6 0C. S uhu campuran tindak balas
meningkat kepada 30.5 °c. Hitungkan haba pemendakan plumbum(II) nitrat.
[Muatan haba tentu larutan, C: 4.2 J g- 1°C- 1; ketumpatan larutan, 1 g cm-3]
50 cm3 of 0.5 mol dm-3 lead(II) nitrate, Pb(NOs}2 solution was mixed with 50 cm3 of 0.5 mol
dm-3 sodium sulphate, Na2S04 solution. The initial temperature of lead(II) nitrate, Pb(NOs}2
solution is 27.4 °C and sodium sulphate, Na2S04 solution is 27.6 °C. The temperature of the
reaction mixture rise to 30.5 °C. Calculate the heat of precipitation oflead(II) sulphate, PbS04.
[Specific heat capacity of solution, c: 4.2 J g-1 °C- 1; density of solution, 1 g cm-3}

Penyelesaian/ Solution:

2. Haba pemendakan ferum(III) hidroksida, Fe(OH)3 ialah- l0kJ, Apabila larutan natrium
hidroksida, NaOH 2 mol dm-3 dicampurkan dengan larutan ferum(III) klorida, FeCb,
200 J haba dibebaskan. Hitung isi padu larutan natrium hidroksida, NaOH yang
digunakan.
The heat of precipitation of iron(III) hydroxide, Fe(OH)s is -1 0 kJ. When a 2 mol dm-3 sodium
hydroxide, NaOH solution was mixed together with a solution ofiron(III) chloride, FeCls, 200
J ofheat was released. Calculate the volume of sodium hydroxide, NaOH solution that was
used.

Penyelesaian/ Solution:

8
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

3. Hitungkan perubahan haba apabila 200 cm3 larutan kalsium klorida, CaCl2 0.5 mol
dm-3 ditambahkan kepada 200 cm3 larutan natrium karbonat, Na2CO 3 0.5 mol dm-3 jika
haba pemendakan kalsium karbonat, CaCO 3 ialah + 12.6 kJ mol- 1•
[Muatan haba tentu larutan, c= 4.2 J g- 1 "C- 1• Ketumpatan larutan = 1 g cm-3 ]
Calculate the heat change when 200 cm3 of 0. 5 mol dm-3 calcium chloride, CaCb solution is
added to 200 cm3 of 0. 5 mol dm-3 sodium carbonate, Na2COs solution if the heat of
precipitation of calcium carbonate, CaCOs is + 12. 6 kJ mol-1
[Specific heat capacity of solution,c = 4.2 J g-1 °C- 1. Density of solution= 1 g cm-3 J

Penyelesaian/ Solution:

4. Persamaan termokimia untuk pemendakan argentum klorida adalah seperti berikut:

The thermochemical equation for the precipitation of silver chloride is as follows:
Ag+ (ak) + Cl- (ak) ➔ AgCl ; t.H - 65.5 kJ mol- 1
=

Ag+ (aq) + Cl- (aq) ➔ AgCl; l1H = - 65.5 kJ mol-1

Hitungkan perubahan suhu apabila 100 cm3 larutan argentum nitrat, AgNO 3 0.5 mol
dm-3 ditambahkan kepada 100 cm3 larutan kalium klorida, KCl 0.5 mol dm-3.
Calculate the temperature change when 1 00 cm3 of 0. 5 mol dm-3 silver nitrate, AgNOs,
solution is added to 1 00 cm3 of 0. 5 mol dm-3 potassium chloride, KCl solution.

Penyelesaian/ Solution:

9
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

2. Latihan Haba Penyesaran/ Exercise of Heat of Displacement:

1. Seorang murid telah memperolehi data berikut dalam mengenal pasti haba
penyesaran bagi Argentum, Ag daripada larutan argentum nitrat, AgNO 3.
Student obtained the following data to determine the heat of displacement to silver, Ag
from silver nitrate, AgNOs solution.
Pengukuran Suhu ( ° C)
Measurements Temperature ( °C)
Suhu awal larutan argentum nitrat(V), AgNO3
29.5
Initial temperature of silver nitrate(V), AgN0s solution
Suhu tertinggi campuran tindak balas
33.1
Highest temperature of reaction mixture
Jisim Kuprum yang digunakan = 1 g
Mass of copper, Cu powder used
Isi padu larutan argentum nitrat 0.1 mol dm-3, =100 cm3
Volume of 0.1 mol dm-3 silver nitrate(V), AgN0s solution

Penyelesaian/ Solution:

2. Dalam satu eksperimen, serbuk magnesium berlebihan ditambahkan ke dalam 50

cm3 larutan ferum(II) sulfat 0.25 mol dm-3 pada 29 °C. Persamaan termokimianya adalah
seperti di bawah:
In an experiment, excess magnesium powder is added to 50 cm3 of 0.25 mol dm-3 iron(II)
sulphate solution at 29.0 °C. The thermochemical equation is shown below,
Mg(p) + Fe2+ (ak) ➔ Mg2+ (ak) + Fe (p); �H = -80.6 kJ mol - 1
Mg(s) + Fe2+ (aq) ➔ Mg2+ (aq) + Fe (s); l1H = -80.6 kJ mol - 1
Berapakah suhu tertinggi yang tercapai di dalam eksperimen ini?
What is the highest temperature reached in this experiment?

