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SPM State Trial Papers Form 5 Chapter 1: Rate of Reaction QUESTION 1 – 2013 SBP Diagram 5.1 and Diagram 5.2 show the apparatus set-up to investigate the factor that affects the rate of reaction for Experiment I and Experiment II respectively. Rajah 5.1 dan Rajah 5.2 menunjukkan susunan radas untuk mengkaji faktor yang mempengaruhi kadar tindak balas masing-masing bagi Eksperimen I dan Eksperimen II.

Table 5.1 and Table 5.2 show the results obtained from Experiment I and Experiment II respectively. Jadual 5.1 dan jadual 5.2 menunjukkan keputusan yang diperoleh masing-masing daripada eksperimen I dan eksperimen II.

Time (s) Masa (s)

0 20 40 60 80 100 120 140 160 180

Volume of gas released (cm³) Isi padu gas yang terbebas (cm³)

0.00 6.50 12.50 17.80 23.50 27.20 31.80 35.00 35.00 35.00

Table 5.1 / Jadual 5.1

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Time (s) Masa (s)

0 20 40 60 80 100 120 140 160 180

Volume of gas released (cm³) Isi padu gas yang terbebas (cm³)

0.00 8.50 15.50 21.00 26.80 31.50 35.00 35.00 35.00 35.00


(a) (i) Name gas Y. Namakan gas Y. [1 mark] (ii) Describe a chemical test to verity gas Y. Jelaskan satu ujian kimia untuk mengesahkan gas Y. [2 marks]

(b) (i) Calculate the average rate of reactions for Experiment I and Experiment II. Hitung kadar tindak balas purata untuk tindak balas bagi Eksperimen I dan Eksperimen II. Experiment I: Eksperimen I: Experiment II: Eksperimen II: [2 marks] (ii) Compare the rate of reaction of Experiment I and Experiment II. Bandingkan kadar tindak balas bagi Eksperimen I dan Eksperimen II. [1 mark] (iii) Explain the answer in 5(b)(ii) with reference to the collision theory. Terangkan jawapan di 5(b)(ii) dengan merujuk kepada teori perlanggaran. [3 marks] (c) Sketch the graph of volume of gas Y produced against time for both experiments on the same axis.

Lakar graf isi padu gas Y yang dihasilkan melawan masa bagi kedua-dua eksperimen di atas paksi yang sama. [2 marks]

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Suggested Answer:

(a) (i) Hydrogen (ii) Place / Insert a lighted splinter into a test tube. "Pop" sound produced.

(b) (i) Experiment I:

Average rate of reaction 1313 25.0

140

35 scmscm

Experiment II:

Average rate of reaction 1313 29.0

120

35 scmscm

(ii) Rate of reaction in Experiment II is higher than Experiment I. (iii) In Experiment II: 1. Size of zinc is smaller // Total surface area of zinc in Experiment II is bigger. 2. Frequency of collision between zinc atom and hydrogen ion is higher. 3. Frequency of effective collisions between zinc atom and hydrogen ion is higher.

(c)

QUESTION 2 – 2014 SBP Experiments I, II and III are carried out to investigate the factors affecting the rate of reaction. Table 6.1 shows the reactants and temperature used in each experiment. Eksperimen I, II dan III dijalankan untuk mengkaji faktor-faktor yang mempengaruhi kadar tindak balas. Jadual 6.1 menunjukkan bahan tindak balas dan suhu yang digunakan dalam setiap eksperimen.

Experiment Eksperimen

Reactants Bahan Tindak Balas

Temperature Suhu (°C)

I Excess zinc powder + 25 cm3 of 0.1 mol dm-3 hydrochloric acid Serbuk zink berlebihan + 25 cm3 asid hidroklorik 0.1 mol dm-3

30

II Excess zinc powder + 25 cm3 of 0.1 mol dm-3 hydrochloric acid Serbuk zink berlebihan + 25 cm3 asid hidroklorik 0.1 mol dm-3

40

III Excess zinc powder + 25 cm3 of 0.1 mol dm-3 sulphuric acid Serbuk zink berlebihan + 25 cm3 asid sulfurik 0.1 mol dm-3

30


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(a) Write the ionic equation for the reaction in Experiment I. Tuliskan persamaan ion untuk tindak balas dalam Eksperimen I. [2 marks]

(b) Based on the experiments, state two factors that affect the rate of reaction. Merujuk kepada eksperimen, nyatakan dua faktor yang mempengaruhi kadar tindak balas.

