Chapter 12 Chemical Kinetics

Kinetics Study of Speed at which reactions take place Effect of each reactant Effect of concentration Mechanisms Reaction process Energy requirements for reactions

Reaction Rate Change in concentration of reactant or product per unit of time Rate = Δ[A]/ Δt Rates decrease as time increases Rates are given positive values If you know the rate of one species in a reaction you can calculate the others

2NO2  2NO + O2 Page 558 Based on the graph What do you notice? Calculate the rate of NO2 0s to 50.s NO2 50.s to 100.s O2 0s to 50.s 2NO2  2NO + O2

Rate Laws Only depends on the forward reaction An expression that shows how the rate depends on the concentrations of reactants A way of determining rate or concentration Depends on what you know.

Form of the Rate Law Assuming the reaction A  B + C Rate = k[A]n k and n are both experimentally determined k is a constant called the rate constant Units vary depending on n n is the order Does NOT depend on balanced equation Whole numbers including zero and fractions

Units of k Depends on what number n is k has units that gives the rate units of mol/L*s Consider - Rate = k[A]n What are the units of k if n is 1? 2? 0?

Multiple Reactants Consider the reaction A + B  C What is the rate law? Rate = k[A]n[B]m What are the units for k when n is 1 & m is 1

Specifically These types of rate laws are differential rate laws Just called rate laws There are also integrated rate laws The form you use depends on the data you have

Homework P. 598 #’s 17,18,19,20

Method of Initial Rates

Integrated Rate Laws A way of determining the order and rate constant of a reaction when time and concentration data is known. Must be for a single reactant. Or have one in excess Equations for 0th, 1st, and 2nd order Use equations as tests for order

1st Order ln[A] = -kt + ln[A]o [A] concentration of reactant at time t [A]o concentration of reactant at time t=0 k is the rate constant This is the equation of a ________? LINE Y axis is ln[A], X axis is t The slope is the rate constant

1st Order If a graph of ln[A] vs t is a straight line then the reaction is 1st order A straight line is the check for all of these

2N2O5 4NO2 + O2 Example Use the data to determine the rate law. Determine the rate constant. Determine the [N2O5] at 2000.s Time (s) [N2O5] .100 100. .0614 300. .0233 600. .00541 900. .00126

1st Order Half Life The half life is the time required for a reactant to reach half of its previous concentration Meaning [A] = [A]o/2 Derive Half Life Equation t1/2=ln2/k Half life is always the same for first order kinetics

2N2O5 4NO2 + O2 Example Use the data to determine the rate law. Determine the rate constant. Determine the [N2O5] at 2000.s Determine the half life Time (s) [N2O5] .100 100. .0614 300. .0233 600. .00541 900. .00126

2nd Order Integrated Rate Law 1/[A] = kt + 1/[A]o This is the equation of a ________? LINE Y axis is 1/[A], X axis is t The slope is the rate constant Half Life t1/2 = 1/(k[A]o) Each half life is double the previous

2C4H6  C8H8 Example #33 p. 600 Determine the rate law Determine the rate constant. Determine the half-life Time (s) [C4H8] .017 195 .016 604 .015 1246 .013 2180 .011 6210 .0068

0th Order Integrated Rate Law [A] = -kt + [A]o This is the equation of a ________? LINE Y axis is [A], X axis is t The slope is the rate constant Half Life t1/2 = [A]o/2k Each half life is the same

2N2O  2N2 + O2 Determine the rate law. Determine the rate constant. Determine the half-life Time (s) [C4H8] .44 10. .33 20. .22 30. .11 40.

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Homework Page 599 #’s 27,29,30,32

Reaction Mechanism A series of elementary steps that must satisfy two requirements The sum of the steps must equal the overall balanced equation Must agree with the rate law Elementary Step – Steps in the mechanism. (Think back to organic)

Cont. Intermediates – Species that are produced in one elementary step and consumed in another Not part of the overall balanced equation Rate-determining step – Slowest step in a reaction that determines the rate Molecularity – Number of species that must collide to produce an elementary step

Molecularity Table Elementary Step Molecularity Rate Law A  Products Unimolecular Rate=k[A] 2 A  Products Bimolecular Rate=k[A]2 A + B  Products Rate=k[A][B]2 2A + B  Products Termolecular Rate=k[A]2[B] A+B+C Products Rate=k[A][B][C]

NO2 + CO  NO + CO2 The reaction above has an experimentally determined rate law of Rate=k[NO2]2 Is the proposed mechanism possible? Explain Step 1 = NO2 + NO2  NO3 + NO Step 2 = NO3 + CO  NO2 + CO2

Collision Model for Kinetics For a reaction to occur particles must collide to react. Must collide in the proper orientation Must collide with enough energy Activation Energy

Proper Orientation +  Images from: http://www.sparknotes.com/chemistry/kinetics/mechanisms/section1.rhtml

Improper Orientation +  Images from: http://www.sparknotes.com/chemistry/kinetics/mechanisms/section1.rhtml

Activation Energy (Ea) The energy required to convert atoms or molecules into their transition state Minimum energy required for effective collisions Can be found by running the same reaction at different temps. Use Arrhenius Equation

Arrhenius Equation K = rate constant Ea = Activation Energy R = Energy Gas Constant 8.314 J/mol*K T = Temp in K A = Frequency Factor Equation of a line Y is ln k X is 1/T Slope is –Ea/R

Find the activation energy 2N2O5  4NO2 + O2 Rate Constant Temp (ºC) 2.0x10-5 20 7.3x10-5 30 2.7x10-4 40 9.1x10-4 50 2.9x10-3 60

2 Temp. Ea Equation You can manipulate the Arrhenius Equ if you only have to temps and rate constants by subtracting the lower temp equation from the higher temp equation

Catalysts Chemical that speeds up a chemical reaction but is not consumed in the reaction Do this by Lowering activation energy Provide alternate reaction pathways Increase effective collisions

Cont. Below are two steps in the destruction of ozone. What is the overall rxn? What is the catalyst? What is the intermediate? Cl + O3 ClO + O2 O + ClO  Cl + O2

Homework Page 602 #’s 45,48,49,54