NILAM-FM5-Chemistry-Form-5

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 149 © Nilam Publication Sdn. Bhd. Energy change in endothermic reaction / Perubahan tenaga dalam tindak balas endotermik LS / SP 3.1.2 Energy profile diagram for endothermic reactions. Rajah profil tenaga untuk tindak balas endotermik. Energy Tenaga Heat energy is absorbed (+ve) Tenaga haba diserap (+if) Reactants Bahan tindak balas Products Hasil tindak balas ∆H is positive ∆H adalah positif Heat energy is released (–ve) Tenaga haba dibebaskan (–if) Compare the quantity of heat energy absorbed and released in the reaction. / Bandingkan kuantiti tenaga haba yang diserap dan dibebaskan dalam tindak balas tersebut. The quantity of heat energy absorbed for bonds breaking in the reactants is higher than heat energy released for the formation of bonds in the products. Kuantiti tenaga haba diserap untuk pemecahan ikatan dalam bahan tindak balas lebih tinggi daripada tenaga haba dibebaskan untuk pembentukan ikatan dalam hasil tindak balas. Compare strength of bonds in the reactants and products. Bandingkan kekuatan ikatan dalam bahan tindak balas dan hasil tindak balas. Strong bonds are broken in the reactants and weak bonds are formed in the products. Ikatan kuat dipecahkan dalam bahan tindak balas dan ikatan lemah dibentuk dalam hasil tindak balas. 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. 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 positive . / Tanda untuk ∆H adalah positif . 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. Tanda positif ∆H menunjukkan haba diserap dari persekitaran. Why is the temperature decreases? Mengapakah suhu menurun? Heat is absorbed from the surroundings, temperature of the surrounding decreases (Surrounding include the reaction solution, container and the air). / Haba diserap dari persekitaran, suhu persekitaran menurun (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: Heat energy → Chemical energy Tenaga haba → Tenaga kimia 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. Draw the energy level diagram for exothermic reaction. Lukis rajah aras tenaga untuk tindak balas eksotermik. Energy / Tenaga Reactants Bahan tindak balas Products / Hasil tindak balas ∆H is positive (heat is absorbed) ∆H adalah positif (haba diserap)

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 150 Exercise / Latihan TAHAP PENGUASAAN (TP) Menguasai Belum menguasai TP1 Mengingat kembali pengetahuan dan kemahiran asas mengenai haba tindak balas. TP2 Memahami haba tindak balas serta dapat menjelaskan kefahaman tersebut dengan contoh. 1 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. (a) Construct energy level diagram for the thermochemical equation. Bina rajah aras tenaga untuk persamaan termokimia tersebut. Energy / Tenaga HCl + NaOH NaCl + H2O ∆H = –57 kJ mol–1 (b) 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) The reaction between hydrochloric acid and sodium hydroxide is exothermic . (ii) During reaction, the temperature of the mixture increases . (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 is 57 kJ. (iv) The total energy of 1 mole of hydrochloric acid and 1 mole of sodium hydroxide is more than the total energy of 1 mole of sodium chloride and 1 mole of water. The difference in energy is 57 kJ. (i) Tindak balas antara asid hidroklorik dengan natrium hidroksida adalah eksotermik . (ii) Semasa tindak balas, suhu campuran meningkat . (iii) Apabila satu mol asid hidroklorik bertindak balas dengan satu mol natrium hidroksida menghasilkan satu mol natrium klorida dan satu mol air, kuantiti haba yang dibebaskan ialah 57 kJ. (iv) Jumlah tenaga bagi 1 mol asid hidroklorik dan 1 mol natrium hidroksida lebih daripada jumlah tenaga 1 mol natrium klorida dan 1 mol air. Perbezaan tenaga adalah 57 kJ. 2 The following is the thermochemical equation for the dissolving of sodium nitrate in water: Berikut ialah persamaan termokimia bagi melarutkan natrium nitrat dalam air: NH4NO3(s/p) H2O NH4 + (aq/ak) + NO3 – (aq/ak)            ∆H = +26 kJ mol–1 TP2 TP1 TP2

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 151 © Nilam Publication Sdn. Bhd. (a) Construct energy level diagram for the thermochemical equation. Bina rajah aras tenaga untuk persamaan termokimia tersebut. Energy / Tenaga NH4 NO3 (s/p) NH4 + (aq/ak) + NO3 – (aq/ak) ∆H = +26 kJ mol–1 (b) 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 endothermic . (ii) Temperature of the solution decreases . (iii) When one mole of ammonium nitrate dissolves in water, the quantity of heat absorbed is 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. (i) Melarutkan ammonium nitrat dalam air adalah endotermik . (ii) Suhu larutan menurun . (iii) Apabila satu mol ammonium nitrat larut dalam air, kuantiti haba yang diserap ialah 26 kJ. (iv) Jumlah tenaga bagi 1 mol pepejal ammonium nitrat adalah kurang daripada jumlah tenaga 1 mol larutan ammonium nitrat. Perbezaan tenaga adalah 26 kJ. 3 The diagram below shows the energy level of reactions I and II. Gambar rajah di bawah menunjukkan aras tenaga bagi tindak balas I dan II. N2 (g/g) + 2O2 (g/g) 2NO2 (g/g) Energy / Tenaga ∆H = +66 kJ mol–1 Zn + CuSO4 ZnSO + Cu 4 ∆H = –210 kJ mol–1 Energy / Tenaga Reaction I / Tindak balas I Reaction II / Tindak balas II (a) 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 while reaction II is exothermic . (ii) Heat is absorbed from the surrounding in reaction I while heat is released to the surrounding in reaction II. (iii) The total energy content of 1 mole of nitrogen gas and 2 moles of oxygen gas is lower 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. (iv) The quantity of heat absorbed during reaction I is 66 kJ (heat of reaction is +66 kJ mol–1) while the quantity of heat released during reaction II is 210 kJ (heat of reaction is –210 kJ mol–1). (i) Tindak balas I adalah endotermik manakala tindak balas II adalah eksotermik . (ii) Haba diserap dari persekitaran dalam tindak balas I manakala haba dibebaskan ke persekitaran dalam tindak balas II. TP2 TP1 TP2

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 152 (iii) Jumlah kandungan tenaga 1 mol gas nitrogen dan 2 mol gas oksigen lebih rendah daripada jumlah kandungan tenaga 2 mol nitrogen dioksida dalam tindak balas I. Jumlah kandungan tenaga 1 mol zink dan 1 mol kuprum(II) sulfat lebih tinggi daripada jumlah tenaga 1 mol zink sulfat dan 1 mol kuprum dalam tindak balas II. (iv) Kuantiti haba yang diserap semasa tindak balas I adalah 66 kJ (haba tindak balas ialah +66 kJ mol–1) manakala kuantiti haba yang dibebaskan semasa tindak balas II adalah 210 kJ (haba tindak balas ialah –210 kJ mol–1). HEAT OF REACTION HABA TINDAK BALAS CS / SK 3.2 3.2 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 “Redox Equilibrium” (Form 5) Tindak balas penyesaran dalam tajuk “Keseimbangan Redoks” (Tingkatan 5) 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. Heat of reaction Haba tindak balas Definition Definisi Example Contoh Heat of precipitation Haba pemendakan LS / SP 3.2.1 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. 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+ + SO4 2– → 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: Pb2+(aq/ak) + SO4 2–(aq/ak) PbSO4 (s/p) ΔH = –50.4 kJ mol–1 Energy / Tenaga

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 153 © Nilam Publication Sdn. Bhd. Heat of reaction Haba tindak balas Definition Definisi Example Contoh Heat of displacement Haba penyesaran LS / SP 3.2.2 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. 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 • 217 kJ heat energy is released when 1 mole of copper is displaced from copper(II) sulphate solution by zinc . 217 kJ tenaga haba dibebaskan apabila 1 mol kuprum disesarkan dari larutan kuprum(II) sulfat oleh zink. Energy level diagram: / Gambar rajah aras tenaga: Energy / Tenaga Cu2+(ak/aq) + Zn(s/p) Zn2+(ak/aq) + Cu(s/p) ∆H = –217 kJ mol–1 Heat of neutralisation Haba peneutralan LS / SP 3.2.3 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. 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 • 57 kJ heat energy is released when 1 mol of water formed from neutralisation of potassium hydroxide with nitric acid . 57 kJ haba dibebaskan apabila 1 mol air terbentuk dari peneutralan kalium hidroksida dengan asid nitrik . Energy level diagram: / Gambar rajah aras tenaga: H+(aq/ak) + OH– (aq/ak) H2 O(l/ce) ΔH = –57 kJ mol –1 Energy / Tenaga Heat of combustion Haba pembakaran LS / SP 3.2.4 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 • 1 366 kJ heat energy is released when one mole of ethanol is burnt completely in excess oxygen . / 1 366 kJ tenaga haba dibebaskan apabila 1 mol etanol dibakar lengkap dalam oksigen berlebihan . Energy level diagram: / Gambar rajah aras tenaga: C2 H5 OH + 3O2 2CO2 + 3H2 O ∆H = –1 366 kJ mol–1 Energy / Tenaga

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 154 Heat of reaction ( DH) is the energy change when one mole of reactant reacts or when one mole of product is formed. Haba tindak balas (DH) ialah perbezaan haba apabila 1 mol bahan bertindak balas atau apabila 1 mol hasil tindak balas dihasilkan. B CALCULATION: / PENGIRAAN: Heat of reaction, / Haba tindak balas, Heat released/Heat absorbed/Heat change / Haba dibebaskan/Haba diserap/Perubahan haba, DH = ± mcq X Q = mcq B B B B ∆ H = Heat of precipitation: Heat change 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) ∆H = Heat of displacement: Heat change 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 neutralisation: Heat released 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 combustion: Heat released when 1 mole 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) B B B B 43 42 41 40 39 38 37 36 35 34 Sodium chloride solution Larutan natrium klorida Silver nitrate solution Larutan argentum nitrat 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 Apparatus: / Radas: Polystyrene cup, measuring cylinder, thermometer / Bekas polistirena, silinder penyukat, termometer Procedure: / Prosedur: 1 Measure 25 cm3 of 0.5 mol dm–3 silver nitrate solution with measuring cylinder and poured into a polystyrene cup. 43 42 41 40 39 38 37 36 35 34 Zinc powder Serbuk zink Copper(II) sulphate solution Larutan kuprum(II) sulfat 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 Measure 25 cm3 of 0.5 mol dm–3 copper(II) sulphate solution with cylinder and poured into a polystyrene cup. 43 42 41 40 39 38 37 36 35 34 Hydrochloric acid Asid hidroklorik Sodium hydroxide solution Larutan natrium hidroksida Manipulated variable: / Pemboleh ubah dimanipulasikan: Type of acid / Jenis asid 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 43 42 41 40 39 38 37 36 35 34 Thermometer Termometer Copper can Tin kuprum Wind shield Pengadang angin Water Air Fuel Bahan api 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 molecule alcohols, the higher the heat of combustion. Semakin bertambah bilangan atom karbon dan hidrogen Activity/Experiment to Determine Heat of Reaction / Aktiviti/Eksperimen untuk Menentukan Haba Tindak Balas

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 155 © Nilam Publication Sdn. Bhd. 2 Measure 25 cm3 of 0.5 mol dm–3 sodium chloride solution with another measuring cylinder and poured into another polystyrene cup. 3 Place a thermometer into each solution and record the initial temperature T1 and T2 of each solution. 4 Pour the sodium chloride solution quickly and carefully into the silver nitrate solution. 5 Stir the reaction of mixture with the thermometer and record the highest temperature, T3. 1 Sukat 25 cm3 larutan argentum nitrat berkepekatan 0.5 mol dm–3 dengan silinder penyukat dan tuang ke dalam cawan polistirena. 2 Sukat 25 cm3 larutan natrium klorida berkepekatan 0.5 mol dm–3 dengan silinder penyukat dan tuang ke dalam cawan polistirena yang lain. 3 Letakkan termometer di dalam setiap larutan dan catat suhu awal setiap larutan T1 dan T2. 4 Tuang larutan natrium klorida dengan cepat dan cermat ke dalam larutan argentum nitrat. 5 Kacau campuran tindak balas dengan termometer dan catat suhu tertinggi, T3. 2 Place a thermometer into the solution and record the initial temperature T1 of the solution. 3 Add half spatula of zinc powder quickly and carefully into the copper(II) sulphate solution. 4 Stir the reaction mixture with the thermometer and record the highest temperature, T2. 1 Sukat 25 cm3 larutan kuprum(II) sulfat berkepekatan 0.5 mol dm–3 dengan silinder penyukat dan tuang ke dalam cawan polistirena. 2 Letakkan termometer di dalam larutan tersebut dan catat suhu awal T1. 3 Tambah separuh spatula serbuk zink dengan cepat dan cermat ke dalam larutan kuprum(II) sulfat. 4 Kacau campuran tindak balas dengan termometer dan catat suhu tertinggi, T2. 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 Measure 50 cm3 of 2 mol dm–3 sodium hydroxide solution with measuring cylinder and pour into a polystyrene cup. 2 Measure 50 cm3 of 2 mol dm–3 hydrochloric acid with another measuring cylinder and pour into another polystyrene cup. 3 Place a thermometer into each solution and record the initial temperature of sodium hydroxide solution and hydrochloric acid, T1 and T2. 4 Pour hydrochloric acid quickly and carefully into the sodium hydroxide solution. 5 Stir the reaction mixture with the thermometer and record the highest temperature, T3. 6 Repeat steps 1 to 5 using sodium hydroxide solution and ethanoic acid. 1 Sukat 50 cm3 larutan natrium hidroksida 2 mol dm–3 dengan silinder penyukat dan tuang ke dalam cawan polistirena. 2 Sukat 50 cm3 asid hidroklorik berkepekatan 2 mol dm–3 dengan silinder penyukat yang lain dan tuang ke dalam cawan polistirena yang lain. 3 Letakkan termometer di dalam setiap larutan dan rekod suhu awal larutan natrium hidroksida dan asid hidroklorik, T1 dan T2. 4 Tuang asid hidroklorik dengan cepat dan cermat ke dalam larutan natrium hidroksida. 5 Kacau campuran tindak balas dengan termometer dan catat suhu tertinggi, T3. 6 Ulang langkah 1 hingga 5 dengan menggunakan larutan natrium hidroksida dan asid etanoik. setiap molekul alkohol, semakin bertambah haba pembakaran. 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 Procedure: / Prosedur: 1 Measure 100 cm3 of water with measuring cylinder and pour the copper can. 2 Place a thermometer into the water and record the initial temperature, T1. 3 Place the copper can on a tripod stand. 4 Fill a lamp with methanol. Weigh the lamp and record the initial mass, m1. 5 Place a wind shield as shown in the diagram to minimise heat loss to the moving air in the surrounding. 6 Place the lamp near the base of the copper can to maximise the heat transfer and the wick is lighted. 7 Stir the water continuously with the thermometer until its temperature increased by 30°C, put off the flame and record the highest temperature, T2 reached by the water. 8 Weigh the final mass of the lamp, m2 and its content immediately and record the mass. 9 Repeat steps 1 to 8 with ethanol, propanol and butanol. 1 Sukat 100 cm3 air dengan silinder penyukat dan tuang ke dalam bekas kuprum. 2 Letakkan termometer di dalam air dan catat suhu awal, T1. 3 Letakkan bekas kuprum di atas tungku kaki tiga. 4 Isi pelita dengan metanol dan timbang. Catat jisim awalnya, m1. 5 Letakkan penghadang angin seperti ditunjukkan dalam rajah untuk mengurangkan kehilangan haba ke udara persekitarannya. 6 Letak pelita dekat dengan bekas kuprum untuk memaksimakan pemindahan haba. Nyalakan sumbu pelita tersebut. 7 Kacau air tersebut dengan termometer sehingga suhunya meningkat sebanyak 30°C, padamkan api dan catat suhu tertinggi, T2 dicapai oleh air. 8 Timbang jisim terakhir pelita m2 dan kandungannya segera dan catat jisimnya. 9 Ulang langkah 1 hingga 8 dengan menggunakan etanol, propanol dan butanol.