[2 marks] (c) Compare the rate of reaction between Experiments I and II. Explain the difference by using

collision theory. Bandingkan kadar tindak balas antara Eksperimen I dan II. Terangkan perbezaan itu dengan menggunakan teori perlanggaran. [4 marks]

(d) Diagram 6.2 shows the curve of the graph of total volume against time for Experiment I. Sketch the curve obtained for Experiment III on the same axes. Rajah 6.2 menunjukkan garis lengkung bagi graf jumlah isi padu gas melawan masa bagi Eksperimen I. Lakarkan garis lengkung yang diperolehi bagi Eksperimen III pada paksi yang

sama. [1 mark]

(e) During a master chef competition, an apprentice found that a piece of meat is still not tender after

cooking for one hour. Semasa satu pertandingan 'master chef’, seorang pelatih mendapati ketulan daging yang dimasak masih tidak lembut selepas satu jam.

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State one method that should be taken to make the meat become tender in a shorter time. Explain you answer. Nyatakan satu kaedah yang boleh diambil supaya daging itu menjadi lembut dalam masa lebih singkat. Terangkan jawapan anda. [2 marks]

Suggested Answer:

(a) 2

22 HZnHZn

(b) Temperature and concentration

(c) 1. The rate of reaction in Experiment II is higher than Experiment I. 2. The temperature in Experiment II is higher. 3. The kinetic energy of hydrogen ions is higher. 4. The frequency of collision between zinc atoms and hydrogen ions is higher. // The frequency of effective collision between zinc atoms and hydrogen ions is higher.

(d)

(e) 1. Cut the meat into smaller size. 2. Larger total surface area of meat will absorb more heat. OR 3. Cook in pressure cooker. 4. High pressure in pressure cooker increase the temperature.

QUESTION 3 – 2015 PULAU PINANG Experiment 1 is carried out to determine the rate of reaction between 25 cm3 of 0.2 mol dm-3 hydrochloric acid and excess zinc powder. Diagram 4 shows the apparatus set-up for Experiment 1. Eksperimen 1 dijalankan untuk menentukan kadar tindak balas antara 25 cm3 asid hidroklorik 0.2 mol dm-3 dan serbuk zink berlebihan. Rajah 4 menunjukkan susunan radas bagi Eksperimen 1.

Diagram 4 / Rajah 4

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(a) Write a chemical equation for the reaction and calculate the maximum volume of hydrogen gas released.

[Molar volume of gas at room conditions is 24 mol dm-3] Tulis persamaan kimia bagi tindak balas itu dan hitung isi padu maksimum gas hidrogen yang terbebas.

[Isi padu molar gas pada keadaan bilik ialah 24 mol dm-3] [5 marks] (b) The rate of reaction between zinc and hydrochloric acid can be changed by using catalyst.

Kadar tindak balas antara zink dan asid hidroklorik dapat diubah dengan menggunakan mangkin.

(i) Name one suitable catalyst that can be used in this reaction. Namakan satu mangkin yang sesuai digunakan dalam tindak balas ini. [1 marks] (ii) Draw the energy profile diagram for the reaction. Show in the diagram the activation energy without the use of catalyst, Ea and with the use of catalyst, Ea ’.

Lukis gambarajah profil tenaga bagi tindak balas itu. Tunjukkan pada rajah tersebut tenaga pengaktifan tanpa menggunakan mangkin, Ea dan dengan menggunakan mangkin, Ea '.

