Chapter 12 Chemical Kinetics. Kinetics Study of Speed at which reactions take place –Effect of...

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

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

1. Use the data to determine the rate law.

2. Determine the rate constant.

3. Determine the [N2O5] at 2000.s

Time (s) [N2O5]

0 .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

1. Use the data to determine the rate law.

2. Determine the rate constant.

3. Determine the [N2O5] at 2000.s

4. Determine the half life

Time (s) [N2O5]

0 .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

1. Determine the rate law

2. Determine the rate constant.

3. Determine the half-life

Time (s) [C4H8]

0 .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

1. Determine the rate law.

2. Determine the rate constant.

3. Determine the half-life

Time (s) [C4H8]

0 .44

10. .33

20. .22

30. .11

40. 0

Page 578

Homework

• Page 599 #’s 27,29,30,32

Reaction Mechanism

• A series of elementary steps that must satisfy two requirements

1. The sum of the steps must equal the overall balanced equation

2. 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 Bimolecular Rate=k[A][B]2

2A + B Products Termolecular Rate=k[A]2[B]

A+B+C Products Termolecular 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

ATR

Ek a ln

1ln

• 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

211

2 11ln

TTR

Ea

k

k

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