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Chemistry The Molecular Science Moore, Stanitski

and Jurs

Chapter 13 Chemical Kinetics Rates of Reactions

Inquiring Minds Want to Know.

- Will a reaction happen? (Ch. 18)
- How fast or slow does a reaction happen? (Ch. 13)
- When will the reaction stop or reach

equilibrium? (Ch. 14 and 17) - Drugs in the body.
- Ozone in the atmosphere.
- Graphite to diamond.

Chemical Kinetics

- Chemical kinetics studies the rate at which a

chemical reaction occurs and the pathway taken.

Rate is the change of something per unit time. - Examples distance/time
- /time
- concentration/time
- C (graphite) C (diamond) excruciatingly

slow rate -?C(graphite) / ?time - CH4 (g) 2 O2 (g) CO2 (g) 2 H2O (g)

rapid - rate -?CH4(g) /?time

Chemical Kinetics

Homogeneous - reactants products in one

phase. ex. molecules more likely to come together

in gas or liquid phase Heterogeneous -

species in multiple phases. ex. solid catalyst

and air (catalytic converter)

Reaction Rate

- Factors affecting the speed of a reaction
- (Homogeneous)
- X, M or P
- Concentration of reactants
- k, units include s-1
- Properties (phase, structure)
- Temperature
- Catalysts

Rate -?Cv/?time k Cv n

Reaction Rate

- Change in reactant (or product) per unit time.

Cresol violet (Cv a dye) decomposes in NaOH(aq)

Cv(aq) OH- (aq) ? CvOH(aq)

Reaction Rates and Stoichiometry

- For any general reaction
- a A b B c C d D
- The overall rate of reaction is

Cv(aq) OH-(aq) ? CvOH(aq)

Loss of 1 Cv ? Gain of 1 CvOH Rate of Cv loss

Rate of CvOH gain Rate -?Cv/?time

?CvOH/?time

Reaction Rates and Stoichiometry

H2 (g) I2 (g) ? 2 HI (g) The rate of loss of

I2 is 0.0040 mol L-1 s-1. What is the rate of

formation of HI ?

Reaction Rate

The rate of the Cv reaction can be calculated

- Time, t Cv Average rate
- (s) (mol / L) (mol L-1 s-1)
- 0.0 5.000 x 10-5
- 10.0 3.680 x 10-5
- 20.0 2.710 x 10-5
- 30.0 1.990 x 10-5
- 40.0 1.460 x 10-5
- 50.0 1.078 x 10-5
- 60.0 0.793 x 10-5
- 80.0 0.429 x 10-5
- 100.0 0.232 x 10-5

13.2 x 10-7 9.70 x 10-7 7.20 x 10-7 5.30

x 10-7 3.82 x 10-7 2.85 x 10-7 1.82 x 10-7

0.99 x 10-7

trend? Rate depends on conc. Rate decrease with

conc. React.

Average Rate and Instantaneous Rate

Graphical view of Cv reaction

5.0E-5 4.0E-5 3.0E-5 2.0E-5 1.0E-5 0

Rate -?Cv/?time

Cv (mol/L)

t (s)

0 20 40 60 80 100

Average rate slope of the blue or grey

triangle but the avg. rate depends on interval

chosen.

Average Rate and Instantaneous Rate

5.0E-5 4.0E-5 3.0E-5 2.0E-5 1.0E-5 0

Cv (mol/L)

t (s)

0 20 40 60 80 100

- Instantaneous rate slope of a line tangent to

the curve. - t 5 s and t 75 s have different instantaneous

rates - Cannot predict Cv at a time too far away since

the rate changes.

The Rate Law

- Rate may change when reactant changes.

- Cv example shows this.
- For Cv the rate is proportional to concentration.

k

(10 second interval)

rate k Cvn

y m x

The Rate Law

Rate Law and Order of Reaction

A general reaction will usually have a rate

law rate k Am Bn . . .

where k rate constant (T, properties,

catalyst) m, n order for A B, respectively m

n overall order of the reaction

- The orders are usually integers (-2, -1, 0, 1,

2), but may also be fractions (½, ?) - The concentration of a reactant with a zero-order

dependence has no effect on the rate of the

reaction. (There just needs to be some reactant

around.