[4 marks] (iii) Referring to the collision theory, explain how catalyst can affect the rate of the reaction. Dengan merujuk kepada teori perlanggaran, terangkan bagaimana mangkin boleh mempengaruhi kadar tindak balas itu. [4 marks] (iv) Catalysts are widely used in industries for the manufacturing of chemical based products. State one chemical process and name the catalyst used in that industrial process. Mangkin digunakan secara meluas dalam industri pembuatan produk berasaskan bahan kimia. Nyatakan satu proses kimia dan nama mangkin yang digunakan dalam proses industri tersebut. [2 marks] (c) Experiment II is also carried out under the same conditions except using 10 cm3 of 0.5 mol dm-3

hydrochloric acid to replace 25 cm3 of 0.2 mol dm-3 hydrochloric acid used in Experiment I. Compare the rate of reaction and the maximum volume of hydrogen gas released for both of the experiments. Explain your answer. Eksperimen II dijalankan pada keadaan-keadaan yang sama kecuali menggunakan 10 cm3 asid hidroklorik 0.5 mol dm-3 untuk menggantikan 25 cm3 asid hidroklorik 0.2 mol dm-3 yang digunakan dalam Eksperimen I. Bandingkan kadar tindak balas serta isi padu maksimum gas hidrogen yang dibebaskan bagi kedua-dua eksperimen. Jelaskan jawapan anda. [4 marks]

Suggested Answer:

(a) 2HCI + Zn ZnCI2 + H2

No. of moles of HCI 005.01000

252.0

No. of moles of H2 2

1 No. of moles of HCI

= 0.0025 Volume of H2 = 0.0025 X 24 = 0.06 dm3

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(b) (i) Copper (II) sulphate solution (ii)

(iii) 1. Catalyst increases the rate of reaction 2. Copper(ll) sulphate / Catalyst lowers the activation energy // Copper(ll) sulphate / Catalyst provides an alternative path with a lower activation energy 3. More zinc atoms and hydrogen ions are able to achieve the lower activation energy 4. The frequency of effective collision increases (iv) 1. Haber process, iron 2. Contact process, vanadium(V) oxide 3. Ostwald process, platinum [Anyone]

(c) 1. Rate of reaction in Experiment II is higher than in Experiment I. 2. The concentration of hydrochloric acid / H+ ions used in Experiment II is higher than in Experiment I. 3. The maximum volume of hydrogen gas released from both experiments is the same. 4. The number of moles of hydrochloric acid used in both experiments is the same.

QUESTION 4 – 2014 KEDAH MODUL 1 A group of pupils carried out three experiments to investigate the factors affecting the rate of reaction. Table 8 shows information about the reaction in each experiment. Sekumpulan pelajar menjalankan tiga eksperimen untuk mengkaji faktor yang mempengaruhi kadar tindak balas. Jadual 8 menunjukkan maklumat yang digunakan dalam setiap eksperimen.



I 60 cm3 of 0.25 mol dm -3 hydrogen peroxide solution 30 cm3 larutan hidrogen peroksida 0.5 mol dm -3

II 30 cm3 of 0.5 mol dm -3 hydrogen peroxide solution 30 cm3 larutan hidrogen peroksida 0.5 mol dm -3

III 30 cm3 of 0.5 mol dm -3 hydrogen peroxide solution + manganese (IV) oxide 30 cm3 larutan hidrogen peroksida 0.5 mol dm -3 + mangan (IV) oksida

Table 8 / Jadual 8

The graph in diagram 8 shows the results of these experiments. Graf pada rajah 8 menunjukkan keputusan eksperimen tersebut.

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(a) Calculate the average rate of reaction for Experiment I. Hitungkan kadar tindak balas purata bagi eksperimen I. [2 marks] (b) Hydrogen peroxide decomposes to oxygen gas and water. Sketch the energy profile diagram for

Experiment II and Experiment III on the same axes. Indicates clearly the activation energy for both experiments. Hidrogen peroksida terurai kepada gas oksigen dan air. Lakarkan gambar rajah profil tenaga bagi Eksperimen II dan Eksperimen lll pada paksi yang sama. Tunjukkan dengan jelas tenaga pengaktifan bagi setiap eksperimen. [4 marks]

