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Chemical Kinetics

- Area of chemistry that deals with rates or speeds

at which reactions occur. - Examples of why the study of kinetics is

important - --allows determination of expiration dates on

food. - --allows determination of how fast a medicine

will work. - --increases efficiency of chemical processes

more product, less waste.

Some Reactions Occur at Very Fast Rate

Some Reactions are Very Slow

Collision Theory In order for molecules to

react, they must be properly oriented. More

collisions provides a better chance of correct

orientations. Higher rate of properly oriented

collisions equals faster reaction.

Proper Orientation

Notice the Orientation

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Not Properly Oriented

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Not Properly Oriented

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Properly Oriented

REACTION !

Several factors influence the rate of

collisionsand therefore, the rate of a reaction.

Factors Affecting Rates of Reaction

- Concentration of reactants
- Temperature at which the reaction occurs
- The presence of a catalyst
- Physical State / Surface area of reactants.

a brief overview.

Increase Temperature

- Temperature is a measure of the kinetic energy of

molecules. - Kinetic Energy ½ mv2
- Temperature related to the SPEED of the molecule

Increased temperature increased collision rate

increase rate of reaction.

Concentration

High Concentration of Reactants Much interaction

Low Concentration of Reactants Little interaction

Oxygen in Atmosphere

Pure Oxygen

Physical States and Surface Area

Gas

Gas

Gas

Gas

Different States Less Interaction

Different States Less Interaction Possible

Liquid

Liquid

Liquid

Gas

Same States More Interaction Possible

Gas

Mg metal reacts with high temperature steam, but

not water.

Mg (s) H2O (l) ? No Reaction Mg (s) H2O (g) ?

MgO H2

When solids and gases react, the reaction is

limited to the surface of the solid.

Grain Elevator Fire

Grinding chemicals with a mortar and pestle

increases reaction rate.

Catalysts

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Reactants

Catalyst (Enzyme)

Enzyme reactants Complex

Enzyme products Complex

Products

Unchanged Enzyme

Reactants

Catalyst Enzyme

Enzyme reactants Complex

Enzyme products Complex

Products

Unchanged Enzyme

Activation Energy

Activation Energy

Activation Energy-- energy need to get a reaction

started

Activation Energy

Activation Energy

Activation Energy

Activation Energy

Activation Energy

Activation Energy

Activation Energy

Activation Energy

Activation Energy

Activation Energy

Activation Energy

As a review the different factors that help

increase the probability that reactants will

align in the correct way and with enough energy

for a reaction to occur are...

- Increase temperature
- Increase concentration
- Physical State and Surface Area
- The presence of a catalyst

Now for the details and calculations!

The Rate of a Chemical Reaction is the rate of

change in the concentration of reactants and/or

products per unit time.

Reaction rates are determined by measuring the

concentration of one or more chemicals involved

at different times during the chemical reaction.

Reaction rates are expressed as the number of

moles per liter that react each second during the

reaction.

The units of a reaction rate aremol L-1 s-1 M

s-1.

Considering Reaction A ? B

The reaction rate is a measure of how quickly A

is consumed or B is produced.

- Average rate of reaction can be written
- (remember ? final initial)

This is a measure of the average rate of

appearance of B.

- Average rate can also be written in terms of A

This is the rate of disappearance of A (negative

value makes final rate positive).

Data for reaction A?B

Notice the decrease in rate as A is used up.

? Moles A ? Moles B

Start with one mole of A at time zero, measure

amounts of A and B at given time intervals.

Data for Reaction A?B

Time A B

0 seconds 1.0 mol 0.0 mol

20 seconds 0.54 mol 0.46 mol

40 seconds 0.30 mol 0.70 mol

Calculate the average rate of appearance of B

over the time interval from 0 to 40 seconds.

You Try One!