The Rate Law Reaction Order

The reaction is second order in NO first order

in H2 third order overall.

The Rate Law Reaction Order

- 22.
- 2 NO (g) Br2 (g) 2 NOBr (g)
- Experiment shows that the reaction is first-order

in Br2 and second-order in NO. - Write the rate law for the reaction.
- Rate kBr2NO2
- If the concentration of Br2 is tripled, how will

the reaction rate change? - 3X
- What happens to the reaction rate when the

concentration of NO is doubled? - 4X

Determining Rate Laws from Initial Rates

Rate laws must be measured experimentally. They

cannot be predicted from reaction stoichiometry.

- Initial Rate Method
- To find the order for a reactant
- Run the experiment with known reactant0.
- Measure the initial rate of reaction (slope at t

0, lt2 consumed.) - Change reactant0 of 1 reactant keep all others

constant. - Re-measure the initial rate.
- The ratio of the two rates gives the order for

the chosen reactant.

Determining Rate Laws from Initial Rates

- Data for the reaction of methyl acetate with base

change concentration of each reactant in turn

Initial concentration (M) Expt. CH3COOCH30 OH

-0 Initial rate (M/s) 1 0.040 0.040

2.2 x 10-4 2 0.040 0.080 4.5 x 10-4

3 0.080 0.080 9.0 x 10-4

gt x2

gt x2

Start with generic rate law rate k

CH3COOCH3m OH-n

Determining Rate Laws from Initial Rates

Initial concentration (M) Expt. CH3COOC

H3 OH- Initial rate (M/s) 1

0.040 0.040 2.2 x 10-4 2 0.040

0.080 4.5 x 10-4 3 0.080 0.080

9.0 x 10-4

- Dividing the first two data sets (OH changing)
- 4.5 x 10-4 M/s k (0.040 M)m(0.080 M)n
- 2.2 x 10-4 M/s k (0.040 M)m(0.040 M)n

Thus 2.05 (2.00)n 2.05 (2.00)n and n

1

rate k CH3COOCH3m OH-1

Determining Rate Laws from Initial Rates

- Use experiments 2 3 to find m (dependance on

CH3COOCH3)

9.0 x 10-4 M/s k (0.080 M)m(0.080 M)n 4.5 x

10-4 M/s k (0.040 M)m(0.080 M)n

So 2.00 (2.00)m (1)n 2.00

(2.00)m and m 1 Also 1st order with respect

to CH3COOCH3.

rate k CH3COOCH31 OH-1

Determining Rate Laws from Initial Rates

- The rate law is
- rate k CH3COOCH3OH-
- Overall order for the reaction is
- m n 1 1 2
- The reaction is 2nd order overall.

1st order in OH- 1st order in CH3COOCH3

Determining Rate Laws from Initial Rates

If a rate law is known, k can be

determined rate k CH3COOCH3OH-

Using Exp. 1

k 0.1375 M-1 s-1 0.1375 L mol-1 s-1

Could repeat for each run, take an average But a

graphical method is better.

Practice

Rate kNOnO2m

- 2NO(g) O2(g) ? 2NO2
- Initial rates were measured at 25 ºC starting

with various concentrations of reactants.

What is the rate law? (m,n) Calculate the rate

constant (k).

Practice What is n?

Rate kNOnO2m

4x

The reaction is second order with respect to NO.

Why cant we compare experiment 1 and 3 ???

n 2

Practice What is m?

Rate kNO2O2m

2x

2x

The reaction is first order with respect to

oxygen.

m 1

Practice What is k?

Rate kNO2O2

Using Exp. 1

Units?... rate always has units of M/s.

Practice What is k?