(c) Based on Table 8 and Graph 8, compare the rate of reaction between: Berdasarkan Jadual 8 dan Graf 8, bandingkan kadar tindak balas antara:

• Experiment I and Experiment II / Eksperimen I and Eksperimen II • Experiment II and Experiment III / Eksperimen II and Eksperimen III

In each case explain the difference rate of reaction with reference to the collision theory. Terangkan perbezaan dalam kadar tindak balas bagi setiap kes dengan merujuk pada teori pelanggaran. [10 marks]

(d) The chemical equation below shows the decomposition of hydrogen peroxide. Persamaan kimia di bawah menunjukkan penguraian hidrogen peroksida.

2222 22 OOHOH

Calculate the maximum volume of oxygen gas produced in Experiment II. Hitungkan isipadu maksimum gas oksigen yang terhasil dalam Eksperimen II. [1 mole of gas occupied 24 dm -3 at room condition] [1 mol gas menempati 24 dm -3 pada keadaan bilik] [4 marks]

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Suggested Answer:

(a) Average rate =

55

50 = 0.909 // 0.91 cm3 s-1

(b)

1. Y axis labeled energy 2. Both curve drawn correctly with label 3. Energy level labeled with reactant and products [a: exo or endo] 4. Activation energy labeled correctly

(c) Experiment I and Experiment II: 1. Rate of reaction of experiment II Is higher than experiment I. 2. Experiment II use higher concentration of hydrogen peroxide. 3. The number of Hydrogen peroxide molecules / reactant particles per unit volume in experiment II is higher. 4. Frequency of collision between Hydrogen peroxide molecules of experiment II become Higher. 5. Frequency of effective collision between Hydrogen peroxide molecules/reactant particles

of experiment II become higher. Experiment II and Experiment III : 6. Rate of reaction of experiment III is higher than experiment II. 7. Catalyst that is Manganese (IV) oxide present in experiment III. 8. Manganese(IV) oxide lower the activation energy of the reaction in experiment III. 9. More colliding hydrogen peroxide molecules easily overcome the lower activation energy in experiment III. 10. Frequency of effective collision between Hydrogen peroxide molecules of experiment III become higher.

(d) 1. Number of mole of 22OH 015.0//

1000

305.0

2. 2 mole of 22OH produce 1 mole of 2O

3. Number of mole of 2O 0075.0//2

015.0

4. Volume of 2O = 0.0075 x 24 dm3 //0.18 dm3 //180 cm3 [r: answer without unit]

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QUESTION 5 – 2014 KEDAH MODUL 2 (a) A student carried out two experiments to investigate the effect of catalyst on the rate of reaction of

decomposition of hydrogen peroxide. Table 8 shows the result of the experiments. Seorang pelajar menjalankan dua eksperimen untuk mengkaji kesan mangkin ke atas kadar tindak balas pengaraian larutan hidrogen peroksida. Jadual 8 menunjukkan keputusan eksperimen tersebut.


Reactant Bahan tindak balas

Time/s Masa/s

A 50 cm 3 of 0.05 mol dm -3 hydrogen peroxide solution 50 cm 3 larutan hidrogen peroksida 0.05 mol dm -3

40

B

50 cm 3 of 0.05 mol dm -3 hydrogen peroxide solution and 1g manganese (IV) oxide 50 cm 3 larutan hidrogen peroksida 0.05 mol dm -3 dan 1g mangan(IV) oksida

20

Table 8 / Jadual 8

(i) Write the equation for the decomposition of hydrogen peroxide solution. Tuliskan persamaan bagi penguraian larutan hidrogen peroksida. [2 marks] (ii) Sketch the graph of the volume of the gas liberated against time for Experiment A and Experiment

B on the same axes. Lakarkan graf isipadu gas terbebas melawan masa pada paksi yang sama bagi Eksperimen A dan Eksperimen B. [3 marks] (iii) Explain briefly one method to manipulate another factor for experiment A in order to increase the rate of reaction same as in experiment B and then give one application of that factor in everyday life. Terangkan secara ringkas satu kaedah untuk memanipulasikan faktor lain bagi eksperimen A untuk meningkatkan kadar tindak balas sama seperti eksperimen B dan kemudian berikan satu penggunaan faktor tersebut dalam kehidupan harian. [5 marks]

(iv) Draw a labelled diagram of the apparatus set- up used in the experiment.

