1. CHAPTER 14 Chemical Kinetics

2. Chemical Reactions There are two things that we are interested in concerning chemical reactions: 1) Where is the system going (chemical equilibrium, covered in chapter 15). 2) How long will it take for the system to get to where it is going (chemical kinetics, covered in chapter 14). Kinetics is the study of the rate at which a chemical reaction takes place, and the mechanism by which reactants are converted into products.

3. Example Consider the following irreversible chemical reaction: A(g)  B(g)

4. Example (continued) We may follow the rate of the reaction by either observing the disappearance of reactant A or the appearance of product B. The average rate of the reaction over some time period from t 1 to t 2 is: Ave. rate =- ( [A] 2 - [A] 1 ) = -  [A] t 2 - t 1  t =( [B] 2 - [B] 1 ) =  [B] t 2 - t 1   t where  [A] = [A] 2 - [A] 1  [B] = [B] 2 - [B] 1  t = t 2 - t 1 We insert a negative sign when find the rate of disappearance of a reactant to make the rate of reaction positive.

5. Example (continued) Average rate between 10 s and 20 s is (looking at disappearance of A) Ave. rate = - (0.022 M - 0.030 M) = 0.00080 mol/L. s (20. s - 10. s)

6. Instantaneous Rate of Reaction The instantaneous rate of reaction at time t is equal to the value for the slope of the tangent line in a plot of concentration vs time (negative if a reactant, positive if a product).

7. Stoichiometry and Reaction Rate Different reactants (and products) may disappear (or appear) at different rates depending on the stoichiometry of the reaction. For example: H 2 (g) + 2 ICl(g)  2 HCl(g) + I 2 (g) In the above reaction ICl disappears twice as fast as H 2, and HCl appears twice as fast as I 2. To take this into account we can define the average rate of reaction as the change in concentration of a reactant or product divided by the stoichiometric coefficient used to balance the reaction. Ave. rate = -  [H 2 ] = - 1  [ICl] = 1  [HCl] =  [I 2 ]  t 2  t 2  t  t Note we still insert a negative sign when looking at the disappearance of a reactant. By this method we will obtain the same value for average rate no matter which reactant or product we observe.

8. Rate Law A rate law is an expression that gives the rate of a chemical reaction in terms of the concentrations of the reactants (and occasionally other concentrations as well). For example, consider the following general chemical reaction a A + b B  “products” where a and b are stoichiometric coefficients, and we have assumed the reaction is irreversible (proceeds only in the forward direction). The rate law for reactions of this type can often be written as rate = k [A] m [B] n m = order of reaction with respect to A n = order of reaction with respect to B m + n = overall reaction order k = rate constant, value depends only on temperature. Units determined by dimensional analysis.

9. For example, for the reaction 2 NO 2 (g) + F 2 (g)  2 NO 2 F(g) rate = -  [F 2 ] = k [NO 2 ] [F 2 ]  t Reaction is 1 st order in NO 2, 1 st order in F 2, and 2 nd order overall. Note the following: 1) The reaction orders are usually small whole numbers (0, 1, or 2; occasionally 1 / 2 or -1; rarely any other values). 2) There is no general relationship between the reaction orders and the stoichiometric coefficients for the reaction. For example, in the above reaction the stoichiometric coefficients for NO 2 and F 2 are 2 and 1, but the reaction orders (determined by experiment) are 1 and 1.

10. Experimental Determination of the Rate Law There are several methods that have been developed for finding the rate law for a chemical reaction. The easiest and most common method used is the initial rates method. Consider a general reaction of the form a A + b B  “products” where we assume the rate law for the reaction has the form rate = k [A] m [B] n For a particular set of initial concentrations, the initial rate of reaction is the rate of the reaction measured before the initial concentrations of reactants have had a chance to change significantly. When we measure the initial rate of reaction, we can then use the initial concentrations of our reactants in the rate law.

