Chemical Kinetics. Reaction Rate: The change in the concentration of a reactant or a product with time (M/s). Reactant Products A B


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1 Reaction Rates: Chemical Kinetics Reaction Rate: The change in the concentration of a reactant or a product with time (M/s). Reactant Products A B change in number of moles of B Average rate = change in time = ( moles of B) t = [B] t Since reactants go away with time: Rate = t
2 Consider the decomposition of N O 5 to give NO and O : N O 5 (g) 4NO (g) + O (g) reactants decrease with time products increase with time
3 From the graph looking at t = 300 to 400 s M 6 Rate O = = 9 0 Ms Why do they differ? 00s M 5 Rate NO = = Ms Recall: 00s 0.009M 5 N O 5 (g) 4NO (g) + O (g) Rate NO 5 = =.9 0 Ms 00s To compare the rates one must account for the stoichiometry. 6 6 Rate O = 9 0 Ms = 9 0 Ms Rate NO = Ms = 9. 0 Ms 4 Rate NO 5 =.9 0 Ms = Ms Now they agree! Reaction Rate and Stoichiometry In general for the reaction: aa + bb cc + dd [ A] [ B] [ C] [ D] Rate = = = = a t b t c t d t
4 Rate Law & Reaction Order The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants. Each concentration is expressed with an order (exponent). The rate constant converts the concentration expression into the correct units of rate (Ms ). (It also has deeper significance, which will be discussed later) For the general reaction: aa+ bb cc+ dd x y Rate = k [B] x and y are the reactant orders determined from experiment. x and y are NOT the stoichiometric coefficients.
5 The Overall Order of a reaction is the sum of the individual orders: Rate (Ms ) = k[b] / [C] Overall order: + ½ + = 3.5 = 7/ or seven halves order note: when the order of a reaction is (first order) no exponent is written. Units for the rate constant: The units of a rate constant will change depending upon the overall order. The units of rate are always M/s or Ms To find the units of a rate constant for a particular rate law, simply divide the units of rate by the units of molarity in the concentration term of the rate law. Rate (Ms ) = k st order Ms k(units) = = s M
6 Reaction Rates Instantaneous rate (tangent line) concentration (M) Average Rate Y [concentration] = X time time (s) Rules of logarithms log () = 0 log (0) = log (00) = Log (0 x ) = x log A x = xlog A ln () = 0 ln (e) = ln (e x ) = x ln A x = xln A x x A A A = = x B B B log log x log log(ab) = log A + log B A log = log A log B B
7 Determining Reaction Order: The Method of Initial Rates The reaction of nitric oxide with hydrogen at 80 C is: NO(g) + H (g) N (g) + H O(g) From the following data, determine the rate law and rate constant. Run [NO] o (M) [H ] o (M) Initial Rate (M/min) NO(g) + H (g) N (g) + H O(g) The rate law for the reaction is given by: Rate(M/min) = k [NO] x [H ] y Taking the ratio of the rates of runs 3 and one finds: Rate(3) Rate() = k [NO] [H ] x y (3) (3) x y () () k [NO] [H ] x y 0.00M / min M / min = k [0.000] [0.000] = x y k [0.000] [0.000] [0.000] = [0.000] y y [0.000] [0.000] y y = y
8 log y = y log = log ylog( ) = log y = Now that y is known, we may solve for x in a similar manner: x y Rate() Rate() = k [NO] [H ] x y k [NO] [H ] () () () () x = k [0.000] [0.000] x k [0.000] [0.0300] 4 = x 3 x = 8 xlog = log 8 x = 3 The Rate Law is: Rate(M/min) = k [NO] 3 [H ] To find the rate constant, choose one set of data and solve: M 0.00 k 0.000M 0.000M min = 3 ( ) ( ) k = M 0.00 min ( 0.000M) 3 ( 0.000M) = M 0.00 min M ( ) ( ) M k = min
9 Integrated Rate Laws: time dependence of concentration For a first order process, the rate law can be written: A products Rate(Ms ) = = k t This is the average rate If one considers the infinitesimal changes in concentration and time the rate law equation becomes: d Rate(Ms ) = = k This is the instantaneous rate dt 0 o Integrating in terms of d and dt: d t = k dt where = o at time t = 0 and = at time t = t 0 o d t = k dt ln o = kt Taking the exponent to each side of the equation: o = e kt or = e o kt Conclusion: The concentration of a reactant governed by first order kinetics falls off from an initial concentration exponentially with time.
