1. Chemical kinetics

2. Speed or rates of reactions Affecting factors: Concentration of reactants Temperature at which reaction occurs Presence of a catalyst Surface area of reactants (solid, liquid, gas) or catalysts Pressure with gases

3. Reaction rate- speed of a chem. Reaction Measures how quickly reactants are used up or products are produced Avg. rate = change in # moles of product  time =  (moles Product)  t

4. Rates in terms of concentration A + B  C + D Avg. rate =  [A]  t use [ ] to express conc. in M molarity = [A] final time – [A] initial time = M/s units t f - t i

5. Plot data on graph Graph curve used to determine instantaneous rate at any point Instantaneous rate is the slope of the straight line tangent that touches the curve at the point of interest

6. Rates and Stoichiometry AgNO 3 + NaCl  AgCl + NaNO 3 Each reactant and product is 1:1 ratio Therefore rate =  [AgNO3] =  [AgCl]  t  t Not all balanced equations are 1:1 2HI  H 2 + I 2 In this case the disappearance of HI is twice the rate of the appearance of both H 2 and I 2

7. Rate = -1  [H 2 ] =  [I 2 ] 2  t  t  t General equation is aA + bB  cC + dD Rate= 1  [A] = 1  [B] = 1  [C] = 1  [D] a  t b  t c  t d  t

8. Do following: 2N 2 O 5  4NO 2 + O 2 Rate of decomposition of N 2 O 5 is 4.2 x 10 -7 M/s, what is the rate of appearance of NO 2 and O 2 1  [NO 2 ] = 1  [N 2 O 5 ] = 4(4.2x10 -7 M/s) = 4  t 2  t 2 8.4 x 10 -7 M/s  [O 2 ] = 1  [N 2 O 5 ] = 1(4.2x10 -7 M/s) =  t 2  t 2 2.1x10 -7 M/s

9. Dependence of rate on concentration Rates diminish as concentration of reactants is decreased Rates generally increase when reactants concentration increases Rates of reaction are dependent on the concentrations of the reactants Called rate law Rate = k[reactant 1] m [reactant 2] n k is the rate constant

10. Given a rate law for a reaction and its rate for a set of reactant concentrations, the value of k at a specific temp. and pressure can be calculated Magnitude of k provides valuable info about reactions Exponents m and n are called reaction orders Sum of exponents is the overall reaction order

11. Reaction order exponents are determined experimentally and do not necessarily correspond to the coefficients of a balanced equation Most rate laws have reaction orders of 0, 1 or 2 Some rate laws have fractional orders or even negative Example: H 2 + I 2  2HI Rate = k[H 2 ][I 2 ] Since altering conc. of either reactant has same effect on rate, they are both first order and have exponent values of 1 Overall reaction order is 1+1=2 ;second order overall

12. Units of rate constants Depend on overall reaction order of rate law For a second order overall, the units must be: Units of rate = (units of rate constant)(units of concentration) 2 Units of rate constant = units of rate (units of concentration) 2 = M/s = M -1 s -1 M 2 Units of rate constant will vary depending on overall order of the reaction

13. Rate law for a reaction must be determined experimentally Observe the effect of changing initial concentration of reactants on initial rate of a reaction Example: if changing the conc. of a reactant has no effect on the reaction rate as long as some of the reactant is present, it has a zero order If reaction is first order in a reactant, changes in conc. of that reactant will produce proportional changes in the rate Rate law second order, doubling conc. will increase rate by a factor of 2 2 =4 Important to note; rate of reaction depends on conc., rate constant does not

14. Determine rate law, magnitude of rate constant and rate of reaction Given equation A + B  C Exp.#[A]M[B]M Init.rate (M/s) 10.1000.1004.0x10 -5 20.1000.2004.0x10 -5 30.2000.10016.0x10 -5 Rate law: rate= k[A] m [B] n Need to solve for m and n

15. Look at rate comparing [A] to [B] If [B] is changed(exp.2 doubled) it had no effect([A] held constant) (remember only change one variable at a time) Therefore rate law is zero order in B If [A] is doubled: rate increases fourfold Rate is proportional to [A] 2 Rate law of A is 2 nd order Rate = k[A] 2 [B] 0