11. Consider the following two experiments carried out at the same temperature. trial 1, initial concentrations of A and B are [A] 1 and [B] 1, initial rate of reaction is R 1. trial 2, initial concentrations of A and B are [A] 2 and [B] 2, initial rate of reaction is R 2. R 1 = k [A] 1 m [B] 1 n R 2 = k [A] 2 m [B] 2 n Then R 2 = k [A] 2 m [B] 2 n = ( [A] 2 /[A] 1 ) m ( [B] 2 /[B] 1 ) n R 1 k [A] 1 m [B] 1 n The above is true in general. Now, consider the case where the initial concentration of B is the same in both trials.

12. If [B] 2 = [B] 1, then ( [B] 2 /[B] 1 ) n = 1 n = 1 and R 2 = ( [A] 2 /[A] 1 ) m R 1 We can usually find the value for m by inspection, but for a rigorous method for determining m we can simply take the natural logarithm of both sides of the above equation ln (R 2 /R 1 ) = ln ( [A] 2 /[A] 1 ) m = m ln ([A] 2 /[A] 1 ) m = ln (R 2 /R 1 ) ln ([A] 2 /[A] 1 ) The value for n (order of reaction with respect to B) can be found by comparing trials where the initial concentration of A is the same in both trials.

13. Example The following data were obtained for the reaction A + B  “products” trialinitial Ainitial Binitial rate (mol/L)(mol/L)(mol/L. s) 10.1000.1003.1 x 10 -5 20.1000.2003.0 x 10 -5 30.2000.2001.2 x 10 -4 We may assume the rate of reaction is given by the expression rate = k [A] m [B] n Based on the above data, find m, n, and k.

14. Finding m. Compare trial 3 and trial 2. R 3 = k [A] 3 m [B] 3 n = ( [A] 3 /[A] 2 ) m ([B] 3 /[B] 2 ) n R 2 k [A] 2 m [B] 2 n 1.2 x 10 -4 = (0.200/0.100) m (0.200/0.200) n 3.0 x 10 -5 4.0 = 2 m, and so m = 2 Finding n. Compare trial 2 and trial 1. R 2 = k [A] 2 m [B] 2 n = ( [A] 2 /[A] 1 ) m ([B] 2 /[B] 1 ) n R 1 k [A] 1 m [B] 1 n 3.0 x 10 -5 = (0.100/0.100) m (0.200/0.100) n 3.1 x 10 -5 0.97 = 2 n, and so n = 0 So the rate law is: R = k [A] 2

15. To find k we can now use any of the trials. If we use trial 1, then R 1 = k [A] 1 2 k = R 1 /[A] 1 2 = (3.1 x 10 -5 mol/L. s) = 3.1 x 10 -3 L/mol. s (0.100 mol/L) 2 Notice that the units for k are determined by dimensional analysis. For a real set of experimental data we would find a value for k from each data set, and then average to find the best value for k. Also note for real data the values for the reaction orders would likely not work out to be exactly integers due to experimental error.

16. For example: Experimental valueReaction order 1.94 = 2 n n = 1 9.4 = 3 m m = 2 1.45 = 2 p p = 1 / 2 We have assumed in the above that the true values for the reaction orders should be integer or half-integer values, and that the small differences we observe are due to random error in the experimental data.

17. Typical Types of Rate Laws As previously discussed, the rate law for a reaction can often be written as rate = -  [A] = k [A] m [B] n  t Common rate laws. Zero order rate = k First order homogeneous rate = k [A] Second order homogeneous rate = k [A] 2 Second order heterogeneous rate = k [A] [B]

18. First Order Homogeneous Rate Law For a first order homogeneous rate law rate = k [A] We may show that concentration vs time is given by the expression [A] t = [A] 0 e -kt where [A] t = concentration of A at time t [A] 0 = concentration of A at t = 0 k = rate constant (units of 1/time) A plot of concentration vs time will exhibit what is called an exponential decay in the concentration of A.

