Chemical Kinetics

Chemical Kinetics The “Speed” of the Reaction Or Reaction Rates

Reaction Kinetics Reactants Products Rate of Change Initial rates Method Reaction Order Average Instantaneous Integrated Rate Forms of Rate Law Rate Law Graphical Analysis of Rates Half-Life

Map Reactants  • What happens to the reactants in a reaction? • If we measure the concentration of reactants as a reaction proceeds, what would the graph look like? Or • How does the concentration vary with time? • Is it linear? • Exponential? • Random? • Do all reactions only move forward? • Assume for now there is no reverse reaction • Or the reverse reaction proceeds so slowly, we are willing to ignore it graph

Map  Products • Where do the products come from? • If we measure the concentration of the products from the first second the reaction starts, • how would the concentration vary with time? graph

Map Rate of Change • In the reaction A  B • How does the [A] vary with time? Or • What is the rate of the reaction? • Rate = [A]time2 – [A]time1Time 2 – Time 1 [A] = the molarity of A

Rate of Change • The symbol delta, , means “the change in” • So the reaction rate can be written • Rate = [A]t2 – [A]t1ortime 2 – time 1 • Rate = [A]t graph

Average Rate of Change • The concentration varies as time goes on. • Because we use up reactants • We can calculate an average rate of product consumption • over a period of time. • Rate is like velocity of a reaction. • the rate of change of meters vs rate of change of [A] • Instead of meters per second, it is concentration per second. • To calculate speed/velocity, divide the distance traveled by the time it took. Meters = m2 – m1 =  m t2 – t1t sec graph

Average Rate of Change • What is the change in concentration? (how far did it go? If you are calculating speed) [A] @ time2 – [A] @time1 = [A] • How much time has elapsed? • [A] @ time2 – [A] @time1 =[A]t2 – t1t graph

Instantaneous Rate • An average rate describes what reaction rate over a time, but does not tell us the rate at any particular moment. • The rate at any moment is the instantaneous rate

Map Instantaneous Rate • If we take the average rate over a period of time and continuously make the time period smaller • When the time period is infinitesimally small, you approach the instantaneous rate • Graphically, it is the slope of the tangent line at the instant. • That’s why graphing programs have that tangent line function! Rates are important in bio, physics and chem.

Map (differential) Rate Law • Expresses how rate depends on concentration. • Rate = - [Reactant] = k [reactant]n t • k is the rate constant • The bigger the k value, the faster the reaction • The smaller the k value, ….? • n = the order of the reaction and must be determined experimentally.

Reaction Rates • Reaction rates are considered positive • Rateinstantaneous = kinstantaneous = - slope of tangent line • Rateaverage = k average = - ([A]2 - [A]1) t2-t1 • So the rate constant, k, is always negative • Rate = - [Product] = k [Product]n t • Assuming no reverse reaction!

Back to: Reactant/product Average Rate Reaction Rate Instantaneous Rate Product Formation Average Rate of Change From 0 to 300 s = 0.01 –0.0038 = 0.000021 M/s 300 s

2NO2 2NO + O2 • Let’s consider the above reaction • How can we measure the rate? • What data do we need? • Measure the time • Measure the concentration • We will take advantage of color in our lab • If we are measuring light, we are doing….. SPECTROSCOPY

Spectrometer One frequency Of light Many frequencies of light Monochrometer, LED Or Filter Source Sample Detector The sample absorbs the light. The detector determines how much.

Beer’s Law • Beer’s law states that the amount of light absorbed depends on: • The material • molar absorbtivity (physical property) • How much is there? • molarity • And how big the sample holder • The light spends more “time” in contact with a longer sample

Spectrometer One frequency Of light Many frequencies of light Monochrometer, LED Or Filter Source Sample Detector The sample absorbs the light. The detector determines how much.

Spectroscopy • Assume the concentration is directly proportional to absorbance of light • The more stuff there is that absorbs the light • the less light that goes through …. or • More light is absorbed Beer’s Law a = e l c = k M a = absorbtivity e = molar absorbtivity (physical property) l = length of light path c = molarity or the solution Molarity and absorbtivity Are directly proportional

2NO2 2NO + O2 Compare the [NO2] in the first 50 secs and the last 50 secs Why does the rate slow down?

Formation of Products 2NO2 2NO + O2 Rate of Consumption NO2 = Rate Formation NO Rate = k[NO2 ] = - k[NO] Because For every two NO2 consumed two NO formed

Formation of Products 2NO2 2NO + O2 Rate of Consumption NO2 = 2 x Rate Formation O2 Rate = k[NO2 ] = - k/2 [O2] Because For every two NO2 consumed one O2 formed

Compare the Instantaneous Rates • At any moment in time [NO2] = - [NO] = 2 - [O2]t t t Or k [NO2] = - k [NO] = - k/2 [O2] graph

Form of the Rate Law For aA + bBcC +dD • Rate = k [A]n [B]m • Where k is the rate constant n = order of reactant A m = order of reactant B • n and m must be determined experimentally • n +m = order of the reaction

Experimental Order • the order in the integrated rate law Rate = - [Reactant] = k [Reactant]n t n = 0, zero order n = 1, first order n = 2, second order Determine order

Order of Reaction A + B → C • Rate = k[A]n [B]m • (n + m) = order of the reaction = 1 unimolecular =2 bimolecular =3 trimolecular This means how many particles are involved in the rate determining step

Method of Initial Rates • A series of experiments are run to determine the order of a reactant. • The reaction rate at the beginning of the reaction and the concentration are measured • These are evaluated to determine the order of each reactant and the overall reaction order

If you plot the concentration versus time of [N2O5], you can determine the rate at 0.90M and 0.45M. What is the rate law for this reaction? Rate = k [N2O5]nn = the order. It is determined experimentally.