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 156 Calculating heat of reaction, / Pengiraan haba tindak balas, ∆H: What are the quantities needed to calculate heat change / heat absorbed / heat given out or heat release, H in a substance? Apakah kuantiti yang diperlukan untuk mengira perubahan haba / haba yang diserap atau 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, for example: Ketumpatan larutan akueus sama dengan ketumpatan air = 1 g cm–3, contoh: • 1 cm3 of aqueous solution has a mass of 1 g 1 cm3 larutan akueus mempunyai jisim 1 g • 2 cm3 of aqueous solution has a mass of 2 g 2 cm3 larutan akueus mempunyai jisim 2 g • m cm3 of aqueous solution has a mass of m g m cm3 larutan akueus mempunyai jisim m g (ii) Specific heat capacity of solution = Specific heat capacity of water = 4.2 J g–1 °C–1 Muatan haba tentu bahan larutan = Muatan haba tentu bahan air = 4.2 J g–1 °C–1 (iii) No heat lost to the surroundings: Tiada haba hilang ke persekitaran: – all heat released in an exothermic reaction = heat absorbed by the solution (temperature increases) semua haba dibebaskan dalam satu tindak balas eksotermik = haba yang diserap oleh larutan (suhu meningkat) – all heat absorbed in endothermic reaction = heat lost by the solution (temperature decreases) semua haba diserap dalam satu tindak balas endotermik = haba yang hilang oleh larutan (suhu menurun) How to calculate heat change / heat absorbed / heat given out or heat released, H in a reaction? Bagaimana mengira perubahan haba / haba yang diserap / haba yang dikeluarkan atau haba dibebaskan, H dalam tindak balas? The heat change, Q in a reaction can be calculated with the following formula Perubahan haba, Q dalam tindak balas boleh dikira dengan formula berikut: Heat change (Q) / Perubahan haba (Q) = 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 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 Q J of heat energy X mol bahan/hasil menyerap/membebaskan Q J tenaga haba 1 mol of reactant/ product absorbs/releases / 1 mol of bahan/hasil menyerap/membebaskan Q X J mol–1 ⇒ ∆H (heat of reaction) / ∆H (Haba tindak balas) = +/– Q J X mol , X = number of moles of reactant/product / bilangan mol bahan/hasil Remark: / Catatan: (i) The sign of ∆H is negative for exothermic reaction (temperature increases). Tanda ∆H adalah negatif untuk tindak balas eksotermik (suhu menaik). (ii) The sign of ∆H is positive for endothermic reaction (temperature decreases). Tanda ∆H adalah positif untuk tindak balas endotermik (suhu menurun). What is the unit heat of reaction? / Apakah unit haba bagi tindak balas? The unit for heat of reaction is kJ mol–1. Unit untuk haba tindak balas ialah kJ mol–1.

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 157 © Nilam Publication Sdn. Bhd. 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, Q Hitung *perubahan haba tindak balas, Q Q = mcθ Step 3: / Langkah 3: Calculate *heat of reaction, ΔH Hitung *haba tindak balas, ΔH ΔH = +/– Q X J Remark: / Catatan: For specific reaction in this chapter: / Bagi tindak balas tertentu dalam bab ini: 1 *Heat change, Q is heat released for exothermic reaction or heat absorbed for endothermic reaction. *Perubahan haba, Q ialah haba yang dibebaskan untuk tindak balas eksotermik atau haba yang diserap untuk tindak balas endotermik. 2 *Heat of reaction, ΔH is / * Haba tindak balas, ΔH ialah (i) Heat of precipitation for precipitation reaction / Haba pemendakan bagi tindak balas pemendakan (ii) Heat of displacement for displacement reaction / Haba penyesaran bagi tindak balas penyesaran (iii) Heat of neutralisation for neutralisation reaction / Haba peneutralan bagi tindak balas peneutralan (iv) Heat of combustion for combustion reaction / Haba pembakaran bagi tindak balas pembakaran Example: / Contoh: 60 cm3 of 0.25 mol dm–3 silver nitrate solution reacts with 60 cm3 of 0.25 mol dm–3 potassium bromide solution with an average temperature of 29°C. A yellow precipitate was formed and the highest temperature reached is 32°C. Determine the heat of reaction, ∆H. / 60 cm3 larutan argentum nitrat 0.25 mol dm–3 bertindak balas dengan 60 cm3 larutan kalium bromida 0.25 mol dm–3 dengan suhu purata 29°C. Mendakan kuning terbentuk dan suhu tertinggi dicapai ialah 32°C. Tentukan haba tindak balas, ∆H. Steps / Langkah-langkah Calculation / Pengiraan Step 1: / Langkah 1: Calculate the number of moles of silver bromide precipitated (X) Hitung bilangan mol mendakan argentum bromida (X) 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+ = 0.25 mol dm–3 × 60 1 000 dm3 = 0.015 mol Number of moles of Br– / Bilangan mol Br – = 0.25 mol dm–3 × 60 1 000 dm3 = 0.015 mol From the equation: / Daripada persamaan: 1 mole of Ag+ ions reacts with 1 mole of Br– ions to form 1 mole of AgBr 1 mol ion Ag+ bertindak balas dengan 1 mol ion Br – membentuk 1 mol AgBr 0.015 mole of Ag+ ions reacts with 0.015 mole Br– ions to form 0.015 mole AgBr 0.015 mol ion Ag+ bertindak balas dengan 0.015 mol ion Br – membentuk 0.015 mol AgBr X = 0.015 Step 2: / Langkah 2: Calculate the heat released, Q Hitung haba yang dibebaskan, Q (Specific heat capacity of solution Muatan haba tentu larutan = 4.2 J g–1 °C–1) Q = mcθ J Q = 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 = – Q X (negative because heat is released to the surrounding or temperature increases) (negatif sebab haba dibebaskan ke persekitaran atau suhu menaik) = – 1 512 0.015 mol = –100.8 kJ mol–1 Draw the energy level diagram: Lukis rajah aras tenaga: Energy / Tenaga AgNO3 (aq/ak) + KBr(aq/ak) AgBr(s/p) + KNO3 (aq/ak) ΔH = –100.8 kJ mol–1

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 158 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 Zn + CuSO4 → ZnSO4 + Cu Number of moles of CuSO4 / Bilangan mol CuSO4 = 50 × 0.1 1 000 = 0.005 mol 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, Q Langkah 2: Hitung haba dibebaskan, Q Heat released in the experiment, Haba dibebaskan dalam eksperimen, Q = 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 ∆H = – 1 050 J 0.005 mol = –210 kJ mol–1 Example 2 / Contoh 2: The following is the thermochemical equation for a reaction. Berikut adalah persamaan termokimia untuk suatu tindak balas. Zn + CuSO4 →ZnSO4 + Cu ∆H = –210 kJ mol–1 Calculate the heat released when 50 cm3 of 1.0 mol dm–3 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, Q Langkah 2: Hitung haba dibebaskan, Q ∆H = Q X 210 kJ mol–1 = Q 0.05 Q = 210 kJ mol–1 × 0.05 mol = 10.5 kJ Numerical Problems Involving Heat of Displacement Pengiraan Melibatkan Haba Penyesaran

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 159 © Nilam Publication Sdn. Bhd. 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 TP1 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, Q Langkah 2: Hitung haba dibebaskan, Q TP1 ∆H = Q X 189 kJ mol–1 = Q 0.032 mol ; Q = Heat released in the experiment Haba dibebaskan dalam eksperimen Q = 189 kJ mol–1 × 0.032 mol = 6.048 kJ = 6 048 J Step 3: Calculate the increase in temperature, θ Langkah 3: Hitung peningkatan suhu, θ TP2 6 048 J = mcθ = 80 × 4.2 × θ θ = 18°C 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, Q = 50 g × 4.2 J g–1 ºC–1 × 3ºC = 630 J Numerical Problems Involving Heat of Displacement Pengiraan Melibatkan Haba Penyesaran Exercise / Latihan

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 160 1 100 cm3 of 2.0 mol dm–3 sodium hydroxide solution is added into 100 cm3 of 2.0 mol dm–3 ethanoic acid. The initial temperature for both solutions is 28.0ºC and the highest temperature is 41.0ºC. Calculate heat of neutralisation. 100 cm3 larutan natrium hidroksida 2.0 mol dm–3 ditambah kepada 100 cm3 asid etanoik 2.0 mol dm–3. Suhu awal keduadua 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 1 000 = 0.2 mol Number of mole of CH3COOH / Bilangan mol CH3COOH = 100 × 2 1 000 = 0.2 mol 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, Q Langkah 2: Hitung haba dibebaskan, Q Heat released / Haba dibebaskan, Q = (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, ΔH = 10.92 kJ 0.2 mol = –54.6 kJ mol–1 2 The thermochemical ionic equation below represents the reaction between lead(II) nitrate solution and potassium sulphate solution. / Persamaan ion termokimia di bawah mewakili tindak balas antara larutan plumbum(II) nitrat dengan larutan kalium sulfat. Pb2+ + SO4 2– → PbSO4 ∆H = –50.4 kJ mol–1 Calculate the increase in temperature when 25 cm3 of 1 mol dm–3 of lead(II) nitrate solution is added into 25 cm3 of 1 mol dm–3 of potassium sulphate solution. [Specific heat capacity of solution = 4.2 J g–1 °C–1, density of solution = 1 g cm–3] Hitungkan kenaikan suhu apabila 25 cm3 larutan plumbum(II) nitrat 1 mol dm–3 ditambah kepada 25 cm3 larutan kalium sulfat 1 mol dm–3. [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3] Step 1: Calculate number of mole of lead(II) sulphate precipitate formed Langkah 1: Hitung bilangan mol mendakan plumbum(II) sulfat yang terbentuk Number of moles of Pb2+ = 25 × 1 1 000 = 0.025 mol, Bilangan mol Pb2+ Number of moles of SO4 2– = 25 × 1 1 000 = 0.025 mol Bilangan mol SO4 2– From the equation: / Dari persamaan • 1 mol of Pb2+ ions reacts with 1 mol of SO4 2– ions to form 1 mole of PbSO4 1 mol ion Pb2+ bertindak balas dengan 1 mol ion SO4 2– membentuk 1 mol PbSO4 • 0.025 mole of Pb2+ ions reacts with 0.025 mole SO4 2– ions to form 0.025 mole of PbSO4 0.025 mol ion Pb2+ bertindak balas dengan 0.025 mol ion SO4 2– membentuk 0.025 mol PbSO4 Step 2: Calculate the heat released, Q Langkah 2: Hitung haba dibebaskan, Q X = 0.025 mol 50.4 kJ mol–1 = Q 0.025 mol Heat released / Haba dibebaskan, Q = 1.26 kJ = 1 260 J 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 Numerical Problems Involving Heat of Neutralisation Pengiraan Melibatkan Haba Peneutralan

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 161 © Nilam Publication Sdn. Bhd. 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 ∆ H = –57 kJ mol–1 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. – 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. – 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. – 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. – Asid hidroklorik dan asid nitrik adalah asid monoprotik kuat. Satu mol asid hidroklorik atau asid nitrik mengion lengkap dalam air untuk menghasilkan satu mol ion hidrogen. – Natrium hidroksida dan kalium hidroksida adalah alkali kuat. Satu mol natrium hidroksida atau kalium hidroksida mengion lengkap dalam air untuk menghasilkan satu mol ion hidroksida. – Haba peneutralan bagi larutan natrium hidroksida/kalium hidroksida dengan asid hidroklorik/asid nitrik ialah –57 kJ mol–1 kerana semua tindak balas menghasilkan satu mol air. HCl + KOH → KCl + H2O HCl + NaOH → NaCl + H2O HNO3 + KOH → KNO3 + H2O H+ + OH– → H2O, ∆H = –57 kJ mol–1 HNO3 + NaOH → NaNO3 + H2O ⇒ 1 mol of hydrogen ions react with 1 mol of hydroxide ions to form 1 mol of water to release 57 kJ of 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. – 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 2 The reaction between 25 cm3 of hydrochloric acid and 25 cm3 of sodium hydroxide solution releases the heat of 2 100 J. What is the temperature change of the mixture? Tindak balas antara 25 cm3 asid hidroklorik dan 25 cm3 larutan natrium hidroksida membebaskan haba sebanyak 2 100 J. Apakah perubahan suhu campuran tindak balas? [Specific heat capacity of a solution = 4.2 J g–1 °C–1] / [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1] Answer: / Jawapan: Heat released / Haba dibebaskan = mcθ = 2 100 J (25 + 25) g × 4.2 J g–1 °C–1 × θ = 2 100 J θ = 10°C Comparing the Heat of Neutralisation Membandingkan Haba Peneutralan TP2

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 162 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 ∆H = –55 kJ mol–1 NaOH + HCN → NaCN + H2O ∆H = –12 kJ mol–1 Explanation: / Penerangan: (i) Weak acids ionise partially in water to produce hydrogen ions. Asid lemah mengion separa dalam air menghasilkan ion hidrogen . CH3COOH CH3COO– + H+ (ii) Some of the particles still remain in the form of molecules . Sebahagian zarah masih kekal dalam bentuk molekul . (iii) Heat energy is absorbed to ionise molecules of the weak acid that have not been ionised so that they ionise completely. / Tenaga haba diserap untuk mengionkan molekul asid lemah yang masih belum mengion supaya mengion sepenuhnya. (iv) Part of the heat that is released is used/absorbed to ionise the molecules of weak acid that has not been ionised. / Sebahagian haba yang dibebaskan digunakan/diserap untuk mengionkan molekul asid lemah yang masih belum mengion. Calculation guide: / Panduan pengiraan: # Calculation guide I: / Panduan pengiraan I: If any reaction is repeated by changing the volume without changing the concentration, change in temperature is the same. Jika sebarang tindak balas diulangi dengan menukarkan isi padu tanpa menukar kepekatan, perubahan suhu adalah sama. Example 1: / Contoh 1: • Reaction I: / Tindak balas I: 50 cm3 of 2 mol dm–3 hydrochloric acid is added to 50 cm3 of 2 mol dm–3 potassium hydroxide solution. The temperature rises by 13°C. / 50 cm3 asid hidroklorik 2 mol dm–3 ditambah dengan 50 cm3 larutan kalium hidroksida 2 mol dm–3. Suhu naik sebanyak 13°C. • Reaction II: / Tindak balas II: 100 cm3 of 2 mol dm–3 hydrochloric acid is added to 100 cm3 of 2 mol dm–3 potassium hydroxide solution. What is the temperature change in this reaction? / 100 cm3 asid hidroklorik 2 mol dm–3 ditambah dengan 100 cm3 larutan kalium hidroksida 2 mol dm–3. Apakah perubahan suhu dalam tindak balas ini? Answer: / Jawapan: ∆H = Q X where / di mana X = Number of moles of water / Bilangan mol air Q = 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: ∆H = 100 × 4.2 × 13 J 0.1 = 54 600 J Reaction II: / Tindak balas II: 54 600 J = 600 × 4.2 × T 0.2 , where / di mana T = temperature change in reaction II perubahan suhu dalam tindak balas II T = 13ºC

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 163 © Nilam Publication Sdn. Bhd. Explanation: – When Reaction II is repeated by doubled volume of acid and alkali, the number of moles of water produced in Reaction II is double, hence amount of heat energy released is double. – The amount of heat energy which is doubled is used to increase total volume of solution which is also double. – Therefore, the increase in temperature remains the same. Penerangan: – Apabila Tindak balas II diulang dengan menggandakan isi padu asid dan alkali, bilangan mol air yang dihasilkan dalam Tindak balas II adalah dua kali ganda, oleh itu jumlah tenaga haba yang dibebaskan adalah dua kali ganda. – Jumlah tenaga haba yang berganda digunakan untuk meningkatkan jumlah isi padu larutan yang juga dua kali ganda. – Oleh itu, peningkatan suhu kekal sama. # Calculation guide II: / Panduan pengiraan II: If the reaction is repeated by changing the concentration of the solution by n times without changing the volume, the temperature change is n times. / Jika sebarang tindak balas diulangi dengan menukarkan kepekatan larutan sebanyak n kali tanpa menukar isi padu, perubahan suhu adalah n kali. Example 2: / Contoh 2: • Reaction I: / Tindak balas I: 50 cm3 of 0.2 mol dm–3 lead(II) nitrate is added to 50 cm3 of 0.2 mol dm–3 sodium carbonate solution. The temperature of the mixture rises by 2.4°C. / 50 cm3 larutan plumbum(II) nitrat 2 mol dm–3 ditambah dengan 50 cm3 larutan natrium karbonat 0.2 mol dm–3. Suhu naik sebanyak 2.4°C. • Reaction II: / Tindak balas II: 50 cm3 of 0.6 mol dm–3 lead(II) nitrate solution is added to 50 cm3 of 0.6 mol dm–3 sodium carbonate solution. What is the temperature rise in this experiment? / 50 cm3 larutan plumbum(II) nitrat 0.6 mol dm–3 ditambah dengan 50 cm3 larutan natrium karbonat 0.6 mol dm–3. Apakah kenaikan suhu dalam eksperimen ini? Answer: / Jawapan: ∆H = Q 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 Q = Heat change / Perubahan haba = mcθ Ionic equation for both reactions: / Persamaan ion untuk kedua-dua tindak balas: Pb2+ + CO3 2– → PbCO3 Reaction I: / Tindak balas I: ∆H = 100 × 4.2 × 2.4 J 0.01 = 100 800 J mol–1 Reaction II: / Tindak balas II: 100 800 J = 100 × 4.2 × T 0.03 , where / di mana T = temperature change in reaction II 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 isrepeated 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.