- Iodide ion is oxidized in acidic solution to the

triiodide ion, I3- , by hydrogen peroxide.

A series of four experiments were run at

different concentrations, and the initial rates

of I3- formation were determined. From the

following data, obtain the reaction orders with

respect to H2O2, I-, and H. Calculate the

numerical value of the rate constant.

Practice What is m,n,p?

Rate kH2O2mI-nHp

Rate kH2O21I-2H0

Because H 0 1, the rate law is

Rate kH2O2I-2

We can now calculate the rate constant by

substituting values from any of the experiments.

Using Experiment 1

Change of Concentration with Time

- A rate law simply tells you how the rate of

reaction changes as reactant concentrations

change.

The integrated rate law shows how a reactant

concentration changes over a period of time.

The Integrated Rate Law (Calculus)

Consider a 1st-order reaction

A products

A big change A little change (differential

equation)

Integrates to ln At -k t ln A0

y m x b (straight line)

If a reaction is 1st-order, a plot of ln A vs.

t will be linear.

Graphing Kinetic Data

- In addition to the method of initial rates, rate

laws and k can be deduced by graphical methods.

For first-order reaction

y mx b

This means if we plot lnAt versus t, we will

get a straight line for a first-order reaction.

Concentration-Time Equations

- Second-Order Rate Law

Using calculus, you get the following equation.

Here At is the concentration of reactant A at

time t, and Ao is the initial concentration.

Graphing Kinetic Data

For second-order reaction

y mx b

- This means if we plot 1/At versus time,
- we will get a straight line for a second-order

reaction.

Concentration-Time Equations

For a zero-order reaction we could write the rate

law in the form

Rearrange get the following equation.

Here At is the concentration of reactant A at

time t, and Ao is the initial concentration.

Graphing Kinetic Data

For zero-order reaction

y mx b

- This means if we plot At versus time,
- we will get a straight line for a second-order

reaction.

The Integrated Rate Law

0 rate k At -kt A0 -k 1 rate kA

lnAt -kt lnA0 -k

y mx b

The most accurate k is obtained from the slope of

a plot.

The Integrated Rate Law

Rate data for the decomposition of

cyclopentene C5H8(g) ? C5H6(g) H2(g) were

measured at 850C. Determine the order of the

reaction from the following plots of those data

NOT a line line! NOT a line

- The reaction is first order (the only linear

plot) - k -1 x (slope) of this plot.

Example 1

- The decomposition of N2O5 to NO2 and O2 is first

order with a rate constant of 4.800 x 10-4 s-1.

If the initial concentration of N2O5 is 1.65 x

10-2 mol/L, what is the concentration of N2O5

after 825 seconds?

The first-order time-concentration equation for

this reaction would be

Practice- using integrated rate laws

- The decomposition of N2O5 to NO2 and O2 is first

order with a rate constant of 4.800 x 10-4s-1. If

the initial concentration of N2O5 is 1.65x10-2

mol/L, what is the concentration of N2O5 after

825 seconds?

eln(x) x

Example 2

- Consider the following first-order reaction
- CH3CH2Cl(g)? C2H4(g) HCl(g)
- The initial concentration of ethyl chloride was

0.00100M. After heating at 500.0ºC for 155.00 s

the concentration was reduced to 0.000670 M.

What was the concentration of ethyl chloride

after a total of 256.00 s?

Example 1

- The initial concentration of ethyl chloride was

0.00100M. After heating at 500.0ºCfor 155.00 s

the concentration was reduced to 0.000670 M.

What was the concentration of ethyl chloride

after a total of 256.00 s?

C2H5Cl0 0.00100M C2H5Clt1 0.000670M t1

155.00s t2 256.00s, C2H5Clt2?

2.58 10-3 s-1

Example 1

The initial concentration of ethyl chloride was

0.00100M. After heating at 500.0ºCfor 155.00 s

the concentration was reduced to 0.000670 M.

What was the concentration of ethyl chloride

after a total of 256.00 s?

