Chapter 14: Kinetics Wasilla High School

What is Kinetics? In kinetics we study the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

Factors That Affect Reaction Rates Physical State of the Reactants – In order to react, molecules must come in contact with each other. – The more homogeneous the mixture of reactants, the faster the molecules can react. Concentration of Reactants – As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. Temperature – At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. Presence of a Catalyst – Catalysts speed up reactions by changing the mechanism of the reaction. – Catalysts are not consumed during the course of the reaction.

Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

Reaction Rates In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times. C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq)

The average rate of the reaction over each interval is the change in concentration divided by the change in time: Average rate =  [C 4 H 9 Cl]  t

Note that the average rate decreases as the reaction proceeds. This is because as the reaction goes forward, there are fewer collisions between reactant molecules.

A plot of [C 4 H 9 Cl] versus time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction.

Reaction Rates In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) Rate =  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t

aA + bB  products [A]Initial Rate[B]Initial Rate 1.00 M0.01 M/s1.00 M0.01 M/s 2.00 M0.02 M/s2.00 M0.04 M/s 3.00 M0.03 M/s3.00 M0.09 M/s Increase [A] by 2  Initial Rate increases by 2 1 = 2 Increase [A] by 3  Initial Rate increases by 3 1 = 3 R  [A] 1 Increase [B] by 2  Initial Rate increases by 2 2 = 4 Increase [B] by 3  Initial Rate increases by 3 2 = 9 R  [B] 2 Rate Law: R = k[A] x [B] y So…. R = k[A] 1 [B] 2 Reaction is first order in relation to A and second order in relation to B Overall Order = 3

Consider the previous reaction… R = k[A] 1 [B] 2 What would be the units for k? R = M/s M/s = kM 1 M 2 k = 1/s M 2

A  products [A]Initial Rate 1.00 M0.01 M/s 2.00 M0.01 M/s Increase [A] by 2  Initial Rate stays the same 2 o = 1 Zero Order in A So… R = k[A] o R = k Units for k = M/s

2NO (g) + 2H 2(g)  N 2(g) + 2H 2 O (At C) Experiment[NO][H 2 ]Initial Rate M/s X X X Task #1: Find the rate law R = k [NO] x [H 2 ] y Start by examining Exp 1 and 2 because [H 2 ] is constant Increase [NO] by 2  Initial Rate increases by 2 2 = 4 To figure out that 5.00 x is an increase by a factor of 4 Divide 5.00 X /1.25 X Next Exp 2 and 3 because [H 2 ] is constant Increase [H 2 ] by 2  Initial Rate increases by 2 1 = 2 To figure out that 1.00 x is an increase by a factor of 2 Divide 1.00 X /5.00 X R = k[NO] 2 [H 2 ] 1 Overall order = 3

Using Our Previous Data…. Determine the rate constant (k) Choose any 1 experiment and use its data to plug into our equation R = k[NO] 2 [H 2 ] M/s = k[0.0050M] 2 [0.0020M] M/s = k(5 x M 2 M) k = 250 1/sM C

Using Our Previous Data…. Determine the rate of reaction when [NO] = 0.012M and [H 2 ] = M R = k[NO] 2 [H 2 ] 1 R = 250 1/sM 2 [0.012M] 2 [0.0060M] 1 R = 2.2 X C

Concentration and Rate If we compare Experiments 1 and 2, we see that when [NH 4 + ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l) Likewise, when we compare Experiments 5 and 6, we see that when [NO 2 − ] doubles, the initial rate doubles.

Rate Laws The exponents tell the order of the reaction with respect to each reactant. Since the rate law is Rate = k [NH 4 + ] [NO 2 − ] the reaction is First-order in [NH 4 + ] First-order in [NO 2 − ]

Rate Laws Rate = k [NH 4 + ] [NO 2 − ] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall.

First-Order Processes For a first-order reaction, the graph of the natural logarithm of the concentration versus time is a straight line. ln [A] t = -kt + ln [A] 0 Notice this equation is in the form of y=mx+b

First-Order Processes

Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NCCH 3 CN

First-Order Processes This data were collected for this reaction at  C. CH 3 NCCH 3 CN

First-Order Processes When ln P is plotted as a function of time, a straight line results. Therefore, – The process is first-order. – k is the negative of the slope: 5.1  10  5 s  1.

Second-Order Processes 1 [A] t = kt + 1 [A] 0 So if a process is second-order in A, a plot of vs. t will be linear, and the slope of the line is k. 1 [A]

Second-Order Processes

The decomposition of NO 2 at 300 °C is described by the equation NO 2 (g)NO(g) + O 2 (g) and yields data comparable to this table: Time (s)[NO 2 ], M

Plotting ln [NO 2 ] vs. t yields the graph at the right. Time (s)[NO 2 ], Mln [NO 2 ]      The plot is not a straight line, so the process is not first-order in [A].

Graphing ln vs. t, however, gives this plot Time (s)[NO 2 ], M1/[NO 2 ] Because this is a straight line, the process is second- order in [A]. 1 [NO 2 ]

Half-Life Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0

Half-Life For a first-order process, this becomes 0.5 [A] 0 [A] 0 ln = −kt 1/2 ln 0.5 = − kt 1/2 −0.693 = − kt 1/2 = t 1/ k

Half-Life For a second-order process, © 2009, Prentice-Hall, Inc [A] 0 = kt 1/2 + 1 [A] 0 2 [A] 0 = kt 1/2 + 1 [A] 0 2 − 1 [A] 0 = kt 1/2 1 [A] 0 == t 1/2 1 k[A] 0

Temperature and Rate

The Collision Model Molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

Activation Energy

Maxwell–Boltzmann Distributions Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. At any temperature there is a wide distribution of kinetic energies.

As the temperature increases, the curve flattens and broadens. Thus, at higher temperatures, a larger population of molecules has higher energy. If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation-energy barrier.

Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and E a : k = A e - E a RT

Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes ln k =  ( ) + ln A 1T1T y = mx + b Therefore, if k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs.. EaREaR 1T1T

Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step.

Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs. © 2009, Prentice-Hall, Inc.