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 164 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. 1 Methanol burns in oxygen in a reaction as shown in the thermochemical equation below. Metanol terbakar dalam oksigen seperti persamaan termokimia di bawah. CH3OH(s/p) + 3 2 O2(g/g)→ CO2 (g/g) + 2H2O ∆H = –725 kJ mol–1 What isthe 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 = Q X , X = Number of moles of methanol / Bilangan mol metanol 725 kJ mol–1 = 145 kJ X Number of moles of methanol / Bilangan mol metanol = 145 kJ 725 kJ mol–1 = 0.2 mol Calculate mass of methanol Hitung jisim metanol Mass of methanol / Jisim metanol = 0.2 × [12 × 1 + 4 × 1 + 16] = 6.4 g 2 22 g of butanol is burnt completely in excess of oxygen. The heat released is used to heat up 500 cm3 of water from 27.5°C to 55.8°C. Calculate the heat of combustion of butanol. / 22 g butanol terbakar lengkap dalam oksigen berlebihan. Haba yang dibebaskan memanaskan 500 cm3 air dari 27.5°C ke 55.8°C. Hitungkan haba pembakaran butanol. [Specific heat capacity of a solution = 4.2 J g–1 °C–1, relative atomic mass: H, 1; C, 12; O, 16 ] [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, jisim atom relatif: H, 1; C, 12; O, 16] Calculate number of mole of butanol Kira bilangan mol butanol Number of moles of butanol / Bilangan mol butanol = 22 g 74 g mol–1 Calculate heat released, Q Hitung haba dibebaskan, Q Q = 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 = 59.43 kJ 22 74 mol = 199.9 kJ mol–1 Numerical Problems Involving Heat of Combustion Pengiraan Melibatkan Haba Pembakaran Comparing the Heat of Combustion of Various Fuels Membandingkan Haba Pembakaran Pelbagai Bahan Api

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 165 © Nilam Publication Sdn. Bhd. I Hot packs: Pek panas: 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 exothermic reaction. Mengandungi bahan kimia yang membebaskan haba, aplikasi bagi tindak balas eksotermik . 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 release heat, thus causing the temperature to increase . / Kalsium klorida kontang larut dalam air dan membebaskan haba yang seterusnya menyebabkan suhu meningkat . CaCl2(s/p) H2O Ca2+(aq/ak) + 2Cl– (aq/ak) ∆H = –83 kJ mol–1 APPLICATION OF EXOTHERMIC AND ENDOTHERMIC REACTIONS IN EVERYDAY LIFE / APLIKASI TINDAK BALAS EKSOTERMIK DAN ENDOTERMIK DALAM KEHIDUPAN SEHARIAN CS / SK 3.3 3.3 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. 0 1 000 1 2 3 4 2 000 3 000 Heat of combustion of alcohol (kJ mol –1) Haba pembakaran alkohol (kJ mol –1) Number of carbon atom per molecule Bilangan atom karbon per molekul State relationship between the number of carbon atom per molecule of alcohol with the heat of combustion. Explain. Nyatakan hubungan antara bilangan atom karbon per molekul alkohol dengan haba pembakaran. Terangkan. Answer: / Jawapan: – When the number of carbon atom per molecule of alcohol increases, the heat combustion increases . – When the number of carbon atom per molecule of alcohol increases , the number of carbon dioxide and water molecules produced as products increases . – More bonds between atoms in carbon dioxide and water molecules are formed, more heat is released. – Apabila bilangan atom karbon dalam setiap molekul alkohol bertambah, haba pembakaran bertambah . – Apabila bilangan atom karbon dalam setiap molekul alkohol bertambah , bilangan molekul karbon dioksida dan air yang dihasilkan bertambah . – Lebih banyak ikatan antara atom dalam molekul air dan karbon dioksida terbentuk, lebih banyak haba dibebaskan.

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 166 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. 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 endothermic reaction. Mengandungi bahan kimia yang menyerap haba, aplikasi bagi tindak balas endotermik . 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 absorb heat, thus causing the temperature to decrease . / Pepejal ammonium nitrat larut dalam air menyerap haba yang seterusnya menyebabkan suhu menurun . NH4NO3(s/p) H2O 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. 1 The diagram below shows two different situations occurred in daily lives. Rajah di bawah menunjukkan dua situasi berbeza yang terjadi dalam kehidupan seharian. Situation I: Face mask causes the glass to fog up Situasi I: Pelitup muka menyebabkan cermin mata berkabus Situation I: A girl is sweating. When she is exposing herself in front of the fan, she feel cold. Situasi I: Seorang perempuan berpeluh. Apabila dia mendedahkan dirinya di hadapan kipas, dia berasa sejuk. Exercise / Latihan

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 167 © Nilam Publication Sdn. Bhd. (a) State the reaction involved in situation I and II either exothermic or endothermic. Nyatakan tindak balas yang terlibat bagi situasi I dan II sama ada eksotermik atau endotermik. Situation I: / Situasi I : Exothermic / Eksotermik Situation II: / Situasi II : Endothermic / Endotermik (b) Explain how heat change has occurred during the process in situation I and II. Terangkan mengapa perubahan haba berlaku di sepanjang proses dalam situasi I dan II. Situation I: Hot air that we exhale escapes through the face mask will go directly to the cooler lenses of the glasses. This causes the formation of small droplets of water. As the condensation process occur, heat energy is released. Situation II: The fan helps to evaporate sweat from skin. As evaporation occur, heat energy is absorbed from the body causes the body feel cold. 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 in excess oxygen, the unit is kJ g–1. / Nilai bahan api adalah jumlah haba yang dibebaskan apabila 1 g bahan api terbakar lengkap dalam lebihan oksigen, unitnya adalah kJ g–1. The unit for heat of combustion is in kJ mol–1. Why the unit for value is in kJ g–1? / Unit bagi haba pembakaran adalah dalam kJ mol –1. Mengapa unit bagi nilai ini dalam kJ g–1? In daily life, it is not possible to measure the quantity of fuel in mole. The quantity of fuel normally measured by its mass in gram. Dalam kehidupan seharian, ia tidak mungkin praktikal untuk mengukur kuantiti bahan api dalam mol. Kuantiti bahan api kebiasaannya diukur oleh jisimnya dalam gram. How to calculate the fuel value of a fuel from its heat of combustion? Bagaimana untuk menghitung nilai bahan api bagi bahan api daripada haba pembakaran? Fuel value = Heat of combustion in kJ mol–1 Molar mass of the fuel in g mol–1 Nilai bahan api = Haba pembakaran dalam kJ mol–1 Jisim molar bahan api dalam g mol–1 Example: / Contoh: Fuel Bahan api Molecular formula Formula molekul Heat of combustion, ∆H Haba pembakaran, ∆H (kJ mol–1) Molar mass Jisim molar (g mol–1) Fuel value Nilai bahan api (kJ g–1) Carbon Karbon C –389 12 32.4 Propane Propana C3H8 –2202 44 50.0 Methane Metana CH4 –394 16 24.6 What are the aspects to be considered when choosing a fuel in industry? Apakah aspek yang perlu dipertimbangkan ketika memilih bahan api dalam industri? – Fuel value of the fuel. / Nilai bahan api bagi bahan api. – Cost of energy/cost of fuel. / Harga tenaga/bahan api. – Availability and sources of the fuel. / Ketersediaan dan sumber bahan api. – Effect of the fuel to the environment. / Kesan bahan api kepada persekitaran. State two advantages of using hydrogen as a fuel. Nyatakan dua kelebihan penggunaan hidrogen dalam bahan api. 1 Hydrogen produces the highest energy per gram compared to other fuels. Hidrogen menghasilkan tenaga tertinggi per gram berbanding bahan api lain. 2 Hydrogen burns in air to produce steam which does not pollute the environment. Pembakaran hidrogen di udara menghasilkan wap yang tidak mencemarkan alam sekitar. Why coke is used as fuel in iron extraction? Mengapa kok digunakan sebagai bahan api dalam pengambilan besi? The blast furnace uses coke as a fuel to meet the energy requirements of the process as well as a reducing agent for the extraction of iron ore to iron. Coke is cheaper than other fuel. Relau bagas menggunakan kok sebagai bahan api untuk memenuhi keperluan tenaga proses tersebut dan juga agen pengurangan pengekstrakan bijih besi ke besi. Kok lebih murah daripada bahan api lain. TP2 TP4

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 168 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. 1 The table below shows some properties of various chemicals used as fuels. Jadual di bawah memberikan beberapa ciri pelbagai bahan kimia yang digunakan sebagai bahan bakar. Fuel Bahan api Sootiness of flame Kejelagaan api Relative cost Kos relatif Fuel value / kJ g–1 Nilai bahan api / kJ g–1 Hydrogen / Hidrogen Clean / Bersih Very expensive / Sangat mahal 142.5 Ethanol / Etanol Clean / Bersih Expensive / Mahal 30 Kerosene / Kerosin Moderate / Sederhana Moderate / Sederhana 37 Petrol / Petrol Clean / Bersih Expensive / Mahal 34 Natural gas / Gas asli Clean / Bersih Moderate / Sederhana 50 (a) What is meant by fuel value? / Apakah yang dimaksudkan dnegan 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. (b) Hydrogen has fuel value of 142.5 kJ g–1 and ethanol has fuel value of 30 kJ g–1. Both ethanol and hydrogen are fuel that is used in transportation. Between hydrogen and ethanol, which fuel can be used as rocket propellant into the space? Give reason for your answer. / Hidrogen mempunyai nilai bahan api 142.5 kJ g–1 manakala etanol mempunyai nilai bahan api 30 kJ g–1. Bahan api yang manakah sesuai digunakan bagi pelancar roket ke angkasa lepas? Berikan alasan bagi jawapan anda. Hydrogen. Hydrogen produces the highest fuel value compared to other fuels. Hydrogen burns in air to produce steam which does not pollute the environment. Hidrogen. Hidrogen menghasilkan nilai bahan api paling tinggi berbanding bahan api yang lain.Hidrogen terbakar dalam udara membentuk wap air yang tidak mencemarkan alam sekitar. (c) The diagram below shows a process to produce another type of fuel. Rajah di bawah menunjukkan proses bagi menghasilkan sejenis bahan api. Ocean 300 – 400 million years ago Laut 300 – 400 juta tahun lalu Ocean 50 – 100 million years ago Laut 50 – 100 juta tahun lalu Petroleum Petroleum Sand, rocks Tanah, batu Sand / Tanah Dead organisms Organisma mati (i) Name three types of fuel that can be produced from the process above. Namakan tiga bahan api yang boleh dihasilkan daripada proses di atas. 1 Petrol / Petrol 2 Kerosene / Kerosin 3 Natural gas / Gas asli Exercise / Latihan TP2 TP3 TP3 TP4

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 169 © Nilam Publication Sdn. Bhd. (ii) State two disadvantages of using this type of fuel. Nyatakan dua kekurangan menggunakan jenis bahan api ini. 1 Non-renewable / Tidak boleh diperbaharui 2 Production process of fuel costs a lot of money / Proses penghasilan bahan api menggunakan kos yang tinggi (iii) Fuel that has been extracted from plants and crops is known as biofuel. Is biofuel production better than the fuel above? Justify your answer. Bahan api yang diekstrak daripada tumbuhan dan tanaman dikenali sebagai bahan api bio. Adakah bahan api bio lebih baik berbanding bahan api di atas? Berikan hujah anda. Biofuel is better. Renewable// environmental-friendly. OR Fossil fuel is better. Production of biofuel destroy crops and plant to obtain sugar. This will affect the climate change, land and many more. 1 The diagram below shows the apparatus set-up for an experiment to determine the heat of displacement of silver. Rajah di bawah menunjukkan susunan radas untuk eksperimen menentukan haba penyesaran argentum. 34 35 36 37 38 39 40 41 42 43 Excess of copper powder Serbuk kuprum berlebihan 100 cm3 of 0.5 mol dm–3 silver nitrate solution 100 cm3 larutan argentum nitrat 0.5 mol dm–3 Plastic cup Cawan plastik The following data was obtained: / Berikut adalah data yang diperoleh: 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 (ii) Colourless solution turns blue because copper(II) ion is produced (iii) The thermometer reading rises or the container becomes hot or warm because heat is released to the surroundings/ the reaction is exothermic (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 Structured Questions / Soalan Struktur TP1 TP2 TP2

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 170 (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 Q = (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 1 000 = 0.05 mol From the equation, / Dari persamaan, 2 mol of AgNO3 produce 2 mol of Ag / 2 mol AgNO3 menghasilkan 2 mol Ag 0.05 mol of AgNO3 produce 0.05 mol of Ag / 0.05 mol AgNO3 menghasilkan 0.05 mol Ag Heat of displacement of silver / Haba penyesaran argentum = –5 250 J 0.05 mol = –105 kJ mol–1 (f) (i) The experiment is repeated using 100 cm3 of 1.0 mol dm–3 silver nitrate solution and excess copper powder. Calculate the temperature change in this experiment. Eksperimen itu diulangi menggunakan 100 cm3 larutan argentum nitrat 1.0 mol dm–3 dan serbuk kuprum yang berlebihan. Hitungkan perubahan suhu dalam eksperimen ini. Number of moles of Ag+ / Bilangan mol Ag+ = 1 × 100 1 000 = 0.1 mol, Number of moles of Ag displaced / Bilangan mol Ag disesarkan = 0.1 mol Temperature change, / Perubahan suhu, θ = 0.1 × 105 000 100 × 4.2 = 25 ºC (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. 2 Experiment I is carried out to determine the heat of neutralisation between strong acid and strong alkali. 50 cm3 of 0.5 mol dm–3 sodium hydroxide solution is poured into a plastic cup and the initial temperature is recorded. 50 cm3 of 0.5 mol dm–3 nitric acid is then poured into the cup containing the sodium hydroxide solution. The mixture is stirred and heat produced raises the temperature by 3°C. [Specific heat capacity of the solution = 4.2 J g–1 °C–1] Eksperimen I dijalankan untuk menentukan haba peneutralan antara asid kuat dengan alkali kuat. 50 cm3 larutan natrium hidroksida 0.5 mol dm–3 dituangkan dalam cawan plastik dan suhu awal dicatat. 50 cm3 asid nitrik 0.5 mol dm–3 kemudian dituangkan ke dalam cawan mengandungi larutan natrium hidroksida. Campuran tindak balas dikacau dan haba yang terbebas menaikkan suhu sebanyak 3°C. [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1] TP3 TP3 TP1 TP2

