1. RATE LAW Experiment #4

2. What are we doing in this experiment? Measure the rate of a chemical reaction between potassium iodide (KI) and hydrogen peroxide (H 2 O 2 ) and use this information to obtain the order and rate constant of the reaction.

3. What is rate? Rate refers to speed. In this experiment we are trying to measure the speed or the rate of a chemical reaction. What is rate of a reaction? It describes how fast reactants are used up and how fast products are formed.

4. What is rate of a reaction? Change in concentration of the reactant or product with change in time Mathematically expressed, change Square brackets denotes concentration

5. Average and instantaneous rate If  (time) is large----- Average rate If  (time) is small ----- instantaneous rate Large and small can be defined based on the time scale of the reaction

6. Unit of rate

8. Rate of a reaction A + 3B 2C+ 2D - sign signifies decrease in concentration + sign signifies decrease in concentration

9. A + 3B 2C+ 2D Rate of consumption of B is 3 times the rate of consumption of A Rate of formation of C is 2 times the rate of consumption of A Rate of formation of D is 2 times the rate of consumption of A Rate of formation of C is equal to the rate of formation of D Rate of consumption of B is 1.5 times the rate of formation of C and D These fractions makes sure that the rates are equal

10. Conclusion I If want to measure the rate of a reaction we can follow the decrease in concentration of a reactant or increase in the concentration of a product over a period of time.

11. Rate law The rate of a homogenous reaction at any instant is proportional to the product of molar concentrations of the reactants, each molarity raised to some power or exponent that has to be found by experiment. A + B products According to the above state, its rate can be expressed as Instantaneous rate

12. Rate law k is called the rate constant This equation is called the rate law m and n are the order of the reaction due to A and B respectively Overall order of the reaction = m + n

13. RATE CONSTANTS HAVE UNITS THAT DEPEND ON THE ORDER OF THE REACTION Rate constant of a first order reaction: A products

15. Rate constant of a second order reaction: A products

16. Rate constant of a third order reaction: A products

17. Conclusion II A + B products For any reaction, Rate law is given by, k is called the rate constant m and n are the order of the reaction due to A and B respectively. The orders have to be determined experimentally. Overall order of the reaction = m + n

18. Conclusion II A + B products For any reaction, The order of the reaction due to each reactant is usually not the same as the stoichiometric coefficients of the reactants. The order of the reaction due to each reactant could be the same as the stoichiometric coefficients of the reactants, only if experimentally proved so. The unit of the rate constant changes with the order of the reaction.

19. Example problem A + B products For any reaction, Concentration-Rate Data for the hypothetical reaction A + B Products Initial concentration Exp [A] [B]Initial rate of formation of P (mol L -1 ) (mol L -1 s -1 ) 1 0.10 0.100.20 2 0.20 0.100.40 3 0.30 0.100.60 4 0.30 0.202.40 5 0.30 0.305.40

20. Example problem What are the rate law, orders and rate constant of the reaction? Solution: The given reaction is, A + B products Since we do not know the rate law for the reaction, we have to start with a general rate law. The general rate law is given by Where k is the unknown rate constant that we have to find. Where m, n are the unknown orders of A and B respectively, that we have to determine based on the experimental data.

21. Initial concentration Exp [A] [B]Initial rate of formation of P (mol L -1 ) (mol L -1 s -1 ) 1 0.10 0.100.20 2 0.20 0.100.40 Substituting the concentrations and rate in the general rate law ------------- (1) ------------- (2)

22. ------------- (1) ------------- (2) Dividing equation (1) by equation (2) 1 2

23. Since the bases are equal, for left hand side (LHS) of the equation to be equal to the right hand side (RHS) of the equation, the exponents must be equal. Method 1 Base Exponent

24. Method 2 Take log on both sides of the equation Applying

25. Initial concentration Exp [A][B]Initial rate of formation of P (mol L -1 ) (mol L -1 s -1 ) 3 0.30 0.100.60 4 0.30 0.202.40 Substituting the concentrations and rate in the general rate law ------------- (3) ------------- (4)

26. ------------- (3) ------------- (4) Dividing equation (3) by equation (4) 1 4