16. Can also find the values by taking the ratio of the rates from two experiments Solving for n: Rate 2 = 4.0x10 -5 M/s =1 Rate 1 4.0x10 -5 M/s Using rate law: 1= rate 2 = k[0.100M] m [0.200M] n = [0.200] n =2 n rate 1 k[0.100M] m [0.100M] n [0.100] n 2 n = 1 if n=0

17. Solving for m: Rate 3 = 16.0 x 10 -5 M/s = 4 Rate 1 4.0 x 10 -5 M/s Using rate law 4 = rate3 = k[0.200M] m [0.100M] n = [0.200] m = 2 m rate1 k[0.100M] m [0.100M] n [0.100] m 2 m = 4 so therefore m = 2

18. Solving for magnitude of k k = rate = 4.0x10 -5 M/s = 4.0x10 -3 M -1 s -1 [A] 2 (0.100M) 2

19. Solving for rate of reaction if [A]=0.050M and [B]=0.100M Rate = k[A] 2 = (4.0x10 -3 M -1 s -1 )(.050M) 2 = 1.0x10 -5 M/s Since B is not part of the rate law, it is immaterial to the rate, provided there is at least some B to react with A

20. Change of Concentration with Time Rate laws can be converted into equations that show concentration of reactants or products at any given time during a reaction First-Order Reactions Reactions whose rate depends on the concentration of a single reactant raised to the first power Example: where A  products Rate =  [A] = k[A]  t

21. Relating the conc. of A at start of reaction, [A] 0, to its conc. at any other time, [A] t : In[A] t – In[A] 0 = -kt In[A] t = -kt [A] 0 In[A] t = -kt + In[A] 0 form of straight line equation y = m x + b A graph of a 1 st order react. of In[A] t vs. time gives a straight line with a slope of –k and y intercept of In[A] 0

22. Using the equation for a first order reaction can determine: Concentration of a reactant remaining at any time after the reaction has started The time required for a given fraction of a sample to react The time required for a reactant concentration to reach a certain level

23. Do following problem: The 1 st order constant for the decomposition on an insecticide in water is 1.45yr -1. On July 1 a quantity is washed in to a lake, giving a concentration of 5.0x10 -7 g/cm 3 of water. What is the insecticide conc. 1 yr. later? How long before the concentration drops to 3.0x10- 7 g/cm 3 ?

24. k= 1.45 -1 yr,t=1.00yr [insecticide] 0 =5.0x10 -7 g/cm 3 In[insecticide] t=1yr = -(1.45yr -1 )(1.00yr) + In(5.0x10 -7 g/cm 3 ) = -1.45 + (-14.51) = -15.96 Need same units so: To obtain [insecticide] t=1yr use inverse natural logarithm (e x ) [insecticide] t=1yr = e -15.96 = 1.2x10 -7 g/cm 3 In(3.0x10 -7 ) = -(1.45yr -1 )(t) + In(5.0x10 -7 ) t = -[In(3.0x10 -7 ) - In(5.0x10 -7 )]/1.45yr -1 = -(-15.02 + 14.52)/1.45yr -1 = 0.35yr

25. Half-life of a reaction t 1/2 is amount of time for conc. Of a reactant to drop one half its value [A] 1/2 = ½[A] 0 To determine half life of 1 st order reaction: In1/2[A] 0 =-kt 1/2 [A] 0 In1/2 = -kt 1/2 t 1/2 = - In1/2 = 0.693 k k T 1/2 for 1 st order rate law is independent of initial conc. of reactant Therefore half-life is same at any time during the reaction

26. Second order reaction kts Rate depends on reactant concentration raised to the second power or on concentrations of two different reactants each raised to first power Rate law is Rate = k[A] 2 1 = kt + 1 [A] t [A] 0 (straight line equation) 2 nd order half-life: t 1/2 = 1 k[A] 0

27. Effect of temperature on rate Rates of most chemical reactions increases as temp increases Temperature does not usually affect concentration, therefore: The rate constant must increase with increasing temperature Reactions rate approximately double for each 10 o C rise in temperature

28. Collision model of chemical kinetics explains effect of temp and conc on reactions Central idea: molecules must collide in order to react The greater the number of collisions: the greater the rate So increase concentration increase # collisions increase rate

29. Increase temp Increase molecular velocities Increase frequency of collisions Molecules hit harder (greater energy) Increase rate