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 171 © Nilam Publication Sdn. Bhd. 34 35 36 37 38 39 40 41 42 43 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. Number of moles / Bilangan mol = 50 × 0.5 1 000 = 0.025 mol (ii) the heat released in the experiment. / haba yang dibebaskan dalam tindak balas itu. Heat released / Haba dibebaskan = Heat changed / Perubahan haba = 100 × 4.2 × 3 = 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 Heat of neutralisation / Haba peneutralan = ∆H = – 1 260 J 0.025 mol = –50.4 kJ mol–1 (c) Write the thermochemical equation for the reaction in the experiment. Tulis persamaan termokimia untuk tindak balas dalam eksperimen. NaOH + HNO3 → NaNO3 + H2O ∆H = –50.4 kJ mol–1 (d) (i) Construct energy level diagram for the reaction. / Bina rajah aras tenaga bagi tindak balas tersebut. Energy / Tenaga NaOH + HNO3 NaNO3 + H2O ∆H = –50.4 kJ TP1 TP2 TP3 TP2

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 172 (ii) State two information about the reaction which can be obtained from the above energy level diagram. Nyatakan dua maklumat tentang tindak balas yang boleh didapati dari rajah aras tenaga di atas. – It is an exothermic reaction // heat energy is released to the surrounding – The total energy of reactants is higher than the products – 50.4 kJ of heat energy is released when 1 mole of water is formed (any 2) (e) 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. 34 35 36 37 38 39 40 41 42 43 Plastic cup Cawan plastik 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 1 000 = 1 × (50) 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 Heat of neutralisation / Haba peneutralan = – 2 310 J 0.05 mol = – 46 200 J mol–1 = 46.2 kJ mol–1 (f) 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. (g) Experiment II is repeated by adding 100 cm3 of 1 mol dm–3 ethanoic acid to 100 cm3 of 1 mol dm–3 sodium hydroxide solution, the increase in temperature is still 5.5ºC. Explain why. Eksperimen II diulang dengan menambah 1 mol dm–3 asid etanoik berisi padu 100 cm3 kepada 1 mol dm–3 larutan natrium hidroksida berisi padu 100 cm3 , peningkatan suhu masih 5.5ºC. Terangkan mengapa. The number of moles of water produced is doubled, hence amount of heat energy released is doubled. The amount of heat energy which is double is used to increase the total volume of solution which is also doubled. Therefore, the temperature increase remains the same. TP2 TP3 TP3 TP4 TP4

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 173 © Nilam Publication Sdn. Bhd. 3 An experiment was carried out to determine the heat of precipitation for the reaction between lead(II) nitrate and potassium sulphate. 50.0 cm3 of 0.5 mol dm–3 lead(II) nitrate solution was added to 50.0 cm3 of 0.5 mol dm–3 of potassium sulphate solution in a plastic cup. / Satu eksperimen dijalankan untuk menentukan haba pemendakan antara plumbum(II) nitrat dan kalium sulfat. 50.0 cm3 larutan plumbum(II) nitrat 0.5 mol dm–3 ditambahkan kepada 50.0 cm3 larutan kalium sulfat 0.5 mol dm–3 di dalam cawan plastik. The thermochemical equation for the reaction is shown below: / Persamaan termokimia untuk tindak balas seperti berikut: Pb(NO3)2 + K2SO4 → PbSO4 + 2KNO3 ∆H = –50.4 kJ mol–1 [Specific heat capacity of the solution = 4.2 J g–1 °C–1, density solution = 1 g cm–3] [Muatan haba tentu bahan larutan = 4.2 J g–1 °C–1, ketumpatan larutan = 1 g cm–3] (a) What is meant by ‘heat of precepitation’ in the experiment? Apakah yang dimaksudkan dengan ‘haba pemendakan’ dalam eksperimen itu? Heat is released when 1 mole of lead(II) sulphate is precipitated from mixing the aqueous solution of the Pb2+ ions and SO4 2– ions. (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 or / atau 50.4 kJ = Q 0.025 Q = 50.4 × 0.025 = 1.26 kJ (iii) Temperature change / Perubahan suhu 1 260 J = 100 × 4.2 × θ θ = 1 260 100 × 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 + K2 SO4 PbSO4 + 2KNO3 ∆H = –50.4 kJ mol–1 (e) Write an ionic equation for the above reaction. Tulis persamaan ion untuk tindak balas di atas. Pb2+ + SO4 2– → PbSO4 TP1 TP2 TP2 TP2 TP3

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 © Nilam Publication Sdn. Bhd. 174 (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 SO4 2– ions. 3°C. Pemendakan plumbum(II) sulfat hanya melibatkan ion Pb2+ dan ion SO4 2–. (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. (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 CaCl2 + Na2 CO3 CaCO3 + 2NaCl 4 The apparatus set-up below was used to determine the heat of combustion of butanol. Susunan radas di bawah telah digunakan untuk menentukan haba pembakaran butanol. 34 35 36 37 38 39 40 41 42 Thermomete 43 r Termometer Metal can / Tin logam Water / Air Lamp + Butanol Pelita + Butanol The results are as follows: / Keputusan adalah seperti berikut: 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] (a) Write the equation for the combustion of butanol, C4H9OH. Tulis persamaan untuk pembakaran butanol, C4H9OH. C4H9OH + 6O2 → 4CO2 + 5H2O TP3 TP3 TP3 TP2 TP2

MODULE • Chemistry FORM 5 UNIT 3 UNIT 3 175 © Nilam Publication Sdn. Bhd. (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, Q = 500 × 4.2 × 30 = 63 000 J/63 kJ (c) Calculate the number of moles of butanol that was burnt. / Hitungkan bilangan mol butanol yang telah terbakar. [Relative atomic mass: / Jisim atom relatif: C = 12, H = 1] Relative molecular mass / Jisim molekul realtif = 4(12) + 10(1) + 16 = 74 Number of moles / Bilangan mol = 2.28 74 = 0.03 mol (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 1 mol of butanol releases / 1 mol butanol membebaskan = 63 kJ 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 • Make sure the flame touches the bottom of the metal can • Stir the water in the metal can continuously (any 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. (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 Molecular formula Formula molekul Heat of combustion/ kJ mol–1 Haba pembakaran/ kJ mol–1 Methanol / Metanol CH3OH 725 Ethanol / Etanol C2H5OH 1 376 Propan-1-ol / Propan-1-ol C3H7OH 2 015 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. QUIZ KUIZ TP3 TP3 TP3 TP2 TP2 TP4

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 © Nilam Publication Sdn. Bhd. 176 POLYMER POLIMER UNIT 4 Concept Map / Peta Konsep Addition Polymerisation: / Pempolimeran Penambahan: – Monomers which have double covalent bonds between carbon, C=C join together to form long chain molecules called polymer. / Monomer-monomer yang mempunyai ikatan kovalen ganda dua antara karbon, C=C bergabung bersama untuk membentuk molekul rantai panjang yang disebut polimer. – Example of polymers produced: Polythene, polyvinyl chloride (PVC), polystyrene and polypropene. Contoh polimer yang terbentuk: Politena, polivinil klorida (PVC), polistirena dan polipropena. Condensation Polymerisation: / Pempolimeran Kondensasi: – Monomers join together to form large molecule with the elimination of small molecules such as water and hydrogen chloride. Monomer-monomer bergabung bersama untuk membentuk molekul besar dengan penyingkiran molekul kecil seperti air dan hidrogen klorida. – Example of polymers produced: Synthetic fiber such as terylene (polyester) and nylon. / Contoh polimer terbentuk: Gentian sintetik seperti terilena (poliester) dan nilon. – Activity to produce nylon in the laboratory. Aktiviti untuk menghasilkan nilon dalam makmal Natural Polymers Polimer Semula Jadi Example: Contoh: – Protein Protein – Starch Kanji – Natural Rubber Getah Asli Synthetic Polymers Polimer Sintetik Example: Contoh: – Polyethene Polietena – Polystyrene Polistirena – Styrenebutadiene rubber (SBR) Getah stirenabutadiena Themoplastic / Termoplastik – Becomes soft when heated and hardened when cold / Menjadi lembut apabila dipanaskan dan keras apabila disejukkan – Can be recycled / Boleh dikitar semula – Example: Polythene, perspex and polyvinyl chloride (PVC) / Contoh: Politena, perspeks dan polivinil klorida (PVC) Thermoset / Termoset – Decomposed instead of melt when heated to high temperature / Terurai, bukannya mencair apabila dipanaskan pada suhu yang tinggi – Cannot be recycled / Tidak boleh dikitar semula – Example: Bakelite, resin epoxy and melamine Contoh: Bakelit, epoksi resin dan melamin. Elastomer / Elastomer – Elastic / Kenyal – Not easily recycled / Tidak mudah dikitar semula – Example: Silicone rubber, styrene butadiane rubber (SBR), neoprene and natural rubber. Contoh: Getah silikone, getah stirena-butadiena, neoprena dan getah asli. Justify the uses of polymers in daily life Wajarkan kegunaan polimer dalam kehidupan seharian Formed by / Terbentuk oleh Characteristics of polymer / Ciri-ciri polimer Source of polymer Sumber polimer Compare natural rubber and vulcanised rubber Bandingkan getah asli dan getah tervulkan POLYMERS: A long chain molecules made up of large number of small repeating identical unit of monemer. POLIMER: MOLEKUL BERANTAI PANJANG YANG TERBENTUK DARIPADA GABUNGAN BANYAK UNIT KECIL YANG SAMA DIPANGGIL MONOMER. POLYMER: A long chain molecule made up of large number of small repeating identical unit of monomers. POLIMER: Molekul berantai panjang yang terbentuk daripada gabungan banyak unit kecil yang sama dipanggil monomer. Synthetic Rubber Getah Sintetik – Meaning: Synthetic polymers that are elastic and can regain their original length after being stretched. Makna: Polimer sintetik yang elastik dan dapat memperoleh kembali panjang asalnya setelah diregangkan. – Example: Thiokol, slicone rubber, styrene-butadiene rubber (SBR) and neoprene. Contoh: Tiokol, getah silikone, getah stirena-butadiena dan neoprena. – Properties and uses Sifat dan kegunaan – Effects of synthetic rubber to the environment / Kesan getah sintetik pada persekitaran Synthetic Rubber Getah Sintetik Natural Rubber Getah Asli Natural Rubber Getah Asli – Structural formula Formula struktur – Latex coagulation prevention Pencegahan penggumpalan lateks – Latex coagulation process / Proses penggumpalan lateks – Characterics and uses Ciri-ciri dan kegunaan Vulcanised rubber Getah tervulkan – Vulcanisation process Proses pemvulkanan – Alternative vulcanisation Pemvulkanan alternatif

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 177 © Nilam Publication Sdn. Bhd. Define: (i) Polymer (ii) Monomer (iii) Polymerisation Terangkan: (i) Polimer (ii) Monomer (iii) Pempolimeran (i) Polymer is a long chain molecule made up of a large number of small repeating identical unit of monomers. Polimer ialah molekul berantai panjang yang terbentuk daripada gabungan banyak unit kecil yang sama dipanggil monomer. (ii) Monomer is small identical repeating units in the polymer. Monomer adalah unit kecil yang berulang dalam polimer. (iii) Polymerisation is the process of joining together the large number of monomers to form a polymer. / Pempolimeran ialah proses penggabungan bersama monomer-monomer untuk membentuk polimer. Polymerisation Pempolimeran Polymer Polimer Monomers Monomer Remark: / Catatan: Addition polymerisation reactions have been studied in the topic of Carbon Compounds (Unit 2) on page 108. Tindak balas penambahan pempolimeran telah dipelajari dalam topik Sebatian Karbon (Unit 2) pada muka surat 108. What are the types of polymers based on their source? / Apakah jenis polimer berdasarkan sumber mereka? (i) Natural polymers / Polimer semula jadi (ii) Synthetic polymers / Polimer sintetik What are the natural polymers? Give examples. Apakah polimer semula jadi? Berikan contoh. – Natural polymers are usually found in plants and in animals. Polimer semula jadi biasanya dijumpai di dalam tumbuhan dan haiwan. – Example of naturally occurring polymers and their monomers are: Contoh polimer semula jadi dan monomernya adalah: Natural polymer Polimer semula jadi Monomer Monomer Protein / Protein Amino acid / Asid amino Starch / Kanji Glucose / Glukosa Rubber / Getah Isoprene / Isoprena What are synthetic polymers? Give examples. Apakah itu polimer sintentik? Berikan contoh. – Synthetic polymers are manufactured polymers. The monomers are usually obtained from petroleum after refining and cracking. / Polimer sintetik adalah polimer buatan. Monomer biasanya diperoleh daripada petroleum yang telah mengalami proses penapisan dan peretakan. – Examples of synthetic polymers are plastic, synthetic rubber and synthetic fibers. Contoh polimer sintetik adalah plastik, getah sintetik dan gentian sintetik. – Example of synthetic polymers and their monomers are: Contoh polimer sintetik dan monomernya adalah: Synthetic polymer Polimer sintetik Monomer Monomer Polythene Politena Ethene, C2H4 Etena, C2H4 Polypropene Polipropena Propene, C3H6 Propena, C3H6 Polyvinyl chloride (PVC) Polivinil klorida (PVC) Chloroethene, C2H3Cl Kloroetena, C2H3Cl Polystyrene Polistirena Styrene, C2H3C6H5 Stirena, C2H3C6H5 Styrene-butadiene rubber (SBR) Getah stirena-butadiena Styrene and butadiene Stirena dan butadiena Remark: / Catatan: The word "plastic" comes from the Greek word “plastikos”, which means it can be shaped or moulded. Perkataan “plastik” datang daripada perkataan Greek “plastikos”, yang bermaksud boleh dibentuk. POLYMER POLIMER CS / SK 4.1 4.1 Explain Polymer / Menerangkan tentang Polimer LS / SP 4.1.2

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 © Nilam Publication Sdn. Bhd. 178 Thermoplastics, thermosets or elastomers are three types of polymers with different characteristic. Termoplastik, termoset dan elastomer adalah tiga jenis polimer dengan ciri yang berbeza. Describe the characteristics of these three types of polymers: / Huraikan ciri-ciri bagi ketiga-tiga jenis polimer ini: TP2 Thermoplastics / Termoplastik Thermosets / Termoset Elastomers / Elastomer Linear chains without cross-link Rantai linear tanpa rangkai silang Cross-link Rangkai silang Linear chains with cross-link Rantai linear dengan rangkai silang – A polymer that becomes soft when heated and hardened when cold . This process can be repeated. / Polimer yang menjadi lembut apabila dipanaskan dan keras apabila sejuk . Proses ini boleh berulang. – Can be moulded many times. Therefore, thermoplastic polymers can be recycled. Boleh dibentuk banyak kali. Maka, polimer termoplastik boleh dikitar semula. – Example: Polythene , perspex and polyvinyl chloride (PVC) Contoh: Politena , perspeks dan polivinil klorida (PVC) – A polymer that does not melt when heated. The polymer decomposed instead of melt when heated to high temperature. / Polimer yang tidak mencair apabila dipanaskan. Polimer ini akan terurai , bukan mencair apabila dipanaskan pada suhu yang tinggi. – Cannot be re-moulded. Therefore, thermoset polymers cannot be recycled . Tidak boleh dibentuk. Maka, polimer termoset tidak boleh dikitar semula . – Example: Bakelite, resin epoxy and melamine . / Contoh: Bakelit, epoksi resin dan melamin . – A polymer with high elastic properties. It will return to its original shape after being stretched or compressed. Polimer dengan sifat kenyal yang tinggi. Ia akan kembali kepada bentuk asal selepas diregangkan atau dimampatkan. – Not easily recycled Tidak mudah untuk dikitar semula – Example: Silicone rubber, styrenebutadiene rubber (SBR) and neoprene Contoh: Getah silikone, getah stirena -butadiena dan neoprena Remark: / Catatan: Thermoplastics and thermoset polymers are the two types of plastic. / Polimer termoplastik dan termoset adalah dua jenis plastik. Give examples of consumer goods made from thermoplastics, thermoset and elastomers polymer in manufacture of consumer goods. Berikan contoh barang pengguna yang diperbuat dari termoplastik, termoset dan polimer elastomer dalam pembuatan barang pengguna. Thermoplastics Termoplastik • Pails, plastic bags and bottles are made from polythene Baldi, beg plastik dan botol diperbuat daripada politena • Packaging materials are made from polystyrene Bahan pembungkusan diperbuat daripada polistirena • Electric-wire insulation and pipes are made from polyvinyl chloride(PVC) Penebat wayar elektrik dan paip diperbuat daripada polivinil klorida (PVC) Thermosets Termoset • Electrical plugs, car bumpers and handles of iron or cookware are made from bakelite / Palam elektrik, bumper kereta dan pemegang seterika atau peralatan memasak diperbuat daripada bakelite • Adhesive is made from resin epoxy Pelekat diperbuat daripada epoksi resin • Dinnerware is made from melamine Peralatan pinggan mangkuk diperbuat daripada melamin