10 Recognizing a first order process: A products Whenever the conc. of a reactant falls off exponentially, the kinetics follow first order. = e o kt Determining the Rate constant for a first order process Taking the log of the integrated rate law for a first order process we find: kt ( = oe ) ln ln = ln o k t A plot of ln versus time (t) is a straight line with slope k and intercept ln o
11 Example: The rate of decomposition of azomethane (C H 6 N ) was studied by monitoring the partial pressure of the reactant as a function of time. Determine if the data below support a first order reaction. Calculate the rate constant for the reaction. Time (s) P (mmhg ln (P) k =.6x03 s  ln P Plot of lnp vs. time y = x R = time (s) Reaction HalfLife
12 Reaction Halflife: Halflife is the time taken for the concentration of a reactant to drop to half its original value. = o For a first order process the half life (t ½ ) is found mathematically from: () ln[ A] = kt + ln[ A] 0 [ ] [ ] 0 () ln A ln A = kt [ A] (3) ln = kt 0 [ A] 0 (4) ln = kt (5) ln = kt t [ A] 0 0 ln = = k k A certain reaction proceeds through t first order kinetics. The halflife of the reaction is 80 s. What percent of the initial concentration remains after 900s? Step : Determine the magnitude of the rate constant, k. ln ln ln t = = k = = = s k k t 80s Using the integrated rate law, substituting in the value of k and 900s we find: o = e kt o s 900 s = e = 0.03 Since the ratio of to 0 represents the fraction of o that remains, the % is given by: = 3.%
13 For a Second Order Process: A Products Rate = k Integrating as before we find: d Rate(Ms ) = = k dt = kt + [ A] [ A] 0 A plot of / versus t is a straight line with slope k and intercept / 0 For a second order reaction, a plot of ln vs. t is not linear. = kt + [ A] [ A] 0 Non linearity indicates that the reaction is not first order. Slope = k (rate constant)
14 Halflife for a secondorder reaction Unlike a first order reaction, the rate constant for a second order process depends on and the initial concentration of a reactant. = kt + [ A] [ A] t 0 at the half life, [ A] = [ A] t 0 Substituting and solving, t / = ka [ ] 0 EXAMPLE: The reaction NOBr (g) NO (g) + Br (g) is a second order reaction with respect to NOBr. k = 0.80 M  s  at 0 o C. If [NOBr] o = M, how much NOBr will be left after a reaction time of 0 minutes? Determine the halflife of this reaction.
15 SOLUTION: One can solve for the amount of NOBr after 0 minutes by substituting the given data into the integrated rate law for a secondorder reaction. = kt + [ NOBr] [ NOBr] t 0 (Second Order) [ NOBr] t =   (0.80 M s ) (600 s) M [ NOBr] t = M  3 [ ] = NOBr.6 0 M t To determine the halflife for this reaction, we substitute the initial concentration of NOBr and the rate constant for the reaction into the equation for the halflife of a secondorder reaction. t / = ka [ ] 0 t / = M s (7.5 0 M) =60 s
16 The Arrhenius Equation: Temperature dependence of the Rate Constant Most reactions speed up as temperature increases. (example: food spoils when not refrigerated.) k = f(t) Where f is some function. The magnitude of a first order rate constant is seen to increase exponentially with an increase in temperature. Therefore one can conclude that: k(t) e f(t)
17 Why would k (along with the rate) increase with temperature? Let s go back to Kinetic Molecular Theory to understand The Collision Model: In order for molecules to react they must collide. As temperature increases, the molecules move faster and the collision frequency increases. Therefore, the greater the number of collisions the faster the rate. Thus reaction rate should increase with an increase in temperature. Also, the more molecules present, the greater the probability of collision and the faster the rate. Thus reaction rate should increase with an increase in the concentration of reactant molecules. Activation Energy Arrhenius: Molecules must posses a minimum amount of energy to react. Why? () In order to form products, bonds must be broken in the reactants. () Bond breakage requires energy. (3) Molecules moving too slowly, with too little kinetic energy, don t react when they collide. The Activation energy, E a, is the minimum energy required to initiate a chemical reaction. E a is specific to a particular reaction.
18 Consider the reaction: N O(g) + O (g) N (g) + NO (g) The progress of a reaction can be described by a Reaction Coordinate Reactants Products Recall for KMT that the temperature for a system of particles is described by a distribution. E < E a E > E a Since the probability of a molecule reacting increases, the rate increases. At higher temps, more particles have enough energy to go over the barrier.
19 Orientation factors into the equation The orientation of a molecule during collision can have a profound effect on whether or not a reaction occurs. Some collisions do not lead to reaction even if there is sufficient energy. The Arrhenius Equation Arhenius discovered that most reactionrate data obeyed an equation based on three factors: () The number of collisions per unit time. () The fraction of collisions that occur with the correct orientation. (3) The fraction of the colliding molecules that have an energy greater than or equal to E a. From these observations Arrhenius developed the aptly named Arrhenius equation.
20 Arrhenius equation k is the rate constant E a is the activation energy E a k = Ae RT T is the temperature in K R is the idealgas constant (8.34 J/Kmol) A is known the frequency or pre exponential exponential factor In addition to carrying the units of the rate constant, A relates to the frequency of collisions and the orientation of a favorable collision probability Both A and E a are specific to a given reaction. Determining the Activation Energy E a may be determined experimentally. First take natural log of both sides of the Arrhenius equation: ln k ln k = Ae E a RT /T E a ln k = + ln A RT Golly gee, what do we have once again A plot of ln k vs /T will have a slope of E a /R and a yintercept of ln A.
21 Determining the Activation Energy One can determine the activation energy of a reaction by measuring the rate constant at two temperatures: Writing the Arrhenius equation for each temperature: ln k E E = + ln A a a = + ln A ln k RT RT If one takes the natural log of the ratio of k over k we find that: k = ln k ln k ln k Substituting in the values for E a into the equation: ln k a ln k = + ln A RT E E RT a + ln A Lookie what happens presto! Knowing the rate at two temps yields the rate constant. or Knowing the E a and the rate constant at one temp allows one to find k(t ) k E = a ln k R T T
22 Example: The activation energy of a first order reaction is 50. kj/mol at 5 o C. At what temperature will the rate constant double? () k = k () k = ln k k E a ln k = ln() = R T T (3) (4) E a R = 3 0 J 50. kj/mol kj 8.34 J mol K ln() = = K 3 = K 98K T A 0 o C change of temperature doubles the rate!! = 308 K 3 (5) = K T T = 308 K
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