# SPM State Trial Papers Form 5 Chapter 1: Rate of Reaction

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1 SPM State Trial Papers Form 5 Chapter 1: Rate of Reaction QUESTION 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) Volume of gas released (cm³) Isi padu gas yang terbebas (cm³) Table 5.1 / Jadual 5.1 SPMSureSkor.com 1

2 Time (s) Masa (s) Volume of gas released (cm³) Isi padu gas yang terbebas (cm³) Table 5.2 / Jadual 5.2 (a) (i) Name gas Y. Namakan gas Y. (ii) Describe a chemical test to verity gas Y. Jelaskan satu ujian kimia untuk mengesahkan gas Y. [1 mark] (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: (ii) Compare the rate of reaction of Experiment I and Experiment II. Bandingkan kadar tindak balas bagi Eksperimen I dan Eksperimen II. (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. [1 mark] [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. SPMSureSkor.com 2

3 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 35 cm s 0.25cm s 140 Experiment II: Average rate of reaction 35 cm s 0.29cm s 120 (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 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 I II III Reactants Bahan Tindak Balas Excess zinc powder + 25 cm 3 of 0.1 mol dm -3 hydrochloric acid Serbuk zink berlebihan + 25 cm 3 asid hidroklorik 0.1 mol dm -3 Excess zinc powder + 25 cm 3 of 0.1 mol dm -3 hydrochloric acid Serbuk zink berlebihan + 25 cm 3 asid hidroklorik 0.1 mol dm -3 Excess zinc powder + 25 cm 3 of 0.1 mol dm -3 sulphuric acid Serbuk zink berlebihan + 25 cm 3 asid sulfurik 0.1 mol dm -3 Table 6.1 / Jadual 6.1 Temperature Suhu ( C) SPMSureSkor.com 3

4 (a) Write the ionic equation for the reaction in Experiment I. Tuliskan persamaan ion untuk tindak balas dalam Eksperimen I. (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. (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. SPMSureSkor.com 4

5 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. Suggested Answer: 2 (a) Zn 2H Zn H2 (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 PULAU PINANG Experiment 1 is carried out to determine the rate of reaction between 25 cm 3 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 cm 3 asid hidroklorik 0.2 mol dm -3 dan serbuk zink berlebihan. Rajah 4 menunjukkan susunan radas bagi Eksperimen 1. Diagram 4 / Rajah 4 SPMSureSkor.com 5

7 (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 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. Experiment Eksperimen I II III Reactants Bahan Tindak Balas 60 cm 3 of 0.25 mol dm -3 hydrogen peroxide solution 30 cm 3 larutan hidrogen peroksida 0.5 mol dm cm 3 of 0.5 mol dm -3 hydrogen peroxide solution 30 cm 3 larutan hidrogen peroksida 0.5 mol dm cm 3 of 0.5 mol dm -3 hydrogen peroxide solution + manganese (IV) oxide 30 cm 3 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. SPMSureSkor.com 7

9 Suggested Answer: (a) 50 Average rate = = // 0.91 cm 3 s-1 55 (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) Number of mole of H 2O2 // mole of H 2O2 produce 1 mole of O Number of mole of O // Volume of O 2 = x 24 dm 3 //0.18 dm 3 //180 cm 3 [r: answer without unit] SPMSureSkor.com 9

11 Suggested Answer: (a)(i) (ii) 2H H O O 2O (iii) (iv) 1. Increase the concentration of hydrogen peroxide mol dm -3 / any suitable higher concentration 3. Volume 25cm 3 / any suitable volume to get that match the concentration 4. Any suitable application of concentration 5. How the application is used (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) SPMSureSkor.com 11

17 Suggested Answer: (a) (b) (c) 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 = mol Maximum volume of CO2 = X 24 = 0.6 dm 3 = 600cm 3 (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 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 cm cm3 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 The present of catalyst copper (II) sulphate will increase the rate of reaction reduced tune taken to collect 50 cm' of gas. SPMSureSkor.com 17

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