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 179 © Nilam Publication Sdn. Bhd. Compare the characteristics of thermoplastic and thermoset. Bandingkan ciri termoplastik dan termoset. Characteristics Ciri-ciri Differences / Perbezaan Thermoplastics /Termoplastik Thermosets / Termoset Flammability / Kebolehbakaran Easy / Mudah Difficult / Susah Resistance to heat / Ketahanan pada haba Low / Rendah High / Tinggi Hardness / Kekerasan Soft / Lembut Hard / Keras Melting point / Takat lebur Low / Rendah High / Tinggi Can be remoulded / Boleh dibentuk Yes / Ya No / Tidak Can be recycled / Boleh dikitar semula Yes / Ya No / Tidak Presence of cross-links / Kehadiran rangkai silang No / Tidak Yes / Ya Similarities / Persamaan Thermoplastic and thermoset polymers are good heat and electrical insulator, not easily oxidised , waterproof and oil proof. Polimer termoplastik dan polimer termoset adalah penebat haba dan elektrik yang baik, tidak mudah dioksidakan , kalis air dan kalis minyak. POLYMERISATION / PEMPOLIMERAN (i) Addition Polymerisation Reaction / Tindak Balas Pempolimeran Penambahan What is addition polymerisation? Apakah pempolimeran penambahan? Addition polymerisation is a reaction that occurs when many monomers which have double covalent bonds between carbon, C = C join together to form long chain molecules called polymer. Pempolimeran penambahan adalah tindak balas yang wujud apabila banyak monomer yang mempunyai ikatan kovalen ganda dua antara karbon, C = C bergabung bersama untuk membentuk rantai panjang yang dipanggil polimer. Example: Complete the polymerisation equation of ethene, propene and chloroethene. Contoh: Lengkapkan persamaan pempolimeran bagi etena, propena dan kloroetena. (a) Polymerisation of ethene / Pempolimeran etena: Ethene / Etena Polythene / Politena H H n C = C H H H H – C – C – H H where n is large number up to a few thousands n di mana n ialah bilangan yang sangat besar sehingga beberapa ribu Elastomers Elastomer • Medical devices are made from silicone rubber Peralatan perubatan diperbuat daripada getah silikone • Tyres are made from styrene-butadiene rubber(SBR) Tayar diperbuat daripada getah stirena-butadiena • Boots, small boats and wetsuits are made from neoprene Kasut, perahu kecil dan pakaian selam diperbuat daripada neoprena

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 © Nilam Publication Sdn. Bhd. 180 Remark: / Catatan: Refer the exercise below to construct an addition polymerisation equation (alkene addition polymerisation reaction). Rujuk latihan di bawah untuk membina persamaan pempolimeran penambahan (tindak balas pempolimeran penambahan alkena). (b) Polymerisation of propene / Pempolimeran propena: Propene / Propena Polypropene / Polipropena H CH3 n C = C H H H CH3 – C – C – H H n (c) Polymerisation of chloroethene / Pempolimeran kloroetena: Polyvinyl chloride Polivinil klorida Chloroethene Kloroetena H H n C = C H CI H H – C – C – H CI n Complete the following table for polymers produced by addition polymerisation. Lengkapkan jadual berikut bagi polimer yang dihasilkan oleh pempolimeran penambahan. TP1 TP2 TP3 Polymerisation equation / Persamaan pempolimeran Properties & uses / Sifat & kegunaan Polymer: Polyethene / Polimer: Polietena C C H H H H n C C H H H H n Ethene / Etena Polyethene / Polietena • Durable, light and easily melted and moulded. Tahan lama, ringan dan mudah dicairkan dan dibentuk. • Plastic bags, shopping bags, plastic containers and plastic toys. Beg plastik, beg membeli belah, bekas plastik dan permainan plastik. Polymer: Polypropene / Polimer: Polipropena C C CH3 H H H n C C CH3 H H H n Propene / Propena Polypropene / Polipropena • Strong, durable and easily melted and moulded. Kuat, tahan lama dan mudah dicairkan dan dibentuk. • Plastic bottles, plastic tables and chairs, car batteries casing and ropes Botol plastik, meja dan kerusi plastik, bekas bateri kereta dan tali. Polymer: Polyvinyl chloride (PVC) / Polimer: Polivinil klorida (PVC) C C Cl H H H n C C Cl H H H n Chloroethene Kloroetena Polyvinyl chloride (PVC) Polivinil klorida (PVC) • Strong, lightweight, durable, flame retardant and excellent electrical insulation properties. Kuat, ringan, tahan lama, tahan api dan mempunyai sifat penebat elektrik yang sangat baik. • Pipes, raincoats, bags, footwear, artificial leather and cable casing. Paip, baju hujan, beg, kasut, kulit tiruan dan sarung kabel. Polymer: Polystyrene / Polimer: Polistirena C C H H H n C C H H H n Styrene / Stirena Polystyrene / Polistirena • Heat insulator, light and can be moulded. Penebat haba, ringan dan mudah dibentuk. • Packaging materials, disposable cups and plates. Bahan pembungkusan, cawan dan pinggan pakai buang. Exercise / Latihan TAHAP PENGUASAAN (TP) Menguasai Belum menguasai TP1 Mengingat kembali pengetahuan dan kemahiran asas mengenai polimer. TP2 Memahami polimer serta dapat menjelaskan kefahaman tersebut.

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 181 © Nilam Publication Sdn. Bhd. Polymer: Polytetrafluoroethene (PPTFE or Teflon) Polimer: Politetrafluoroetena (PPTFE atau Teflon) C C F F F F n C C F F F F n Tetrafluoroethene Tetrafluoroetena Polytetrafluoroethene (PPTFE or Teflon) Politetrafluoroetena (PPTFE atau Teflon) • Durable, non-stick, chemically inert and strong. Tahan lasak, tidak melekat, lengai secara kimia dan kuat. • Non-stick coating for pans and other cookware, electrical insulators. Lapisan tidak melekat untuk kuali dan peralatan memasak lain, penebat elektrik. (ii) Condensation Polymerisation Reaction / Tindak Balas Pempolimeran Kondensasi What is condensation polymerisation? Apakah pempolimeran kondensasi? It is a process in which monomers join together to form large molecule with the elimination of small molecules such as water and hydrogen chloride. Ia adalah proses di mana monomer-monomer bergabung bersama untuk membentuk molekul yang besar dengan penyingkiran molekul-molekul kecil seperti air dan hidrogen klorida. Remark: / Catatan: Condensation polymerisation involves at least two different types of monomers. The polymerisation reaction occurs at the functional group of these monomers. Pempolimeran kondensasi melibatkan sekurang-kurangnya dua jenis monomer yang berbeza. Tindak balas pempolimeran ini berlaku pada kumpulan berfungsi monomer ini. What are the two polymers made from this reaction? Apakah dua polimer yang dihasilkan daripada tindak balas ini? (i) Terylene / Terilena (ii) Nylon / Nilon Remark: / Catatan: Terylene and nylon are synthetic fiber. Terilena dan nilon adalah gentian sintetik. Terylene / Terilena What is another name of terylene? Apakah nama lain bagi terilena? Polyester / Poliester Remark: / Catatan: There are ester linkages in terylene. Terdapat rangkaian ester di dalam terilena. How is terylene produced? / Bagaimanakah terilena dihasilkan? – The monomer used to make polyester are diol molecules and diacid molecules. – A diol molecule has two -OH groups and a diacid molecules has two -COOH groups. – When many diacid molecules condense with diol molecules, a polyester and water are formed. – Terylene is manufactured from ethane-1, 2-diol and benzoic-1, 4-dicarboxylic acid. – Monomer yang digunakan untuk membuat poliester adalah molekul-molekul diol dan diasid. – Molekul diol mempunyai kumpulan –OH dan molekul-molekul diasid mempunyai dua kumpulan –COOH. – Apabila banyak molekul-molekul diasid terkondensasi dengan molekul-molekul diol, poliester dan air dihasilkan. – Terilena diperbuat daripada etana-1, 2-diol dan asid benzena-1, 4-dikarboksilik. H O C C O H H O (CH2)2 O H O O n Benzene-1, 4-dicarboxylic acid Asid benzena-1, 4-dikarboksilik Ethane-1, 2-diol Etana-1, 2-diol C C O (CH2)2 O nH2O O O Terylene Terilena n Remark: / Catatan: Benzene is an aromatic molecule. The molecular formula for benzene is C6H6. The diagram on the right shows three ways of representing benzene. / Benzena adalah molekul aromatik. Formula molekul untuk benzena ialah C6H6. Rajah di sebelah menunjukkan tiga cara untuk mewakili benzena. What are the uses of terylene? Apakah kegunaan terilena? Terylene is suitable for making textile, socks, parachutes and fishing nets because it is elastic, chemically inert, can be coloured and easily made into fiber. Terilena sesuai untuk membuat tekstil, stoking, payung terjun dan jaring ikan kerana kenyal, lengai secara kimia, dapat diwarnai dan mudah dijadikan gentian. C C C H H H H H H C C C (1) (2) (3)

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 © Nilam Publication Sdn. Bhd. 182 Nylon / Nilon How is nylon produced? Bagaimanakah nilon dihasilkan? LS / SP 4.1.2 Nylon is a general term given to syntenic polymer made from two types of monomers: Nilon adalah istilah umum diberikan kepada polimer sintetik diperbuat daripada dua jenis monomer: (i) Amines (–NH2) and carboxylic acids (–COOH) produce nylon and water. Amina (–NH2) dan asid karboksilik (–COOH) menghasilkan nilon dan air. (ii) Amines (–NH2) and acyl chlorides (–COCl) produce nylon and hydrogen chloride. Amina (–NH2) dan klorida asil (–COCI) menghasilkan nilon dan hidrogen klorida. Example: / Contoh: (i) Condensation between 1, 6-hexanediamine molecules and 1, 6-hexanedioic acid to form nylon-6, 6. Kondensasi antara molekul 1, 6-heksanadiamina dan asid 1, 6-heksanadioik untuk membentuk nilon-6, 6. H O C (CH2)4 C O H H N (CH2)6 N H O O H H n 1, 6-hexanedioic acid Asid 1, 6-heksanadioik 1, 6-hexanediamine 1, 6-heksanadiamina C (CH2)4 C N (CH2)6 N nH2O O O H H n Nylon-6, 6 Nilon-6, 6 (ii) Condensation between 1, 6-hexanediamine molecules and decanedioyl dichloride to form nylon-6,10. Kondensasi antara molekul 1, 6-heksanadiamina dan dekanadiol diklorida untuk membentuk nilon-6,10. C (CH2)8 C N (CH2)6 N nHCl O O H H n Nylon-6, 10 Nilon-6, 10 CI C (CH2)8 C CI H N (CH2)6 NH O O H H n Decanedioyl dichloride Dekanadiol diklorida 1, 6-hexanediamine 1, 6-heksanadiamina How to produce nylon in the laboratory? Bagaimanakah untuk menghasilkan nilon di dalam makmal? Nylon is prepared in the laboratory through the reaction between amines with acyl chlorides instead of acid because acyl chloride is more reactive than carboxylic acid. Nilon disediakan di dalam makmal melalui tindak balas antara amina dan klorida asil, bukannya asid kerana klorida asil lebih reaktif daripada asid karboksilik. State the properties of nylon. Nyatakan sifat-sifat nilon. Nylon is a strong and very light in weight, tough and waterproof polymer and can be easily made into fibers, and are easy to wash and to dye in a wide range of colours. Nilon adalah polimer yang kuat dan ringan, tahan lasak dan kalis air serta mudah dijadikan gentian, senang dicuci dan diwarnakan dalam pelbagai warna. What is the uses of nylon? Apakah kegunaan nilon? Nylon is used to make toothbrushes, ropes, fishing lines, parachutes, carpets, textile, threads and electric insulators. Nilon digunakan untuk membuat berus gigi, tali, tali pancing, payung terjun, karpet, tekstil, benang dan penebat elektrik. Procedure of producing polymer Prosedur penghasilan polimer

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 183 © Nilam Publication Sdn. Bhd. Justify The Use Of Polymers to Environment. / Wajarkan Kegunaan Polimer kepada Alam Sekitar. LS / SP 4.1.3 State the advantages of synthetic polymers. Nyatakan kelebihan-kelebihan sintetik polimer. (a) Light and strong / Ringan dan kuat (b) Cheap / Murah (c) Easily shaped and coloured / Mudah dibentuk dan diwarna (d) Inert and resistant to chemicals / Lengai dan tahan kepada bahan kimia (e) Very stable and do not corrode / Sangat stabil dan tidak terkakis Explain the environmental pollution from synthetic polymers. Terangkan pencemaran persekitaran dari polimer sintetik. TP2 (a) Most of synthetic polymers are non-biodegradable (cannot be decomposed by microorganisms). Disposal of synthetic polymers such as plastic bottles and containers cause blockage of drainage systems and river thus causing flash floods . Sebilangan besar polimer sintetik tidak terbiodegradasi (tidak dapat diuraikan oleh mikroorganisma). Pembuangan polimer sintetik seperti botol plastik dan bekas menyebabkan sistem saliran dan sungai tersumbat sehingga menyebabkan banjir kilat . (b) Open burning of polymers will release acidic and poisonous gas that will cause air pollution: Pembakaran terbuka polimer akan membebaskan gas berasid dan beracun yang akan menyebabkan pencemaran udara: – Burning most of synthetic polymers will produce: Pembakaran kebanyakan sintetik polimer akan menghasilkan: (i) Carbon dioxide gas which cause greenhouse effect Gas karbon dioksida yang menyebabkan kesan rumah hijau (ii) Carbon monoxide which is poisonous Karbon monoksida yang beracun – Burning of PVC will release hydrogen chloride gas which will cause acid rain Pembakaran PVC akan membebaskan gas hidrogen klorida yang menyebabkan hujan asid – Burning of synthetic polymers contains carbon and nitrogen such as nylon will produce highly poisonous gas such as hydrogen cyanide Pembakaran polimer sintetik yang mengandungi karbon dan nitrogen seperti nilon akan menghasilkan gas yang sangat beracun seperti hidrogen sianida (c) Plastic containers that are left in open area will collect rain water and become breeding ground for mosquito which will cause diseases such dengue fever. Bekas plastik yang tertinggal di tempat terbuka boleh menyebabkan air hujan bertakung dan boleh menyebabkan pembiakan nyamuk yang membawa penyakit demam denggi. (d) Plastics materials that are carried to rivers and into sea may be mistaken by marine animals as food. Plastics are not digestible. Eating them will cause death to the marine animals. Bahan plastik yang dibawa ke sungai dan ke laut boleh disalah anggap oleh hidupan laut sebagai makanan. Plastik tidak boleh dicerna. Memakannya akan menyebabkan kematian hidupan laut. Suggest ways to reduce the pollution of synthetic polymers. Cadangkan cara untuk mengurangkan pencemaran polimer sintetik. TP3 (a) Reduce, recycle and reuse the synthetic polymers. Kurang, kitar semula dan gunakan semula polimer sintetik. (b) Using biodegradable plastics. Menggunakan plastik terbiodegradasi. (c) On-going research to produce cheap biodegradable polymers. Penyelidikan berterusan untuk menghasilkan polimer biodegradasi yang murah. (d) Disintegrate plastics by pyrolysis : Plastic can be disintegrated by heating at the temperature between 400°C – 800°C without oxygen. Hancurkan plastik dengan pirolisis : Plastik boleh hancur dengan pemanasan pada suhu antara 400°C – 800°C tanpa oksigen.

