1. 16.1 Rate Expression

2. Assessment Statements 16.1.1 Distinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant. 16.1.2 Deduce the rate expression for a reaction from experimental data. 16.1.3 Solve probems involving the rate expression. 16.1.4 Sketch, identify and analyse graphical representations for zero-, first-, and second- order reactions.

3. Assessment Statements 16.1.1 Distinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant.

4. Keep In Mind Kinetics is an experimentally determined science. There are no theories that can determine the rate expression by merely considering the chemicals reacting. By observation we can see that varying certain conditions changes the rate with which the reactants are used up (or the products formed). Kinetics investigates this and in doing so gives us clues as to the mechanism of the reaction.

5. Experimentally When all other factors are kept constant we can investigate how the rate of a reaction depends on the concentration of the reactants. [A] + [B] --> [Products]

6. The Rate Expression To allow for any possibility we can express the effect of the concentrations of A and B on the rate of the reaction by the expression: Rate = k [A] x [B] y Order of the reaction. With respect to that species Rate Constant Concentrations of the species

7. Notes: If changing the concentration of A has no effect on the rate  x=0, If the rate is directly proportional to the concentration of A  x=1. If there is some other effect (e.g rate is directly proportional to the square of the concentration) then x will take another number. This value is called THE ORDER OF REACTION with respect to that reactant.

8. Rate Constant The 'k' is called the rate constant It has a constant value for any specific reaction at constant temperature. The value of the rate constant for a reaction gives us a measure of how fast the reaction is. A very small value for k means a very slow reaction and vice versa.

9. Order of The Reaction The total order of the reaction (the OVERALL ORDER OF REACTION) is the sum of the individual orders, in this case X + Y. The order of the reaction with respect to a specific reagent provides information about the number of particles of this reagent involved in the rate determining step of the mechanism. The rate determining step is the slowest step. Note: If an equilibrium feeds into the rate determining step then this will also influence the order with respect to any particles that appear on the left hand side (reactants) of the equilibrium.

10. Units for the Overall Rate Constant Overall Reaction Order Units of k (t in seconds) omol.dm -3.s -1 1s -1 2dm 3.mol -1.s -1 3dm 6.mol -2.s -1

11. First Order The units of the rate constant depend on the total order of the reaction. For order = 1 (first order reaction) The rate expression is: Rate = k [A] 1 Substituting the units in each part we get: mol dm -3 s -1 = k [mol dm -3 ] 1 Cancelling out from both sides gives: s -1 = k

12. 2nd Order Rate = k [A] 2 Substituting the units in each part we get: mol dm -3 s -1 = k [mol dm -3 ] 2 Rearranging and cancelling out from both sides gives: dm 3 mol -1 s -1 = k

13. Assessment Statements 16.1.2 Deduce the rate expression for a reaction from experimental data. 16.1.3 Solve problems involving the rate expression.

14. Experimental Measurements These can involve following the disappearance of reactants or the generation of products. There are many techniques that may be used: Measuring the evolution of a gas volume with time Measuring the decreasing mass of a reaction vessel as a gas is evolved Measuring the changing characteristics of a solution in terms of colour (using a colourimeter), conductivity, pH etc.

15. Experimental Procedure to determine the rate expression Measure the rate for a known series of initial reactant concentrations Change one of the concentrations keeping the other constant and measure the rate again Change the concentration of the one previously kept constant, and measure the initial rate again. Repeat this procedure until enough data is obtained

16. Example Experiment[A] mol dm -3 [B] mol dm -3 [C] mol dm -3 Rate mol dm -3 s -1 10.1 6.2 x 10 -4 20.10.20.11.2 x 10 -3 30.1 0.26.2 x 10 -4 40.20.10.22.5 x 10 -3 When a component is kept constant between two experiments we can effectively remove it from the rate expression by combining its constant value with the rate constant to make a kind of super constant (that we are not concerned with any way). In this way we know that any change in rate is due to the concentration that changes.

17. Changing B (Expt 1 and 2) This means that the rate is directly proportional to the concentration of B. Rate = k'[B] So in this case the order with respect to [B] is equal to 1 Rate = k'[B] 1 Experiment[A] mol dm -3 [B] mol dm -3 [C] mol dm -3 Rate mol dm -3 s -1 10.1 6.2 x 10 -4 20.10.20.11.2 x 10 -3 30.1 0.26.2 x 10 -4 40.20.10.22.5 x 10 -3

18. Experiments 1 and 3 Repeat the process Concentrations of A and B are constant, so any change in the rate is due to the change in concentration of C Rate does not change. So changing C doesn't affect the rate. The order with respect to the concentration of C must be zero. Rate = k''[C] 0

19. Experiments 3 and 4 The concentrations of B and C are kept constant while [A] doubles. Rate increases fourfold between these two experiments. That means that any change in the concentration of A produces a squared change in the rate. The order with respect to [A] is therefore 2. Rate = k''[A] 2