Half-Life

- t1/2 time for reactant to fall to

½reactant0.

- For 1st -order reactions
- t1/2 is independent of the starting

concentration. - only true for 1st order reactions (not 0th,

2nd) - t1/2 is constant for a given 1st -order reaction.

Half-Life

For a 1st-order reaction lnAt -kt

lnA0

When t t1/2, At ½A0

Then ln(½A0) -kt1/2 lnA0

ln(½A0/A0) -kt1/2 note ln x ln

y ln(x/y) ln(½) -ln(2) -kt1/2 note

ln(1/y) ln y

Half Life

t1/2 of a 1st-order reaction can be used to find

k.

- For cisplatin (a chemotherapy agent)
- 0.0100 M ? 0.0050 M, after 475 min
- (t½475minutes)

graphically

t½

t½

t½

Calculating or t

- Use an integrated rate equation or half-life

equation.

Example In a 1st-order reaction, reactant0

0.500 mol/L and t1/2 400.s. Calculate

- reactant,1600.s after initiation.
- t for reactant to drop to 1/16th of its initial

value. - t for reactant to drop to 0.0500 mol/L.

Calculating or t

- In a 1st-order reaction, reactant0 0.500

mol/L and t1/2 400.s - (a) Calculate reactant ,1600.s after initiation.

1st order k ln 2/ t½ 0.6931/(400. s)

1.733x10-3 s-1 lnAt -(0.001733

s-1)(1600 s) ln(0.500) lnAt -2.773

-0.693 -3.466 At e-3.466 0.0312 mol/L

ln At -kt ln A0

OR1600 s 4 t1/2 so 0.500 ? 0.250 ? 0.125 ?

0.0625 ? 0.0313 M.

Calculating or t

- In a 1st-order reaction, reactant0 0.500

mol/L and t1/2 400.s - (b) Calculate t for reactant to drop to 1/16th

of its initial value.

4 (400 s) 1600 s

Calculating or t

In a 1st-order reaction, reactant0 0.500

mol/L and t1/2 400.s (c) Calculate t for

reactant to drop to 0.0500 mol/L ?

- From part (a) k 1.733 x 10-3 s-1
- ln At -kt ln A0
- then ln (0.0500) -(0.001733 s-1) t

ln(0.500) - -2.996 -(0.001733 s-1) t 0.693
- t 1.33 x 103 s

Elementary and Complex Reactions

- On the nanoscale level, a reaction may be
- unimolecular a single particle (atom, ion,

molecule) rearranges into 1 or 2 different

particles. - (example some decomposition, geometric isomer)
- bimolecular two particles collide and rearrange

- (example composition)
- Observed reactions, on the macroscale level, may

be - elementary directly occur by one of these two

processes, or - complex occur as a series of elementary steps

Elementary Reactions

Example Label these elementary reactions as

unimolecular or bimolecular

unimolecular

bimolecular

unimolecular

unimolecular

Elementary Reactions - Unimolecular

2-butene isomerization is unimolecular

- cis-trans conversion twists the CC bond.
- This requires a lot of energy (Ea

4.35x10-19J/molecule 262 kJ/mol) - Even more (4.42x10-19J/molecule) to convert back.

cis-2-butene trans-2-butene

Unimolecular Reactions

transition state or activated complex

500 400 300 200 100 0

Ea is the activation energy, the minimum E to go

over the barrier.

Potential energy (10-21 J)

Ea 435 x 10-21 J

?E -7 x 10-21 J

Initial state

Final state

-30 0 30 60 90

120 150 180 210

Exothermic overall

Reaction Progress (angle of twist)

- kinetics if lower Ea, increase the speed of the

reaction - thermodynamics if EproductltEreactant, the

reaction is exothermic

Transition State

- Occurs at the top of the activation barrier
- Exists for very short time (few fs, 1 fs 10-15

s). - Falls apart to form products or reactants.
- the molecules must have enough energy

activation energy to form the transition state

and convert it into products.