Penyelesaian/ Solution:

10
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

Langkah 5 - Lukiskan gambar rajah aras tenaga

Step 5 - Draw the energy level diagram

3. Latihan untuk Haba Peneutralan/ Exercise of Heat of Neutralisation

1. Seorang murid mencampurkan 60 cm3 larutan natrium hidroksida 2 mol dm-3

bersama 60 cm3 larutan asid etanoik 2 mol dm-3. Dia merekodkan datanya di dalam
jadual di bawah:
A student mixed togeth er 60 cm3 of 2 mol dm-3 sodium h ydroxide, NaOH solution and 60
cm3 of 2 mol dm-3 ethanoic acid, CHsCOOH solution. She recorded her data in Table below:
Pengukuran Suhu ( ° C)
Measurements Temperature ( °C)
Initial temperature of sodium hydroxide, NaOH solution 27.0
Initial temperature of ethanoic acid, CH3 COOH solution 28.0
Final temperature of reaction mixture 40.5
Dengan menggunakan data berikut, hitungkan haba peneutralan.
By using the given data, calculate the heat of neutralization.
[Muatan haba tentu larutan, C: 4.2 J g- 1 ° C- 1; ketumpatan larutan, 1 g cm-3 ]
[Specific heat capacity, c: 4.2 J g- 1 C-1,· density of solution, 1 g cm-3}
0

Penyelesaian/ Solution:

11
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

2. Persamaan termokimia bagi tindak balas di antara asid etanoik dan natrium
hidroksida diberikan seperti di bawah:
The thermochemical equation for the reaction between ethanoic acid and sodium hydroxide
is given below,
CH3COOH (ak) NaOH (ak) ➔ NaCH3COO (ak) + H2O (ce) ; �H = - 55 kJ mol- 1
+
CHsCOOH (aq) + NaOH (aq) ➔ NaCHsCOO (aq) + H20 (l); l1H = - 55 kJ mol-1
Hitungkan haba yang dibebaskan, apabila 200 cm3 asid etanoik 0.5 mol dm-3
ditambahkan kepada 20 cm3 natrium hidroksida 0.5 mol dm-3.
Calculate the heat given out, when 200 cm3 of ethanoic acid 0. 5 mol dm-3 is added to 200
cm3 of sodium hydroxide 0.5 mol dm-3.

Penyelesaian/ Solution:

3. Rajah aras tenaga di bawah merupakan tindak balas peneutralan.

The energy level diagram of a neutralization reaction is shown in figure below
Tenaga/ Energy
H2SO4 + 2NaOH

�H = - 114 kJ
Na2SO4 + 2H2O

Apabila 100 cm3 larutan asid sulfurik 1.0 mol dm-3ditambahkan ke dalam 100 cm3
larutan natrium hidroksida 1.0 mol dm-3. Berapakah perubahan suhu?
When 100 cm3 of 1.0 mol dm-3 sulphuric acid is added to 100 cm3 of 1.0 mol dm-3 sodium
hydroxide solution. What is the change in temperature?

Penyelesaian/ Solution:

12
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

4. Latihan untuk haba pembakaran/ Exercise ofHeat ofCombustion

1. C5H12O5(p) + 6 O2(g) ➔ 6 CO2(g) + 6 H2O(ce); 11H = - 2 400 kJ

Cd--I12O6(s) + 6 O2(g) ➔ 6 CO2(g) + 6 H2O(l); LJH = - 2 400 kJ
36 g glukosa, C5H12O6 terbakar lengkap digunakan untuk memanaskan 800 cm3 air.
Hitungkan peningkatan suhu bagi air.
36 g of glucose, Cd--!12O6 is burnt completely to heat up 800 cm3 of water. Calculate the rise
in temperature of the water.
[JMR glukosa = 180; Muatan haba tentu larutan,C: 4.2 J g-1° C-1; ketumpatan air, 1 g cm-3 ]
[RMM of glucose = 180; specific heat capacity of water, c: 4.2 J g-1 °C- 1; density of water, 1 g
cm-3}

Penyelesaian/ Solution:

2. Seorang murid memanaskan 250 cm3 air dengan menggunakan sebuah lampu kecil
yang mengandungi propan-1-ol, C3 H7OH. Beliau telah memperoleh data seperti dalam
jadual di bawah:
A student heated up 2 50 cm3 of water using a small lamp filled with propan-l-ol, CsH7OH.
He obtained the data in table below.
Pengukuran Bacaan termometer ( ° C)
Measurements Thermometer reading ( °C)
Suhu awal air
27.5 ° C
Initial temperature of water
Suhu akhir air
50.6 ° C
Final temperature of water
Pengukuran Penimbang (g)
Measurements Weighing balance (g)
Jisim awal lampu
21.40 g
Initial mass of lamp
Jisim akhir lampu
20.68 g
Final mass of lamp
Dengan menggunakan data yang diberikan, hitungkan haba pembakaran bagi propan-1-
ol, C3 H7OH.
By using the given data, calculate the heat of combustion of propan-l-ol, CsH7OH.
[JAR: H = l; C= 12; 0 = 16; Muatan haba tentu larutan, C: 4.2 J g- 1° C-1; ketumpatan
larutan, 1 g cm- 3]
[RAM : H= l; C= 12; 0 = 16; specific heat capacity ofwater,c: 4.2 J g-1 ° C-1; density ofwater,
1 g cm-3]