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 © Nilam Publication Sdn. Bhd. 184 1 The diagram below shows the examples of synthetic polymer products used in daily lives. Complete the table below. Rajah di bawah menunjukkan contoh produk polimer sintetik yang digunakan dalam kehidupan seharian. Lengkapkan jadual di bawah. Produk Product Plug Palam Polythene bag Beg politena Cooking utensil Peralatan memasak Polymer Polimer Bakelit Bakelit Polythene Politena Silicon rubber Getah silikone Type of polymer Jenis polimer Thermoset Termoset Thermoplastic Termoplastik Elastomer Elastomer State two characteristics Nyatakan dua ciri-ciri polimer – Does not melt and decomposed when heated to high temperature. / Tidak mencair dan terurai apabila dipanaskan pada suhu yang tinggi. – Cannot be recycled. Tidak boleh dikitar semula. – Becomes soft when heated and hardened when cold. Menjadi lembut apabila dipanaskan dan keras apabila disejukkan. – Can be recycled. Boleh dikitar semula. – High elastic properties. Sifat kenyal yang tinggi. – Heat resistant. Tahan haba. (a) State two effects of improper disposal of the polymer to the environment. Nyatakan dua kesan pelupusan secara tidak teratur bagi polimer itu terhadap alam sekitar. – Non-biodegradable, cause blockage of drainage systems and river thus causing flash flood. Tidak terbiodegredasi, boleh menyebabkan longkang tersumbat dan banjir kilat. – Burning of polymer will release acidic gas which will cause acid rain. Pembakaran polimer membebaskan gas berasid yang menyebabkan hujan asid. (b) State two proper ways to dispose polymer. Nyatakan dua cara yang betul untuk melupuskan polimer itu. – Recycle / Mengitar semula – Disintegrate plastics by pyrolysis / Diuraikan secara pirolisis 2 The diagram shows Dacron, or polyethylene terephthalate (PET), a synthetic polymer which is used for making food container. / Rajah menunjukkan Dacron, atau polietilena tereftalat (PET), satu polimer sintetik yang digunakan untuk membuat bekas makanan. It is made by polymerisation of a mixture of the monomers terephthalic acid, C6H4(COOH)2 and ethylene glycol, (CH2OH)2. Ia dibuat dengan pempolimeran campuran monomer asid tereftalat, C6H4(COOH)2 dengan etilena glikol, (CH2OH)2. O O HO OH C C H O C C O H H H H H Terephthalic acid Asid tereftalat Ethylene glycol Etilena glikol TAHAP PENGUASAAN (TP) Menguasai Belum menguasai TP3 Mengaplikasikan pengetahuan mengenai polimer untuk menerangkan kejadian atau fenomena alam dan dapat melaksanakan tugasan mudah. TP4 Menganalisis pengetahuan mengenai polimer dalam konteks penyelesaian masalah mengenai kejadian atau fenomena alam. Exercise / Latihan TP3 TP3

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 185 © Nilam Publication Sdn. Bhd. (a) What type of polymerisation does the two monomers undergo? Give reason. Apakah jenis pempolimeran yang dialami oleh dua monomer ini? Berikan sebab. Condensation polymerisation. The reaction eliminates a small molecule from the monomers to form polymer. Pempolimeran kondensasi. Tindak balas ini menyingkirkan molekul kecil dari monomer untuk membentuk polimer. (b) Draw the structural formula of the polymer that is formed by these two monomers. Lukis formula struktur bagi polimer yang terbentuk dari monomer-monomer tersebut. C C O C C O O O H H H H (c) Suggest another name of this polymer. Give a reason for your answer. Cadangkan nama lain bagi polimer ini. Berikan sebab bagi jawapan anda. Polyester. It contains ester linkage. Poliester. Ia mengandungi rangkaian ester. (d) Dacron is non-biodegradable. / Dacron tidak terbiodegradasi. (i) What is meant by non-biodegradable? Apakah yang dimaksudkan dengan tidak terbiodegradasi? A substance which cannot be broken down by microorganisms such as bacteria. Bahan yang tidak boleh diuraikan oleh mikroorganisma seperti bakteria. (ii) Explain why the properties of non-biodegradable of Dacron have advantage and disadvantage when using it as food container. Terangkan mengapa sifat tidak terbiodegradasi Dacron mempunyai kelebihan dan kekurangan apabila digunapakai sebagai bekas makanan. The container can store food for a very long time but the waste can cause land pollution when landfill sites are full. Bekas makanan boleh menyimpan makanan dalam masa yang lama tetapi bahan buangannya menyebabkan pencemaran tanah kerana tapak pelupusan menjadi penuh. (e) Another polymer is also made from the same reaction as Dacron. The two monomers are shown below: Suatu polimer lain juga dibuat dengan cara yang sama seperti Dacron. Dua monomernya adalah seperti di bawah: HO OH CI C CI O Hydroquinone Hydroquinone Phosgene Fosgen (i) Draw the structural formula to represent the polymer produced from these two monomers. Lukis formula struktur untuk mewakili polimer yang dihasilkan daripada dua monomer ini. O O C O (ii) Name a product from this reaction. Namakan hasil daripada tindak balas ini. Hydrogen chloride / Hidrogen klorida TP2 TP2 TP2 TP2 TP4

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 © Nilam Publication Sdn. Bhd. 186 Explain the structure of natural rubber. Terangkan struktur getah asli. LS / SP 4.2.1 (a) Rubber is natural polymer. It is formed from the monomer isoprene . Getah ialah polimer asli. Ia terbentuk daripada monomer isoprena . (b) Molecular formula of isoprene is C5H8. / Formula molekul isoprena ialah C5H8. (c) Isoprene molecules are joined together by addition polymerisation process to form the polymer of natural rubber, polyisoprene: / Molekul isoprena terikat bersama oleh proses pempolimeran penambahan untuk membentuk polimer getah asli, poliisoprena: Isoprene (2-methylbut-1, 3-diene) Isoprena (2-metilbut-1, 3-diena) Polyisoprene Poliisoprena n is large numbers n ialah nombor yang besar ( C C C C ) n H H H H H n ( C C C C ) H H H H H CH3 CH3 What is latex? Apakah lateks? Latex is milk like liquid obtained from tapped rubber tree. Latex is a colloid which contains suspension of rubber particles in water. Lateks ialah cecair seperti susu yang diperoleh daripada pokok getah yang ditoreh. Lateks ialah koloid yang mengandungi zarah-zarah getah yang tersebar dalam air. Explain why latex exist as liquid. Terangkan mengapa lateks wujud sebagai cecair. – The rubber particles are made up of long chain rubber polymers, (C5H8)n surrounded by a protein membrane . / Zarah-zarah getah terdiri daripada polimer getah berantai panjang, (C5H8)n yang dikelilingi membran protein . – The protein membrane is negatively charged. Membran protein adalah bercas negatif . – The forces of repulsion between negatively charged particles prevent them from combining or coagulating. / Daya tolakan di antara zarah-zarah bercas negatif menghalang zarah-zarah tersebut daripada bergabung atau bergumpal. Negative charge Bercas negatif Rubber polymer Polimer getah Water Air Repulsion Tolakan What is coagulation of latex? Apakah penggumpalan lateks? Latex in liquid state change to semi-solid. Lateks dalam keadaan pepejal bertukar menjadi separa pepejal. What are the causes for coagulation of latex? Apakah sebab penggumpalan lateks? Latex coagulates when: / Lateks tergumpal apabila: (i) Acid is added to it such as methanoic acid (formic acid), ethanoic acid (acetic acid) or any other weak acids. / Asid seperti asid metanoik (asid formik), asid etanoik (asid asetik) atau asid lemah lain ditambah ke dalamnya. (ii) Left aside for 1 – 2 days. / Dibiarkan selama 1 – 2 hari. Explain how latex coagulates when acid is added into it. Terangkan bagaimana lateks menggumpal apabila asid ditambah kepadanya. (i) Positively charged hydrogen ions from the acid neutralises the negative charges on the surface of the protein membrane. A neutral rubber particle is formed. / Ion hidrogen bercas positif daripada asid meneutralkan cas-cas negatif pada permukaan membran protein. Zarah getah yang neutral terbentuk. H+ H+ H+ H+ H+ H+ H+ H+ H+ H+ H H+ + H+ H+ H+ H+ Collide Perlanggaran Breaking of protein membrane Membran protein pecah Neutralised rubber particle Zarah getah yang neutral H+ H+ H+ H+ H+ H+ H+ H+ H+ H+ H H+ + H+ H+ H+ H+ Repulsion Tolakan NATURAL RUBBER GETAH ASLI CS / SK 4.2 4.2

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 187 © Nilam Publication Sdn. Bhd. (ii) The neutral particles no longer repel each other. These neutral particles collide with each other, causing the membrane to break . Zarah-zarah neutral tidak lagi menolak di antara satu sama lain. Zarah-zarah neutral ini berlanggar di antara satu sama lain, menyebabkan membran pecah . (iii) The rubber polymers are freed and they coagulate by combining together to form large lump of rubber polymer. The latex has coagulated. Polimer getah terbebas dan bergumpal dengan bergabung untuk membentuk gumpalan getah yang besar. Lateks telah tergumpal. Breaking of protein membrane Membran protein pecah Rubber polymers coagulate Polimer getah menggumpal Explain how latex coagulates when latex is left aside for 1 to 2 days. Explain. Jelaskan bagaimana penggumpalan lateks apabila dibiarkan 1 hingga 2 hari. Terangkan. (i) Bacteria from the air enter latex. (ii) Activity of bacteria in the latex produce lactic acid that contains hydrogen ions (H+) which causes coagulation of latex. Coagulated latex is semi-solid. (iii) Positively charged hydrogen ions from the acid neutralises the negative charges on the surface of the protein membrane. A neutral rubber particle is formed. (iv) The neutral particles no longer repel each other. These neutral particles collide with each other, causing the membrane to break . (v) The rubber polymers are freed and they coagulate by combining together to form large lump of rubber polymer. The latex has coagulated. (i) Bakteria dari udara masuk ke dalam lateks. (ii) Aktiviti bakteria di dalam lateks menghasilkan asid laktik yang mengandungi ion hidrogen (H+) yang menyebabkan penggumpalan lateks. Lateks tergumpal adalah separa pepejal. (iii) Ion hidrogen bercas positif daripada asid meneutralkan cas-cas negatif pada permukaan membran protein. Zarah getah yang neutral terbentuk. (iv) Zarah-zarah neutral tidak lagi menolak di antara satu sama lain. Zarah-zarah neutral ini berlanggar di antara satu sama lain, menyebabkan membran pecah . (v) Polimer getah terbebas dan bergumpal dengan bergabung untuk membentuk gumpalan getah yang besar. Lateks telah tergumpal. How to prevent coagulation of latex? Bagaimanakah menghalang penggumpalan lateks? Coagulation of latex can be prevented by adding ammonia (alkali) to it. Penggumpalan lateks boleh dihalang dengan menambah ammonia (alkali) kepadanya. Explain how the presence of an alkali can prevent the coagulation process of latex. Terangkan bagaimana kehadiran alkali boleh menghalang proses penggumpalan lateks. (a) The ammonia solution (containing OH– ions) will neutralise any acids that may be produced by the bacteria. (b) Hydroxide ions, OH– from alkali neutralise hydrogen ions, H+ produced by acid as a result of bacterial attack on protein. (c) The protein membrane remains negatively charge because there is no hydrogen ions. (d) The rubber particles repel each other. (e) The rubber polymers cannot combine and coagulate . (a) Larutan ammonia (mengandungi ion OH– ) akan meneutralkan sebarang asid yang mungkin dihasilkan oleh bakteria. (b) Ion hidroksida, OH– daripada alkali meneutralkan ion hidrogen, H+ yang dihasilkan oleh asid disebabkan serangan bakteria ke atas protein. (c) Membran protein kekal bercas negatif kerana tiada ion-ion hidrogen. (d) Zarah-zarah getah menolak di antara satu sama lain. (e) Polimer-polimer getah tidak boleh bergabung dan menggumpal .

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 © Nilam Publication Sdn. Bhd. 188 Laboratory activity to investigate the coagulation of latex. / Aktiviti makmal untuk menyiasat penggumpalan lateks. LS / SP 4.2.2 Aim: / Tujuan: To investigate the coagulation of latex. / Untuk menyiasat penggumpalan lateks. Materials: / Bahan: Latex, 2 mol dm–3 ethanoic acid, 2 mol dm–3 ammonia solution, red and blue litmus paper. Lateks, 2 mol dm–3 asid etanoik, 2 mol dm–3 larutan ammonia, kertas litmus merah dan biru. Apparatus: / Radas: Beakers, dropper, glass rod. / Bikar, penitis, rod kaca. Procedure: / Prosedur: (i) Measure and pour 20 cm3 of latex into each of the three beakers labelled A, B and C. (ii) Add ethanoic acid drop by drop into beaker A until the latex becomes acidic (blue litmus paper turns red). Stir the mixture after each addition of acid. (iii) Add ammonia solution drop by drop into beaker B until latex becomes alkaline (red litmus paper turns blue). Stir the mixture after each addition of alkali. (iv) Leave the three beakers overnight. (v) Observe and record any changes that occur. (i) Ukur dan tuang 20 cm3 lateks ke dalam tiga bikar yang berlabel A, B dan C. (ii) Tambah asid etanoik titis demi titis ke dalam bikar A sehingga lateks menjadi berasid (kertas litmus biru bertukar merah). Kacau campuran bagi setiap penambahan asid. (iii) Tambah larutan ammonia titis demi titis ke dalam bikar B sehingga lateks menjadi beralkali (kertas litmus merah bertukar biru). Kacau campuran bagi setiap penambahan alkali. (iv) Biarkan ketiga-tiga bikar semalaman. (v) Perhatikan dan rekod apa-apa perubahan yang berlaku. Observations: / Pemerhatian: Beaker / Bikar Observation / Pemerhatian A The latex coagulates very quickly / Lateks menggumpal dengan cepat B The latex does not coagulate / Lateks tidak menggumpal C The latex coagulates slowly / Lateks menggumpal dengan perlahan Properties and Uses of Natural Rubber. / Ciri-ciri dan Kegunaan Getah Asli. Property / Sifat Description / Penerangan Uses / Kegunaan Elasticity Kekenyalan When it is stretched , it straighten out. It returns back to its original shape once the stretching force is released. Apabila diregangkan , ia menjadi lurus. Ia kembali kepada bentuk asal apabila daya regangan dilepaskan. Rubber tube, gloves, rubber bands and shoe soles. Tiub getah, sarung tangan, getah pengikat dan tapak kasut. Resistance to oxidation Ketahanan terhadap pengoksidaan The natural rubber polymers are easily oxidised due to the presence of double bonds. / Polimer getah asli teroksida dengan mudah kerana kehadiran ikatan ganda dua. These properties make the usage of natural rubber limited. Sifat-sifat ini menjadikan kegunaan getah asli terhad. Effect of heat Kesan haba When it is heated, it soften and becomes sticky . Apabila dipanaskan, getah menjadi lembut dan menjadi melekit . When it is cooled, it becomes hard and brittle . Apabila disejukkan, ia menjadi keras dan rapuh . Effect of solvent Kesan pelarut Natural rubber is soluble in organic, alkaline and acidic solution. Getah asli larut dalam larutan organik, beralkali dan berasid. Remark: / Catatan: (a) Natural rubber is elastic (it will return to its original shape after stretching force is released). Getah asli adalah kenyal (ia akan kembali kepada bentuk asal apabila daya regangan dilepaskan). (b) When the rubber is over stretched, the rubber molecules do not return to their original positions. The rubber has lost its elasticity. Apabila getah diregang secara berlebihan, molekul getah tidak kembali kepada kedudukan asal. Getah telah hilang sifat kekenyalannya.