19. Finding the Rate Constant It is difficult to find the value for k (rate constant) from a plot of concentration vs time since we get nonlinear behavior. We can find a linear relationship as follows [A] t = [A] 0 e -kt Take the ln of both sides ln [A] t = ln [A] 0 - kt (y) = b + m(x)y = ln[A] t x = t This predicts that for a first order reaction a plot of ln [A] t vs time will give a straight line, with slope = -k (and intercept = ln [A] 0 )

20. One way to test whether or not a reaction is first order is to plot the logarithm of concentration vs time. If you get a linear result in the plot, then you know the reaction is first order. If you do not get a linear result in the plot, then you know the reaction is not first order.

21. Half-life By definition, the half-life for a chemical reaction, t 1/2, is the time it takes for the concentration of a reactant to decrease to 1 / 2 of its initial value. For a first order reaction [A] t = [A] 0 e -kt At t = t 1/2, [A] t = [A] 0 /2, so [A] 0 /2 = [A] 0 e -kt½ Divide both sides by [A] 0 then 1/2 = e -kt½ Take the ln of both sidesthen ln(1/2) = -kt 1/2 Divide by -k then t 1/2 = - ln(1/2) k But - ln(1/2) = ln(2), so t 1/2 = ln(2)  0.693 k k

22. Note the following: 1) Since t 1/2 = ln(2)/k for a first order reaction, the value for the half-life is independent of the initial concentration of the reactant. 2) First order reactions are the only reactions where the half-life is independent of concentration. 3) If we wait two half-lives, the concentration will decrease to 1 / 4 of the initial value. For three half-lives the concentration will decrease to 1 / 8 of the initial value, and so forth.

23. Second Order Homogeneous Rate Law For a second order homogeneous rate law rate = k [A] 2 we may show that concentration vs time is given by the expression [A] t = [A] 0 (1 + kt[A] 0 ) where [A] t = concentration of A at time t [A] 0 = concentration of A at t = 0 k = rate constant (units of 1/(concentration)(time)) Concentration will not decrease as quickly in a second order reaction as it does for a first order reaction.

24. Finding the Rate Constant Since [A] t = [A] 0 (1 + kt[A] 0 ) then if we invert both sides of this equation we get 1 = (1 + kt[A] 0 ) = 1 + kt [A] t [A] 0 [A] 0 (y) = b + m(x)y = 1/[A] t x = t This predicts that for a second order reaction a plot of 1/[A] t vs time will give a straight line, with slope = k (and intercept = 1/[A] 0 ).

25. One way to test whether or not a reaction is second order is to plot 1/concentration vs time. If you get a linear result in the plot, then you know the reaction is second order. If you do not get a linear result in the plot, you know the reaction is not second order.

26. Half-life We may use the definition of half-life to find an expression for t 1/2 for a second order reaction. It is easiest to do this starting with the expression 1 = 1 + kt [A] t [A] 0 If we substitute [A] t = [A] 0 /2 at t = t 1/2, we get (after some algebra) t 1/2 = 1 k [A] 0 Unlike a first order reaction, the half-life for a second order reaction depends on concentration. As the initial concentration decreases the half life becomes longer.

27. Zero Order Rate Law For a zero order rate law rate = k [A] 0 = k we may show that concentration vs time is given by the expression [A] t = [A] 0 - kt, t < [A] 0 /k = 0, t  [A] 0 /k where [A] t = concentration of A at time t [A] 0 = concentration of A at t = 0 k = rate constant (units of concentration/time) Concentration decreases at a constant rate until all of the reactant has disappeared.

28. Finding the Rate Constant and Half-Life Since [A] t = [A] 0 - kt, t < [A] 0 /k a plot of concentration vs time will be a straight line, with slope = -k, (and intercept = [A] 0 ). The half life for a zero order reaction will be t 1/2 = [A] 0 /2k

29. Summary of Results First order[A] t = [A] 0 e- kt Second order[A] t = [A] 0 (1 + kt[A] 0 )

30. Sample Problem Consider the reaction A  “products” The reaction obeys first order homogeneous kinetics. The initial concentration of A in the system is [A] 0 0.418 M. After 100. s the concentration of A is 0.322 M. Find k (the rate constant), and the concentration of A after 500. s.

