Nilam Workbook - Ch.4 F5 by izzahedi

Unit 4 ChemF5 (BIl) 2017 (CSY5p).indd 156

Heat change when 1 mole of metal is displaced from its solution by a more electropositive metal. Perubahan haba apabila 1 mol logam disesarkan daripada larutannya oleh logam yang lebih elektropositif.

Meaning / Maksud

1 Evaluate energy changes in chemical reactions Menilai perubahan tenaga dalam tindak balas kimia 2 Understand the heat of precipitation Memahami haba pemendakan 3 Understand the heat of displacement Memahami haba penyesaran

Meaning / Maksud

Heat released when 1 mole of fuel is burnt completely in excess oxygen / Haba yang dibebaskan apabila 1 mol bahan api terbakar sepenuhnya dalam oksigen berlebihan

Meaning / Maksud

Heat of combustion Haba pembakaran

4 Understand the heat of neutralisation Memahami haba peneutralan 5 Understand the heat of combustion Memahami haba pembakaran 6 Appreciate the existence of various sources of energy Menghargai kewujudan pelbagai sumber tenaga

Heat released when 1 mole of water is formed from neutralisation of acid with an alkali / Haba yang dibebaskan apabila 1 mol air terbentuk daripada peneutralan asid dengan alkali

H = mcθ H ΔH = +/– — — X

Hot pack Pek panas

Sign of ΔH “–” Simbol ΔH “–”

Products / Hasil tindak balas

∆ H is negative (heat is released) ∆ H adalah negatif (haba dibebaskan)

H = Heat released/Heat absorbed/Heat change Haba dibebaskan/haba diserap/perubahan haba ΔH = Heat of precipitation/ Heat of neutralisation/Heat of displacement/ Heat of combustion // Haba pemendakan/haba peneutralan/haba penyesaran/heba pembakaran m = total mass of solution / jumlah jisim larutan θ = change in temperature / perubahan suhu c = specific heat capacity of water / muatan haba tentu air x = mol of precipitate/metal displaced/water/alcohol mol mendakan/logam disesarkan/air/alkohol

Application Aplikasi

Heat of neutralisation Haba peneutralan

Involves / Melibatkan

Bond Formation Pembentukan ikatan

Heat of displacement Haba penyesaran

Calculation Pengiraan

Learning objective / Objektif pembelajaran

Heat change when 1 mole of precipitate is formed from its ions in aqueous solution. Perubahan haba apabila 1 mol mendakan terbentuk daripada ionnya dalam larutan akueus.

Meaning / Maksud

Heat of precipitation Haba pemendakan

Experiment to determine Eksperimen untuk menentukan

Heat of reaction, ΔH Haba tindak balas, ΔH

Resulting / Mengakibatkan

Bond Breaking Pemutusan ikatan

Cold pack / Pek sejuk

Exothermic Eksotermik

Reactants Bahan tindak balas

Energy / Tenaga

Application Aplikasi

Endothermic Endotermik

In / Dalam

Energy change, H Perubahan tenaga, H

Involves / Melibatkan

TERMOKIMIA

UNIT

Sign of ΔH “+” Simbol ΔH “+”

∆H is positive (heat is absorbed) ∆H adalah positif (haba diserap)

Products / Hasil tindak balas

4 THERMOCHEMISTRY

U N I T

Reactants Bahan tindak balas

Energy / Tenaga

MODULE • Chemistry Form 5

THERMOCHEMISTRY

TERMOKIMIA Concept Map / Peta Konsep

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MODULE • Chemistry Form 5

Energy Changes in Chemical Reactions Perubahan Tenaga dalam Tindak Balas Kimia Almost all chemical reactions involve energy change. Why is there energy change? Hampir semua tindak balas kimia melibatkan perubahan tenaga. Mengapakah ada perubahan tenaga?

During chemical reaction, heat energy is either released to the surrounding or absorbed from the surrounding. Semasa tindak balas kimia, tenaga haba dibebaskan ke persekitaran atau diserap dari persekitaran.

What are the two types of energy changes in chemical reaction? Apakah dua jenis perubahan tenaga dalam tindak balas kimia?

Two types of reactions that occur are: Dua jenis tindak balas yang berlaku adalah:

How do we know energy changes during chemical reaction? Bagaimanakah kita mengetahui perubahan tenaga semasa tindak balas kimia?

Measure the temperature change in the surrounding. Ukur perubahan suhu persekitaran.

Define exothermic reaction. Nyatakan maksud tindak balas eksotermik. How do we know if a reaction is exothermic? / Bagaimanakah kita mengetahui jika tindak balas adalah eksotermik?

(b) endothermic reaction / tindak balas endotermik

Remark / Catatan: Other possible ways to detect energy change in a reaction quickly is by touching. Cara lain yang untuk mengesan perubahan tenaga dalam suatu tindak balas ialah dengan meyentuh.

releases

It is a chemical reaction that

heat to the surrounding.

Tindak balas kimia yang membebaskan haba ke persekitaran. rises

The temperature in the surrounding Suhu persekitaran

meningkat

Reaction between zinc and copper(II) sulphate / Tindak balas antara zink dengan kuprum(II) sulfat

Excess zinc powder Serbuk zink berlebihan

U N I T

Temperature increases Suhu meningkat

Examples of exothermic reactions. Contoh tindak balas eksotermik.

(a) exothermic reaction / tindak balas eksotermik

Heat / Haba

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

Heat energy is released to the surroundings Tenaga haba dibebaskan kepada persekitaran

Give other examples of chemical equations which are exothermic reaction. Berikan contoh lain persamaan kimia yang merupakan tindak balas eksotermik.

Type of reaction Jenis tindak balas Neutralisation / Peneutralan

Example of chemical equation for the reaction Contoh persamaan kimia untuk tindak balas 2KOH + H2SO4 → K2SO4 + 2H2O

Reaction between acids and metals Tindak balas antara asid dengan logam

Mg + 2HCl → MgCl2 + H2

Reaction between acids and metal carbonate Tindak balas antara asid dengan karbonat logam

CaCO3 + 2HCl → CaCl2 + H2O + CO2

Combustion of alcohol / Pembakaran alkohol

C2H5OH + 3O2 → 2CO2 + 3H2O

Remark / Catatan: – * Neutralisation reaction, reaction between metal with acid, reaction between carbonate and acids were studied in topic “Acid and base” (Form 4). * Tindak balas peneutralan, tindak balas antara logam dengan asid, tindak balas antara karbonat dan asid dipelajari dalam tajuk “Asid dan Bes” (Tingkatan 4) – *Combustion of alcohols was studied in topic “Carbon Compound” (Form 5). *Pembakaran alkohol dipelajari dalam tajuk “Sebatian Karbon” (Tingkatan 5).

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MODULE • Chemistry Form 5

Can physical changes also be exothermic? Give examples. Bolehkah perubahan fizikal juga menjadi eksotermik? Berikan contoh.

Yes / Ya Physical changes Perubahan fizikal

Example Contoh Dissolving sodium hydroxide/potassium hydroxide in water Melarutkan natrium hidroksida/kalium hidroksida dalam air H 2O NaOH(s/p) Na+ (aq/ak) + OH–(aq/ak) Dissolving anhydrous salt such as copper(II) sulphate in water Melarutkan garam kontang seperti kuprum(II) sulfat dalam air H 2O CuSO4(s/p) Cu2+(aq/ak) + SO42–(aq/ak)

Dissolving substance in water Melarutkan bahan dalam air

Adding water to concentrated acid Menambahkan air kepada asid pekat H 2O H2SO4(aq/ak) 2H+ (aq/ak) + SO42–(aq/ak) – Condensation / Kondensasi – Freezing / Pembekuan – Sublimation (gas changes to solid) Pemejalwapan (gas bertukar kepada pepejal)

Change of state of matter Perubahan keadaan jirim

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

Define endothermic reaction. Nyatakan maksud tindak balas endotermik. How do we know if a reaction is endothermic? / Bagaimanakah kita mengetahui jika suatu tindak balas adalah endotermik?

Tindak balas kimia yang

menyerap

The temperature in the surrounding Suhu persekitaran

menurun

heat energy from the surrounding.

tenaga haba dari persekitaran. decreases

Dissolving ammonium nitrate in water / Melarutkan ammonium nitrat dalam air

Ammonium nitrate Ammonium nitrat

U N I T

Temperature decreases Suhu menurun

Examples of endothermic reaction. Contoh tindak balas endotermik.

absorbs

It is a chemical reaction that

Water / Air

Heat / Haba

Heat energy is absorbed from the surroundings Tenaga haba diserap dari persekitaran

Give other examples of chemical equations which are endothermic reaction: Berikan contoh lain persamaan kimia yang merupakan tindak balas endotermik:

Type of reaction Jenis tindak balas Thermal decomposition of nitrate salts reaction. Penguraian terma tindak balas garam nitrat. Thermal decomposition of carbonate salts reaction. Penguraian terma tindak balas garam karbonat. Reaction of acids with hydrogen carbonate. Tindak balas asid dengan hidrogen karbonat. Photosynthesis Fotosintesis

Example of chemical equation for the reaction Contoh persamaan kimia tindak balas 2Cu(NO3)2

ZnCO3

2CuO + 4NO2 + O2 Δ

HCl + KHCO3 6CO2 + 6H2O

ZnO + CO2 KCl + CO2 + H2O

C6H12O6 + 6O2

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

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Can physical changes also be endothermic? Give examples. Bolehkah perubahan fizikal juga menjadi endotermik? Berikan contoh.

Yes / Ya Physical changes Perubahan fizikal

Example Contoh Dissolving ammonium salts/nitrate salts in water Melarutkan garam ammonium/garam nitrat dalam air

Dissolving substance in water Melarutkan bahan dalam air

NH4NO3(s/p) KNO3(s/p)

H 2O H 2O

NH4+ (aq/ak) + NO3– (aq/ak) K+ (aq/ak) + NO3– (aq/ak)

Decomposition of hydrated salt to anhydrous salt and water Penguraian garam terhidrat kepada garam kontang dan air

Heating of hydrated salt Pemanasan garam terhidrat

D CuSO4.5H2O(s/p) CuSO4(s/p) + 5H2O(l/ce) (blue/biru) (white/putih) – Melting Peleburan – Boiling/evaporation Pendidihan/penyejatan – Sublimation (solid changes to gas) Pemejalwapan (pepejal bertukar kepada gas)

Change of state of matter Perubahan keadaan jirim

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

Energy change during formation and breaking of bonds Perubahan tenaga semasa pembentukan dan pemutusan ikatan The energy changes are always caused by two processes taking place during chemical reaction when reactants change to products. What are these two processes? Perubahan tenaga sering disebabkan oleh dua proses yang berlaku semasa tindak balas kimia apabila bahan tindak balas berubah kepada hasil. Apakah dua proses ini? How is breaking of bonds in the reactants occur? Bagaimanakah pemutusan ikatan dalam bahan tindak balas berlaku? How is formation of bond in the product occur? Bagaimana pembentukan ikatan dalam hasil berlaku?

(i) (ii)

Breaking of bonds in the reactant Pemutusan ikatan dalam bahan tindak balas Formation of bonds in the products Pembentukan ikatan dalam hasil

Heat energy is Tenaga haba

absorbed diserap

U N I T

to break the bonds in the reactants.

untuk memecahkan ikatan dalam bahan tindak balas.

released dibebaskan

during bond formation in the products. semasa pembentukan ikatan dalam hasil tindak balas.

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

It depends on the strength of bonds: Ia bergantung kepada kekuatan ikatan: (i) More energy is absorbed to break a strong bond compared to a weak bond. Lebih tenaga diserap untuk memutuskan ikatan yang kuat berbanding ikatan lemah. (ii) More energy is released when a strong bond is formed compared to a weak bond. Lebih tenaga dibebaskan apabila ikatan yang kuat terbentuk berbanding ikatan yang lemah.

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

Heat of reaction, ∆H is the difference between heat energy absorbed and heat energy released when 1 mole of reactant react or 1 mole of product is formed. Haba tindak balas, ∆H ialah perbezaan antara tenaga haba yang dibebaskan dan tenaga haba yang diserap apabila 1 mol bahan bertindak balas atau 1 mol hasil terbentuk.

