This slide show for undergraduates SM Sains Raja Tun Azlan Shah. Use this for perfect score in SPM

1. Rate of Reaction

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3. using deep fryer

4. cut into small part

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6. FACTORS THAT AFFECT THE RATE OF REACTION

7. Rate of reaction is affected by : size of SOLID concentration of SOLUTION temperature of SOLUTION Presence of catalyst pressure of GAS reactant

8. experiments Investigate factors that affect the rate of reaction Factor 1: size of SOLID Factor 2: concentrationof SOLUTION Factor 3: temperature of SOLUTION Factor 4: presence of catalyst

9. experiments Factor 1: size of SOLID Reactant : Differents SIZE : calcium carbonate, CaCO3 Fixed VOLUME and CONCENTRATION : 20 cm3 of 0.5 mol dm-3hydrochloric acid, HCℓ

10. 20 cm3 of 0.5 mol dm-3 hydrochloric acid, HCℓ CO2 burette water EXPERIMENT I : excess of calcium carbonate small chips EXPERIMENT II : excess of calcium carbonate large chips

11. size of SOLID calcium carbonate, CaCO3 LARGE SMALL POWDER

12. Balanced equation : CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g) Observable changes: Volume of carbon dioxide collected every 30 seconds by water displacement in the burette. Volume of CO2/cm3 Experiment I : small CaCO3 V Experiment II : large CaCO3 Time/s

13. From graph sketch: The gradient of the curve for experiment I is greater than the curve for experiment II. The rate of reaction in experiment I is higher than experiment II. Volume of CO2/cm3 Gradient Experiment I : small CaCO3 Volume of Carbon dioxide,CO2 gas collected are equal V Gradient Experiment II : large CaCO3 Time/s

14. The gradient of the curve for experiment I is greater than the curve for experiment II. The rate of reaction in experiment I is higherthan experiment II. The rate of reaction of the small calcium carbonate chips is higher. Since calcium carbonate used in excess, all hydrochloric acid has reacted. [ HCℓ reacted completed  CaCO3excess]. The number of mole of hydrochloric acid in both experiments: = 0.5 mol dm-3 x 20 cm3 1000 = 0.01 mol The volume of carbon dioxide gas collected for both experiments are equal because number of mol of hydrochloric acid experiment I and experiment II are equal = MV 1000

15. explaination…! break into small size 1 unit 2 unit Total surface Area = 6 sides x 2 unit x 2 unit = 24 unit2 Total surface Area = 8 cubes x 6 sides x 1 unit x 1 unit = 48 unit2 Size of reactant (solid) decrease, total surface area for reaction occurs increase, reaction occurs faster, rate of reaction is higher.

16. experiments Factor 2: concentration of SOLUTION Reactant : Differents CONCENTRATION : 45 cm3 of 0.2 mol dm-3sodium thiosulphate, Na2S2O3 [diluted with different volume of distilled water] Fixed VOLUME and TEMPERATURE : 5 cm3 of 1.0 mol dm-3sulphuric acid, H2SO4

17. Record the time as soon as the cross mark cannot be seen Mencatatmasasebaiksahajatandapangkahtidakkelihatan Experiment repeated, four more times using 2.0 mol dm-3 sodium thiosulphate solution dilute with different volume of distilled water. Sodium thiosulphate solution + sulphuric acid + distilled water Larutannatriumtiosulfat + asidsulfurik + air suling White paper Kertasputih Cross mark Tandapangkah V x 0.2 50

18. concentration of SOLUTION Experiment 1: Concentration Na2S2O3 (high concentration) 1. Measure 50 cm3 of 0.2 mol dm3 sodium thiosulphate solution and pour into conical flask. 2. Measure 5 cm3sulphuric acid and mix into solution in conical flask. At the same time, start the stopwatch. 3. Swirl and see “X” sign disappear (no longer visible) Concentration of Na2S203 used is changed when diluted with distilled water in Experiment 2 until Experiment 5. Total volume of Na2S203 and distilled are 50 cm3 Experiment 2-5: Concentration Na2S2O3 (lower concentration) 1. Measure 40 cm3 of 0.2 mol dm3 sodium thiosulphate solution and pour into conical flask and diluted with 10 cm3 distilled water. (total volume = 50 cm3) 2. Measure 5 cm3sulphuric acid and mix into solution in conical flask. At the same time, start the stopwatch. 3. Swirl and see “X” sign disappear (no longer visible)

19. Balanced equation : Na2S2O3(aq) + H2SO4(aq)  Na2SO4(aq) + H2O(l) + SO2(g) + S(s) Yellow precipitate Observable changes: Time taken for the ‘X’ sign placed under the conical flask to disappear from view. Fixed quantity of solid sulphur is formed in every experiment.