30. Most collisions however do not lead to a reaction Why? Two major factors Correct orientation or geometry of collision Molecules must hit a right place Sufficient energy to stretch, bend and break bonds for reaction to occur Activation Energy E a Minimum energy required to initiate chem reaction Based on K.E. of molecules (temperature)

31. Activated complex or transition state Shown on Potential Energy Diagram

33. Arrhenius Equation Increase in rate with increasing temperature is nonlinear in most reactions Rates obey equation: k = Ae -Ea/RT K=rate constant, Ea=activation energy R=gas constant(8.314J/mol-K) T=absolute temperature A= constant called frequency factor

34. Reaction Mechanism- process by which a reaction occurs Each process occurs in a single event or step called an ELEMENTARY STEP Multi-step mechanisms consist therefore of a series of elementary steps Number of molecules that participate in an elementary step is the molecularity of the step Single molecule: unimolecular Two molecule collision: bimolecular Three molecules: termolecular

35. The elementary steps in a multistep mechanism must always add to give the chemical equation of the overall process Substances in the elementary steps that are not original reactants or final product are called intermediates

36. Elementary steps have a rate law Rate of unimolecular process is 1 st order A  product rate = k[A] Bimolecular is 2 nd order A + B  product rate = k[A][B] A + A  product rate= k[A] 2 Termolecular is 3 rd order A + B + C  product rate= k[A][B][C] A + A + B  product rate= k[A] 2 [B] A + A + A  product rate= k[A] 3

37. In multiple step mechanisms, the overall rate of the reaction is determined by the slowest occurring elementary step Called rate-determining step Rate-determining step governs the rate law for the overall reaction

38. Determining rate law of reaction with an intermediate as a reactant in the rate- determining step is difficult Especially so if the rate determining step is not first step Example: 2NO + Br 2  2NOBr

39. The experimentally determined rate law for the reaction is 2 nd order in NO and 1 st order Br 2 Rate = k[NO] 2 [Br 2 ] Need to find mechanism that is consistent with exp. rate law Since termolecular processes are rare the reaction of NO + NO + Br 2  2NOBr is unlikely as a single step mechanism Consider bimolecular 2 step mechanism

40. k 1 Step 1 NO + Br 2  NOBr 2 (fast) k -1 k 2 Step 2 NOBr 2 + NO  2NOBr (slow) Because 2 nd step is rate-determining overall rate is governed by rate law of that step Rate = k 2 [NOBr 2 ][NO] But NOBr 2 is an intermediate. Intermediates are unstable and typically have low unknown concentrations. So the following assumptions must be considered

41. Need to express concentration of [NOBr 2 ] in terms of [NO] and [Br 2 ] Since NOBr 2 is unstable it can be consumed in two ways Can combine with NO to form NOBr Can fall apart back to NO and Br 2 Look at original equation, the first scenario is Step 2, slow process 2 nd scenario is unimolecular process and reverse of Step 1 k -1 NOBr 2  NO + Br 2

42. Since Step 2 is slow: assume most of NOBr 2 falls back The forward and reverse reactions of Step 1 occur faster than Step 2 An equilibrium is established and the rates of the reverse and forward reactions are equal k 1 [NO][Br 2 ] = k -1 [NOBr] Rate of forward reaction Rate of reverse reaction Solving for [NOBr 2 ]

43. [NOBr 2 ] = k 1 [NO][Br 2 ] k -1 Substitute the relationship into the rate law of the rate-determining step gives: Rate = k 2 k 1 [NO][Br 2 ][NO] = k[NO] 2 [Br 2 ] k -1 This is consistent with experimental rate law Rate constant k= k 2 k 1 /k -1 In general, whenever a fast step precedes a slow step, the concentration of the intermediate may be solved by assuming that an equilibrium is established in the fast step.

44. Catalysis Use of a catalyst- a substance that changes the speed of a chem. reaction without undergoing a permanent chemical change itself in the process Catalyst lowers the overall activation energy for a reaction Homogeneous Catalysis Catalyst present in same phase as reactants

45. Heterogeneous catalysis Catalyst exists in different phase than reacting molecules Initial step with this type is typically adsorption of reactants Binding of molecules to surface Example: enzymes,catalytic converters, Pt Inhibitors- slow down or cause reactions to cease Bind to active sites #53, 55, 57