Lukiskan gambar rajah berlabel bagi susunan radas yang digunakan dalam eksperimen tersebut. [2 marks]

(c)(i) Compare the rate of reaction between Experiment A and Experiment B. Your explanation must refer to the collision theory. Bandingkan kadar tindak balas eksperimen A dan eksperimen B. Huraian anda mestilah merujuk kepada teori pelanggaran. [5 marks]

(ii) Draw an energy profile diagram to show the effect of catalyst on the rate of decomposition of hydrogen peroxide. Lukiskan gambar rajah profil tenaga untuk menunjukkan kesan mangkin ke atas penguraian hidrogen peroksida. [3 marks]

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Suggested Answer:

(a)(i) 2222 22 OOHOH

(ii)

(iii) 1. Increase the concentration of hydrogen peroxide 2. 0.1 mol dm-3 / any suitable higher concentration 3. Volume 25cm3 / any suitable volume to get that match the concentration 4. Any suitable application of concentration 5. How the application is used

(iv)

(c)(i) 1. Rate of reaction of experiment B is higher 2. The presence of catalyst in experiment B 3. The reaction through the alternative path // lower activation energy 4. More colliding particle can achieve the lower activation energy 5. Frequency of effective collision is higher

(ii)

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QUESTION 6 – 2015 PERLIS A student carried out three sets of experiment to investigate the factors that affect the rate of reaction. The time taken to collect 40 cm3 of hydrogen gas is recorded in Table 9. Seorang pelajar menjalankan tiga set eksperimen untuk mengkaji faktor yang mempengaruhi kadar tindak balas. Masa yang diambil untuk mengumpul 40 cm3 gas hidrogen direkodkan dalam Jadual 9.

Set Reactants

Bahan Tindak Balas

Temperature Of The Mixture

Suhu Campuran

(C)

Time Taken To Collect 40 Cm3 Of Hydrogen Gas Masa Yang Diambil Untuk Mengumpul 40 Cm3 Gas

Hydrogen/S

I

25 cm3 of 0.2 mol dm -3 nitric acid 25 cm3 asid nitrik 0.2 mol dm -3

+ Excess zinc powder Serbuk zink berlebihan

30 90

II



30 55

III



40 30

Table 9 / Jadual 9

(a) Zinc, Zn reacts with nitric acid, HNO3 to produce zinc nitrate, Zn(NO3 )2 and hydrogen gas, H2. Write a balanced chemical equation for the reaction and calculate the maximum volume of hydrogen gas produced in set 1. [Relative atomic mass: Zn = 65; 1 mol of any gas occupies 24 dm3 mol-1 at room conditions] Zink, Zn bertindak balas dengan asid nitrik, HNO3 menghasilkan zink nitrat, Zn(NO3)2 dan gas hidrogen, H2. Tulis persamaan kimia yang seimbang bagi tindak balas itu dan hitung isipadu maksimum gas hidrogen yang dihasilkan dalam set 1. [Jisim atom relatif: Zn = 65; 1 mol bagi sebarang gas menempati 24 dm3 mol-1 pada keadaan bilik]

[4 marks] (b) Based on Table 9, compare the rates of reaction. Berdasarkan Jadual 9, bandingkan kadar tindak balas.

(i) Between set I and set II // Antara set I dengan set II (ii) Between set I and set III // Antara set I dengan set III By referring to collision theory, explain your answer in 9(b)(i) or 9(b)(ii). Dengan merujuk kepada teori perlanggaran jawapan, terangkan jawapan anda di 9(b)(i) atau 9(b)(ii). [6 marks]

(c) Size of reactants and catalyst can also affect the rate of reaction between acid and zinc. Choose one of these two factors and describe an experiment to show how this factor affects the rate of reaction.