31. Consider the reaction A  “products” The reaction obeys first order homogeneous kinetics. The initial concentration of A in the system is [A] 0 0.418 M. After 100. s the concentration of A is 0.322 M. Find k (the rate constant), and the concentration of A after 500. s. [A] t = [A] 0 e -kt ln([A] t ) = ln([A] 0 ) - kt ln([A] t /[A] 0 ) = - kt k = - (1/t) ln([A] t /[A] 0 ) = + (1/t) ln([A] 0 /[A] t ) Sok = (1/100. s) ln(.418/.322) = 2.61 x 10 -3 s -1. At 500. s, [A] 500 = (0.418 M) exp[-(2.61 x 10 -3 s -1 )(500. s)] = 0.1134 M

32. Temperature Dependence of the Rate Constant For a reaction that obeys the rate law rate = k [A] m [B] n the rate, and therefore the rate constant, for the reaction usually increases as temperature increases. Experimentally it is found that the temperature dependence of the rate constant for the reaction often follows a simple expression called the Arrhenius equation: k = A e -Ea/RT wherek - the rate constant for the reaction A - pre-exponential factor E a - activation energy for the reaction If we take the ln of both sides of the Arrhenius equation, we get ln(k) = ln(A) - (E a /R)(1/T)

33. Example: 2 HI(g)  H 2 (g) + I 2 (g) If we assume the data fit the Arrhenius equation, then ln k = ln A - (E a /R) (1/T) and so for each experimental value for k and T we must 1) Convert T into units of K, then find 1/T 2) Find the value for ln k and then plot the results.

34. The value of E a is found from the slope of the above plot. The value for A is then found by substituting one of the data points into the Arrhenius equation.

35. x y 0.0013 K -1 - 4.0 0.0018 K -1 - 15.0 At T = 556. K we have k = 3.52 x 10 -7 L/mol. s

36. x y 0.0013 K -1 - 4.0 0.0018 K -1 - 15.0 slope =  y = [ (- 15.0) - (-4.0) ] = - 22000. K  x [0.0018 - 0.0013]K -1 So E a = - R (slope) = - (8.314 x 10 -3 kJ/mol.K) (- 22000. K) = 182.9 kJ/mol

37. Since k = A e -Ea/RT, then A = k e +Ea/RT At T = 556. K we have k = 3.52 x 10 -7 L/mol. s, and so A = (3.52 x 10 -7 L/mol. s) e (182900. J/mol)/(8.314 J/mol.K)(556.K) = 5.37 x 10 10 L/mol. s

38. Other Forms of the Arrhenius Equation Beginning with the Arrhenius equation k = A e -Ea/RT we can derive the following equation ln (k 2 /k 1 ) = - (E a /R) 1 1 T 2 T 1 In this equation k 1 is the rate constant at T 1 and k 2 is the rate constant at T 2. By knowing the value for the rate constant at two different temperatures we may use this equation to find E a, and then use the value of k at either temperature to find A.

39. Collision Theory The theoretical model on which the Arrhenius equation is based is called collision theory. The theory makes the following assumptions: 1) For a reaction to take place a collision between reactant molecules must occur. 2) The reactants must collide with sufficient kinetic energy to overcome the energy barrier separating reactants and products. 3) The reactants must have a favorable orientation for reaction to occur. For example, consider the reaction AB + C  A + BC Transition state - The species that forms as reactants are converted into products. It has a structure intermediate between that of the reactants and that of the products of the reaction.