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

Bonds are formed in the products: Ikatan terbentuk dalam hasil tindak balas:

Heat is absorbed Haba diserap

Heat is released Haba dibebaskan

(–1 856 kJ heat energy released) (–1 856 kJ tenaga haba dibebaskan)

(+1 370 kJ heat energy absorbed) (+1 370 kJ tenaga haba diserap)

➢ The heat released from bond formation is

greater

tinggi

Tenaga haba yang dibebaskan dari pembentukan ikatan lebih released

➢ A negative sign for ∆H shows that heat is

than heat absorbed for bond breaking. . dibebaskan

Tanda negatif untuk ∆H menunjukkan bahawa haba

daripada haba yang diserap semasa pemutusan ikatan. .

Energy change in exothermic reaction / Perubahan tenaga dalam tindak balas eksotermik Energy profile diagram for exothermic reactions. Rajah profil tenaga untuk tindak balas eksotermik.

Energy / Tenaga Heat energy is absorbed (+ve) Tenaga haba diserap (+if)

Heat energy is released (–ve) Tenaga haba dibebaskan (–if)

Reactants Bahan tindak balas ∆H is negative ∆H adalah negatif

U N I T

Products Hasil tindak balas

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

Compare the quantity of heat energy absorbed and released in the reaction. Bandingkan kuantiti tenaga haba yang diserap dan dibebaskan dalam tindak balas tersebut.

heat energy absorbed

The quantity of

heat energy released rendah

daripada

tindak balas.

Ikatan

What is the sign of ΔH? Apakah tanda ΔH?

The sign of ∆H is

than

for the formation of bonds in the products.

Weak

tenaga haba yang dibebaskan

bonds are broken and

lemah

strong

dipecahkan dan ikatan

A negative sign for ∆H shows that heat is

Why is the temperature increases? Mengapakah suhu meningkat?

Heat energy is

Tanda negatif ∆H menunjukkan haba

increases

kuat

released

untuk pembentukan ikatan dalam hasil

bonds are formed. dibentuk.

negative . / Tanda bagi ∆H adalah

Why is the sign of ΔH negative? Mengapa tanda ΔH adalah negatif?

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lower

Kuantiti tenaga haba yang diserap untuk pemecahan ikatan dalam bahan tindak balas adalah lebih

Compare strength of bonds in the reactants and products. Bandingkan kekuatan ikatan dalam bahan tindak balas dan hasil tindak balas.

for bonds breaking in the reactants is

released dibebaskan

negatif

. to the surrounding.

ke persekitaran.

to the surroundings, temperature of the surroundings

. (Surroundings include the reaction solution, container and the air).

dibebaskan Tenaga haba ke persekitaran, suhu persekitaran termasuklah larutan bahan tindak balas, bekas dan udara)

naik

. (Persekitaran

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MODULE • Chemistry Form 5

What is the energy change in the reaction? Apakah perubahan tenaga dalam tindak balas?

Energy change: / Perubahan tenaga: Chemical

Compare the total energy content of the reactants and products. Bandingkan jumlah kandungan tenaga bahan dan hasil tindak balas.

Total energy content of the products is less than total energy of the reactants. Jumlah kandungan tenaga hasil kurang daripada jumlah kandungan tenaga bahan tindak balas.

kimia

Tenaga

Draw the energy level diagram for exothermic reaction. Lukis rajah aras tenaga untuk tindak balas eksotermik.

Heat

energy →

haba

→ Tenaga

energy

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

Energy change in endothermic reaction: Perubahan tenaga dalam tindak balas endotermik: Energy profile diagram for endothermic reactions. Rajah profil tenaga untuk tindak balas endotermik.

Energy Tenaga

Heat energy is released (–ve) Tenaga haba dibebaskan (–if) Heat energy is absorbed (+ve) Tenaga haba diserap (+if)

Products Hasil tindak balas

∆H is positive ∆H adalah positif

Reactants Bahan tindak balas

Compare the quantity of heat energy absorbed and released in the reaction. Bandingkan kuantiti tenaga haba yang diserap dan dibebaskan dalam tindak balas tersebut. Compare strength of bonds in the reactants and products. Bandingkan kekuatan ikatan dalam bahan tindak balas dan hasil tindak balas. How heat of reaction, ΔH is obtained? Bagaimana haba tindak balas, ΔH diperolehi?

The quantity of

heat energy absorbed

for bonds breaking in the reactants is

higher

than

heat energy released from the formation of bonds in the products. Kuantiti

tenaga haba diserap

daripada

tenaga haba dibebaskan

Strong Ikatan

bonds are broken and kuat

untuk pemecahan ikatan dalam bahan tindak balas lebih tinggi dari pembentukan ikatan dalam hasil tindak balas. weak

dipecahkan dan ikatan

bonds are formed.

lemah

dibentuk.

Heat of reaction, ∆H is the difference between heat energy absorbed and heat energy released. Haba tindak balas, ∆H adalah perbezaan antara tenaga haba yang diserap dengan tenaga haba yang dibebaskan.

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

The sign of ∆H is

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

A positive sign for ∆H shows that heat is absorbed from the surrounding.

positive . / Tanda untuk ∆H adalah

Tanda positif ∆H menunjukkan haba

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U N I T

diserap

positif

dari persekitaran.

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MODULE • Chemistry Form 5

Why is the temperature decreases? Mengapakah suhu menurun?

Heat is

absorbed

from the surroundings, temperature of the surrounding decreases (Surrounding diserap

include the reaction solution, container and the air). / Haba persekitaran

menurun

dari persekitaran, suhu

(Persekitaran termasuklah larutan bahan tindak balas, bekas dan udara).

What is the energy change in the reaction? Apakah perubahan tenaga dalam tindak balas tersebut?

Energy change: / Perubahan tenaga:

Compare the total energy content of the reactants and products. Bandingkan jumlah kandungan tenaga bahan dan hasil tindak balas.

Total energy content of the products is more than total energy of the reactants. Jumlah kandungan tenaga hasil tindak balas lebih daripada jumlah kandungan tenaga bahan tindak balas.

Heat haba

Tenaga

Draw the energy level diagram for exothermic reaction. Lukis rajah aras tenaga untuk tindak balas eksotermik.

Chemical

energy →

energy kimia

→ Tenaga

Energy / Tenaga Products / Hasil tindak balas

Reactants Bahan tindak balas

∆H is positive (heat is absorbed) ∆H adalah positif (haba diserap)

Application of knowledge of exothermic and endothermic reactions in everyday life. Aplikasi tindak balas eksotermik dan endotermik dalam kehidupan seharian. I Hot packs: Pek panas:

U N I T

State the type of reaction take place in the hot packs. Nyatakan jenis tindak balas yang berlaku di dalam pek panas.

Contain chemicals that release heat, application of

How is the structure of the hot packs? / Bagaimanakah struktur pek panas?

It is a plastic bag containing separate compartments of water and anhydrous calcium chloride. Ia adalah beg plastik yang mengandungi ruang berasingan air dan kalsium klorida kontang.

Explain how the reaction take place in the hot pack. Terangkan bagaimana tindak balas berlaku dalam pek panas.

The anhydrous calcium chloride dissolves in water to

exothermic

reaction.

Mengandungi bahan kimia yang membebaskan haba, aplikasi bagi tindak balas

increase

release

. / Kalsium klorida kontang larut dalam air dan

menyebabkan suhu

meningkat

CaCl2(s/p)

H 2O

eksotermik

heat, thus causing the temperature membebaskan

haba yang seterusnya

. Ca2+(aq/ak) + 2Cl–(aq/ak)

∆ H = –83 kJ mol–1

What are other substances that can be used in the hot packs? Apakah bahan lain yang boleh digunakan dalam pek panas?

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

How is the structure of reusable hot packs? Bagaimanakah struktur pek panas yang boleh diguna semula?

A reusable hot pack uses supersaturated solution of sodium ethanoate crystallisation and resolution. Pek panas yang boleh dipakai semula menggunakan larutan tepu natrium etanoat yang akan menghablur.

State the uses of hot packs. Nyatakan kegunaan pek panas.

– Mountain climbers use hot packs to keep their hands and feet warm. Pendaki gunung menggunakan pek panas untuk menghangatkan tangan dan kaki. – Heat can increase blood flow and help restore movement to injured tissue. Haba dapat meningkatkan pengaliran darah dan membantu mengembalikan pergerakan tisu yang tercedera. – Heat can also reduce joint stiffness, pain, and muscle spasm. Haba juga mengurangkan sengal sendi, kesakitan dan kejang otot.

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MODULE • Chemistry Form 5

II Cold packs: Pek sejuk:

State the type of reaction take place in the cold packs. Nyatakan jenis tindak balas yang berlaku dalam pek sejuk.

Contain chemicals that absorb heat, application of

How is the structure of the cold packs? Bagaimanakah struktur pek sejuk?

It is a plastic bag containing separate compartments of water and solid ammonium nitrate. Ia adalah beg plastik yang mengandungi ruang yang berasingan air dan pepejal ammonium nitrat.

Explain how the reaction take place in the cold pack. Terangkan bagaimana tindak balas berlaku dalam pek sejuk.

The solid ammonium nitrate dissolves in water to

endothermic

reaction.

Mengandungi bahan kima yang menyerap haba, aplikasi bagi tindak balas

decrease

heat, thus causing the temperature to

Pepejal ammonium nitrat larut dalam air menurun

absorb

endotermik

. NH4NO3(s/p)

H 2O

menyerap

haba yang seterusnya menyebabkan suhu

NH4+(aq/ak) + NO3– (aq/ak)

∆H = + 26 kJ mol–1

What are other substances can be used in the cold packs? Apakah bahan lain yang boleh digunakan dalam pek sejuk?

Other substances that can be used in a cold pack are ammonium chloride, potassium nitrate and sodium thiosulphate. Bahan lain yang boleh digunakan dalam pek sejuk adalah ammonium klorida, kalium nitrat dan natrium tiosulfat.

State the uses of cold packs. Nyatakan kegunaan pek sejuk.

– To reduce body temperature for a patient who has fever. Untuk mengurangkan suhu badan pesakit yang demam. – To ease pain from both acute and chronic injuries, such as sprains and arthritis. Untuk melegakan kesakitan kecederaan akut dan kronik, seperti terseliuh dan artritis.

Exercise / Latihan 1

U N I T

The following is the thermochemical equation of neutralisation: Berikut ialah persamaan termokimia peneutralan: HCl + NaOH

NaCl + H2O ∆H = –57 kJ mol–1

Remark: / Catatan: Thermochemical equation is an equation that shows the heat of reaction together with the balanced equation. Persamaan termokimia ialah satu persamaan yang menunjukkan haba tindak balas bersama dengan persamaan seimbang. Construct energy level diagram for the thermochemical equation. Bina rajah aras tenaga untuk persamaan termokimia tersebut.

Energy / Tenaga

HCl + NaOH

∆H = –57 kJ mol–1 NaCl + H2O

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MODULE • Chemistry Form 5

Give four statements to interpret the energy level diagram that you have constructed. Berikan empat pernyataan untuk mentafsir rajah aras tenaga yang telah anda bina.

exothermic

(i) The reaction between hydrochloric acid and sodium hydroxide is Tindak

balas

antara

eksotermik

asid

hidroklorik

dengan

hidroksida

adalah

. increases

(ii) During reaction, the temperature of the mixture

natrium

meningkat

Semasa tindak balas, suhu campuran

(iii) When one mole of hydrochloric acid reacts with one mole of sodium hydroxide to produce one mole of sodium chloride and one mole of water, the quantity of heat

released

57 kJ.

Apabila satu mol asid hidroklorik bertindak balas dengan satu mol natrium hidroksida menghasilkan satu mol natrium klorida dan satu mol air, kuantiti haba yang dibebaskan

(iv) The

total

ialah 57 kJ. energy of 1 mole of hydrochloric acid and 1 mole of sodium hydroxide is

than the

total

energy of 1 mole of sodium chloride and 1 mole of water.

The difference in energy is 57 kJ.