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Mangkin dan saiz bahan tindak balas juga dapat mempengaruhi kadar tindak balas antara asid dengan zink. Pilih satu daripada dua faktor ini dan huraikan satu eksperimen untuk menunjukkan bagaimana faktor ini dapat mempengaruhi kadar tindak balas. [10 marks]

Suggested Answer:

(a) Zn + 2HNO3 Zn (NO3)2 + H2

No. of moles of HNO3 = 0.2 X 25 = 0.005mol 1000

2 mol of HNO3 1 mol H2

0.005 mol 0.0025 mol H2

Max volume of H2 = 0.0025 X 24 = 0.06 dm3 = 60cm 3

(b) (i)

Rate of reaction set II higher than set I The concentration of nitric acid / HNO3 in set II higher than set I // No. of particles per unit volume in set II is higher Frequency of effective collision between hydrogen ions / H+ and zinc atom / Zn higher in set II

(ii) Rate of reaction set III higher than set I The temperature in set III higher than set I // Kinetic energy of particles in set in is higher Frequency of effective collision between hydrogen ions/ H+ and zinc atom / Zn higher in set II

(c) Size of Reactants: 1. (25-50) cm 3 of (0.1-1.0) mol dm3 of hydrochloric acid is measured and poured into a conical flask. 2. About 5.0 g of zinc granules is weigh. 3. A burette is filled with water and inverted into a basin containing water. 4. The water level in the burette is adjusted to 50 cm mark. 5. The granulated zinc is added into the conical flask. 6. Immediately the conical flask is closed and connect it using delivery tube to the burette. 7. The stopwatch is started. 8. The conical flask is shaken steadily. 9. Record volume of hydrogen gas every 30 seconds interval. 10. The experiment is repeated using 5.0 g of zinc powder to replace 5.0 g of zinc granules. Catalyst 1. (25-50) cm 3 of (0.1-1.0) mol dm3 of hydrochloric acid is measured and poured into a conical flask. 2. About 5.0 g of zinc granules is weigh. 3. A burette is filled with water and inverted into a basin containing water. 4. The granulated zinc is added into the conical flask. 5. 5 cm 3 of 0.5 mol dm3 copper (II) sulphate solution is added into the Conical flask. 6. Immediately the conical flask is closed and connect it using delivery tube to the burette. 7. The stopwatch is started. 8. The conical flask is shaken steadily. 9. Record volume of hydrogen gas every 30 seconds interval. 10. The experiment is repeated without adding copper (II) sulphate Solution.

QUESTION 7 – 2015 KELANTAN (a) A group of students carried out experiments to investigate the factor affecting the rate of reaction

between metal P and an acid Q. Sekumpulan pelajar telah menjalankan eksperimen untuk mengkaji kesan faktor yang mempengaruhi kadar tindak balas antara logam P dan asid Q.

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Table 9 shows the information about the reactants and the time taken to collect 30 cm 3 of hydrogen gas. Jadual P menunjukkan maklumat tentang bahan tindak balas dan masa diambil untuk mengumpul 30 cm 3 gas hidrogen.



Time Taken (s) Masa diambil (s)

I Powdered metal P and 50 cm 3 of 1.0 mol dm3 acid Q Serbuk logam P dan 50 cm 3 asid Q 1.0 mol mol dm3

10

II Powdered metal P and 100 cm 3 of 0.5 mol dm3 J acid Q Serbuk logam P dan 100 cm 3 asid Q 0.5 mol dm3

20

Table 9 / Jadual 9

(i) Suggest the name of metal P and acid Q. Cadangkan nama logam P dan asid Q. [2 marks] (ii) By using the named metal P and acid Q. Write the chemical equation. Menggunakan logam P dan asid Q yang dinamakan, tulis persamaan kimia. [4 marks]

(iii) Explain the difference in the rate of reaction between Experiment I and Experiment II. Use the collision theory in your explanation. Terangkan perbezaan kadar tindak balas antara Eksperimen I dan Eksperimen II. Gunakan teori perlanggaran dalam penerangan anda. [4 marks]