20. Putting It All Together Now all of the orders can be placed in the rate expression. Rate = k [A] 2 [B] 1 [C] 0

21. Calculating Rate Constant Choosing 1 Experiment 6.2 x 10 -4 = k [0.1] 2 [0.1] 1 [0.1] 0 k = 6.2 x 10 -4 0.1 x 0.1 x 0.1 k = 6.2 x 10 -1 dm 6 mol -2 s -1

22. Writing a (differential) Rate Law 2 NO(g) + Cl 2 (g)  2 NOCl(g) Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: Experiment[NO] (moldm - ) (moldm -3 ) [Cl 2 ] (moldm - ) (moldm -3 )Rate moldm - s -1 moldm -3 s -1 10.2500.250 1.43 x 10 -6 20.5000.250 5.72 x 10 -6 30.2500.500 2.86 x 10 -6 40.5000.500 11.4 x 10 -6

23. Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: Experiment[NO] (moldm -3 ) [Cl 2 ] (moldm -3 ) Rate moldm -3 s -1 10.2500.250 1.43 x 10 -6 20.5000.250 5.72 x 10 -6 30.2500.500 2.86 x 10 -6 40.5000.500 1.14 x 10 -5 In experiment 1 and 2, [Cl 2 ] is constant while [NO] doubles. The rate quadruples, so the reaction is second order with respect to [NO] R = k[NO] x [Cl 2 ] y R = k[NO] 2 [Cl 2 ] y

24. Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law : Experiment[NO] (mol/L) [Cl 2 ] (mol/L) Rate Mol/L·s 10.250 1.43 x 10 -6 20.5000.2505.72 x 10 -6 30.2500.5002.86 x 10 -6 40.500 1.14 x 10 -5 R = k[NO] 2 [Cl 2 ] y In experiment 2 and 4, [NO] is constant while [Cl 2 ] doubles. The rate doubles, so the reaction is first order with respect to [Cl 2 ] R = k[NO] 2 [Cl 2 ]

25. Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data: Experiment[NO] (mol/L) [Cl 2 ] (mol/L) Rate Mol/L·s 10.250 1.43 x 10 -6 R = k[NO] 2 [Cl 2 ]

26. Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO] 2 [Cl 2 ] Overall order is the sum of the exponents, or orders, of the reactants 2 +1= 3 The reaction is 3 rd order

27. Practice Determine the orders of reactions for each species Determine the rate constant.

28. Practice 2NO + 2H 2  N 2 + 2H 2 O is the reaction we’re studying, this is the data found during our experimentation. Determine the orders of reactions for each species Determine the rate constant

29. Practice Use the kinetics data to write the rate law for the reaction. What overall reaction order is this? 2NO + O 2  2NO 2

30. Sample Exercise 3 Rate data for the reaction: CH 3 Br + OH -  CH 3 OH + Br - Use the data to find the experimental rate law, the rate constant and the overall order of reaction

31. Distinctions Between Rate And The Rate Constant k The rate of a reaction is the change in concentration with time, whereas the rate constant is the proportionality constant relating reaction rate to the concentrations of reactants. The rate constant remains constant throughout a reaction, regardless of the initial concentrations of the reactants. The rate and the rate constant have the same numerical values and units only in zero-order reactions. For reaction orders other than zero, the rate and rate constant are numerically equal only when the concentrations of all reactants are 1 M. Even then, their units are different.

32. Collecting Data

33. Assessment Statements 16.1.4 Sketch, identify and analyse graphical representations for zero-, first-, and second- order reactions.

34. Other Methods Order of a reaction can also be found from a graph of concentration against time.

35. Graphing conc. Vs time Shows the effect of reactants being used up on the rate of reaction If constant = zero order Decreasing concentration is not affecting the rate

36. 0 th Order Reactions There is no correlation between the rate and the concentration of a zeroth order reaction. A flat line is obtained. This is in keeping with the rate law expression: Rate = k[A] 0

37. Graphing first order If the reaction rate is halved when the concentration is halved, then the reaction is first order

38. 1 st Order Reactions There is direct proportionality between the rate and the concentration. A straight line is obtained passing through the origin. This is in keeping with the rate law expression: Rate = k[A] 1 The gradient (slope) of the graph gives the rate constant k, whose units are now s -1

39. Graphing Second Order If halving the concentration causes the rate to decrease by a factor of 4, the reaction is second order (1/2 ) 2 = ¼

40. 2 nd Order Reactions When rate is plotted against concentration a curve is obtained. However, when the order is any other than 0th or 1st a curve is also given. The graph is of limited value in this case and the data must be further processed. Inspection of the rate expression for a second order reaction Rate = k[A] 2 shows that a plot of rate against concentration would give a curve.

41. Rate vs. Conc Graphs

42. Using Logs If logs are taken throughout however this gives: log Rate = log k + 2log[A] This has the same form as a straight line graph y = mx +c A plot of log rate against log[A] gives a straight line graph whose intercept is the value for log k and the gradient is equal to the order of the reaction. This treatment is valid for any order values. The gradient (slope) of the line is always equal to the order.

43. Graph Representations Zero OrderFirst Order Second Order

44. Using Half Life Half Life is Halving Constant half Life (= 0.693/rate constant k) Half Life is Doubling

47. An Overview of zero, first and simple second order reactions