Jumlah

lebih

tenaga bagi 1 mol asid hidroklorik dan 1 mol natrium hidroksida

daripada

jumlah

tenaga 1 mol natrium klorida dan 1 mol air. Perbezaan tenaga

adalah 57 kJ. 2

The following is the thermochemical equation for the dissolving of sodium nitrate in water: Berikut ialah persamaan termokimia bagi melarutkan natrium nitrat dalam air: NH4NO3(s/p)

U N I T

H 2O

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

Construct energy level diagram for the thermochemical equation. Bina rajah aras tenaga untuk persamaan termokimia tersebut.

Energy / Tenaga +

–

NH4 (aq/ak) + NO3 (aq/ak)

∆H = +26 kJ mol–1 NH4NO3(s/p)

Give four statements to interpret the energy level diagram that you have constructed. Berikan empat pernyataan untuk mentafsir rajah aras tenaga yang telah anda bina.

(i) Dissolving ammonium nitrate in water is

Suhu larutan

menurun

decreases

endotermik

Melarutkan ammonium nitrat dalam air adalah

(ii) Temperature of the solution

endothermic

(iii) When one mole of ammonium nitrate dissolves in water, the quantity of heat absorbed

is 26 kJ.

Apabila satu mol ammonium nitrat larut dalam air, kuantiti haba yang

diserap

ialah 26 kJ. (iv) The

total

energy of 1 mole of solid ammonium nitrate is less than the

total

energy of ammonium nitrate solution. The difference in energy is 26 kJ.

Unit 4 ChemF5 (BIl) 2017 (CSY5p).indd 164

Jumlah tenaga bagi 1 mol pepejal ammonium nitrat adalah kurang daripada jumlah tenaga 1 mol larutan ammonium nitrat. Perbezaan tenaga adalah 26 kJ.

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MODULE • Chemistry Form 5

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

Energy / Tenaga

2NO2(g/g)

Zn + CuSO4

∆H = –210 kJ mol–1 ∆H = +66 kJ mol–1

N2(g/g) + 2O2(g/g)

ZnSO4 + Cu

Reaction II / Tindak balas II

Reaction I / Tindak balas I

Based on the diagram above, compare the energy level diagram in reaction I and reaction II. Berdasarkan gambar rajah di atas, bandingkan rajah aras tenaga tindak balas I dan tindak balas II.

(i) Reaction I is

endothermic

endotermik

Tindak balas I adalah eksotermik

exothermic

while reaction II is

manakala tindak balas II adalah

(ii) Heat is absorbed from the surrounding in reaction I while heat is

released to

the

surrounding in reaction II.

Haba diserap dari persekitaran dalam tindak balas I manakala haba dibebaskan ke persekitaran dalam tindak balas II. lower

(iii) The total energy content of 1 mole of nitrogen gas and 2 moles of oxygen gas is

than the total energy content of 2 moles of nitrogen dioxide in reaction I. The total energy of the content of 1 mole of zinc and 1 mole of copper(II) sulphate is higher than the total energy content of 1 mole of zinc sulphate and 1 mole of copper in reaction II.

Jumlah kandungan tenaga 1 mol gas nitrogen dan 2 mol gas oksigen lebih rendah daripada jumlah kandungan tenaga 2 mol nitrogen dioksida dalam tindak balas tinggi

I. Jumlah kandungan tenaga 1 mol zink dan 1 mol kuprum(II) sulfat lebih

daripada jumlah tenaga 1 mol zink sulfat dan 1 mol kuprum dalam tindak balas II. absorbed

(iv) The quantity of heat

during reaction I is

+66 kJ mol ) while the quantity of heat –1

released

66 kJ

(heat of reaction is

during reaction II is

210 kJ

(heat of reaction is –210 kJ mol–1).

Kuantiti haba yang

diserap

semasa tindak balas I adalah

tindak balas ialah +66 kJ mol ) manakala kuantiti haba yang –1

semasa

tindak

balas

adalah

210 kJ

(haba

tindak

66 kJ

U N I T

(haba

dibebaskan

balas

ialah

–210 kJ mol ). –1

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MODULE • Chemistry Form 5

The Heat of Reaction (∆H) Haba Tindak Balas (∆H) What is heat of reaction, ΔH? Apakah haba tindak balas, ΔH?

The amount of heat energy released or absorbed during chemical reaction. Jumlah tenaga haba dibebaskan atau diserap semasa tindak balas kimia. Or / Atau The difference between total energy content of the reactants and the products. Perbezaan antara jumlah kandungan tenaga bahan tindak balas dan hasil.

Different types of chemical reactions will give different value of heat of reaction. What are the heat of reactions for specific reactions in this chapter? Jenis tindak balas kimia yang berbeza akan memberikan nilai haba tindak balas yang berbeza. Apakah haba tindak balas untuk tindak balas tertentu dalam bab ini?

Type of reaction Jenis tindak balas

Specific heat of reaction Haba tentu tindak balas

Remark Catatan

Precipitation Pemendakan

Heat of precipitation Haba pemendakan

Precipitation reaction in topic “Salt” (Form 4) Tindak balas pemendakan dalam tajuk “Garam” (Tingkatan 4)

Displacement Penyesaran

Heat of displacement Haba penyesaran

Displacement reaction in topic “Electrochemistry” (Form 4) Tindak balas penyesaran dalam tajuk “Elektrokimia” (Tingkatan 4)

Neutralisation Peneutralan

Heat of neutralisation Haba peneutralan

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

Combustion Pembakaran

Heat of combustion Haba pembakaran

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

Complete the following table. Lengkapkan jadual di bawah.

U N I T

Heat of reaction Haba tindak balas

Definition Definisi

Heat of precipitation Haba pemendakan

Heat of precipitation is heat change when 1 mole of precipitate is formed from its ions in aqueous solution. Haba pemendakan ialah perubahan haba apabila 1 mol mendakan terbentuk dari ion-ionnya dalam larutan akueus. Remark / Catatan: Precipitation reaction occurs when two solutions containing cations and anions of insoluble salts are added together. This reaction is used to prepare any insoluble salt. Tindak balas pemendakan berlaku apabila dua larutan mengandungi kation dan anion garam tak terlarutkan dicampur bersama. Tindak balas ini digunakan untuk menyediakan garam tak terlarutkan.

Example Contoh Thermochemical equation: Persamaan termokimia: Pb(NO3)2(aq/ak) + Na2SO4(aq/ak) → PbSO4(s/p) + 2NaNO3(aq/ak) ∆H = –50.4 kJ mol–1 Ionic equation: Persamaan ion: Pb2+ + SO42– → PbSO4 • 50.4 kJ heat energy is released when 1 mole of lead(II) ions reacted with 1 mole of sulphate ions to form 1 mole of lead(II) sulphate . 50.4 kJ tenaga haba dibebaskan apabila 1 mol ion plumbum(II) bertindak balas dengan 1 mol ion sulfat untuk membentuk 1 mol plumbum(II) sulfat . Energy level diagram:

Gambar rajah aras tenaga: Energy / Tenaga Pb2+(aq/ak) + SO42–(aq/ak) ΔH = –50.4 kJ mol–1 PbSO4(s/p)

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MODULE • Chemistry Form 5

Heat of reaction Haba tindak balas Heat of displacement Haba penyesaran

Definition Definisi Heat of displacement is heat change when 1 mole of a metal is displaced from its solution by a more electropositive metal. Haba penyesaran ialah perubahan haba apabila 1 mol logam disesarkan dari larutan garamnya oleh logam yang lebih elektropositif. Remark / Catatan: Displacement reaction occurs when a metal which is situated at a higher position (higher tendency to release electron) in the electrochemical series displaces a metal below it from its salt solution. Tindak balas penyesaran berlaku apabila logam yang berada di kedudukan yang lebih tinggi (lebih cenderung melepaskan elektron) dalam siri elektrokimia menyesar logam di bawahnya dari larutan garamnya.

Heat of neutralisation Haba peneutralan

Heat of neutralisation is heat released when 1 mole of water is formed from neutralisation of acid with an alkali. Haba peneutralan ialah perubahan haba yang dibebaskan apabila 1 mol air terbentuk dari peneutralan asid dan alkali. Remark / Catatan: Neutralisation is the reaction between an acid and a base to form only salt and water. Peneutralan ialah tindak balas antara asid dan bes menghasilkan garam dan air sahaja.

Example Contoh Thermochemical equation: / Persamaan termokimia: Zn(s/p) + CuSO4(aq/ak) → ZnSO4(aq/ak) + Cu(s/p) ∆H = –217 kJ mol–1 Ionic equation: / Persamaan ion:

Cu2+ + Zn → Zn2+ + Cu

released

• 217 kJ heat energy is from

copper(II) sulphate

zinc

solution by

dibebaskan

217 kJ tenaga haba

copper

when 1 mole of

is displaced

. kuprum

apabila 1 mol

disesarkan

dari larutan kuprum(II) sulfat oleh zink. Energy level diagram: / Gambar rajah aras tenaga: Energy / Tenaga Cu2+(ak/aq) + Zn(s/p) ∆H = –217 kJ mol–1 Zn2+(ak/aq) + Cu(s/p)

Thermochemical equation: / Persamaan termokimia:

KOH(aq/ak) + HNO3(aq/ak) → KNO3(aq/ak) + H2O(l/ce) ∆H = –57 kJ mol–1

Ionic equation: / Persamaan ion:

H+ + OH– → H2O

released

• 57 kJ heat energy is neutralisation of

potassium hydroxide

dibebaskan

57 kJ haba

when 1 mol of

formed from

with nitric acid . air

apabila 1 mol

kalium hidroksida dengan

water

asid nitrik

terbentuk dari peneutralan

Energy level diagram: / Gambar rajah aras tenaga:

U N I T

Energy / Tenaga H (aq/ak) + OH (aq/ak) +

–

ΔH = –57 kJ mol–1

H2O(l/ce)

Heat of combustion Haba pembakaran

Heat of combustion is heat released when 1 mole of fuel is burnt completely in excess oxygen. Haba pembakaran ialah haba yang dibebaskan apabila 1 mol bahan api terbakar lengkap dalam oksigen berlebihan. Remark / Catatan: Combustion is a reaction when a substance burns completely in the excess oxygen. Tindak balas pembakaran adalah tindak balas yang berlaku apabila bahan terbakar lengkap dalam oksigen berlebihan.

Thermochemical equation: / Persamaan termokimia: C2H5OH + 3O2 → 2CO2 + 3H2O ∆H = –1 366 kJ mol–1 released

• 1 366 kJ heat energy is completely apabila 1 mol

when one mole of ethanol is burnt

excess oxygen . / 1 366 kJ tenaga haba dibebaskan

etanol

dibakar

lengkap

dalam oksigen berlebihan .

Energy level diagram: / Gambar rajah aras tenaga: Energy / Tenaga

C2H5OH + 3O2 ∆H = –1 366 kJ mol–1 2CO2 + 3H2O

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H = mcq

when 1 mole of metal is displaced from its solution by a more electropositive metal. / Haba penyesaran: Perbezaan haba apabila 1 mol logam disesarkan daripada larutannya oleh logam yang lebih elektropositif. m = volume of salt solution isi padu larutan garam X = number of moles of metals displaced (from the balanced equation) bilangan mol logam yang disesarkan (dari persamaan seimbang)

∆H = Heat of displacement: Heat change when 1 mole of water is formed from neutralisation of acid with an alkali. Haba peneutralan: Haba yang terbebas apabila 1 mol air dihasilkan daripada peneutralan asid dengan suatu alkali. m = total volume of acid and alkali jumlah isi padu asid dan alkali X = number of moles of water (from the balanced equation) bilangan mol air (dari persamaan seimbang)

∆H = Heat of neutralisation: Heat released

Sodium chloride solution Larutan natrium klorida

Apparatus: / Radas: Polystyrene cup, measuring cylinder, thermometer / Bekas polistirena, silinder penyukat, termometer Procedure: / Prosedur: 1 25 cm3 of 0.5 mol dm–3 silver nitrate solution is measured with measuring cylinder and poured into a polystyrene cup.

Material: / Bahan: 0.5 mol dm–3 of sodium chloride solution, 0.5 mol dm–3 of silver nitrate solution Larutan natrium klorida 0.5 mol dm–3, larutan argentum nitrat 0.5 mol dm–3

Silver nitrate solution Larutan argentum nitrat

Material: / Bahan: 0.5 mol dm–3 of copper(II) sulphate, zinc powder / Larutan kuprum(II) sulfat 0.5 mol dm–3, serbuk zink Apparatus: / Radas: Polystyrene cup, measuring cylinder, thermometer Bekas polistirena, silinder penyukat, termometer Procedure: / Prosedur: 1 25 cm3 of 0.5 mol dm–3 copper(II) sulphate solution is measured with measuring cylinder and poured into a polystyrene cup.