(b) By using either size of reactant or temperature, describe an experiment how this factor affecting the rate of reaction. Dengan menggunakan faktor saiz bahan tindak balas atau suhu, huraikan satu eksperimen bagaimana faktor berkenaan mempengaruhi kadar tindak balas. [10 marks]

Suggested Answer:

(a) (i)

P : [mana-mana logam di alas Cu dalam SEK] Contoh: Magnesium / Zink / Aluminium [r: kalium / natrium] Q: [mana-mana asid] Contoh: Asid hidroklorik / Asid sulfurik / Asid nitric [a: asid lemah]

(ii) Mg + 2HC1 MgC12 + H2

(iii) Eksperimen I = 30 // 3 cm 3 S-1

10 Eksperimen II = 30 // 1.5 cm 3 S-1

20 1. Kadar tindak balas dalam eksperimen I lebih tinggi berbanding eksperimen II. 2. Kepekalan asid dalam eksperimen I lebih tinggi berbanding eksperimen II. // Bilangan ion hydrogen per unit isipadu dalam eksperimen I lebih tinggi berbanding eksperimen II. 3. Frekuensi perlanggaran antara ion hydrogen dan atom [P] dalam eksperimen I lebih tinggi berbanding eksperimen II.

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4. Frekuensi perlanggaran berkesan antara zarah-zarah dalam eksperimen I lebih tinggi berbanding eksperimen II.

(b) Saiz Bahan Tindak Balas 1. Isi sebuah besin dengan air. 2. Isi sebuah buret dengan air dan telangkupkan dalam besin. 3. Rekod bacaan awal buret. 4. Tuangkan [20-200 cm3] asid hidroklorik [0.1 - 2.0 mol dm-3] ke dalam sebuah kelalang kon. 5. Tambah [0.5-5 g] ketulan zink // berlebihan. 6. Rekod bacaan buret pada sela masa 30 s/60 s // kumpul [isi padu tetap gas] dan rekod masa diambil. 7. Ulang langkah 1-6 menggunakan serbuk zink dengan jisim yang sama. 8. ****Jadual 9. ****graf //

Eks. Serbuk zink Kadar tindak balas = [isipadu gas]/masa = a Eks. Ketulan zink Kadar tindak balas = [isipadu gas]/masa = b a>b

10. Kesimpulan: Kadar tindak balas menggunakan serbuk zink lebih tinggi. Suhu 1. Isi sebuah besin dengan air. 2. Isi sebuah buret dengan air dan telangkupkan dalam besin. 3. Rekod bacaan awal buret. 4. Tuangkan [20-200 cm3] asid hidroklorik [0.1 -2 0 mol dm-3] ke dalam sebuah kelalang kon. 5. Rekodkan suhu asid. 6. Tambah [0.5-5 g] ketulan zink // berlebihan. 7. Rekod bacaan buret pada sela masa 30 s/60 s // kumpul [isi padu tetap gas] dan rekod masa diambil. 8. Ulang langkah 1-6 menggunakan asid hidroklorik yang dipanaskan pada suhu. [> 30"C] dengan mengekalkan jisim zink, isipadu dan kepekatan asid. 9. ****Jadual 10. ****graf //

Eks. Serbuk zink Kadar tindak balas = [isipadu gas]/masa = a Eks. Ketulan zink Kadar tindak balas = [isipadu gas]/masa = b a>b

11. Kesimpulan: Kadar tindak balas menggunakan serbuk zink lebih tinggi.

QUESTION 8 – 2015 KEDAH MODUL 2 Table 10 shows the results of three sets of experiments to investigate the factors that affect the rate of reaction. 50 cm3 of carbon dioxide gas is collected in each experiment. Jadual 10 menunjukkan keputusan bagi tiga set eksperimen untuk menyiasat faktor yang mempengaruhi kadar tindak balas. 50 cm3 gas karbon dioksida dikumpulkan dalam setiap eksperimen.

