1. Kinetics How fast does your reaction go?

2. Reaction rates Rate is how fast a process occurs Rates are measured in units of Results Time Example: speed is measured in m/s (or mi/hr). A remote control car covers 125 meters in 15 seconds. What is its rate of speed?

3. Reaction rates Example #2: Lucy wraps chocolates at a rate of 5 pieces/minute. The conveyor belt moves chocolates by at a rate of 9 pieces/minute. At what rate does Lucy have to eat chocolates to keep them from piling up on the conveyor belt? Chemical reaction rates are often mol/s or mol/Ls.

4. Reaction rates Average rate is the (total results)/(total time). MgO + H 2  Mg + H 2 O Average rate = ([H 2 O final ]-[H 2 O initial ])/(t f -t i ) =  [H 2 O]/  t

5. Reaction rates Instantaneous rate is the results for an infinitesimally small amount of time divided by that time Instantaneous rate = d[H 2 O]/dt Rates can also be for the disappearance of reactants Average rate = -  [H 2 ]/  t

6. Reaction rates Measured in units of amount of product formed time The amount of product can be any convenient unit (grams, moles, liters of gas, etc.)

7. Reaction rates Example: MgO + H 2  Mg + H 2 O If the concentration of water vapor in the above reaction increases from 1.0x10 -3 moles/liter to 8.8x10 -3 moles/liter in 3.5 seconds, What would be the average reaction rate? (8.8-1.0)x10 -3 /3.5 = 2.2x10 -3 mol/L∙s

8. Graphing rates Results (dependent variable) are plotted on the y axis, and time (independent variable) is plotted on the x axis. The rate at any time is the slope of the graph at that time. The rate of a straight line plot is just the slope of the whole line (rise/run).

9. Graphing rates Curved plots –Instantaneous rate at any time is the slope of the tangent to the curve at that point. –Average rate for any time period is the slope of the straight line connecting the two desired time points on the graph.

10. Collision theory Particles must collide in order to react. Particles must be oriented correctly when colliding in order to react Correct orientation results in a temporary high-energy arrangement called an activated complex or transition state.

11. Collision theory

12. Activated complex for S N2 substitution of bromomethane with hydroxide to make methanol Transition state may form products or break apart and re-form reactants

13. Activation energy Particles must collide with sufficient energy to cause a reaction Activation energy is the difference in energy between the reactants and the transition state

14. Activation energy

15. If reactants do not have enough kinetic energy, the reaction will not start.

16. Factors affecting reaction rates Free energy and reaction rate are unrelated Surface area –More surface area (smaller pieces or finer powder) means a faster reaction. All reactions involving a solid phase take place at the surface of the solid.

17. Factors affecting reaction rates Concentration –Higher concentration means a faster reaction. Nature of reactants –Rate of reaction depends on what is reacting. Temperature –The activation energy is supplied by the kinetic energy of the particles.

18. Factors affecting reaction rates

19. Higher temperature means that more particles will have enough energy to react.

20. Catalysts A reaction will proceed faster in the presence of a catalyst because the activation energy is lowered.

21. Catalysts A catalyst speeds a reaction without being used up. Transition metals and their oxides often have catalytic properties – used in catalytic converters in automobiles

22. Catalysts For solid catalysts, the reaction usually takes place on the surface of the catalyst, therefore the more surface area the catalyst has, the faster the reaction Biological catalysts are called enzymes

23. Rate Laws Rate varies with concentration of reactants A  B Rate = k[A] where k is the rate constant. The rate constant is temperature dependent. Rate constant is experimentally determined.

24. Reaction order Expresses how rate depends on concentration as an exponent Zero order: rate = k[A] 0 = k First order: rate = k[A] Second order: rate = k[A] 2 Reactions beyond second order in one reactant are extremely rare.

25. Reaction order A + B  AB Rate = k[A][B] Reaction is first order in A, first order in B and second order overall. In a simple reaction the order represents the molecularity of the reaction.

26. Reaction order Molecularity is the number of particles that have to collide to make a reaction Zero order – generally reactions that happen at a surface and depend only on surface area. First order – depends only on concentration of reactant

27. Reaction order Second order Rate = k[A] 2 Two molecules of A must collide. Probability of a collision is n(n-1)/2. If the concentration is doubled to 2n (and n is large) the probability increases by a factor of four, so the rate depends on the square of the concentration.

