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Rate Of Reaction

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Rate Of Reaction

  1. 1. Rate of Reaction<br />
  2. 2. What is the best way to fry chicken quickly?<br />Freestyle Fried Chicken<br />
  3. 3. using deep fryer<br />
  4. 4. cut into small part<br />
  5. 5. EASY… call delivery<br />
  6. 6. FACTORS THAT AFFECT<br />THE RATE OF REACTION<br />
  7. 7. Rate of reaction is affected by :<br />size of SOLID<br />concentration of SOLUTION<br />temperature of SOLUTION<br />Presence of catalyst<br />pressure of GAS reactant<br />
  8. 8. experiments<br />Investigate factors that affect the rate of reaction <br />Factor 1:<br />size of SOLID<br />Factor 2:<br />concentrationof SOLUTION<br />Factor 3:<br />temperature of SOLUTION<br />Factor 4:<br />presence of catalyst<br />
  9. 9. experiments<br />Factor 1:<br />size of SOLID<br />Reactant :<br />Differents SIZE :<br />calcium carbonate, CaCO3<br />Fixed VOLUME and CONCENTRATION :<br />20 cm3 of 0.5 mol dm-3hydrochloric acid, HCℓ<br />
  10. 10. 20 cm3 of 0.5 mol dm-3 hydrochloric acid, HCℓ<br />CO2<br />burette<br />water<br />EXPERIMENT I : excess of calcium carbonate small chips<br />EXPERIMENT II : excess of calcium carbonate large chips<br />
  11. 11. size of SOLID<br />calcium carbonate, CaCO3<br />LARGE<br />SMALL<br />POWDER<br />
  12. 12. Balanced equation :<br />CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)<br />Observable changes:<br />Volume of carbon dioxide collected every 30 seconds by water displacement in the burette.<br />Volume of CO2/cm3<br />Experiment I : small CaCO3<br />V<br />Experiment II : large CaCO3<br />Time/s<br />
  13. 13. From graph sketch:<br />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.<br />Volume of CO2/cm3<br />Gradient Experiment I : small CaCO3<br />Volume of Carbon dioxide,CO2 gas collected are equal<br />V<br />Gradient Experiment II : large CaCO3<br />Time/s<br />
  14. 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.<br />The rate of reaction of the small calcium carbonate chips is higher.<br />Since calcium carbonate used in excess, all hydrochloric acid has reacted. [ HCℓ reacted completed  CaCO3excess].<br />The number of mole of hydrochloric acid in both experiments:<br />= 0.5 mol dm-3 x 20 cm3<br /> 1000<br />= 0.01 mol<br />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<br />= MV<br /> 1000<br />
  15. 15. explaination…!<br />break into small size<br />1 unit<br />2 unit<br />Total surface Area <br />= 6 sides x 2 unit x 2 unit<br />= 24 unit2<br />Total surface Area <br />= 8 cubes x 6 sides x 1 unit x 1 unit<br />= 48 unit2<br />Size of reactant (solid) decrease, <br />total surface area for reaction occurs increase, <br />reaction occurs faster, <br />rate of reaction is higher.<br />
  16. 16. experiments<br />Factor 2:<br />concentration of SOLUTION<br />Reactant :<br />Differents CONCENTRATION :<br />45 cm3 of 0.2 mol dm-3sodium thiosulphate, Na2S2O3 <br />[diluted with different volume of distilled water]<br />Fixed VOLUME and TEMPERATURE :<br />5 cm3 of 1.0 mol dm-3sulphuric acid, H2SO4<br />
  17. 17. Record the time as soon as the cross mark cannot be seen<br />Mencatatmasasebaiksahajatandapangkahtidakkelihatan<br />Experiment repeated, four more times using 2.0 mol dm-3 sodium thiosulphate solution dilute with different volume of distilled water.<br />Sodium thiosulphate solution + sulphuric acid + distilled water<br />Larutannatriumtiosulfat + asidsulfurik + air suling<br />White paper<br />Kertasputih<br />Cross mark<br />Tandapangkah<br />V x 0.2<br />50<br />
  18. 18. concentration of SOLUTION<br />Experiment 1: Concentration Na2S2O3 (high concentration)<br />1. Measure 50 cm3 of 0.2 mol dm3 sodium thiosulphate solution and pour into conical flask.<br />2. Measure 5 cm3sulphuric acid and mix into solution in conical flask. At the same time, start the stopwatch.<br />3. Swirl and see “X” sign disappear (no longer visible) <br />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<br />Experiment 2-5: Concentration Na2S2O3 (lower concentration)<br />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)<br />2. Measure 5 cm3sulphuric acid and mix into solution in conical flask. At the same time, start the stopwatch.<br />3. Swirl and see “X” sign disappear (no longer visible) <br />