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 189 © Nilam Publication Sdn. Bhd. What is vulcanisation of rubber? Apakah pemvulkanan getah? Vulcanisation is a process whereby sulphur atoms form cross-link between adjacent chains of rubber polymer at the double bond between carbon atoms. Pemvulkanan adalah proses di mana atom sulfur getah membentuk rangkaian silang antara rantai polimer getah bersebelahan pada ikatan berganda antara atom karbon. C C C C C C C C C C C C C C C C H H H H H H H H H S S H CH3 CH3 H H H H H S S H H H CH3 CH3 H H Natural rubber Getah asli Vulcanised rubber Getah tervulkan Sulphur cross-link Rangkai silang sulfur How vulcanised rubber is prepared? Bagaimana getah tervulkan disediakan? (i) Sulphur is heated together with natural rubber. Sulfur dipanaskan bersama dengan getah asli. (ii) Rubber stripe is soaked in sulphur monochloride solution in methylbenzene and then dried. Jalur getah direndam dengan larutan sulfur monoklorida dalam metilbenzena dan kemudiannya dikeringkan. Explain how the presence of sulphur atoms changes the properties of vulcanised rubber. Jelaskan bagaimana kehadiran atom sulfur mengubah sifat-sifat getah tervulkan. (i) The sulphur atoms form cross-link between the long rubber molecule. Atom-atom sulfur membentuk rangkai silang di antara molekul panjang getah. (ii) This reduces the ability of the polymers to slide over each other. Ini mengurangkan kebolehan polimer untuk menggelongsor di antara satu sama lain. (iii) The rubber molecules return to their original positions after being stretched. Molekul-molekul getah kembali kepada kedudukan asal selepas diregangkan. Vulcanisation Pemvulkanan Natural rubber / Getah asli Vulcanised rubber / Getah tervulkan Describe laboratory activity to produce vulcanised rubber. Huraikan aktiviti makmal untuk menghasilkan getah tervulkan. LS / SP 4.2.3 Aim: / Tujuan: To prepare vulcanised rubber. / Untuk menyediakan getah tervulkan Materials: / Bahan: Latex, solution of disulphur dichloride in methylbenzene. Lateks, larutan disulfur diklorida dalam metilbenzena. Apparatus: / Radas: White tile, blade, glass rod, beaker. Jubin putih, pisau, rod kaca, bikar. Procedure: / Prosedur: Preparation of vulcanised rubber / Penyediaan getah tervulkan (i) Latex is poured carefully onto a white tile. (ii) A glass rod is used to flatten the latex to form a layer of about 1 to 2 mm thick. (iii) The white tile is left aside for about 1 to 2 days for the latex to coagulate. (iv) The rubber sheet formed is then cut into two strips of equal size. (v) One strip of the rubber sheet is dipped into a solution of disulphur dichloride for about 1 minute. (vi) The vulcanised rubber strip is then removed from the solution and dried in air. (i) Lateks dituang perlahan-lahan ke atas jubin putih. (ii) Rod kaca digunakan untuk meratakan lateks untuk membentuk lapisan setebal sekitar 1 hingga 2 mm. (iii) Jubin putih dibiarkan untuk 1 hingga 2 hari bagi lateks menggumpal. (iv) Kepingan getah yang terbentuk kemudian dipotong menjadi dua jalur dengan ukuran yang sama. (v) Satu jalur kepingan getah dicelup ke dalam larutan disulfur diklorida kira-kira 1 minit. (vi) Jalur getah tervulkan kemudian dikeluarkan dari larutan dan dikeringkan di udara.

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 © Nilam Publication Sdn. Bhd. 190 Experiment to compare elasticity of natural rubber and vulcanised rubber. Eksperimen untuk membandingkan kekenyalan getah asli dan getah tervulkan. LS / SP 4.2.3 Aim: / Tujuan: To compare the elasticity of natural rubber and vulcanised rubber. Untuk membandingkan kekenyalan getah asli dan getah tervulkan. Problem statement: / Pernyataan masalah: Is vulcanised rubber more elastic than unvulcanised rubber (natural rubber)? Adakah getah tervulkan lebih kenyal dari getah tak tervulkan (getah asli)? Hypothesis: / Hipotesis: Vulcanised rubber is more elastic than unvulcanised rubber. / Getah tervulkan lebih kenyal daripada getah tak tervulkan. Variables: / Pemboleh ubah: (a) Manipulated: Vulcanised rubber and unvulcanised rubber. (b) Responding: Length of rubber strip after weight is removed (c) Constant: Size of rubber strip, mass of weight (a) Dimanipulasikan: Getah tervulkan dan getah tak tervulkan (b) Bergerak balas: Panjang jalur getah selepas beban dikeluarkan (c) Dimalarkan: Saiz jalur getah, jisim pemberat Operational definition of elastic rubber: / Definisi secara operasi bagi getah kenyal: Rubber that extends slightly when suspended weight and return to its original size when the weight is released. Getah yang memanjang sedikit ketika digantung dengan beban dan kembali ke ukuran asalnya ketika berat dilepaskan. Materials: / Bahan: Vulcanised rubber and unvulcanised rubber strips / Jalur getah tervulkan dan getah tak tervulkan Apparatus: / Radas: Bulldog clips, metre ruler, weights, hooks, retort stand and clamp. Klip bulldog, pembaris meter, pemberat, cangkuk, kaki retort dan pengapit. Procedure: / Prosedur: Vulcanised rubber strip Jalur getah tervulkan Unvulcanised rubber strip Jalur getah tak tervulkan Weight Pemberat Weight Pemberat (i) Measure and record the initial length of vulcanised and unvulcanised rubber strips. (ii) Hang the vulcanised and unvulcanised rubber strips using bulldog clips. (iii) Hang weight of the same mass to each of the two strips. Measure the stretched lengths of the two strips. (iv) Remove the weight, measure and record the lengths of the two strips again. (i) Ukur dan rekod panjang awal jalur getah tervulkan dan tak tervulkan. (ii) Gantung getah tervulkan dan tak tervulkan dengan menggunakan klip bulldog. (iii) Gantungkan pemberat yang sama jisim pada setiap dua jalur tesebut. Ukur panjang kedua-dua jalur yang diregangkan. (iv) Alihkan beban, ukur dan rekod kembali panjang kedua-dua jalur. Results: / Keputusan: Types of rubber Jenis getah Initial length Panjang awal (cm) Stretched length Panjang yang diregangkan (cm) Length after removal of weight / Panjang selepas pemberat dialihkan (cm) Difference in length Perbezaan panjang (cm) Unvulcanised rubber Getah tak tervulkan Vulcanised rubber Getah tervulkan

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 191 © Nilam Publication Sdn. Bhd. What is the effect of stretching on natural rubber and vulcanised rubber? Apakah kesan meregangkan getah asli dan getah tervulkan? Natural Rubber / Getah Asli Before stretching Sebelum meregang Stretching Semasa meregang After stretching Selepas meregang The rubber polymer entangle massively into lump. / Polimer getah berselirat menjadi gumpalan. Rubber molecule slide over each other when natural rubber is stretch. / Molekul getah menggelongsor di antara satu sama lain apabila getah asli diregang. Rubber molecule do not return to its original position when over-stretched. The natural rubber has lost its elasticity. Molekul getah tidak kembali ke kedudukan asalnya apabila diregangkan secara berlebihan. Getah asli telah kehilangan kekenyalannya. Vulcanised Rubber / Getah Tervulkan Before stretching Sebelum meregang Stretching Semasa meregang After stretching Selepas meregang Represent sulphur cross-links in vulcanised rubber. Mewakili rangkai silang sulfur dalam getah tervulkan. Rubber molecule slide over each other when vulcanised rubber is stretched. / Molekul getah menggelongsor di antara satu sama lain apabila getah tervulkan diregangkan. Sulphur cross-link pull the rubber molecules back to original position. Rangkai silang sulfur menarik molekul getah kembali ke kedudukan asal. Comparison of Properties between Unvulcanised Rubber and Vulcanised Rubber. Perbandingan Sifat antara Getah tak Tervulkan dan Getah Tervulkan. Properties Sifat Unvulcanised Rubber Getah Tak Tervulkan Vulcanised Rubber Getah Tervulkan Elasticity Kekenyalan Less elastic because the polymer chain of rubber can slide over one another easily. Kurang kenyal kerana rantai polimer getah dapat menggelongsor antara satu sama lain dengan mudah. More elastic because the sulphur cross-links prevents the polymer chain of rubber molecules slide over one another. Lebih elastik kerana rangkai silang sulfur menghalang rantai polimer molekul getah saling menggelongsor antara satu sama lain. Strength and hardness Kekuatan dan kekerasan Weaker and softer . When it is stretched beyond the elastic limit, the polymer chain will break . / Lebih lemah dan lebih lembut . Apabila diregangkan melebihi had kenyal, rantai polimer akan putus . Stronger and harder because the presence of sulphur cross-links between the polymer. Lebih kuat dan lebih keras kerana adanya rangkai silang antara polimer. Resistance to heat Ketahanan haba Cannot withstand high temperature. Easily melt when heated. Tidak tahan suhu tinggi. Cair dengan mudah apabila dipanaskan. Can withstand high temperature because the presence of sulphur cross-links makes it more difficult to melt . / Boleh tahan suhu tinggi kerana kehadiran rangkai silang sulfur menjadikannya lebih sukar untuk mencair.

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 © Nilam Publication Sdn. Bhd. 192 Resistance to oxidation Ketahanan terhadap pengoksidaan Easily oxidised by oxygen because the presence of many double bonds in the rubber polymer. / Mudah dioksidakan oleh oksigen kerana adanya banyak ikatan berganda dalam polimer getah. Not easily oxidised by oxygen because the number of double bonds is reduced. Tidak mudah dioksidakan oleh oksigen kerana bilangan ikatan berganda berkurang. Remark: / Catatan: Due to its improved properties, vulcanised rubber is suitable for making tyres and parts in automobile industry. Oleh kerana sifatnya yang bertambah baik, getah tervulkan sesuai untuk membuat tayar dan alat ganti dalam industri automobil. Alternative methods of vulcanisation are the vulcanisations of rubber without the use of sulphur. Among the methods are: Kaedah alternatif pemvulkanan adalah pemvulkanan getah tanpa penggunaan sulfur. Antara kaedahnya adalah: Metal oxide vulcanisation is used for synthetic rubber such as chloroprene rubber and chlorobutyl rubber. Pemvulkanan logam oksida digunakan untuk getah sintetik seperti getah kloroprena dan getah klorobutil. Using metal oxides Menggunakan logam oksida Vulcanisation using peroxides is used for synthetic rubber without –C=C– to form cross-links between polymers. Peroxides produce radicals that form the cross-links –C–C– between adjacent polymers. Pemvulkanan yang menggunakan peroksida digunakan untuk getah sintetik tanpa –C=C– untuk membentuk rangkai silang antara polimer. Peroksida menghasilkan radikal yang membentuk rangkai silang –C–C– antara polimer bersebelahan. Using peroxide Menggunakan peroksida It is an electron beam vulcanisation of natural rubber latex. Vulcanised rubber produced is softer and shows high tensile strength. Ia adalah pemvulkanan sinar elektron dari lateks getah asli. Getah tervulkan yang dihasilkan lebih lembut dan menunjukkan kekuatan tegangan yang tinggi. Irradiation Penyinaran What is synthetic rubber? Apakah getah sintetik? LS / SP 4.3.1 Synthetic rubbers are elastomers that are elastic and can regain their original length after being stretched and pressed. / Getah sintetik adalah elastomer yang kenyal dan dapat kembali ke panjang asalnya setelah diregangkan dan ditekan. Remark: / Catatan: The structure of synthetic rubber polymers is the same as the structure of natural rubber polymers. Struktur polimer sintetik adalah sama dengan struktur polimer getah asli. How is synthetic rubber produced? / Bagaimanakah getah sintetik dihasilkan? Synthetic rubber is produced through the polymerisation process of hydrocarbon monomers obtained from petroleum fractions. / Getah sintetik dihasilkan melalui proses pempolimeran monomer hidrokarbon yang diperoleh daripada pecahan petroleum. Compare the properties of natural rubber with synthetic rubber. / Bandingkan sifat-sifat getah asli dengan getah sintetik. Characteristics Sifat-sifat Differences / Perbezaan Natural rubber / Getah asli Synthetic rubber / Getah sintetik Heat resistance Ketahanan haba Less Kurang More Lebih Hardness Kekerasan Softer Lembut Harder Keras Ease of oxidation by oxygen Mudah dioksidakan oleh oksigen Easy Mudah Difficult Sukar Resistance to chemicals and solvents Ketahanan kepada bahan kimia dan pelarut Not resistant Tidak tahan Resistant Tahan Elasticity Kekenyalan Less elastic Kurang kenyal Elastic Kenyal Similarities / Persamaan Natural rubber and synthetic rubber are good heat and electrical insulator as well as easy to vulcanise to increase its strength. Getah asli dan getah sintetik adalah penebat haba dan elektrik yang baik serta mudah untuk tervulkan untuk meningkatkan kekuatannya. SYNTHETIC RUBBER GETAH SINTETIK CS / SK 4.3 4.3

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 193 © Nilam Publication Sdn. Bhd. 1 (a) Natural rubber is a natural polymer. / Getah asli ialah polimer semula jadi. (i) Give another two examples of natural polymer. / Berikan dua lagi contoh polimer semula jadi. Protein, starch / Protein, kanji (ii) Draw the structural formula and write IUPAC name for the monomer of natural rubber. Lukiskan formula struktur dan tulis nama IUPAC bagi monomer getah asli. C C C C H CH3 H H H H Exercise / Latihan State the type of synthetic rubber, their characteristics and uses. Nyatakan jenis getah sintetik, sifat-sifat dan kegunaannya. Types of synthetic rubber Jenis getah sintetik Special properties Sifat istimewa Uses Kegunaan Neoprene Neoprena Not flammable, heat resistant, has resistance to water, oil and chemicals Tidak mudah terbakar, tahan panas, tahan terhadap air, minyak dan bahan kimia Tank linings for storing petrol and solvent, petrol hose, rubber pipes for gases and chemicals, wetsuits and wire insulation / Pelapik tangki untuk menyimpan petrol dan pelarut, hos petrol, paip getah untuk gas dan bahan kimia, pakaian selam dan penebat wayar Thiokol Tiokol Very resistant to oil and solvent Sangat tahan terhadap minyak dan pelarut Tank linings for storing petrol and solvent and also petrol hose. / Pelapik tangki untuk menyimpan petrol dan pelarut serta hos petrol. Styrene-Butadiene Rubber (SBR) Getah stirena-butadiena High heat resistance and abrasion resistance / Rintangan haba dan ketahanan lelasan yang tinggi Car tyres, shoe soles, conveyor belts and gasket Tayar kereta, tapak kasut, tali sawat dan gasket Silicon rubber Getah silikone Withstand high temperature Tahan suhu yang tinggi Rubber tube, cooking utensils, food storage products, adhesives & sealants, medical devices and automotive components. / Tiub getah, peralatan memasak, bekas simpanan makanan, pelekat dan bahan kedap, peralatan perubatan dan komponen automotif. Nitrile rubber (NBR) Getah nitril (NBR) Very resistant to oil and solvent Sangat tahan terhadap minyak dan pelarut For gasket, engine hoses and glove. Untuk gasket, hos enjin dan sarung tangan. Discuss the use of synthetic rubber and their effects on the environment. Explain. Bincangkan kegunaan getah sintetik dan kesannya kepada persekitaran. Terangkan. LS / SP 4.3.2 (a) Tyres are one of the synthetic rubber products which is the largest waste and difficult to dispose of due to its non-biodegradable nature. (b) Without a proper disposal system, used tyres become breeding grounds for mosquitoes and pollute the environment. (c) Used tyres are also prone to fire risk as well as emitting fumes containing toxic chemicals. (d) The nature of tyres that are resistant to high temperatures is used as decoration, especially in open areas. (e) Used tyres can also be recycled by converting them to rubber powder that can be used to make rubber tiles for playgrounds, rubber mats and stadium tracks. (a) Tayar ialah salah satu produk getah sintetik yang merupakan bahan buangan yang terbesar yang sukar dilupuskan kerana sifatnya yang tidak terbiodegradasi. (b) Tanpa sistem pelupusan yang betul, tayar yang telah digunakan boleh menjadi tempat pembiakan nyamuk serta mencemarkan alam sekitar. (c) Tayar terpakai juga terdedah kepada risiko kebakaran serta mengeluarkan asap yang mengandungi bahan kimia toksik. (d) Sifat tayar yang tahan terhadap suhu tinggi menyebabkannya sesuai untuk dikitar semula menjadi hiasan di kawasan terbuka. (e) Tayar terpakai juga boleh dikitar semula dengan menukarkannya kepada serbuk getah yang boleh digunakan untuk membuat jubin getah untuk taman permainan, tikar getah serta trek stadium. QUIZ KUIZ TP1 TP2 2-methylbut-1, 3-diene 2-metilbut-1, 3-diena