40. How will the energy of the system change as we proceed from reactants to products? We can represent this by an energy diagram. In this diagram E a, the activation energy, is the height of the barrier separating the reactants (AB + C) and products (A + BC).  E represents the change in energy in going from reactants to products. This reaction is exothermic (  E < 0). A---B---C is the transition state. Note that it is a good approximation to say  H   E.

41. Collision Theory and the Arrhenius Equation How does collision theory relate to the Arrhenius equation? We can use the theory to give a physical interpretation to the constants that appear in the equation. k = A e -Ea/RT = pz e -Ea/RT A depends on the collision frequency (z, the number of collisions between reactant molecules per unit time) multiplied by an orientation factor (p, the fraction of collisions that have the correct orientation of reactant molecules). e -Ea/RT represents the fraction of collisions that have sufficient kinetic energy to pass over the barrier separating reactants from products. Notice that as T becomes larger, this term also becomes larger, which makes sense, as the molecules move faster and have higher kinetic energy at high temperature than they do at low temperature. Note: As previously discussed in CHM 1045, v ave = (3RT/M) 1/2

42. Orientation The pre-exponential factor A in the Arrhenius equation depends on two factors - the rate at which collisions take place and the fraction of collisions that have a favorable orientation for a reaction to occur. k = A e -Ea/RT Example:Cl(g) + NOCl(g)  Cl 2 (g) + NO(g) Favorable orientation Unfavorable orientation

43. Reaction Mechanism A reaction mechanism is a sequence of elementary reactions that take place on a molecular level and lead from reactants to products. For example, consider the stoichiometric reaction NO 2 (g) + CO(g)  NO(g) + CO 2 (g) One possible mechanism for this reaction is step 1NO 2 + NO 2  NO 3 + NO step 2NO 3 + CO  NO 2 + CO 2 In this mechanism NO 3 is a reaction intermediate, a substance that is neither a reactant nor a product, but which is produced and consumed as reactants are converted into products.

44. Types of Elementary Reactions There are three common types of elementary reactions that appear in reaction mechanisms. UnimolecularA  “products”rate = k [A] BimolecularA + A  “products”rate = k [A] 2 A + B  “products”rate = k [A] [B] TermolecularA + A + A  “products”rate = k [A] 3 A + A + B  “products”rate = k [A] 2 [B] A + B + C  “products”rate = k [A] [B] [C] Note that the rate for a particular elementary reaction is a rate constant multiplied by the concentrations of the reactants.

45. Requirements For a Reaction Mechanism An acceptable reaction mechanism must satisfy two requirements. 1) The individual elementary steps in the mechanism must add up to the overall reaction. 2) The rate law predicted by the mechanism must agree with the experimentally determined rate law. If the above two requirements are not met, then the reaction mechanism is not acceptable. Unfortunately, the reverse is not true. In principle all we can do with a mechanism is show it is consistent with experiment. Also note that the final expression for the rate law should not involve any reaction intermediates.

46. Finding the Rate Law From the Reaction Mechanism In general, it is difficult to obtain the rate law for a reaction from the reaction mechanism (in fact, for complicated systems one often uses a computer to model the reaction). There are a few simple cases where we can obtain a rate law from a mechanism: 1) One step mechanism. Examples: O(g) + HBr(g)  OH(g) + Br(g)rate = k [O] [HBr] H + (aq) + OH - (aq)  H 2 O( )rate = k [H + ] [OH - ]

47. 2) Mechanisms with a single slow step. For a multistep mechanism with one slow step, the overall rate of the reaction will be the rate of the slow step. This makes it possible to obtain a rate law as long as there are no reaction intermediates involved in the slow step. Example: step 1H 2 (g) + ICl(g)  HI(g) + HCl(g)slow step 2HI(g) + ICl(g)  I 2 (g) + HCl(g)fast overall H 2 (g) + 2 ICl  I 2 (g) + 2 HCl(g) (HI is an intermediate) Since the overall rate of reaction is equal to the rate of the slow step, we may say rate = k 1 [H 2 ] [ICl]1 st order in H 2, 1 st order in ICl 2 nd order overall