13
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

Penyelesaian/ Solution:

2. Apabila 1 mol butanol, C4H9OH terbakar dengan lengkap di dalam oksigen berlebihan,
2600 kJ haba telah terbebas. Hitungkan jisim butanol yang diperlukan untuk terbakar
lengkap di dalam oksigen berlebihan untuk meningkatkan suhu 500 cm3 air sebanyak
30 °C.
[JAR; H = l; C= 12; 0 = 16; Muatan haba tentu larutan, C: 4.2 J g- 1 ° C- 1; ketumpatan
larutan, 1 g cm-3]
When 1 mole of butanol, C4H90H is burnt in excess of oxygen, 2600 kJ of heat is produced.
Calculate the mass of butanol needed to burn completely in oxygen in order to raise the
temperature of 500 cm3 of water by 30 °C.
[RAM: H = l; C= 12; 0 = 16; Specific heat capacity of water, c= 4.2 J g- 1 ° C- 1]

Penyelesaian/ Solution:

14
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

5. Haba pembakaran bagi alkohol/ The heat of combustion of alcohol

1. Haba pembakaran beberapa alkohol yang berbeza, seperti dalamjadual di bawah:

The heat of combustion of various alcohols is different, as list in table below:

alkohol Formula Jisim Molar (g mol-1) Haba Pembakaran (kJ)

Alcohol Formula Molar mass (g mol-1) Heat of combustion (kJ)
Metanol
CH3OH 32 - 726
Methanol
Etanol
C2HsOH 46 - 1376
Ethanol
Propan-1-ol
C3H7OH 60 - 2026
Propan-1-ol
Butan-1-ol
C4H9OH 74 - 2679
Butan-1-ol
Pentan-1-ol
CsH11OH 88 - 3332
Pentan-1-ol

2. Apabilajisim molar alkohol meningkat, haba pembakaranjuga meningkat. Haba

pembakaran meningkat dengan sekata.
When the molar mass of alcohols increases, the heat of combustion also increases.
The heat of combustion increases regularly.

3. Ini keranajisim molarjuga meningkat sebanyak 14 unit bila menuruni siri homolog
alkohol. Peningkatan 14 unit kerana setiap ahli alkohol mempunyai lebihan satu atom
karbon, C dan dua atom hidrogen, H berbanding ahli sebelumnya.
This is because the molar mass also increases regularly by 14 units when going down the
alcohol homologous series. The increase of 14 units is because each alcohol member has
one carbon, C atom and two hydrogen, H atoms more than the member above it.

6. Nilai Bahan Api/ Fuel value

1. Haba dibebaskan apabila bahan api dibakar.

Energy is given off when a Juel bums.

2. Jumlah tenaga (diukur dalam kilo joule) boleh diperolehi apabila 1 g bahan api
dibakar dikenali sebagai nilai bahan api atau nilai haba.
The amount of energy (measured in kilojoules) that can be obtained when l g of.fuel is
burnt is called the Juel value or heat value.

3. Unitnya ialah kJ g-1.

Its unit is kJ g-1.

4. Formula yang boleh digunakan:

The formula can be used:

Nilai bahan api (kJ g-1) =haba pembakaran bahan ( kJ mol-1 )

Jisim molar bahan

Fuel value (kJ g-I) heat of combustion of substance ( kJ mol-1 )

molar mass of substance

15
 PROGRAM SMART SPM 2021 PERINGKAT DKBB

6. Latihan/ Exercise:

a. Haba pembakaran etanol, C3H70H ialah - 2016 kJ mol- 1• Berapakah nilai bahan
apinya?
The heat of combustion of propanol, CsH70H is - 2016 kJ mol-1. What is its.fuel value?

Penyelesaian/ Solution:

(i) Hitungkanjisim molar bagi propanol, C3H70H.

Calculate the molar mass of propanol, CsH70H.

(ii) Hitungkan nilai bahan api bagi propanol, C3H70H.

Calculate the fuel value of propanol, C3H70H.

2. Nilai bahan api bagi arang ialah 35 kJ g- 1• Berapakah jumlah arang diperlukan untuk
mendidihkan 1.8 dm3 air?
The fuel value of charcoal is 35 kJ g-1. How much charcoal must be burnt to boil 1. 8 dm3 of
water?
[Muatan haba tentu larutan, C: 4.2 J g- 1 0
c- 1; ketumpatan larutan, 1 g cm- 3; suhu
air pada suhu bilik, 27 °C]
[Specific heat capacity of water, c: 4.2 J g-1 °C- 1; density of water, 1 g cm-3; room
temperature of water, 27 °Cj

Penyelesaian/ Solution

16