Endothermic reaction

overall up hill EproductsgtEreactants

Ea

Exothermic reaction

overall down hill EproductsgtEreactants

Bimolecular Reactions

I- must collide with enough E and with the right

orientation to cause the inversion.

Bimolecular Reactions

For the reaction to take place two molecules must

effectively collide. One out of four collisions

is successful. (backside of tetrahedron, opposite

Br). This geometric constraint on approach is

called a steric factor.

Bimolecular Reaction

Bimolecular Reaction

O3 NO O2 NO2

Reaction Energy Diagrams

Forward reaction Ea 10kJ/mol Exothermic Backwa

rd reaction Ea? Exo or Endo?

Factors Affecting Reaction Rates

Increasing T will speed up most reactions.

Higher T higher average Ek for the

reactants. larger fraction of the molecules

can overcome the activation barrier.

Many more molecules have enough E to react at

75C, so the reaction goes much faster.

Temperature and Reaction Rate

- Rule of thumb approximately double the rate for

every 10 K.

I- CH3Br ? Br - CH3I

Rate kI-mCH3Brn

Temperature and Reaction Rate

- The Arrhenius equation shows how k varies with T

A Frequency factor. How often a collision

occurs with the correct orientation. (1/4 for I-

CH3Br) e-Ea/RT Fraction of the molecules

with enough Ea to cross the barrier.

T Temperature Must be in Kelvin. R Gas law

constant 8.314 J K-1 mol-1.

Temperature and Reaction Rate

rate kI-mCH3Brn rate

(Ae-Ea/RT)I-mCH3Brn Can rearrange.

OWL 13.5a OWL link

Factors Affecting Reaction Rates

- Presence of a catalyst.

A catalyst is a substance that increases the rate

of a reaction without being consumed in the

overall reaction. The catalyst generally does

not appear in the overall balanced chemical

equation (although its presence may be indicated

by writing its formula over the arrow).

Catalysts

- A catalyst increases the rate of reaction by

lowering the activation energy, Ea. - To avoid being consumed, the catalyst must

participate in at least one step of the reaction

and then be regenerated in a later step. - Ultimately a catalyzed reaction has a different

reaction mechanism than the same reaction

proceeding without the presence of catalyst.

Catalyzed and Uncatalyzed Reactions

Without catalyst. With I- catalyst.

Catalyzed and Uncatalyzed Reactions

Reaction Mechanisms

- To explain how reactions happen on the molecular

scale we need to look into individual steps that

take place during the reaction.

The collection of individual steps that a

reaction goes through is referred to as the

reaction mechanism. The individual steps are

referred to as elementary reactions. The sum of

several elementary reactions is a complex

reaction.

Elementary/Complex Reactions

- Consider the reaction of nitrogen dioxide with

carbon monoxide.

reaction intermediate not in net equation

Elementary/Complex Reactions

The overall chemical equation is obtained by

adding the two steps together and canceling any

species common to both sides.

Rate Laws for Elementary Reactions

- Elementary reactions
- Occur as written and their rate laws are

predictable. - Unimolecular reactions are always 1st -order.
- Bimolecular reactions are always 2nd-order.

A Products rate kA A B Products

rate kAB A A Products rate

kA2

Rate Laws for Elementary Reactions

- For an elementary reaction, the coefficient of

each reactant becomes the power to which it is

raised in the rate law for that reaction.

rate kNOO3 (determined experimentally)

Rate Laws for Complex Reactions

- Complex reactions
- Do not occur as written.
- They occur as a series of elementary steps.

rate kH2O2 (determined experimentally)

Rate Laws and Mechanisms

- Consider the reaction below.

This implies that the reaction above is not an

elementary reaction but rather the result of

multiple steps.

Rate Laws and Mechanisms

2I- H2O2 2 H3O I2 4H2O Rate

I-H2O2

Rate Laws and Mechanisms

What is the net reaction? What is the rate law

for this reaction?

(fast) (slow) (fast)

Rate N2O2H22