48. We can often use experimental data to decide whether a particular reaction mechanism is possible or not. Example: Consider the following two mechanisms for the reaction NO 2 (g) + CO(g)  NO(g) + CO 2 (g) One step mechanism NO 2 (g) + CO(g)  NO(g) + CO 2 (g) predicted rate law: rate = k 1 [NO 2 ][CO] Two step mechanism step 1NO 2 (g) + NO 2 (g)  NO 3 ( g)+ NO(g)slow step 2NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) fast predicted rate law: rate = k 1 [NO 2 ] 2 Experimentally it is found that this reaction is 0 th order in CO and 2 nd order in NO 2. That means the first mechanism is not correct, and that the second mechanism is consistent with the observed rate law.

49. Fast and Reversible Elementary Reactions In some reaction mechanisms there will be a step that is both fast and which goes in both directions. k 1 A + B C + Dfast, reversible k -1 Since the reaction is fast in both directions we can assume that equilibrium is rapidly achieved. At that point, the rate of the forward and reverse reactions must be the same, so k 1 [A] [B] = k -1 [C] [D] We can solve the above expression for the concentration of one reactant in terms of other reactants and the rate constants. This is often useful in finding the rate law from a mechanism

50. Example Consider the following three step mechanism for the stoichio- metric reaction 2 H 2 + 2 NO  2 H 2 O + N 2 overall reaction step 12 NO N 2 O 2 fast, reversible step 2H 2 + N 2 O 2  H 2 O + N 2 Oslow step 3N 2 O + H 2  N 2 + H 2 Ofast (N 2 O 2, N 2 O are intermediates) What is the rate law predicted for the above mechanism?

51. step 12 NO N 2 O 2 fast, reversible step 2H 2 + N 2 O 2  H 2 O + N 2 Oslow step 3N 2 O + H 2  N 2 + H 2 Ofast Overall rate = rate of slow step = k 2 [H 2 ][N 2 O 2 ] N 2 O 2 is a reaction intermediate and so should not appear in the final rate expression. But from step 1 we may say k 1 [NO] 2 = k -1 [N 2 O 2 ] ; so [N 2 O 2 ] = (k 1 /k -1 ) [NO] 2 So rate = k 2 [H 2 ] {(k 1 /k -1 ) [NO] 2 } = (k 1 k 2 /k -1 ) [H 2 ] [NO] 2 = k overall [H 2 ] [NO] 2

52. Catalyst A catalyst is a substance that changes the rate of a chemical reaction without itself being produced or consumed by the reaction. Example: The reaction 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) is slow even at high temperatures. However, if a small amount of solid MnO 2 is added to the reaction mixture the reaction will become fast. The MnO 2 is not consumed in the reaction and can be recovered at the end of the reaction. Homogeneous catalyst - Appears in the same phase as the reactants. Heterogeneous catalyst - Appears in a different phase than the reactants.

53. Many catalysts work by lowering the activation energy for a reaction. Catalysts can also affect the rate of reaction by providing new pathways for converting reactants to products, or by changing the value for the pre-exponential factor (A), usually by forcing the reactants into a favorable orientation. Biological catalysts (enzymes) often work in this way.

54. Example: Hydrogenation of ethene. C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)

55. Enzyme - A biological catalyst. Most biological enzymes are proteins, composed of chains of amino acids, that take on a particular shape, making it possible to catalyze a specific biological reaction.

56. End of Chapter 14 “He (van't Hoff) was a modest and unassuming man who never made priority claims; indeed, in his later books he sometimes gave credit to others for things - like the ‘Arrhenius equation’ and the ‘Le Chatlier principle’ - that he had first discovered himself.” - K. J. Laidler, The World of Physical Chemistry “Nothing has such power to broaden the mind as the ability to investigate systematically and truly all that comes under your observa- tion in life.” - Marcus Aurelius