Copper(II) sulphate solution Larutan kuprum(II) sulfat

Zinc powder Serbuk zink

Responding variable: Pemboleh ubah bergerak balas: Heat of neutralisation / Haba peneutralan Constant variable: / Pemboleh ubah dimalarkan: Volume and concentration of acid and alkali, type of alkali / Isi padu dan kepekatan asid dan alkali, jenis alkali Hypothesis: / Hipotesis: Reaction between hydrochloric acid and sodium hydroxide has higher heat of neutralisation than

Manipulated variable: / Pemboleh ubah dimanipulasikan: Type of acid / Jenis asid

Sodium hydroxide solution Larutan natrium hidroksida

Hydrochloric acid Asid hidroklorik

Fuel Bahan api

Copper Can Tin kuprum

Thermometer Termometer

molecule alcohols, the higher the heat of combustion. Semakin bertambah bilangan atom karbon dan hidrogen

Manipulated variable: Pemboleh ubah dimanipulasikan: Type of alcohols / Jenis alkohol Responding variable: Pemboleh ubah bergerak balas: Heat of combustion / Haba pembakaran Constant variable: / Pemboleh ubah dimalarkan: Volume of water, type of metal can Isi padu air, jenis bekas logam Hypothesis: / Hipotesis: The higher the number of carbon and hydrogen atom per

Water Air

Wind shield Pengadang angin

of fuel is burnt completely in excess oxygen. Haba pembakaran: Haba yang terbebas apabila 1 mol bahan api terbakar lengkap dalam oksigen berlebihan. m = volume of water in copper can isi padu air di dalam tin kuprum X = number of moles of alcohol (from mass fuel in the lamp) bilangan mol alkohol (dari perbezaan jisim pelita)

∆H = Heat of combustion: Heat released when 1 mole

B B B

when 1 mol of precipitate is formed from its ions in aqueous solution. Haba pemendakan: Perubahan haba apabila 1 mol mendakan dihasilkan daripada ion-ionnya di dalam larutan akueus. m = total volume of salt solution jumlah isi padu larutan garam X = number of moles of precipitate (from the balanced equation) bilangan mol mendakan (dari persamaan seimbang)

Heat released/Heat absorbed/ Heat change / Haba dibebaskan/ Haba diserap/Perubahan haba,

B B B

CALCULATION: / PENGIRAAN: Heat of reaction, / Haba tindak balas, mcq DH = ± X

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

∆H = Heat of precipitation: Heat change

43 42 41 40 39 38 37 36 35 34

43 42 41 40 39 38 37

Unit 4 ChemF5 (BIl) 2017 (CSY5p).indd 168 36

4 34

U N I T

Activity/Experiment to Determine Heat of Reaction Aktiviti/Eksperimen untuk Menentukan Haba Tindak Balas

MODULE • Chemistry Form 5

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23/10/2017 3:13 PM

Unit 4 ChemF5 (BIl) 2017 (CSY5p).indd 169

1 25 cm3 larutan argentum nitrat berkepekatan 0.5 mol dm–3 disukat dengan silinder penyukat dan dituang ke dalam cawan polistirena. 2 25 cm3 larutan natrium klorida berkepekatan 0.5 mol dm–3 disukat dengan silinder penyukat dan dituang ke dalam cawan polistirena yang lain. 3 Termometer diletakkan di dalam setiap larutan dan suhu awal setiap larutan T1 dan T2 dicatat. 4 Larutan natrium klorida dituang dengan cepat dan cermat ke dalam larutan argentum nitrat. 5 Campuran tindak balas dikacau dengan termometer dan suhu tertinggi, T3 dicatat.

2 25 cm3 of 0.5 mol dm–3 sodium chloride solution is measured with another measuring cylinder and poured into another polystyrene cup. 3 A thermometer is placed into each solution and the initial temperature T1 and T2 of each solution is recorded. 4 The sodium chloride solution is poured quickly and carefully into the silver nitrate solution. 5 The reaction of mixture is stirred with the thermometer and the highest temperature, T3 is recorded. 1 25 cm3 larutan kuprum(II) sulfat berkepekatan 0.5 mol dm–3 disukat dengan silinder penyukat dan dituang ke dalam cawan polistirena. 2 Termometer diletakkan di dalam larutan tersebut dan suhu awal T1 dicatat. 3 Separuh spatula serbuk zink ditambahkan dengan cepat dan cermat ke dalam larutan kuprum(II) sulfat. 4 Campuran tindak balas dikacau dengan termometer dan suhu tertinggi, T2 dicatat.

2 A thermometer is placed into the solution and the initial temperature T1 of the solution is recorded. 3 Half spatula of zinc powder is quickly and carefully added into the copper(II) sulphate solution. 4 The reaction mixture is stirred with the thermometer and the highest temperature, T2 is recorded.

169

1 50 cm3 larutan natrium hidroksida 2 mol dm–3 disukat dengan silinder penyukat dan dituang ke dalam cawan polistirena. 2 50 cm3 asid hidroklorik berkepekatan 2 mol dm–3 disukat dengan silinder penyukat yang lain dan dituang ke dalam cawan polistirena yang lain. 3 Termometer diletakkan di dalam setiap larutan dan suhu awal larutan natrium hidroksida dan asid hidroklorik, T1 dan T2 dicatat. 4 Asid hidroklorik dituang dengan cepat dan cermat ke dalam larutan natrium hidroksida. 5 Campuran tindak balas dikacau dengan termometer dan suhu tertinggi, T3 dicatat. 6 Langkah 1 hingga 5 diulang dengan menggunakan larutan natrium hidroksida dan asid etanoik.

reaction between ethanoic acid and sodium hydroxide / Tindak balas asid hidroklorik dengan natrium hidroksida menghasilkan haba peneutralan yang lebih tinggi dari haba peneutralan antara asid etanoik dengan natrium hidroksida. Material: / Bahan 2 mol dm–3 of hydrochloric acid, 2 mol dm–3 ethanoic acid, 2 mol dm–3 of sodium hydroxide solution Asid hidroklorik 2 mol dm–3, asid etanoik 2 mol dm–3, larutan natrium hidroksida 2 mol dm–3 Apparatus: / Radas: Polystyrene cup, measuring cylinder, thermometer / Bekas polistirena, silinder penyukat, termometer Procedure: / Prosedur: 1 50 cm3 of 2 mol dm–3 sodium hydroxide solution is measured with measuring cylinder and poured into a polystyrene cup. 2 50 cm3 of 2 mol dm–3 hydrochloric acid is measured with another measuring cylinder and poured into another polystyrene cup. 3 A thermometer is placed into each solution and the initial temperature of sodium hydroxide solution and hydrochloric acid, T1 and T2 are recorded. 4 Hydrochloric acid is poured quickly and carefully into the sodium hydroxide solution. 5 The reaction mixture is stirred with the thermometer and the highest temperature, T3 is recorded. 6 Steps 1 to 5 are repeated using sodium hydroxide solution and ethanoic acid. 1 100 cm3 air disukat dengan silinder penyukat dan dituang ke dalam bekas kuprum. 2 Termometer diletakkan di dalam air dan suhu awal, T1 dicatat. 3 Bekas kuprum diletakkan di atas tungku kaki tiga. 4 Pelita diisikan dengan metanol dan ditimbang. Jisim awalnya, m1 dicatat. 5 Penghadang angin diletakkan seperti ditunjukkan dalam rajah untuk mengurangkan kehilangan haba ke udara persekitarannya. 6 Pelita diletak dekat dengan bekas kuprum untuk memaksimakan pemindahan haba. Sumbu pelita tersebut dinyalakan. 7 Air tersebut dikacau berterusan dengan termometer sehingga suhunya meningkat sebanyak 30°C, api dipadamkan dan suhu tertinggi, T2 dicapai oleh air dicatat. 8 Jisim terakhir pelita m2 dan kandungannya segera ditimbang dan dicatat. 9 Langkah 1 hingga 8 diulangi dengan menggunakan etanol, propanol dan butanol.

Procedure: / Prosedur: 1 100 cm3 of water is measured with measuring cylinder and poured into the copper can. 2 A thermometer is placed into the water and the initial temperature, T1 is recorded. 3 The copper can is placed on a tripod stand. 4 A lamp is filled with methanol. The lamp is weighed and the initial mass, m1 is recorded. 5 A wind shield is placed as shown in the diagram to minimise heat loss to the moving air in the surrounding. 6 The lamp is placed near the base of the copper can to maximise the heat transfer and the wick is lighted. 7 The water is stirred continuously with the thermometer until its temperature increased by 30°C, the flame is put off and the highest temperature, T2 reached by the water is recorded. 8 The final mass of the lamp, m2 and its content is weighed immediately and recorded. 9 Steps 1 to 8 are repeated with ethanol, propanol and butanol.

Material: / Bahan: Methanol, ethanol, propanol, butanol Metanol, etanol, propanol, butanol Apparatus: / Radas: Lamp, copper can, thermometer, wind shield, wooden block, measuring cylinder, tripod stand Pelita, bekas kuprum, termometer, pengadang, bongkah kayu, silinder penyukat, tungku kaki tiga

pembakaran.

setiap molekul alkohol, semakin bertambah haba

MODULE • Chemistry Form 5

U N I T

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MODULE • Chemistry Form 5

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

U N I T

What are the quantities needed to calculate heat change / heat absorbed / heat given out / heat release, H in a substance? Apakah kuantiti yang diperlukan untuk mengira perubahan haba / haba yang diserap / haba yang diberikan / haba yang dibebaskan, H dalam bahan?

(i) Mass of substance (m in grams) Jisim bahan (m dalam gram) (ii) Specific heat capacity of a substance (c in J g–1 °C–1) Muatan haba tentu bahan (c dalam J g–1 °C–1) (iii) Temperature change (θ°C) Perubahan suhu (θ°C)

What are the assumptions made to get mass, specific capacity and temperature of the substance? Apakah andaian yang dibuat untuk mendapatkan jisim, muatan tentu dan suhu bahan tersebut?

For a chemical reaction that occurs in an aqueous solution (precipitation, displacement of metal and neutralisation), assumptions are made during the calculation of heat of reaction: Untuk tindak balas kimia yang berlaku dalam larutan akueus (pemendakan, penyesaran logam dan peneutralan) anggapan dibuat semasa pengiraan haba tindak balas: (i) Density of aqueous solution is equal to the density of water = 1 g cm–3, contoh: Ketumpatan larutan akueus sama dengan ketumpatan air = 1 g cm–3, for example: • 1 cm3 of aqueous solution has a mass of 1 g 1 cm3 larutan akueus mempunyai jisim 1 g • 2 cm3 of aqueous solution has a mass of 2 g 2 cm3 larutan akueus mempunyai jisim 2 g • m cm3 of aqueous solution has a mass of m g m cm3 larutan akueus mempunyai jisim m g (ii) Specific heat capacity of solution = Specific heat capacity of water = 4.2 J g–1 °C–1 Muatan haba tentu bahan larutan = Muatan haba tentu bahan air = 4.2 J g–1 °C–1 (iii) No heat lost to the surroundings: Tiada haba hilang ke persekitaran: – all heat released in an exothermic reaction = heat absorbed by the solution (temperature increases) semua haba dibebaskan dalam satu tindak balas eksotermik = haba yang diserap oleh larutan (suhu meningkat) – all heat absorbed in endothermic reaction = heat lost by the solution (temperature decreases) semua haba diserap dalam satu tindak balas endotermik = haba yang hilang oleh larutan (suhu menurun)

How to calculate heat change / heat absorbed / heat given off / heat released, H in a reaction? Bagaimana mengira perubahan haba / haba yang diserap / haba yang dikeluarkan / haba dibebaskan, H dalam tindak balas?

The heat change, H in a reaction can be calculated with the following formula Perubahan haba, H dalam tindak balas boleh dikira dengan formula berikut:

How to calculate heat of reaction, ΔH in a reaction? Bagaimanakah cara mengira haba tindak balas, ΔH dalam tindak balas?