27. Since the bases are equal, for left hand side (LHS) of the equation to be equal to the right hand side (RHS) of the equation, the exponents must be equal. Method 1 Base Exponent

28. Method 2 We know that the LHS of the equation must be equal to RHS. So we can ask ourselves this question: How many times do I have to multiply half to itself to obtain one-fourth? The answer is 2 times

29. Method 3 Take log on both sides of the equation Applying

30. Now we have, So the rate law becomes, Now we can pick any one of the 5 experiments and plug-in in the above equation, the values of concentration and rate and obtain the value of rate constant for this reaction. The overall order of the reaction= m + n = 1 + 2 = 3

31. Let us say, we pick experiment 5 Initial concentration Exp [A] [B]Initial rate of formation of P (mol L -1 ) (mol L -1 s -1 ) 5 0.30 0.305.40

33. Our experiment 2 I - (aq) + H 2 O 2 (aq) + 2 H + (aq) → I 2 (aq) + 2 H 2 O(l) There are 3 concentrations that change in this experiment. We can simplify the experiment by keeping one of the concentrations a constant. This means that we will never vary the concentration of that particular molecule in all our experiments. Kept Constant by the use of buffer

34. Determination 1:Iodide Solution:Peroxide Solution: 10 mL 0.3 M KI30 mL 0.1 M H 2 O 2 10 mL 0.02 M Na 2 S 2 O 3 25 mL 0.5/0.5 M acetate buffer 2 mL starch solution 123 mL distilled water Determination 2: Iodide Solution:Peroxide Solution: 20 mL 0.3 M KI 30 mL 0.1 M H 2 O 2 10 mL 0.02 M Na 2 S 2 O 3 25 mL 0.5/0.5 M acetate buffer 2 mL starch solution 113 mL distilled water Determination 3: Iodide Solution: Peroxide Solution: 30 mL 0.3 M KI 30 mL 0.1 M H 2 O 2 10 mL 0.02 M Na 2 S 2 O 3 25 mL 0.5/0.5 M acetate buffer 2 mL starch solution 103 mL distilled water

35. Determination 3:Iodide Solution:Peroxide Solution: 30 mL 0.3 M KI30 mL 0.1 M H 2 O 2 10 mL 0.02 M Na 2 S 2 O 3 25 mL 0.5/0.5 M acetate buffer 2 mL starch solution 123 mL distilled water Determination 4: Iodide Solution: Peroxide Solution: 30 mL 0.3 M KI 50 mL 0.1 M H 2 O 2 10 mL 0.02 M Na 2 S 2 O 3 25 mL 0.5/0.5 M acetate buffer 2 mL starch solution 83 mL distilled water Determination 5: Iodide Solution: Peroxide Solution: 30 mL 0.3 M KI 70 mL 0.1 M H 2 O 2 10 mL 0.02 M Na 2 S 2 O 3 25 mL 0.5/0.5 M acetate buffer 2 mL starch solution 63 mL distilled water

36. Example problem A + B products For any reaction, Concentration-Rate Data for the hypothetical reaction A + B Products Initial concentration Exp [A] [B]Initial rate of formation of P (mol L -1 ) (mol L -1 s -1 ) 1 0.10 0.100.20 2 0.20 0.100.40 3 0.30 0.100.60 4 0.30 0.20 2.40 5 0.30 0.30 5.40

37. Determining m: Taking natural log on both sides of the equation

38. When determining m, the rates are measured by varying [I - ] and keeping [H 2 O 2 ] constant. So, nLn[H 2 O 2 ] remains a constant say C 1.

39. Since K’ is a constant, Ln(K’) is also a constant. We can sum Ln(k’) and C1 and assign a new constant C’ [I-]RateLn (Rate)Ln [I - ] [I 1 ] [I 2 ] [I 3 ] Rate1Ln [I 1 ] Rate2Ln [I 2 ] Rate3Ln [I 3 ] Ln (Rate1) Ln (Rate2) Ln (Rate3)

40. Ln [I - ] Ln (Rate) Ln [I - ] p Best-fit line x x p q x y Equation of the best-fit line: Y=mX + z m = slope = Ln [I - ] 2,Ln (Rate 2 ) Ln [I - ] 3,Ln (Rate 3 ) Ln (Rate p ) Ln (Rate q ) Ln [I - ] 1,Ln (Rate 1 ) Ln [I - ] q Intercept, Z

41. Slope = order = m Determining n: When determining n, the rates are measured by varying [H 2 O 2 ] and keeping [I - ] constant. So, mLn[I - ] remains a constant say C 2.