28. Determining reaction order Method of initial rates –Reaction is run for short periods of time and stopped before reactant concentrations change much –Initial concentrations are changed and rates compared –If initial concentration is doubled, the rate will zero order: not change first order: 2x second order: 4x

29. Determining reaction order Rate constant can also be determined Example. A + B  C Determine rate law for this reaction, rate constant, and rate when [A] = 0.050 M and [B] = 0.100 M Experiment[A][B]Initial rate (M/s) 10.100 4.0x10 -5 20.1000.2004.0x10 -5 30.2000.10016.0x10 -5

30. Determining reaction order Rate law for B: Use exp. 1 & 2 Rate 2 /rate 1 = 4.0x10 -5 /4.0x10 -5 = k(0.100) m (0.200) n = 1 k(0.100) m (0.100) n Rate law for A: Use exp. 1 & 3 Rate 2 /rate 1 = 16.0x10 -5 /4.0x10 -5 = 4 = k(0.200) m (0.100) n k(0.100) m (0.100) n rate = k[A] 2 [B] 0 = k[A] 2 n = 0 m = 2 = 2 m = 4

31. Determining reaction order Determining k: use any experiment k = rate/[A] 2 = 4x10 -5 /0.1002 = 4.0x10 -3 M -1 s -1 Rate at given concentrations: Use the rate law equation rate = k[A] 2 = (4x10 -3 )(0.200) 2 = 1.6x10 -4 M/s

32. Determining reaction order Example: 2NO + 2H 2  N 2 + 2H 2 O Determine the rate law and rate constant, and the rate when [NO] = 0.050 M and [H 2 ] = 0.150 M Experiment[NO][H 2 ]Initial rate (M/s) 10.10 1.23x10 -3 20.100.202.46x10 -3 30.200.104.92x10 -3 Answer: rate = k[NO] 2 [H 2 ]; k = 1.2 M -2 s -1 ; rate = 4.5x10 -4 M/s

33. Integrated rate laws Concentration expressed in terms of time Zero order: [A] = -kt + [A] 0 Graph of [A] vs t is a straight line. 1st order: ln[A] = -kt + ln[A] 0 Graph of ln[A] vs t is a straight line.

34. Integrated rate laws 2nd order: 1/[A] = kt + 1/[A] 0 Graph of 1/[A] vs t is a straight line. 3 rd order reactions are practically non-existent.

35. Reaction mechanisms Complex reactions are made of several simple steps. Example: NO 2 + CO  NO + CO 2 occurs in two steps: i. NO 2 + NO 2  NO 3 + NO ii. NO 3 + CO  NO 2 + CO Overall process is sum of steps.

36. Reaction mechanisms Intermediates are short-lived species that are temporarily formed as the reaction proceeds. The rate-determining step is the slowest step of the mechanism. Intermediates are short-lived species that are temporarily formed as the reaction proceeds. The rate-determining step is the slowest step of the mechanism.

37. Changing components of this step changes the reaction rate; changing components of other steps does not. Rate laws for complex reactions include the components of the slow step and critical components of earlier steps.

38. Reaction mechanisms Example: 2NO + 2H 2  N 2 + 2H 2 O occurs in three steps: 2NO  N 2 O 2 (fast) N 2 O 2 + 2H 2  N 2 O + H 2 O (slow) N 2 O + H 2  N 2 + H 2 O (fast) Rate law is rate = k[NO] 2 [H 2 ]. [H 2 ] appears in the slow step, and [N 2 O] depends on [NO] 2, so [NO] 2 appears in the rate law as well.

39. Reaction mechanisms Further examples: Example 1 HBr + O 2  2H 2 O + 2Br 2 Step 1 (slow) HBr + O 2  HOOBr Step 2 HOOBr + HBr  2HOBr Step 3 2HOBr+ 2HBr  2H 2 O + 2 Br 2

40. Reaction mechanisms Choose the most likely rate law: a. rate = [HBr] b. rate = [O 2 ] c. rate = [HBr][O 2 ] d. rate = [HBr] 2

41. Reaction mechanisms Example 2 (CH 3 ) 3 CBr + H 2 O  (CH 3 ) 3 COH + HBr