(i) Heat of reaction (∆H) is the energy change when one mole of reactant reacts or when one mole of product is formed. Haba tindak balas (∆H) ialah perubahan tenaga apabila satu mol bahan bertindak balas atau satu mol hasil terbentuk. (ii) X mol of reactant/product absorbs/releases H J of heat energy X mol bahan/hasil menyerap/membebaskan H J tenaga haba H 1 mol of reactant/ product absorbs/releases / 1 mol of bahan/hasil menyerap/membebaskan J mol–1 X HJ ⇒ ∆H (heat of reaction) / ∆H (Haba tindak balas) = +/– , X mol

Heat change (H) / Perubahan haba (H) = mcθ J where / di mana m = mass of the solution in gram jisim larutan dalam gram c = specific heat capacity of solution in J g–1 °C–1 muatan haba tentu larutan dalam J g–1 °C–1 θ = temperature change in °C perubahan suhu dalam °C

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

Remark / Catatan: (i) The sign of ∆H is Tanda ∆H adalah

untuk tindak balas eksotermik (suhu menaik).

(ii) The sign of ∆H is

positive

for endothermic reaction (temperature decreases).

Unit 4 ChemF5 (BIl) 2017 (CSY5p).indd 170

for exothermic reaction (temperature increases).

negatif

What is the unit heat of reaction? Apakah unit haba bagi tindak balas?

negative

Tanda ∆H adalah

positif

untuk tindak balas endotermik (suhu menurun).

The unit for heat of reaction is kJ mol–1. Unit untuk haba tindak balas ialah kJ mol–1.

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MODULE • Chemistry Form 5

Steps to calculate heat of reaction, ΔH. Langkah-langkah untuk mengira haba tindak balas, ΔH.

Step 1 / Langkah 1: Calculate the number of moles of reactants or products, X Hitung bilangan mol bahan atau hasil tindak balas, X

Step 2 / Langkah 2: Calculate the *heat change of the reaction, H Hitung *perubahan haba tindak balas, H H = mcθ

Step 3 / Langkah 3: Calculate *heat of reaction, ΔH Hitung *haba tindak balas, ΔH H J ΔH = +/– X

Remark / Catatan: For specific reaction in this chapter: / Bagi tindak balas tertentu dalam bab ini: 1 *Heat change, H is heat released for exothermic reaction or heat absorbed for endothermic reaction. *Perubahan haba, H ialah haba yang dibebaskan untuk tindak balas eksotermik atau haba yang diserap untuk tindak balas endotermik. 2 *Heat of reaction, ΔH is / * Haba tindak balas, ΔH ialah (i) Heat of precipitation for precipitation reaction / Haba pemendakan bagi tindak balas pemendakan (ii) Heat of displacement for displacement reaction / Haba penyesaran bagi tindak balas penyesaran (iii) Heat of neutralisation for neutralisation reaction / Haba peneutralan bagi tindak balas peneutralan (iv) Heat of combustion for combustion reaction / Haba pembakan bagi tindak balas pembakaran

Example: / Contoh: 60 cm3 of 0.25 mol dm–3 silver nitrate solution reacts with 60 cm3 of 0.25 mol dm–3 potassium bromide solution with an average temperature of 29°C. A yellow precipitate was formed and the highest temperature reached is 32°C. Determine the heat of reaction, ∆H. / 60 cm3 larutan argentum nitrat 0.25 mol dm–3 bertindak balas dengan 60 cm3 larutan kalium bromida 0.25 mol dm–3 dengan suhu purata 29°C. Mendakan kuning terbentuk dan suhu tertinggi dicapai ialah 32°C. Tentukan haba tindak balas, ∆H. Steps / Langkah-langkah Step 1: / Langkah 1: Calculate the number of moles of silver bromide precipitated (X) Hitung bilangan mol mendakan argentum bromida (X)

Calculation / Pengiraan AgNO3(aq/ak) + KBr (aq/ak) → AgBr(s/p) + KNO3(aq/ak) or / atau Ag+ + Br– → AgBr + Number of moles of Ag / Bilangan mol Ag+ = 60 dm3 × 0.25 mol dm–3 = 0.015 mol 1 000

Number of moles of Br – / Bilangan mol Br – = 60 dm3 × 0.25 mol dm–3 = 0.015 mol 1 000

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

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

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

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

H = mcθ J H = 120 g × 4.2 J g–1 °C–1 × 3°C = 1 512 J

Remark / Catatan: 1 Mass of the solution, m / Jisim larutan, m = (60 cm3 + 60 cm3) × 1 g cm–3 = 120 g 2 Temperature change, q / Perubahan suhu, q = (32 – 29)°C = 3°C Step 3: / Langkah 3: Calculate the heat of precipitation, ΔH Hitung haba pemendakan, ΔH

∆H = – =–

Draw the energy level diagram: Lukis rajah aras tenaga:

U N I T

H (negative because heat is released to the surrounding or temperature increases) X (negatif sebab haba dibebaskan ke persekitaran atau suhu menaik) 1 512 = –100.8 kJ mol–1 0.015 mol

Energy / Tenaga AgNO3(aq/ak) + KBr(aq/ak) ΔH = –100.8 kJ mol–1 AgBr(s/p) + KNO3(aq/ak)

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Numerical Problems Involving Heat of Displacement Pengiraan Melibatkan Haba Penyesaran Example 1 / Contoh 1: Excess of zinc powder is added to 50 cm3 of 0.1 mol dm–3 copper(II) sulphate solution. The temperature of reaction mixture increases by 5°C. Calculate the heat of displacement of copper by zinc from copper(II) sulphate solution. [Specific heat capacity of solution = 4.2 J g–1 o C–1, density of solution = 1 g cm–3] Serbuk zink berlebihan ditambah kepada 50 cm3 larutan kuprum(II) sulfat 0.1 mol dm–3. Suhu campuran tindak balas meningkat sebanyak 5°C. Hitungkan haba penyesaran kuprum oleh zink dari larutan kuprum(II) sulfat. [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3] Step 1: Calculate number of mole of copper displaced Langkah 1: Hitung bilangan mol kuprum yang disesarkan

Excess / Lebih

50 cm3 0.1 mol dm–3 ? mol + CuSO4 → ZnSO4 + Cu

Number of moles of CuSO4 / Bilangan mol CuSO4 50 × 0.1 = = 0.005 mol 1 000

From the equation, / Dari persamaan, 1 mol CuSO4 : 1 mol Cu 0.005 mol CuSO4 : 0.005 mol Cu Step 2: Calculate the heat released, H Langkah 2: Hitung haba dibebaskan, H

Heat released in the experiment, Haba dibebaskan dalam eksperimen,

H = 50 × 4.2 × 5 J = 1 050 J Step 3: Calculate the heat of reaction (ΔH) Langkah 3: Hitung haba tindak balas (ΔH)

Heat of displacement, / Haba penyesaran 1 050 J ∆H = – 0.005 mol

U N I T

= –210 kJ mol–1

Example 2 / Contoh 2: The following is the thermochemical equation for a reaction. Berikut adalah persamaan termokimia untuk suatu tindak balas. Zn + CuSO4 → ZnSO4 + Cu ∆H = –210 kJ mol–1 3 –3 Calculate the heat released when 50 cm of 1.0 mol dm copper(II) sulphate solution reacts with excess zinc. Hitungkan haba yang dibebaskan apabila 50 cm3 larutan kuprum(II) sulfat 1.0 mol dm–3 bertindak balas dengan zink berlebihan. Step 1: Calculate number of mole of copper displaced Langkah 1: Hitung bilangan mol kuprum yang disesarkan

Number of moles of CuSO4 / Bilangan mol CuSO4

50 × 1 1 000

= 0.05 mol From the equation, / Dari persamaan

1 mol CuSO4 : 1 mol Cu 0.05 mol CuSO4 : 0.05 mol Cu Step 2: Calculate the heat released, H Langkah 2: Hitung haba dibebaskan, H

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

∆H =

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Exercise / Latihan The thermochemical ionic equation below represents the reaction between magnesium powder and iron(II) sulphate solution. Persamaan ion termokimia di bawah mewakili tindak balas antara serbuk magnesium dengan larutan ferum(II) sulfat. Mg(s/p) + Fe2+(aq/ak) → Mg2+(aq/ak) + Fe(s/p)

∆H = –189 kJ mol–1

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

Number of moles of FeSO4 / Bilangan mol FeSO4 80 × 0.4 = 1 000 = 0.032 mol From the equation, / Dari persamaan 1 mol Fe 2+ : 1 mol Fe 0.032 mol Fe 2+ : 0.032 mol Fe

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

∆H = 189 kJ mol–1 =

Step 3: Calculate the increase in temperature, θ Langkah 3: Hitung peningkatan suhu, θ

H X

H ; H = Heat released in the experiment 0.032 mol Haba dibebaskan dalam eksperimen

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

6 048 J = mcθ = 80 × 4.2 × θ θ = 18°C

U N I T

4 Numerical Problems Involving Heat of Precipitation Pengiraan Melibatkan Haba Pemendakan 1

When 25 cm3 of 0.25 mol dm–3 silver nitrate solution is added into 25 cm3 of 0.25 mol dm–3 sodium chloride solution, the temperature of the mixture rises by 3°C. What is the quantity of heat released in this experiment? [Specific heat capacity of a solution = 4.2 J g–1 °C–1] Apabila 25 cm3 larutan argentum nitrat 0.25 mol dm–3 ditambah kepada 25 cm3 larutan natrium klorida 0.25 mol dm–3, suhu campuran tindak balas naik sebanyak 3ºC. Berapa kuantiti haba yang dibebaskan dalam eksperimen ini? [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1] Answer: / Jawapan: Heat released in the experiment, / Haba dibebaskan dalam eksperimen, H = 50 g × 4.2 J g–1 ºC–1 × 3ºC = 630 J

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The thermochemical ionic equation below represents the reaction between lead(II) nitrate solution and potassium sulphate solution. Persamaan ion termokimia di bawah mewakili tindak balas antara larutan plumbum(II) nitrat dengan larutan kalium sulfat. Pb2+ + SO42– → PbSO4 ∆H = –50.4 kJ mol–1 3 –3 Calculate the increase in temperature when 25 cm of 1 mol dm of lead(II) nitrate solution is added into 25 cm3 of 1 mol dm–3 of potassium sulphate solution. [Specific heat capacity of solution = 4.2 J g–1 °C–1, density of solution = 1 g cm–3] Hitungkan kenaikan suhu apabila 25 cm3 larutan plumbum(II) nitrat 1 mol dm–3 ditambah kepada 25 cm3 larutan kalium sulfat 1 mol dm–3. [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3]

Step 1: Calculate number of mole of lead(II) sulphate precipitate formed Langkah 1: Hitung bilangan mol mendakan plumbum(II) sulfat yang terbentuk

Number of moles of Pb2+ Bilangan mol Pb2+

25 × 1 = 0.025 mol, 1 000

Number of moles of SO42– 25 × 1 = = 0.025 mol Bilangan mol SO42– 1 000 From the equation: / Dari Persamaan • 1 mol of Pb2+ ions reacts with 1 mol of SO42– ions to form 1 mole of PbSO4 1 mol ion Pb2+ bertindak balas dengan 1 mol ion SO42– membentuk 1 mol PbSO4 • 0.025 mole of Pb2+ ions reacts with 0.025 mole SO42– ions to form 0.025 mole of PbSO4 0.025 mol ion Pb2+ bertindak balas dengan 0.025 mol ion SO42– membentuk 0.025 mol PbSO4

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

Step 3: Calculate the increase in temperature, θ Langkah 3: Hitung peningkatan suhu, θ

1 260 J = mcθ = 50 g × 4.2 J g–1 ºC–1 × θ θ = 6 ºC

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

Numerical Problems Involving Heat of Neutralisation Pengiraan Melibatkan Haba Peneutralan U N I T

100 cm3 of 2.0 mol dm–3 sodium hydroxide solution is added into 100 cm3 of 2.0 mol dm–3 ethanoic acid. The initial temperature for both solutions is 28.0ºC and the highest temperature is 41.0ºC. Calculate heat of neutralisation. / 100 cm3 larutan natrium hidroksida 2.0 mol dm–3 ditambah kepada 100 cm3 asid etanoik 2.0 mol dm–3. Suhu awal kedua-dua larutan ialah 28.0ºC dan suhu tertinggi ialah 41.0ºC. Hitungkan haba peneutralan. [Specific heat capacity of a solution = 4.2 J g–1 °C–1] / [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1]

