1. 8-3-2011

4. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the reaction is first order with respect to O 3 From experiments 2 and 3 keeping the concentration of O 3 constant at 2.0x10 -5 M, decreasing the concentration of NO 2 by one-half results in a decrease of the initial rate by the same value. The reaction is therefore first order with respect to NO 2 The rate law can be written as: Rate = k [NO 2 ][O 3 ]

5. The rate constant can be calculated by substitution for the concentrations of reactants and corresponding rate value, taking the first experiment, for instance, will give: 0.022 mol L -1 s -1 = k * 5.0x10 -5 (mol L -1 )* 1.0x10 -5 (mol L -1 ) k = 4.4x10 7 L mol -1 s -1

6. Example In the reaction: 2 NOCl  2 NO + Cl 2 The following data was collected at 27 o C: What is the rate law of the reaction? What is the value of the rate constant?

8. Determination of Rate Law Experiment[F 2 ][ClO 2 ]Rate (M/s) 10.040.031.0x10 -2 20.04 1.3x10 -2 30.020.046.7x10 -3 40.040.062.0x10 -2 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g)

9. Determination of Rate Law Experiments 1 & 4 As [F 2 ] doubles, so does the rate Experiments 2 & 3 As [ClO 2 ] doubles, so does the rate 2:2 ratio…..1:1 ratio x = 1 and y = 1 Rate = k [F 2 ] [ClO 2 ]

10. Experiment[S 2 O 8 2- ][I - ] Initial Rate (M/s) 10.080.0342.2 x 10 -4 20.080.0171.1 x 10 -4 30.160.0172.2 x 10 -4 Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq)

11. rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 k = rate [S 2 O 8 2- ][I - ] = 2.2 x 10 -4 M/s (0.08 M)(0.034 M) = 0.08/Ms rate = k [S 2 O 8 2- ][I - ]

12. Relation between Reactant Concentration and Time First Order Reaction- a reaction whose rate depends on the reactant concentration raised to the first power. Reaction Type: A→B Rate of: -Δ [A]/Δt or k[A] Combining and simplifying these equations brings us to the following rate equation: ln[A] t = -kt + ln[A 0 ]

13. Relation between Reactant Concentration and Time

14. Reaction Time The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88M to 0.14M? ln[A] = ln[A] 0 - kt kt = ln[A] 0 – ln[A] t = ln[A] 0 – ln[A] k = 66 s [A] 0 = 0.88 M [A] = 0.14 M ln [A] 0 [A] k = ln 0.88 M 0.14 M 2.8 x 10 -2 s -1 =

15. Example At 400 o C, the first order conversion of cyclopropane into propylene has a rate constant of 1.16x10 -6 s -1. What will the concentration of cyclopropane be after 24 hours of reaction if the initial concentration was 0.0100 M. Cyclopropane  propylene

16. ln [cyclopropane]0/[cyclopropane]t = kt ln 0.0100/[cyclopropane] t = 1.16x10 -6 s -1 * (24 hr *3600 s/hr) ln (0.0100 M)/[cyclopropane] t = 0.100 0.0100/[cyclopropane] t = e 0.100 [cyclopropane] t = 9.01x10 -3 M

17. The isomerization reaction CH3NC  CH3CN obeys the first order rate law rate = k [CH3NC] Measurements at 500 K reveal that in 520 seconds the concentration of CH3NC has decreased to 71% of its original value. Calculate the rate constant k of the reaction at 500 K.

18. ln[A t ] = -kt + ln[A 0 ] It is important to remember that for first order reactions, one does not need to know the absolute values of [A t ] and [A 0 ] to determine the rate constant (if time is known) or the time (if the rate constant is given). One can work with relative values since the above equation can be recast in the following way: ln([A t ]/[A 0 ]) = -kt

19. Since the relationship between [A t ] and [A 0 ] is known then the problem can be solved. The problem states 71% of A is left after 520 seconds: ln([A t ]/[A 0 ]) = -kt ln(0.71[A 0 ]/[A 0 ) = -k (520 seconds) k = 0.00066 s -1

20. Reaction Half-life As a reaction proceeds, the concentrations of the reactants decreases. Another way to measure [reactant] over time is to use the half-life. Half-life, t 1/2 – the time required for the concentration of a reactant to decrease to half of its initial concentration.

21. Reaction Half-life Considering the condition where one-half of A is consumed ([A] t = ½ [A] 0 ), the time can be denoted as t ½ ; ln [A] 0 / ½ [A] 0 = kt ½ ln 2 = kt ½ t ½ = (ln 2)/k t ½ = 0.693/k

22. A product First-order reaction # of half-lives [A] = [A] 0 /n 1 2 3 4 2 4 8 16 13.3

23. Reaction Half-life

24. The decomposition of RX has a rate law: rate = k[RX]. If, at 550 o C, k = 0.032 s -1, find: 1.The half life of the reaction 2.The initial concentration of RX provided that the concentration of RX after one min of reaction was 0.010 M.

25. a. Calculation of the half life of the reaction t ½ = (ln 2)/k t ½ = 0.693/0.032 s -1 t ½ = 22 s b. ln [A] 0 /[A] t = kt ln [RX] 0 /(0.010 M) = 0.032 s -1 * (1 min * 60 s/min) ln [RX] 0 /(0.010 M) = 1.92 [RX] 0 = 0.068 M

26. Reaction Half-life What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? t½t½ ln2 k = 0.693 5.7 x 10 -4 s -1 = = 1200 s = 20 minutes t½t½ t½t½ t½t½

27. Second-Order Reactions Second-order reaction- a reaction whose rate depends on the concentration of one reactant raised to the second power OR on the concentrations of two different reactants, each raised to the first power. Simple Type: A→B rate = k[A] 2 Complex Type: A + B→C rate = k[A][B]

28. Second-order Reactions For A→B, the following expression is used: rate = -  [A]/  t = k[A] 2 k = -  [A]/([A] 2 *  t) M/M 2 s = 1/s = M -1 s -1 Integration gives: we can obtain t 1/2, by setting [A] t = [A] o /2, therefore, 1/([A] o /2) = 1/[A] o + kt 1/2 t 1/2 = 1/([A] o * k). 1 [A] t = 1 [A] 0 + kt

29. Half-life of a Second-order Reaction Equation for half-life What is the difference between this equation and the equation for half-life of first-order reactions? 1 t ½ = k[A] 0

30. At 25 o C with CCl 4 as solvent, the reaction: I + I  I 2 is second order with respect to the concentration of iodine atoms. The rate constant has been measured as 8.2 x 10 9 L mol -1 s -1. Suppose the initial concentration of I is 1.00 x 10 -4 mol L -1. Calculate [I] after 2.0 x 10 -6 s. In this particular problem, k is given as well as t and [A 0 ] while [A t ] is unknown. This is usually the case, the integrated form of the second order rate law has 4 variables so three should be given in order to solve for one.

31. The solution to this problem is straightforward, it merely involves substituting the known values for t, k and [A 0 ] and then solving for [A t ]. (1/[A t ]) = kt + (1/[A 0 ]) (1/[A t ]) = (8.2 x 10 9 L mol -1 s -1 ) (2.0 x 10 -6 s) + (1/1.00 x 10 -4 mol L -1 ) (1/[A t ]) = (2.6 x 10 4 L mol-1) [A t ] = 3.8 x 10 -5 mol L-1

32. Zero-order Reactions Very rare reactions (dissociation of ammonia) Usually occur on metallic surfaces Half-life Equation: Reaction rate is described by: Rate = k t ½ = [A] 0 2k2k