  19. 19. Balanced equation :<br />Na2S2O3(aq) + H2SO4(aq)  Na2SO4(aq) + H2O(l) + SO2(g) + S(s)<br />Yellow precipitate<br />Observable changes:<br />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.<br />
  20. 20. Graph concentration of Na2S2O3 against time<br />As the concentration of sodium thiosulphate solution decrease, the longer time is needed for marked cross to disappear.<br />Concentration of sodium thiosulphate solution is inversely proportional to time taken for the marked cross to disappear.<br />The higher the concentration, the short is the time taken for the yellowsulphurprecipitate to appear and the faster for the ‘X’ sign to disappear.<br />
  21. 21. Graph concentration of Na2S2O3 against 1/time<br />As the concentration of sodium thiosulphate solution increase, the value of 1/time increases. 1/time represents the rate of reaction.<br />The higher the concentration of sodium thiosulphate solution, the higher is the rate of reaction<br />
  22. 22. explaination…!<br />sulphur atom<br />concentrated solution<br />diluted solution<br />Number of particle per unit volume higher<br />Number of particle per unit volume lower<br />Concentration of reactant increase,<br />Number of particle per unit volume increase, Products formed increase because reaction occurs faster, <br />Rate of reaction is higher.<br />
  23. 23. experiments<br />Factor 2:<br />temperatureof SOLUTION<br />Reactant :<br />DifferentsTEMPERATURE :<br />Heat sodium thiosulphate, Na2S2O3 solution increase temperature to 35oC, 40oC, 45oC, 50oC respectively<br />Fixed VOLUME and CONCENTRATION :<br />5 cm3 of 1.0 mol dm-3sulphuric acid, H2SO4<br />50 cm3 of 0.2 mol dm-3 sodium thiosulphate, Na2S2O3 <br />
  24. 24. 50 cm3 of 2.0 mol dm-3 sodium thiosulphate solution<br />5cm3 1.0 mol dm-3 sulphuric acid <br />Cross mark<br />Tandapangkah<br />White paper<br />Kertasputih<br />Sodium thiosulphate solution + sulphuric acid<br />Larutannatriumtiosulfat + asidsulfurik<br />
  25. 25. 50 cm3 of 2.0 mol dm-3 sodium thiosulphate solution<br />5cm3 1.0 mol dm-3 sulphuric acid <br />White paper<br />Kertasputih<br />Experiment 1 : Record initial temperature, RT<br />Cross mark<br />Tandapangkah<br />Experiment repeated, four more times with different TEMPERATURE <br />(heat sodium thiosulphate raise temperature before add acid)<br />
  26. 26. Balanced equation :<br />Na2S2O3(aq) + H2SO4(aq)  Na2SO4(aq) + H2O(l) + SO2(g) + S(s)<br />Yellow precipitate<br />Observable changes:<br />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.<br />
  27. 27. Graph temperature of Na2S2O3 against time<br />From the graph;<br />As the temperature of sodium thiosulphate solution decreases, a longer time needed for the marked cross to disappear<br />Temperature of sodium thiosulphate solution is inversely proportional to time taken for the ‘X’ sign to disappear.<br />Temperature of Na2S2O3, °C<br />Exp. 5<br />Exp. 4<br />Exp. 3<br />Exp. 2<br />Exp. 1<br />Time / s<br />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.<br />
  28. 28. Graph temperature of Na2S2O3 against time<br />From the graph;<br />As the temperature of sodium thiosulphate solution increases, the value of 1/time increases. 1/time represents the rate of reaction.<br />The higher the temperature of sodium thiosulphate solution, the higher is the rate of reaction.<br />Temperature of Na2S2O3, °C<br />Exp. 5<br />Exp. 4<br />Exp. 3<br />Exp. 2<br />Exp. 1<br /> 1<br />Time , s-1<br />
  29. 29. explaination…!<br />Temperature : High<br />Energy Content High<br />Higher Kinetic Energy<br />Particles Move Faster<br />Temperature : Low<br />Energy Content Low<br />Lower Kinetic Energy<br />Particles Move Slower<br />Temperature of reactant increase,<br />kinetic energy increase, particles move faster then easily collide each others, <br />Products formed increase because reaction occurs faster, Rate of reaction is higher.<br />
  30. 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).<br />ENHANCEMENT<br />CORNER<br />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).<br />
  31. 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.<br />ENHANCEMENT<br />CORNER<br />PROBLEM STATEMENT<br />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. <br />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?<br />

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