Step 1: Calculate number of mole of water formed Langkah 1: Hitung bilangan mol air yang terbentuk

Number of mole of NaOH / Bilangan mol NaOH 100 × 2 = = 0.2 mol 1 000 Number of mole of CH3COOH / Bilangan mol CH3COOH 100 × 2 = = 0.2 mol 1 000 CH3COOH + NaOH → CH3COONa + H2O From the equation / Daripada persamaan: 1 mol CH3COOH : 1 mol NaOH : 1 mol H2O 0.2 mol CH3COOH : 0.2 mol NaOH : 0.2 mol H2O

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

Heat released / Haba dibebaskan, H = (100 + 100) g × 4.2 J g–1 °C–1 × (41 – 28) °C–1 = 10 920 J = 10.92 kJ

Step 3: Calculate the heat of neutralisation (ΔH) Langkah 3: Hitung haba peneutralan (ΔH)

Heat of neutralisation / Haba peneutralan, 10.92 kJ ΔH = 0.2 mol = –54.6 kJ mol–1

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The reaction between 25 cm3 of hydrochloric acid and 25 cm3 of sodium hydroxide solution releases the heat of 2 100 J. What is the temperature change of the mixture? Tindak balas antara 25 cm3 asid hidroklorik dan 25 cm3 larutan natrium hidroksida membebaskan haba sebanyak 2 100 J. Apakah perubahan suhu campuran tindak balas? [Specific heat capacity of a solution = 4.2 J g–1 °C–1] / [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1] Answer: / Jawapan: Heat released / Haba dibebaskan = mcθ = 2 100 J (25 + 25) g × 4.2 J g–1 °C–1 × θ = 2 100 J θ = 10°C

Comparing the Heat of Neutralisation / Membandingkan Haba Peneutralan What is heat of neutralisation? Apakah haba peneutralan?

Heat of neutralisation is heat released when one mole of hydrogen ions from acid reacts with one mole of hydroxide ions from alkali to produce one mole of water: Haba peneutralan ialah haba yang dibebaskan apabila satu mol ion hidrogen dari asid bertindak balas dengan satu mol ion hidroksida dari alkali menghasilkan satu mol air: H+(aq/ak) + OH–(aq/ak) → H2O

Explain why the value of heat of neutralisation between sodium hydroxide solution / potassium hydroxide solution with hydrochloric acid / nitric acid is 57 kJ mol–1. Terangkan mengapa nilai haba peneutralan antara larutan natrium hidroksida / larutan kalium hidroksida dengan asid hidroklorik / asid nitrik adalah 57 kJ mol–1.

∆ H = –57 kJ mol–1

– Hydrochloric acid and nitric acid are strong monoprotic acids. One mole hydrochloric acid or nitric acid ionise completely in water to produce one mole of hydrogen ions. Asid hidroklorik dan asid nitrik adalah asid monoprotik kuat. Satu mol asid hidroklorik atau asid nitrik mengion sepenuhnya dalam air untuk menghasilkan satu mol ion hidrogen. – Sodium hydroxide and potassium hydroxide are strong alkali. One mole sodium hydroxide or potassium hydroxide ionise completely in water to produce one mole of hydroxide ions. Natrium hidroksida dan kalium hidroksida adalah alkali kuat. Satu mol natrium hidroksida atau kalium hidroksida mengion sepenuhnya dalam air untuk menghasilkan satu mol ion hidroksida. – Heat of neutralisation of sodium hydroxide solution/potassium hydroxide solution with hydrochloric acid/nitric acid is –57 kJ mol–1 because all the reactions produce one mol of water. Haba peneutralan bagi larutan natrium hidroksida/kalium hidroksida dengan asid hidroklorik/asid nitrik ialah –57 kJ mol–1 kerana semua tindak balas menghasilkan satu mol air. HCl + KOH → KCl + H2O HCl + NaOH → NaCl + H2O HNO3 + KOH → KNO3 + H2O HNO3 + NaOH → NaNO3 + H2O

H+ + OH– → H2O, ∆H = –57 kJ mol–1

⇒ 1 mol of hydrogen ions react with 1 mol of hydroxide ions to form 1 mol of water to release 57 kJ of heat energy. 1 mol ion hidrogen bertindak balas dengan ion hidroksida membentuk 1 mol air dan membebaskan 57 kJ tenaga haba. What is the value of heat of neutralisation of sulphuric acid (strong diprotic acid) with strong alkali? Explain. Apakah nilai haba peneutralan asid sulfurik (asid diprotik kuat) dengan alkali kuat? Terangkan.

U N I T

– Thermochemical equation for the neutralisation between sodium hydroxide with sulphuric acid (diprotic acid): Persamaan termokimia untuk peneutralan antara natrium hidroksida dengan asid sulfurik (asid diprotik): 2NaOH + H2SO4 → Na2SO4 + 2H2O – 2 mol hydroxide ions react with 2 mol of hydrogen ions to form 2 mol H2O. Heat released is 2 × 57 kJ = 114 kJ. 2 mol ion hidroksida bertindak balas dengan 2 mol ion hidrogen membentuk 2 mol H2O. Haba yang dibebaskan ialah 2 × 57 kJ = 114 kJ. 2H+ + 2OH– → 2H2O, ΔH = –114 kJ – Heat of neutralisation of sulphuric acid with sodium hydroxide remains at –57 kJ mol–1 because the definition for heat of neutralisation is heat released for the formation of one mol of water. Haba peneutralan bagi asid sulfurik dengan natrium hidroksida masih –57 kJ mol–1 kerana maksud haba peneutralan adalah haba yang dibebaskan bagi pembentukan satu mol air. H+ + OH– → H2O, ΔH = –57 kJ mol–1

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Explain why the heat of neutralisation between weak acid and strong alkali is less than –57 kJ mol–1. Terangkan mengapa haba peneutralan antara asid lemah dan alkali kuat kurang daripada –57 kJ mol–1.

Magnitude of heat of neutralisation for a weak acid with a strong alkali or strong acid with weak alkali is less than 57 kJ mol–1. Magnitud haba peneutralan untuk asid lemah dengan alkali kuat atau asid kuat dengan alkali lemah adalah kurang daripada 57 kJ mol–1. Example: / Contoh: NaOH + CH3COOH → CH3COONa + H2O NaOH + HCN → NaCN + H2O Explanation: / Penerangan: ionise (i) Weak acids Asid lemah

mengion

∆H = –55 kJ mol–1 ∆H = –12 kJ mol–1 hydrogen

partially in water to produce separa dalam air menghasilkan ion

CH3COOH (ii) Some of the particles still remain in the form of

hidrogen

ions. .

CH3COO + H molecules . –

molekul Sebahagian zarah masih kekal dalam bentuk . absorbed (iii) Heat energy is to ionise molecules of the weak acid that have not been ionised so that

they ionise completely. Tenaga haba diserap

mengion sepenuhnya.

untuk mengionkan molekul asid lemah yang masih belum mengion supaya

(iv) Part of the heat that is released is used/absorbed to ionise the molecules of weak acid that has not been ionised. Sebahagian haba yang dibebaskan digunakan/diserap untuk mengionkan molekul asid lemah yang masih belum mengion.

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

U N I T

Example 1: / Contoh 1: • Reaction I: / Tindak balas I: 50 cm3 of 2 mol dm–3 hydrochloric acid is added to 50 cm3 of 2 mol dm–3 potassium hydroxide solution. The temperature rises by 13°C. / 50 cm3 asid hidroklorik 2 mol dm–3 ditambah dengan 50 cm3 larutan kalium hidroksida 2 mol dm–3. Suhu naik sebanyak 13°C. • Reaction II: / Tindak balas II: 100 cm3 of 2 mol dm–3 hydrochloric acid is added to 100 cm3 of 2 mol dm–3 potassium hydroxide solution. What is the temperature change in this reaction? / 100 cm3 asid hidroklorik 2 mol dm–3 ditambah dengan 100 cm3 larutan kalium hidroksida 2 mol dm–3. Apakah perubahan suhu dalam tindak balas ini? Answer: / Jawapan:

H ∆H = X where / di mana X = Number of moles of water / Bilangan mol air H = Heat change (heat released in the reaction) Perubahan haba (haba dibebaskan dalam tindak balas) = mcq HCl + KOH → KCl + H2O 1 mol 1 mol 1 mol

From the equation / Daripada persamaan: Reaction I: / Tindak balas I: 1 mol HCl : 1 mol KOH : 1 mol H2O 0.1 mol HCl : 0.1 mol HCl : 0.1 mol H2O Reaction II: / Tindak balas II: 1 mol HCl : 1 mol KOH : 1 mol H2O 0.2 mol HCl : 0.2 mol HCl : 0.2 mol H2O Reaction I: / Tindak balas I: Reaction II: / Tindak balas II:

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100 × 4.2 × 13 J = 54 600 J 0.1 600 × 4.2 × T 54 600 J = , where / di mana T = temperature change in reaction II 0.2 perubahan suhu dalam tindak balas II T = 13ºC ∆H =

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Explanation: – When Reaction II is repeated by doubled volume of acid and alkali, the number of moles of water produced in Reaction II is doubled, hence amount of heat energy released is doubled. – The amount of heat energy which is doubled is used to increase total volume of solution which is also doubled. – Therefore, the increase in temperature remains the same. Penerangan: – Apabila Tindak balas II diulang dengan menggandakan isi padu asid dan alkali, bilangan mol air yang dihasilkan dalam Tindak balas II adalah dua kali ganda, oleh itu jumlah tenaga haba yang dibebaskan adalah dua kali ganda. – Jumlah tenaga haba yang berganda digunakan untuk meningkatkan jumlah isi padu larutan yang juga dua kali ganda. – Oleh itu, peningkatan suhu kekal sama. # Calculation guide II: / Panduan pengiraan II: If the reaction is repeated by changing the concentration of the solution by n times without changing the volume, the temperature change is n times. / Jika sebarang tindak balas diulangi dengan menukarkan kepekatan larutan sebanyak n kali tanpa menukar isi padu, perubahan suhu adalah n kali. Example 2: / Contoh 2: • Reaction I: / Tindak balas I: 50 cm3 of 0.2 mol dm–3 lead(II) nitrate is added to 50 cm3 of 0.2 mol dm–3 sodium carbonate solution. The temperature of the mixture rises by 2.4°C. / 50 cm3 larutan plumbum(II) nitrat 2 mol dm–3 ditambah dengan 50 cm3 larutan natrium karbonat 0.2 mol dm–3. Suhu naik sebanyak 2.4°C. • Reaction II: / Tindak balas II: 50 cm3 of 0.6 mol dm–3 lead(II) nitrate solution is added to 50 cm3 of 0.6 mol dm–3 sodium carbonate solution. What is the temperature rise in this experiment? / 50 cm3 larutan plumbum(II) nitrat 0.6 mol dm–3 ditambah dengan 50 cm3 larutan natrium karbonat 0.6 mol dm–3. Apakah kenaikan suhu dalam eksperimen ini? Answer: / Jawapan:

H ∆H = X where / di mana ∆H = Heat of precipitation of lead(II) carbonate Haba pemendakan plumbum(II) karbonat X = Number of moles of lead(II) carbonate precipitated Bilangan mol mendakan plumbum(II) karbonat H = Heat change / Perubahan haba = mcθ Ionic equation for both reactions: / Persamaan ion untuk kedua-dua tindak balas: Pb2+ + CO32– → PbCO3 Reaction I: / Tindak balas I: Reaction II: / Tindak balas II:

U N I T

100 × 4.2 × 2.4 J = 100 800 J mol–1 0.01 100 × 4.2 × T 100 800 J = , where / di mana T = temperature change in reaction II 0.03 ∆H =

perubahan suhu dalam tindak balas II T = 7.2°C (The temperature changes is 3 times more than reaction I) (Perubahan suhu adalah 3 kali lebih daripada tindak balas I)

Explanation: – When Reaction II is repeated by increasing the concentration of lead(II) nitrate solution and sodium carbonate solution by 3 times, number of moles lead(II) carbonate produced is also increased by 3 times. Hence, the amount of heat energy released is increased 3 times. – The heat energy is used to increase the same total volume of solution. – Therefore, the increase in temperature is 3 times. Penerangan: – Apabila Tindak balas II diulang dengan meningkatkan kepekatan larutan plumbum(II) nitrat dan larutan natrium karbonat sebanyak 3 kali, bilangan mol plumbum(II) karbonat yang dihasilkan juga meningkat sebanyak 3 kali ganda. Oleh itu, jumlah tenaga haba yang dilepaskan meningkat 3 kali ganda. – Tenaga haba digunakan untuk meningkatkan jumlah larutan yang sama. – Oleh itu, kenaikan suhu adalah 3 kali ganda.