20. Graph concentration of Na2S2O3 against time As the concentration of sodium thiosulphate solution decrease, the longer time is needed for marked cross to disappear. Concentration of sodium thiosulphate solution is inversely proportional to time taken for the marked cross to disappear. The higher the concentration, the short is the time taken for the yellowsulphurprecipitate to appear and the faster for the ‘X’ sign to disappear.

21. Graph concentration of Na2S2O3 against 1/time As the concentration of sodium thiosulphate solution increase, the value of 1/time increases. 1/time represents the rate of reaction. The higher the concentration of sodium thiosulphate solution, the higher is the rate of reaction

22. explaination…! sulphur atom concentrated solution diluted solution Number of particle per unit volume higher Number of particle per unit volume lower Concentration of reactant increase, Number of particle per unit volume increase, Products formed increase because reaction occurs faster, Rate of reaction is higher.

23. experiments Factor 2: temperatureof SOLUTION Reactant : DifferentsTEMPERATURE : Heat sodium thiosulphate, Na2S2O3 solution increase temperature to 35oC, 40oC, 45oC, 50oC respectively Fixed VOLUME and CONCENTRATION : 5 cm3 of 1.0 mol dm-3sulphuric acid, H2SO4 50 cm3 of 0.2 mol dm-3 sodium thiosulphate, Na2S2O3

24. 50 cm3 of 2.0 mol dm-3 sodium thiosulphate solution 5cm3 1.0 mol dm-3 sulphuric acid Cross mark Tandapangkah White paper Kertasputih Sodium thiosulphate solution + sulphuric acid Larutannatriumtiosulfat + asidsulfurik

25. 50 cm3 of 2.0 mol dm-3 sodium thiosulphate solution 5cm3 1.0 mol dm-3 sulphuric acid White paper Kertasputih Experiment 1 : Record initial temperature, RT Cross mark Tandapangkah Experiment repeated, four more times with different TEMPERATURE (heat sodium thiosulphate raise temperature before add acid)

26. Balanced equation : Na2S2O3(aq) + H2SO4(aq)  Na2SO4(aq) + H2O(l) + SO2(g) + S(s) Yellow precipitate Observable changes: Time taken for the ‘X’ sign placed under the conical flask to disappear from view. Fixed quantity of solid sulphur is formed in every experiment.

27. Graph temperature of Na2S2O3 against time From the graph; As the temperature of sodium thiosulphate solution decreases, a longer time needed for the marked cross to disappear Temperature of sodium thiosulphate solution is inversely proportional to time taken for the ‘X’ sign to disappear. Temperature of Na2S2O3, °C Exp. 5 Exp. 4 Exp. 3 Exp. 2 Exp. 1 Time / s The higher the temperature, the lower/shorter is the time taken for the yellowsulphur precipitate to appear and the faster for the ‘X’ sign to disappear.

28. Graph temperature of Na2S2O3 against time From the graph; As the temperature of sodium thiosulphate solution increases, the value of 1/time increases. 1/time represents the rate of reaction. The higher the temperature of sodium thiosulphate solution, the higher is the rate of reaction. Temperature of Na2S2O3, °C Exp. 5 Exp. 4 Exp. 3 Exp. 2 Exp. 1 1 Time , s-1

29. explaination…! Temperature : High Energy Content High Higher Kinetic Energy Particles Move Faster Temperature : Low Energy Content Low Lower Kinetic Energy Particles Move Slower Temperature of reactant increase, kinetic energy increase, particles move faster then easily collide each others, Products formed increase because reaction occurs faster, Rate of reaction is higher.

30. A (more/less) concentrated liquid detergent is more effective to remove dirt from the clothes as compared to the less concentrated detergent. This is because a (lower/higher) concentrated liquid detergent contains more particles per unit (volume/mass). The rate of removing dirt from the clothes is (lower/higher). ENHANCEMENT CORNER Food stored in refrigerator lasts longer than food stored in the kitchen cabinet. This is because in a (cold/hot/warm) condition, the bacteria are not active and only produce a (little /more) toxic. As a result, the rate of food decay is (higher/lower) and the food can be stored (shorter/longer).

31. Chicken in (bigger / smaller) chunks takes a (longer / shorter) time to cook as compared to chicken in (smaller / bigger) chunks. This is because the smaller chunks of chicken have a (larger / smaller) total surface area. (More/Less) heat can be (released / absorbed) resulting in (higher / lower) rate of cooking. ENHANCEMENT CORNER PROBLEM STATEMENT In the industrial manufacture of chemicals, foods and others; very high temperature and pressure condition to run an industrial process to produce higher yield. Higher cost required to run an industrial process. What we will choose to solve this problem and get optimum condition, run an industrial process in shorter time, lower cost and more products can produced?