MODULE • Chemistry FORM 5 UNIT 4 UNIT 4 © Nilam Publication Sdn. Bhd. 194 (b) The tyres of aircrafts are made from vulcanised rubber. Tayar kapal terbang diperbuat daripada getah tervulkan. (i) What is vulcanised rubber? / Apakah getah tervulkan? Vulcanised rubber is the rubber produced when sulphur atoms form a cross-links between adjacent rubber polymers at the carbon-carbon double bond. Getah tervulkan adalah getah yang dihasilkan apabila atom sulfur membentuk rangkai silang antara polimer getah bersebelahan pada ikatan berganda karbon-karbon. (ii) Explain why vulcanised rubber is more elastic than unvulcanised rubber. Terangkan mengapa getah tervulkan adalah lebih kenyal daripada getah tak tervulkan. – The sulphur atoms form cross-link between the long rubber molecules. – This reduces the ability of the polymers to slide over each other. – The rubber molecules return back to its original positions after being stretched. (c) The diagram below shows three type of materials made from different synthetic rubbers. Rajah di bawah menunjukkan tiga bahan yang diperbuat daripada getah sintetik yang berbeza. Material X Bahan X Material Y Bahan Y Material Z Bahan Z (i) What are the type of synthetic rubbers used to make materials X, Y and Z? State the special properties for each synthetic rubber. / Apakah jenis getah sintetik digunakan untuk membuat bahan X, Y dan Z? Nyatakan sifat istimewa bagi setiap getah sintetik. Synthetic rubber to make material X is Thiokol. Thiokol is very resistant to oil and solvent. Synthetic rubber to make material Y is neoprene. Neoprene is not flammable, heat resistant, has resistance to water, oil and chemicals. Synthetic rubber to make material Z is Styrene-Butadiene Rubber (SBR). SBR has high heat resistance and abrasion resistance. (ii) Material Z plays an important part in manufacturing car. What are the effects toward the environment if material Z is not being disposed properly? Suggest recycling activity for material Z. Bahan Z memainkan peranan penting dalam pembuatan kereta. Apakah kesan terhadap alam sekitar jika bahan Z tidak dilupuskan dengan betul? Cadangkan aktiviti kitar semula untuk bahan Z. Tyres are one of the synthetic rubber products which is the largest waste and difficult to dispose of due to its nonbiodegradable nature. Without a proper disposal system, used tyres become breeding grounds for mosquitoes and pollute the environment. Used tyres are also prone to fire risk as well as emitting fumes containing toxic chemicals. The nature of the tyre that is resistant to high temperatures causes it suitable to be recycled into decoration in open area. Used tyres can also be recycled by converting them to rubber powder that can be used to make rubber tiles for playgrounds, rubber mats and stadium tracks. TP1 TP2 TP4 HOTS TP3

MODULE • Chemistry FORM 5 UNIT 5 UNIT 5 195 © Nilam Publication Sdn. Bhd. CONSUMER AND INDUSTRIAL CHEMISTRY KIMIA KONSUMER DAN INDUSTRI UNIT 5 Concept Map / Peta Konsep Explain Terangkan Can be converted to Boleh ditukar kepada Meaning Maksud Types of chemicals used Jenis bahan kimia digunakan Type Jenis Type Jenis Application Aplikasi Application Aplikasi Describe Huraikan Explain Terangkan Describe & Explain Huraikan & Terangkan Biological enzyme Enzim biologi Optimum condition Keadaan optimum CONSUMER AND INDUSTRIAL CHEMISTRY / KIMIA KONSUMER DAN INDUSTRI Green Technology Teknologi Hijau Saturated Fat Lemak Tepu Soap Sabun Hydrogenation Penghidrogenan Cleansing action Tindakan mencuci Antioxidant / Pengantioksida Stabiliser / Penstabil Flavouring / Perisa Preparation Penyediaan Thickening agents / Pemekat Emulsifiers / Pengemulsi Colouring agent / Pewarna Preservative / Pengawet Effectiveness in soft water and hard water Keberkesanan dalam air lembut dan air liat Temperature 180°C Suhu 180°C Catalyst: Nickel Mangkin: Nikel Unsaturated Fat Lemak tak Tepu Medicine Ubat-ubatan Food Additives Bahan Tambah Makanan Cosmetics Kosmetik Detergent Detergen Semiconductor & Electronic Semikonduktor & Elektronik Energy & Electric Tenaga & Elektrik Agriculture Pertanian Anti-allergy Antialergi Psychotic drug Ubat psikotik Corticosteroids Kortikosteroid Fragrances Pewangi Antimicrobials Antimikrob Treatment cosmetic Kosmetik perawatan Analgesic Analgesik Makeup cosmetics Kosmetik rias Examples of food additives and food Contoh bahan tambah makanan dan makanan Food Makanan Textile Tekstil – Biological enzyme / Enzim biologi – Whitening / Pemutih – Fragrance / Pewangi Oils and Fats Minyak dan Lemak Cleaning Agent Agen Pencuci Waste and wastewater management Pengurusan sisa dan sisa air Wastewater treatment Rawatan sisa air Nanotechnology Nanoteknologi

MODULE • Chemistry FORM 5 UNIT 5 UNIT 5 © Nilam Publication Sdn. Bhd. 196 What are oils and fats? Apakah minyak dan lemak? LS / SP 5.1.1 – Oils and fats are natural esters. / Minyak dan lemak adalah ester semula jadi. – Formed by esterification of alcohol glycerol or propan-1, 2, 3-triol with fatty acid. Terbentuk melalui pengesteran alkohol gliserol atau propan-1, 2, 3-triol dengan asid lemak. – Fats are triesters (triglyceride) / Lemak adalah triester (trigliserida) Remark: / Catatan: Oils and fats are chemically very similar but different in physical state. Minyak dan lemak adalah sangat hampir serupa tetapi berbeza dalam keadaan fizikal. What is alcohol glycerol? Apakah alkohol gliserol? Glycerol is an alcohol with three hydroxyl groups, known as propan-1, 2, 3-triol Gliserol ialah alkohol dengan tiga kumpulan hidroksil, dikenali sebagai propan-1, 2, 3-triol H H H H C C C H OH OH OH What is fatty acid? Apakah asid lemak? Fatty acid is a long chain of carboxylic acid with long carbon chain, R-COOH or CnH2n+1COOH. R is alkyl group with general formula CnH2n+1, n is about 10 to 20 Asid lemak adalah rantai panjang asid karboksilik dengan rantaian karbon yang panjang, R-COOH atau CnH2n+1COOH. R ialah kumpulan alkil dengan formula am CnH2n+1, n adalah 10 hingga 20 General equation for esterification reaction between glycerol and fatty acid. Persamaan am tindak balas pengesteran antara gliserol dan asid lemak. Esterification reaction between glycerol and fatty acid: Tindak balas pengesteran antara gliserol dengan asid lemak: 1 mol of glycerol 1 mol gliserol 3 mol of fatty acid 3 mol asid lemak 1 mol of oil or fat 1 mol minyak atau lemak 3 mol of water 3 mol air H O H O H C O H  +  H O C R H C O C R O O H C O H  +  H O C R' H C O C R' + 3 H – O – H O O H C O H  +  H O C R'' H C O C R'' H H R, R’ and R” represent hydrocarbon chains( alkyl groups) that are the same or different. R, R’ dan R’’ mewakili rantaian hidrokarbon (kumpulan alkil) yang sama atau berbeza. ⇒ Glycerol + Fatty acid Oil or fat + Water Gliserol + Asid lemak Minyak atau lemak + Air ⇒ Oil and fat molecules are made up of two parts i.e derived from glycerol and fatty acid . Molekul minyak dan lemak terdiri daripada dua bahagian iaitu diperoleh daripada gliserol dan asid lemak . OILS AND FATS MINYAK DAN LEMAK CS / SK 5.1 5.1 AR

MODULE • Chemistry FORM 5 UNIT 5 UNIT 5 197 © Nilam Publication Sdn. Bhd. What are saturated fats molecules? Apakah molekul lemak tepu? LS / SP 5.1.2 Saturated fats molecules are esters of saturated fatty acids. Saturated fatty acids contain single carbon-carbon (–C–C–) covalent bonds. / Molekul lemak tepu adalah ester bagi asid lemak tepu. Asid lemak tepu mengandungi ikatan kovalen karbon-karbon (–C–C–) tunggal . Example: / Contoh: Glyceryl tristearate / Gliseril tristearat H O H C O C (CH2)16 — CH3 O H C O C (CH2)16 — CH3 O H C O C (CH2)16 — CH3 H Derived from glycerol Diperoleh daripada gliserol Derived from fatty acid (hydrogen chains contain single covalent bonds between carbon atoms) Diperoleh daripada asid lemak (rantai karbon mengandungi ikatan kovalen tunggal antara atom karbon) What are unsaturated fat molecules? / Apakah molekul lemak tak tepu? LS / SP 5.1.2 Unsaturated fat molecules are esters of unsaturated fatty acids that contain single and double covalent bonds between carbon atoms in their hydrocarbon chain. Molekul lemak tak tepu ialah ester bagi asid lemak tak tepu yang mengandungi ikatan kovalen tunggal dan ganda dua di antara atom-atom karbon dalam rantai hidrokarbonnya. Example / Contoh: Glycerol trilinolate / Gliseril trilinolat H O H C O C (CH2)7 — (CH2)7CH = CHCH2CH = CH(CH2)4CH3 O H C O C (CH2)16 — (CH2)7CH = CHCH2CH = CH(CH2)4CH3 O H C O C (CH2)16 — (CH2)7CH = CHCH2CH = CH(CH2)4CH3 H Hydrocarbon chains contain double covalent bonds between carbon atoms Rantai hidrokarbon mengandungi ikatan kovalen ganda dua antara atom-atom karbon Remark: / Catatan: 1. If there is only one double bond in a fatty acid molecule, the fats formed are mono unsaturated. Sekiranya terdapat hanya satu ikatan berganda dalam molekul asid lemak, lemak yang terbentuk adalah mono tak tepu. 2. If there are two or more double bond in a fatty acid molecule, the fats formed are poly unsaturated. Jika terdapat dua atau lebih ikatan berganda dalam molekul asid lemak, lemak yang terbentuk adalah poli tak tepu.

MODULE • Chemistry FORM 5 UNIT 5 UNIT 5 © Nilam Publication Sdn. Bhd. 198 What are saturated and unsaturated oils or fats? Apakah minyak atau lemak tepu dan tak tepu? The oils and fats are mixture of saturated and unsaturated fats molecules: Minyak dan lemak ialah campuran molekul lemak tepu dan tak tepu: (i) An oil or fat is classified as saturated if it has more saturated fat molecules compared to unsaturated fat molecules, for example animal fats. (ii) An oil or fat is classified as unsaturated if it has more unsaturated fat molecules compared to saturated fat molecules, for example vegetable oils except coconut oil. (i) Suatu minyak atau lemak dikelaskan sebagai tepu jika ia mengandungi lebih banyak molekul lemak tepu berbanding molekul lemak tak tepu , contohnya lemak haiwan. (ii) Suatu minyak atau lemak dikelaskan sebagai tak tepu jika ia mengandungi lebih banyak molekul lemak tak tepu berbanding molekul lemak tepu, contohnya minyak sayuran kecuali minyak kelapa. Compare saturated fats and unsaturated fats. Bandingkan lemak tepu dan lemak tak tepu. LS / SP 5.1.2 Aspect / Aspek Saturated fats / Lemak tepu Unsaturated fats / Lemak tak tepu Bonds Ikatan Only C-C bonds in hydrocarbon chains Hanya ikatan C-C dalam rantaian hidrokarbon C=C and C-C bonds in hydrocarbon chains Ikatan C=C dan C-C dalam rantaian hidrokarbon Melting point / Takat lebur Higher / Lebih tinggi Lower / Lebih rendah Relative formula mass Jisim formula relatif Higher Lebih tinggi Lower Lebih rendah Physical state at room temperature Keadaan fizik pada suhu bilik Solid Pepejal Liquid Cecair Source Sumber Animal fats Lemak haiwan Vegetable oils Lemak tumbuhan What is the importance of oils and fats? Apakah kepentingan minyak dan lemak? (a) Oils and fats provide energy for our bodies. (b) Build membrane cell and certain hormones. (c) Dissolve certain vitamins for absorption. (a) Minyak dan lemak membekalkan tenaga untuk badan kita. (b) Membina membran sel dan hormon-hormon tertentu. (c) Melarutkan vitamin-vitamin tertentu untuk penyerapan. What are the sources of oils and fats? Apakah sumber minyak dan lemak? (a) Fats found in animals like cows and goats, are solid at room temperature. Example of animal fats are butter, cheese etc. (b) Fats from plants are liquid at room temperature. They are called oils . The example of oils are peanut oil, soybean oil and corn oil. (a) Lemak dijumpai dalam haiwan seperti lembu dan kambing, adalah pepejal pada suhu bilik. Contoh lemak haiwan ialah mentega, keju dan lain-lain. (b) Lemak daripada tumbuh-tumbuhan adalah cecair pada suhu bilik. Ia dipanggil minyak Contoh minyak ialah minyak kacang, minyak kacang soya dan minyak jagung. What is the process of changing unsaturated fats to saturated fats? Explain. Apakah proses menukarkan lemak tak tepu kepada lemak tepu? Terangkan. – Unsaturated fats can be converted to saturated fats through hydrogenation process such as in the manufacture of margarine. Catalyst used is nickel at 180°C. Lemak tak tepu boleh ditukarkan kepada lemak tepu melalui proses penghidrogenan, contohnya dalam pembuatan marjerin. Mangkin yang digunakan ialah nikel pada 180°C. – Sources of unsaturated fats are palm oil, soybean oil and corn oil. Sumber bagi lemak tak tepu ialah minyak sawit, minyak kacang soya dan minyak jagung. – Each carbon-carbon double bond absorbs one mole of hydrogen: Setiap ikatan ganda dua karbon-karbon menyerap satu mol hidrogen: H H ~ C C ~ + H2 Nickel / Nikel 180ºC H H ~ C C ~ H H Unsaturated fats (liquid) Lemak tak tepu (cecair) Saturated fats (solid) Lemak tepu (pepejal)