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Experiments Eksperimen


Temperature Suhu

°C

Time Taken To Collect Gas/s

Masa Yang Diambil Untuk Mengumpulkan Gas/s

I

50 cm3 of 1.0 mol dm-3 nitric acid + excess marble chips 50 cm3 Asid nitrik 1.0 mol dm-3 + serpihan marmar berlebihan

28 240

II

50 cm3 of 1.0 mol dm-3 nitric acid + excess powdered marble chips 50 cm3 Asid nitrik 1.0 mol dm-3 + serbuk marmar berlebihan

28 60

III

50 cm3 of 1.0 mol dm-3 nitric acid + excess marble chips 50 cm3 asid nitrik 1.0 mol dm-3 + serpihan marmar berlebihan

40 150

Table 10 / Jadual 10

(a) Nitric acid reacts with marble chips to produce calcium nitrate, carbon dioxide gas and water. Write a balanced chemical equation for the reaction and calculate the maximum volume of carbon

dioxide gas produced in experiment I. [Relative atomic mass: C = 12, Ca = 40; 1 mole of gas occupies 24 dm3 at room conditions] Asid nitrik bertindak balas dengan serpihan marmar untuk menghasilkan kalsium nitrat, gas karbon dioksida dan air. Tulis persamaan kimia yang seimbang bagi tindakbalas itu dan hitung isipadu maksimum gas karbon dioksida yang dihasilkan dalam eksperimen I. [Jisim atom relatif: C = 12, Ca = 40; 1 mol gas menempati 24 dm3 pada keadaan bilik]

[5 marks] (b) Based on table 10, compare the rate of reaction between: Berdasarkan Jadual 10. bandingkan kadar tindak balas antara:

Experiment I and experiment II / Eksperimen I dan Eksperimen II Experiment I dan Experiment III / Eksperimen I dan Eksperimen III In each case explain the difference in the rate of reaction with reference to the collision theory. Bagi setiap kes terangkan perbezaan dalam kadar tindak balas dengan merujuk kepada teori perlanggaran. [6 marks]

(c) Catalyst can be used to increase the rate of reaction between metal and acid. By using a suitable named of metal and acid, describe an experiment to show how catalyst affects the rate of reaction. Mangkin boleh digunakan untuk meningkatkan kadar tindak balas antara logam dan asid. Dengan menggunakan satu logam, asid dan mangkin yang dinamakan, huraikan satu eksperimen untuk menunjukkan bagaimana mangkin boleh mempengaruhi kadar tindak balas.

[9 marks]

SPMSureSkor.com 17

Suggested Answer:

(a) CaCO33+ 2HNO3 → Ca(NO3)2 + CO2 + H 2O Number of mole HNO3 = (50 X 0.1) / 1000

= 0.5 mol 2 mol HNO3 : 1 mol CO2

Number of mol of CO2 produced = 0.05 / 2 = 0.025 mol Maximum volume of CO2 = 0.025 X 24 = 0.6 dm3

= 600cm3

(b) (i) - The rate of reaction of Experiment II higher than Experiment I - Powdered marble chip has higher Total Surface Area that exposed to collision with

hydrogen ions - So, the frequency of collisions is higher - So, the frequency of effective collisions is higher (ii) - The rate of reaction of Experiment III higher than Experiment I - The temperature in experiment III is higher than experiment I - The kinetic energy of reactants/hydrogen ions is higher the reactants move faster - So, the frequency of collisions between hydrogen ions and calcium carbonate is higher - So, the frequency of effective collisions is higher

(c) Zinc magnesium Nitric acid hydrochloric acid sulphuric acid Catalyst: copper(II) sulphate 1. Fill a burette with water, invert into a basin with water, clamp with retort stand, adjust the

meniscus to 50 cm3. 2. 50cm3 of 1.0 mol dm° nitric acid is poured into a conical flask. 3. Excess zinc magnesium granules are added into the acid. 4. A stopper with delivery tube is immediately connected to collect the gas released (consider diagram if any). 5. Start the stop watch the time taken to collect 50 cm3 of gas is recorded. 6. Step 1 to 5 are repeated by adding copper (II) sulphate in step 2. 7. The present of catalyst copper (II) sulphate will increase the rate of reaction reduced tune

taken to collect 50 cm' of gas.

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