42. Since K’ is a constant, Ln(K’) is also a constant. We can sum Ln(k’) and C 2 and assign a new constant C’ [H 2 O 2 ]RateLn (Rate)Ln [H 2 O 2 ] [H 2 O 2 ] 1 [H 2 O 2 ] 2 [H 2 O 2 ] 3 Rate 1 Ln [H 2 O 2 ] 1 Rate 2 Rate 3 Ln (Rate 1 ) Ln (Rate 2 ) Ln (Rate 3 ) Ln [H 2 O 2 ] 2 Ln [H 2 O 2 ] 3

43. Ln [H 2 O 2 ] Ln (Rate) Best-fit line x x p q x y Equation of the best-fit line: Y=mX + z m = slope = Ln (Rate p ) Ln (Rate q ) Ln [H 2 O 2 ] 1,Ln (Rate 1 ) Intercept, Z Ln [H 2 O 2 ] 2,Ln (Rate 2 ) Ln [H 2 O 2 ] 3,Ln (Rate 3 ) Ln [H 2 O 2 ] P Ln [H 2 O 2 ] q

44. Slope = order = n

45. How do we measure the rate of the reaction? 2 I - (aq) + H 2 O 2 (aq) + 2 H + (aq) → I 2 (aq) + 2 H 2 O(l) Either by following the change in concentration of a reactant or product In this reaction we follow the change in concentration of I 2 with time.

46. Since there is no I 2 formed before the reaction begins, t initial =0 and [I 2 ]=0 How do we measure the rate of the reaction?

47. How to detect the formation of I 2 ? I 2 on reaction with starch gives a blue black color Is just adding starch enough to perform all the determinations? No Why not? All the determinations will give a blue black color instantaneously if both starch and I 2 are in the same solution.

48. How do we work around the problem? 2 S 2 O 3 -2 (aq) + I 2 (aq) → 2 I - (aq) + S 4 O 6 -2 (aq) 2 I - (aq) + H 2 O 2 (aq) + 2 H + (aq) → I 2 (aq) + 2 H 2 O(l) 2 moles of I - ≡ 1 mole of I 2 2 moles of S 2 O 3 2- ≡ 1 mole of I 2 Moles of Na 2 S 2 O 3 = 0.01 L × 0.02 M= 0.0002 moles Moles of I 2 = ½ × (Moles of Na 2 S 2 O 3 )= 0.0001 moles

49. We need to exceed the initial 0.0001 moles of I 2 to be able to see the blue-black color. The speed of formation of I 2 depends on the [I - ] and [H 2 O 2 ]. Hence the blue-black color will form at different times depending on how quickly the initial 0.001 moles of I 2 are exceeded. How do we work around the problem?

50. Determination 1:Iodide Solution:Peroxide Solution: 10 mL 0.3 M KI30 mL 0.1 M H 2 O 2 10 mL 0.02 M Na 2 S 2 O 3 25 mL 0.5/0.5 M acetate buffer 2 mL starch solution 123 mL distilled water Moles of I- = 0.01 L × 0.3 M= 0.003 moles Moles of H 2 O 2 = 0.03 L × 0.1 M= 0.003 moles Moles of Na 2 S 2 O 3 = 0.01 L × 0.02 M= 0.0002 moles Moles of I 2 = ½ × (Moles of Na 2 S 2 O 3 )= 0.0001 moles

51. Calculating concentrations 100 mL mark 100mL 100 mL mark 100mL 0.2 M of A0.3 M of B 25 mL of A 65 mL of B + 110 mL of H 2 O What is the concentration of A and B in the final solution?

52. 100mL 0.2 M of A

53. 100mL 0.3 M of B