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Numerical Problems Involving Heat of Combustion Pengiraan Melibatkan Haba Pembakaran 1

Methanol burns in oxygen in a reaction as shown in the thermochemical equation below. Metanol terbakar dalam oksigen seperti persamaan termokimia di bawah. 3 CH3OH(s/p) + O (g/g) → CO2 (g/g) + 2H2O ∆H = –725 kJ mol–1 2 2 What is the mass of methanol that must be burnt completely to produce 145 kJ of heat? [Relative atomic mass: C, 12; O, 16] Apakah jisim metanol yang perlu dibakar lengkap untuk menghasilkan 145 kJ haba? [Jisim atom relatif: C, 12; O, 16]

Calculate number of mole of methanol Kira bilangan mol metanol

H , X = Number of moles of methanol / Bilangan mol metanol X 145 kJ 725 kJ mol –1 = X

∆H =

Number of moles of methanol / Bilangan mol metanol 145 kJ = 725 kJ mol–1 = 0.2 mol

Calculate mass of methanol Hitung jisim metanol

U N I T

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

22 g of butanol is burnt completely in excess of oxygen. The heat released is used to heat up 500 cm3 of water from 27.5°C to 55.8°C. Calculate the heat of combustion of butanol. 22 g butanol terbakar lengkap dalam oksigen berlebihan. Haba yang dibebaskan memanaskan 500 cm3 air dari 27.5°C ke 55.8°C. Hitungkan haba pembakaran butanol. [Specific heat capacity of a solution = 4.2 J g–1 °C–1, relative atomic mass: H, 1; C, 12; O, 16 ] [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, jisim atom relatif: H, 1; C, 12; O, 16]

Calculate number of mole of butanol Kira bilangan mol butanol

Number of moles of butanol / Bilangan mol butanol 22 g = 74 g mol–1

Calculate heat released, H Hitung haba dibebaskan, H

H = 500 cm3 × 4.2 J g–1 °C–1 × 28.3°C = 59 430 J = 59.43 kJ

Calculate the heat of combustion, (ΔH) Hitung haba pembakaran, (ΔH)

∆H =

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59.43 kJ 22 mol 74

= 199.9 kJ mol–1

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Comparing the Heat of Combustion of Various Fuels Membandingkan Haba Pembakaran Pelbagai Bahan Api Define heat of combustion. Nyatakan maksud haba pembakaran.

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

What is the difference in heat of combustion between various alcohols? Apakah perbezaan haba pembakaran di antara pelbagai alkohol?

The higher the number of carbon and hydrogen atoms per molecule of alcohols, the higher the heat energy released by the combustion of 1 mole of alcohols. Semakin tinggi bilangan atom karbon dan hidrogen dalam setiap molekul alkohol, semakin banyak tenaga haba dibebaskan dari pembakaran 1 mol alkohol.

Example: / Contoh: The diagram below shows the graph of heat of combustion of alcohols against number of carbon atom per molecule. Rajah di bawah menunjukkan graf haba pembakaran melawan bilangan atom karbon dalam setiap molekul alkohol. Heat of combustion of alcohol (kJ mol–1) Haba pembakaran alkohol (kJ mol–1) 3 000 2 000 1 000 0

Number of carbon atom per molecule Bilangan atom karbon per molekul

State relationship between the number of carbon atom per molecule of alcohol with the heat of combustion. Explain. Nyatakan hubungan antara bilangan atom karbon per molekul alkohol dengan haba pembakaran. Terangkan. Answer: / Jawapan: – When the number of carbon atom per molecule of alcohol increases, the heat combustion increases. Apabila bilangan atom karbon dalam setiap molekul alkohol bertambah, haba pembakaran bertambah. – When the number of carbon atom per molecule of alcohol as products

increases .

increases , the number of

carbon dioxide

and

water

karbon dioksida Apabila bilangan atom karbon dalam setiap molekul alkohol bertambah , bilangan molekul bertambah dihasilkan . bonds – More between atoms in carbon dioxide and water molecules are formed, more heat is released. Lebih banyak

ikatan

antara atom dalam molekul air dan karbon dioksida terbentuk, lebih banyak

haba

molecules produced dan

air

yang

U N I T

dibebaskan.

Fuel value / Nilai bahan api What are fuels? Apakah bahan api?

Fuels are substances that burn in the air to produce heat energy. Bahan api ialah sebatian yang terbakar dalam udara untuk menghasilkan tenaga haba.

What is fuel value? Apakah nilai bahan api?

Fuel value is the amount of heat released when 1 g of fuel burns completely, the unit is kJ g–1. Nilai bahan api adalah jumlah haba yang dibebaskan apabila 1 g bahan api terbakar lengkap, unitnya adalah kJ g–1.

What are the application of fuel value? Apakah aplikasi nilai bahan api?

Fuel value is used to compare the cost of energy for various fuel. A fuel with high fuel value can supply more energy. / Nilai bahan api digunakan untuk membandingkan kos tenaga pelbagai bahan api. Bahan api dengan nilai bahan api yang tinggi boleh membekalkan lebih tenaga. Example: / Contoh:

Fuel / Bahan api Methanol / Metanol

Charcoal / Arang kayu

Crude oil / Minyak mentah

Kerosene / Kerosin

Petrol / Petrol

Natural gas / Gas asli

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Fuel value/ kJ g–1 / Nilai bahan api/ kJ g–1

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What are the aspects to be considered when choosing a fuel in industry? Apakah aspek yang harus dipertimbangkan ketika memilih bahan api dalam industri?

– Fuel value of the fuel. / Nilai bahan api bagi bahan api. – Cost of energy/fuel. / Harga tenaga/bahan api. – Availability and sources of the fuel. Sumber bahan api. – Effect of the fuel to the environment. Kesan bahan api kepada persekitaran.

What is the major sources of energy? Apakah sumber utama tenaga?

Fossil fuels such as coal, petroleum and natural gas Bahan api fosil seperti arang batu, petroleum dan gas asli

Why fossil fuel is eventually will be used up? Mengapa bahan api fosil akhirnya akan habis digunakan?

Fossil fuels are non-renewable source of energy. Bahan api fosil adalah sumber tenaga yang tidak boleh diperbaharui.

State other sources of energy. Nyatakan sumber tenaga lain.

Other sources of energy are the sun, biomass, water and radioactive substances. Sumber tenaga yang lain adalah matahari, biojisim, air dan bahan radioaktif.

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

Excess of copper powder Serbuk kuprum berlebihan

100 cm3 of 0.5 mol dm–3 silver nitrate solution 100 cm3 larutan argentum nitrat 0.5 mol dm–3

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

U N I T

Plastic cup Cawan plastik

Initial temperature of silver nitrate solution / Suhu awal larutan argentum nitrat = 28.0ºC Highest temperature of the mixture of product / Suhu tertinggi campuran hasil tindak balas = 40.5ºC [Given specific heat capacity of solution = 4.2 J g–1 °C–1, density of solution = 1 g cm–3] [Muatan haba tentu larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3]

(a) What is meant by the ‘heat of displacement’ in the experiment? Apakah yang dimaksudkan dengan ‘haba penyesaran’ dalam eksperimen itu? Heat released when one mole of silver is displaced from silver nitrate solution by copper. Haba dibebaskan apabila satu mol argentum disesarkan dari larutan argentum nitrat oleh kuprum.

(b) State three observations in the experiment and the reason for each observation. Nyatakan tiga pemerhatian dalam eksperimen itu dan berikan sebab untuk setiap pemerhatian. (i) Grey solid is deposited because silver metal is displaced by copper from silver nitrate solution / Pepejal kelabu berkilat

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

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

ion dihasilkan

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

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

persekitaran/ tindak balas adalah eksotermik

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(c) Why is a plastic cup used in the experiment? Mengapakah cawan plastik digunakan dalam eksperimen itu? To reduce heat loss to the surrounding. / Untuk mengurangkan kehilangan haba ke persekitaran

(d) Write the ionic equation for the reaction. Tulis persamaan ion untuk tindak balas itu. Cu + 2Ag+ → Cu2+ + 2Ag

(e) Based on the information given in the experiment, calculate: Berdasarkan maklumat yang diberi, hitungkan: (i) change in temperature / perubahan suhu θ = 40.5 – 28.0 = 12.5°C

(ii) the heat given out in the experiment / haba yang dibebaskan dalam eksperimen

H = (100)(4.2)(12.5) = 5 250 J

(iii) the heat of displacement of silver / haba penyesaran argentum Number of moles of AgNO3 / Bilangan mol AgNO3 = 100 × 0.5 = 0.05 mol 1 000 From the equation, / Dari persamaan, 2 mol of AgNO3 produce 2 mol of Ag / 2 mol AgNO3 menghasilkan 2 mol Ag 0.05 mol of AgNO3 produce 0.05 mol of Ag / 0.05 mol AgNO3 menghasilkan 0.05 mol Ag –5 250 J Heat of displacement of silver / Haba penyesaran argentum = 0.05 mol = –105 kJ mol–1

(f) (i)

The experiment is repeated using 100 cm3 of 1.0 mol dm–3 silver nitrate solution and excess copper powder. Calculate the temperature change in this experiment. Eksperimen itu diulangi menggunakan 100 cm3 larutan argentum nitrat 1.0 mol dm–3 dan serbuk kuprum yang berlebihan. Hitungkan perubahan suhu dalam eksperimen ini. 1 × 100 = 0.1 mol, 1 000 Number of moles of Ag displaced / Bilangan mol Ag disesarkan = 0.1 mol 0.1 × 105 000 Temperature change, / Perubahan suhu, θ = = 25 ºC 100 × 4.2

U N I T

Number of moles of Ag+ / Bilangan mol Ag+ =

(ii) Explain why this change of temperature is different from that in (e)(i). Terangkan mengapa perubahan suhu berbeza dengan (e)(i). The number of mol of silver displaced is doubled, hence amount of heat energy released is also doubled. The amount of heat energy which is doubled is used to increase the temperature of the same volume of solution. The increase in temperature of the solution is also doubled. / Bilangan mol argentum disesar adalah dua kali ganda, maka jumlah tenaga haba dibebaskan juga dua kali ganda. Jumlah tenaga haba yang dua kali ganda digunakan untuk meningkatkan suhu larutan yang sama isi padunya. Kenaikan suhu larutan juga menjadi dua kali ganda.

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Experiment I is carried out to determine the heat of neutralisation between strong acid and strong alkali. 50 cm3 of 0.5 mol dm–3 sodium hydroxide solution is poured into a plastic cup and the initial temperature is recorded. 50 cm3 of 0.5 mol dm–3 nitric acid is then poured into the cup containing the sodium hydroxide solution. The mixture is stirred and heat produced raises the temperature by 3°C. [Specific heat capacity of the solution = 4.2 J g–1 °C–1] Eksperimen I dijalankan untuk menentukan haba peneutralan antara asid kuat dengan alkali kuat. 50 cm3 larutan natrium hidroksida 0.5 mol dm–3 dituangkan dalam cawan plastik dan suhu awal dicatat. 50 cm3 asid nitrik 0.5 mol dm–3 kemudian dituangkan ke dalam cawan mengandungi larutan natrium hidroksida. Campuran tindak balas dikacau dan haba yang terbebas menaikkan suhu sebanyak 3°C. [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1]

Plastic cup Cawan plastik

50 cm3 of 0.5 mol dm–3 sodium hydroxide solution and 50 cm3 of 0.5 mol dm–3 nitric acid 50 cm3 larutan natrium hidroksida 0.5 mol dm–3 dan 50 cm3 asid nitrik 0.5 mol dm–3

Experiment I / Eksperimen I

(a) What is meant by ‘heat of neutralisation’ in the experiment? Apakah yang dimaksudkan ‘haba peneutralan’ dalam eksperimen ini? Heat released when 1 mole of water is formed from the neutralisation between nitric acid and sodium hydroxide solution. Haba yang dibebaskan apabila 1 mol air terbentuk dari tindak balas antara asid nitrik dan larutan natrium hidroksida.