2N2O5(soln) 4NO2(soln) + O2(g) • At 45C, O2 bubbles out of solution, so only the forward reaction occurs. Data [N2O5] Rate ( mol/l • s) 0.90M 5.4 x 10-4 0 45M 2.7 x 10-4 The concentration is halved, so the rate is halved

2N2O5(soln) 4NO2(soln) + O2(g) Rate = k [N2O5]n 5.4 x 10-4 = k [0.90]n 2.7 x 10-4 = k [0 45]n after algebra 2 = (2)n n = 1 which is determined by the experiment Rate = k [N2O5]1

Map Method of Initial Rates • Measure the rate of reaction as close to t = 0 as you can get. • This is the initial rate. • Vary the concentration • Compare the initial rates.

NH4+ + NO2- N2 + 2H2O • Rate = k[NH4+1]n [NO2-1]m • How can we determine n and m? (order) • Run a series of reactions under identical conditions. Varying only the concentration of one reactant. • Compare the results and determine the order of each reactant Order

NH4+ + NO2- N2 + 2H2O

NH4+ + NO2- N2 + 2H2O • Compare one reaction to the next 1.35 x 10-7 = k(.001)n(0.050)m 2.70 x 10-7 = k (0.001)n(0.010)m Form

1.35 x 10-7 = k(0.001)n(0.0050)m 2.70 x 10-7 k (0.001)n(0.010)m 1.35 x 10-7 = (0.0050)m 2.70 x 10-7 (0.010)m • 1/2 = (1/2)m • m = 1 In order to find n, we can do the same type of math with the second set of reactions

Compare one reaction to the next 2.70 x 10-7 = k (0.001)n(0.010)m 5.40 x 10-7 = k(.002)n(0.010)m NH4+ + NO2- N2 + 2H2O

2.70 x 10-7 = k (0.001)n(0.010)m 5.40 x 10-7 k(.002)n(0.010)m • 0.5 = (0.5)n • n = 1 n + m = order of the reaction 1 + 1 = 2 or second order Form

Review • Method of initial rate • In the form Rate = k[A]n[B]m • Where k is the rate constant • n, m = the order of the reactant • The order is determined experimentally • Rate law is important so we can gain an insight into the individual steps of the reaction

Map The Integrated Rate Law • Expresses how concentrations depend on time • Depends on the order of the reaction Remember • Rate = k[A]n[B]m Order = n + m • Integrated Rate law takes the form by “integrating” the rate function. (calculus used to determine) • The value of n and m change the order of the reaction • The form of the integrated rate depends on the value of n • You get a different equation for zero, first and second order equations.

Reaction Order • Order of the reaction determines or affects our calculations. • Zero order indicates the use of a catalyst or enzyme. The surface area of catalyst is the rate determining factor. • First or second order is more typical (of college problems)

Integrated Law - Zero Order Rate = - [A] = k t Set up the differential equation d[A] = -kt Integral of 1 with respect to A is [A]

Map Integrated Rate Law – First Order Rate = [A] = k [A] n t If n = 1, this is a first order reaction. If we “integrate” this equation we get a new form. Ln[A] = -kt + ln[A0] where A0 is the initial concentration

Why? If Rate = - [A] = k [A] 1 t Then you set up the differential equation: d[A] = -kdt [A] Integral of 1/[A] with respect to [A] is the ln[A].

Integrated Rate Law ln[A] = -kt + ln[A]0 • The equation shows the [A] depends on time • If you know k and A0, you can calculate the concentration at any time. • Is in the form y = mx +b Y = ln[A] m = -k b = ln[A]0 Can be rewritten ln( [A]0/[A] ) = kt • This equation is only good for first order reactions!

First Order Reaction [N2O5] Time (s) 0.1000 0 0.0707 50 0.0500 100 0.0250 200 0.0125 300 0.00625 400 Ln[N2O5] Time (s)

Graph Data Map

Given the Reaction2C4H6 C8H12 And the data [C4H6] mol/L Time (± 1 s) 0.01000 0 0.00625 1000 0.00476 1800 0.00370 2800 0.00313 3600 0.00270 4400 0.00241 5200 0.00208 6200

Equations 2C4H6 C8H12

Map Data Graphical Analysis ___1___ [C4H6] Ln [C4H6]

Experimental Derivation of Reaction Order • Arrange data in the form 1/[A] or ln [A] or [A] • Plot the data vs time • Choose the straight line y = mx + b • Determine the k value from the slope • Graphical rate laws 1/[A] = kt + b → 2nd ln[A] = kt + b → 1st [A] = kt + b → zero

Half-life • The time it takes 1/2 of the reactant to be consumed • This can be determined • Graphically • Calculate from the integrated rate law

Chemical Kinetics