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

50 × 0.5 1 000 = 0.025 mol

Number of moles / Bilangan mol =

U N I T

(ii) the heat released in the experiment. haba yang dibebaskan dalam tindak balas itu.

Heat released / Haba dibebaskan = Heat changed / Perubahan haba = 100 × 4.2 × 3 = 1 260 J

(iii) the heat of neutralisation for the reaction. haba peneutralan bagi tindak balas. NaOH + HNO3 → NaNO3 + H2O 0.025 mol 0.025 mol 0.025 mol 0.025 mole of NaOH reacts with 0.025 mole of HNO3 to form 0.025 mole of H2O 0.025 mol NaOH bertindak balas dengan 0.025 mol HNO3 membentuk 0.025 mol H2O The heat change is 1 260 J / Haba dibebaskan ialah 1 260 J 1 260 J Heat of neutralisation / Haba peneutralan = ∆H = – 0.025 mol = –50.4 kJ mol–1

(c) Write the thermochemical equation for the reaction in the experiment. Tulis persamaan termokimia untuk tindak balas dalam eksperimen. NaOH + HNO3 → NaNO3 + H2O

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Plastic cup Cawan plastik

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

50 cm3 of 1 mol dm–3 sodium hydroxide solution and 50 cm3 of 1 mol dm–3 ethanoic acid 50 cm3 larutan natrium hidroksida 1 mol dm–3 dan 50 cm3 asid etanoik 1 mol dm–3

Experiment II / Eksperimen II

(i) Calculate the number of moles of sodium hydroxide used. Hitungkan bilangan mol natrium hidroksida digunakan. Number of moles of sodium hydroxide / Bilangan mol natrium hidroksida =

MV 1 × (50) = 1 000 1 000 = 0.05 mol

(ii) Calculate the heat of neutralisation for the reaction between ethanoic acid and sodium hydroxide solution. [Specific capacity for all solutions is 4.2 J g–1 °C–1 and the density of all solutions is 1.0 g cm–3] Hitungkan haba peneutralan bagi tindak balas antara asid etanoik dengan larutan natrium hidroksida. [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3]

Heat change / Haba dibebaskan = mcθ = (50 + 50) × 4.2 × 5.5 = 2 310 J 2 310 J Heat of neutralisation / Haba peneutralan = – 0.05 mol = – 46 200 J mol–1 = 46.2 kJ mol–1

(e) Compare the heat of neutralisation for Experiment I and Experiment II. Explain your answer. Bandingkan haba peneutralan dalam Eksperimen I dan Eksperimen II. Terangkan jawapan anda. The heat of neutralisation for Experiment I is higher than Experiment II. Nitric acid is a strong acid which ionises completely in water. Ethanoic acid is a weak acid which ionises partially in water, some of the ethanoic acid still remain in the form of molecules. Some of heat released in Experiment II during neutralisation is absorbed to ionise the molecules of ethanoic acid. Haba peneutralan dalam Eksperimen I lebih tinggi daripada Eksperimen II. Asid nitrik adalah asid kuat mengion lengkap dalam air. Asid etanoik adalah asid lemah yang mengion separa dalam air, sebahagian asid etanoik wujud dalam bentuk molekul. Sebahagian haba yang dibebaskan dalam Eksperimen II semasa peneutralan diserap untuk mengionkan molekul asid etanoik yang belum mengion.

U N I T

An experiment was carried out to determine the heat of precipitation for the reaction between lead(II) nitrate and potassium sulphate. 50.0 cm3 of 0.5 mol dm–3 lead(II) nitrate solution was added to 50.0 cm3 of 0.5 mol dm–3 of potassium sulphate solution in a plastic cup. Satu eksperimen dijalankan untuk menentukan haba pemendakan antara plumbum(II) nitrat dan kalium sulfat. 50.0 cm3 larutan plumbum(II) nitrat 0.5 mol dm–3 ditambahkan kepada 50.0 cm3 larutan kalium sulfat 0.5 mol dm–3 di dalam cawan plastik. The thermochemical equation for the reaction is shown below: / Persamaan termokimia untuk tindak balas seperti berikut: Pb(NO3)2 + K2SO4 → PbSO4 + 2KNO3 ∆H = –50.4 kJ mol–1

[Specific heat capacity of the solution = 4.2 J g–1 °C–1, density solution = 1 g cm–3] [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3] (a) What is meant by ‘heat of precepitation’ in the experiment? Apakah yang dimaksudkan dengan ‘haba pemendakan’ dalam eksperimen itu? Heat is released when 1 mole of lead(II) sulphate is precipitated from mixing the aqueous solution of the Pb2+ ions and SO42– ions. / Haba yang dibebaskan apabila 1 mol plumbum(II) sulfat termendak dari larutan akueus yang mengandungi ion Pb2+ dan ion SO42–.

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(b) State one observation in the experiment. Nyatakan satu pemerhatian dalam eksperimen. White precipitate is formed. / Mendakan putih terbentuk.

(c) Calculate / Hitungkan (i) number of moles of lead(II) nitrate Bilangan mol plumbum(II) nitrat Number of moles / Bilangan mol =

50 × 0.5 1 000 = 0.025 mol

(ii) Heat change in the experiment. Perubahan haba dalam eksperimen. Pb(NO3)2 + K2SO4 → PbSO4 + 2KNO3 ∆H = –50.4 kJ mol–1 Number of moles of PbSO4 / Bilangan mol PbSO4 = 0.025 mol 1 mole of lead(II) sulphate is precipitated, heat released is 50.4 kJ 1 mol plumbum(II) sulfat termendak, haba terbebas ialah 50.4 kJ 0.025 mol of lead(II) sulphate, heat released is / 0.025 mol of plumbum(II) sulfat termendak, haba terbebas ialah = 50.4 × 0.025 = 1.26 kJ H or / atau 50.4 kJ = 0.025 H = 50.4 × 0.025 = 1.26 kJ

(iii) Temperature change / Perubahan suhu 1 260 J = 100 × 4.2 × θ 1 260

θ = 100 × 4.2 = 3°C

(d) Construct energy level diagram for the reaction. Lukis gambar rajah aras tenaga untuk tindak balas tersebut. Energy / Tenaga Pb(NO3)2 + K2SO4 ∆H = –50.4 kJ mol–1 PbSO4 + 2KNO3

U N I T

(e) Write an ionic equation for the above reaction. Tulis persamaan ion untuk tindak balas di atas. Pb2+ + SO42– → PbSO4

(f) The experiment is repeated by using 50.0 cm3 of 0.5 mol dm-3 lead(II) ethanoate and 50.0 cm3 of 0.5 mol dm–3 sodium sulphate solution. What is the change in temperature for the reaction? Explain your answer. Eksperimen diulangi dengan menggunakan 50.0 cm3 plumbum(II) etanoat 0.5 mol dm–3 dan 50.0 cm3 larutan natrium sulfat 0.5 mol dm–3. Apakah perubahan suhu untuk tindak balas itu? Terangkan jawapan anda. 3°C. The precipitation of lead(II) sulphate only involves Pb2+ ions and SO42– ions. 3°C. Pemendakan plumbum(II) sulfat hanya melibatkan ion Pb2+ dan ion SO42–.

(g) Why is a plastic cup used in this experiment? Mengapakah cawan plastik digunakan dalam eksperimen ini? Plastic is a good heat insulator // to reduce heat loss to the surrounding. Plastik adalah penebat haba yang baik // untuk mengurangkan kehilangan haba ke persekitaran.

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(h) In another experiment where calcium chloride solution is reacted with sodium carbonate solution, the temperature of the mixture decreases. The temperature change is recorded and ∆H is calculated. Dalam eksperimen lain, larutan kalsium klorida ditindak balaskan dengan larutan natrium karbonat, suhu campuran tindak balas berkurang. Perubahan suhu direkod dan ∆H dihitung. (i) Write a balanced equation for the reaction above. Tulis persamaan seimbang untuk tindak balas di atas. CaCl2 + Na2CO3 → CaCO3 + 2NaCl

(ii) Construct an energy level diagram for the reaction. Lukis gambar rajah aras tenaga untuk tindak balas itu. Energy / Tenaga CaCO3 + 2NaCl

CaCl2 + Na2CO3

Metal can / Tin logam

Thermometer Termometer

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

Water / Air

Lamp + Butanol Pelita + Butanol

The results are as follows: / Keputusan adalah seperti betrikut: Initial mass of lamp + butanol / Jisim awal pelita + butanol = 502.28 g Final mass of lamp + butanol / Jisim akhir pelita + butanol = 500.00 g Initial temperature of water / Suhu awal air = 29°C Highest temperature of water / Suhu tertinggi air = 59°C Volume of water / Isi padu air = 500 cm3 [Specific heat capacity of water = 4.2 J g–1 °C–1] / [Muatan haba tentu air = 4.2 J g–1 °C–1]

U N I T

(a) Write the equation for the combustion of butanol, C4H9OH. Tulis persamaan untuk pembakaran butanol, C4H9OH. C4H9OH + 6O2 → 4CO2 + 5H2O

(b) Calculate the heat energy change for the combustion of butanol in the above experiment. Hitungkan perubahan haba untuk pembakaran butanol di dalam eksperimen di atas. Heat change, / Perubahan haba, H = 500 × 4.2 × 30 = 63 000 J/63 kJ

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

Relative molecular mass / Jisim molekul realtif = 4(12) + 10(1) + 16 = 74 2.28 Number of moles / Bilangan mol = 74 = 0.03 mol

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

(e) Give two precautionary steps that should be taken when conducting the experiment above. Berikan dua langkah berjaga-jaga yang harus diambil semasa menjalankan eksperimen di atas. • Use a wind shield / Gunakan pengadang angin • Make sure the flame touches the bottom of the metal can / Pastikan nyalaan pelita menyentuh bahagaian bawah bekas logam. • Stir the water in the metal can continuously (any 2) / Kacau air dalam bekas logam secara berterusan (mana-mana 2)

(f) The theoretical value for the heat of combustion of butanol is –2 877 kJ. Explain why the experimental value for the heat of combustion of butanol is lower than the theoretical value. Nilai teori untuk haba pembakaran butanol ialah –2 877 kJ. Terangkan mengapa nilai dari eksperimen untuk haba pembakaran butanol adalah lebih rendah daripada nilai teori. Heat is lost to the surrounding. Incomplete combustion of butanol. Heat from the flame during the burning of butanol is absorbed by the tin/heats the tin. / Haba hilang ke persekitaran. Pembakaran butanol yang tidak lengkap. Haba dari nyalaan pembakaran butanol diserap oleh bekas logam/memanaskan bekas logam.

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

U N I T

Molecular formula Formula molekul

Heat of combustion/ kJ mol–1 Haba pembakaran/ kJ mol–1

Methanol / Metanol

CH3OH

Ethanol / Etanol

C2H5OH

1 376

Propan-1-ol / Propan-1-ol

C3H7OH

2 015

725

Explain why there are differences in the value of heat of combustion of the alcohols in the table. Terangkan mengapa terdapat perbezaan pada nilai haba pembakaran alkohol dalam jadual di atas. As the number of carbon and hydrogen atoms per molecule increases, the value of heat combustion increases. The higher the number of carbon and hydrogen atoms per molecule, the more carbon dioxide and water molecules products will be formed. More bonds in the product are formed, more heat is released. / Apabila bilangan atom karbon dalam setiap molekul alkohol bertambah, nilai haba pembakaran bertambah. Semakin bertambah bilangan atom karbon dan hidrogen dalam setiap molekul alkohol, bilangan molekul karbon dioksida dan air sebagai hasil juga bertambah. Semakin banyak ikatan dalam hasil terbentuk, semakin banyak haba dibebaskan.

Additional Question Soalan Tambahan

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