Time A B

0 seconds 1.0 mol 0.0 mol

20 seconds 0.54 mol 0.46 mol

40 seconds 0.30 mol 0.70 mol

Calculate the average rate of disappearance of A

over the time interval from 0s to 40s.

Same Value Notice negative sign makes value

positive.

Average Reaction Ratereaction rate over a given

amount of time. Example Change in

concentration during the time period of 20s-40s

during the reaction (as in previous examples).

Instantaneous Reaction Ratereaction rate at a

given moment in time. Notes Initial Rate where

time 0 is an instantaneous rate.

Use kinetic curve to calculate this.

Sometimes the unit on the y axis is moles (Liters

are assumed).

Remember the equation for finding the slope (math

class).

We use the same equation to determine the

instantaneous (for a specific time) rate of a

reaction from a kinetic curve.

(

)

Reaction rates have values - sign makes rate

of disappearance .

Rates in Terms of Concentration

- The volume in the reaction vessel remains

constant. - Therefore, in analyzing the reaction A?B we can

actually consider the rate in terms of Molarity. - This gives us the units M/s (molarity per second).

Consider the reaction between butyl chloride and

water

C4H9Cl(aq) H2O(l)? C4H9OH(aq) HCl(aq)

Brackets indicate concentration.

- Using the curve created from the data, we can

determine the instantaneous rate for any given

point on the curve. (slope of tangent line

tangent line touches curve at point of interest).

Calculate the rate of disappearance of C4H9Cl at

t 0.

0.06 M 0.1M

Rate -

200 s 0.0 s

2 x 10-4 M/s

Calculate the rate of disappearance of C4H9Cl at

t 300.

0.02 M 0.055 M

Rate -

600 s 300 s

1 x 10-4 M/s

Rates and Stoichiometry

- When mole ratios of equations are not 11.
- aA bB?cC dD

This is how your book explains this. ? Too

Complicated!?

The lengthy explanation provided by your text

basically boils down to A PICKET FENCE

- If the rate of decomposition of N2O5 in a

reaction vessel is 4.2 x 10-7 M/s, what is the

rate of appearance of NO2? - 2N2O5(g)? 4NO2(g) O2(g)

Given 4.2 x 10-7 mol L-1 s-1 N2O5

4.2 x 10-7 mol L-1 s-1 N2O5

4 mol NO2

2 mol N2O5

8.4 x 10-7 mol L-1 s-1 NO2

- If the rate of decomposition of N2O5 in a

reaction vessel is 4.2 x 10-7 M/s, what is the

rate of appearance of O2? - 2N2O5(g)? 4NO2(g) O2(g)

Given 4.2 x 10-7 mol L-1 s-1 N2O5

4.2 x 10-7 mol L-1 s-1 N2O5

1 mol O2

2 mol N2O5

2.1 x 10-7 mol L-1 s-1 O2

What are the 4 factors that affect the rate of a

chemical reaction?

- Concentration of reactants
- Temperature at which the reaction occurs
- The presence of a catalyst
- State and Surface area of solid or liquid

reactants.

Concentration and Reaction Rates

- Increasing concentration of reactants increases

reaction rate. Examples Add more wood to fire

step on gas in automobile. - How do you think the reaction rate would change

over time? - Reaction rates usually decrease over time.
- Due to decreasing concentration of reactants.

Compare reaction rates with varying

concentrations of reactants

The effect of concentration on a chemical

reaction is expressed in a rate law.

k rate constantbrackets concentration of

reactantsThe exponents m and n are called

reaction orders.

Reaction Order

- General form for rate laws

Reactions show how changing a concentration on a

given reaction influences the rate of a

reaction. Sum of m and n is the overall reaction

order.

- Reaction orders are determined experimentally and

do not relate to coefficients of balanced

equation (We will see how to determine these in a

bit) - .
- In most rate laws reaction orders are 0, 1, or 2.
- Can be fractional or negative at times.

- In the above equation NH4 NO2- are raised to

the first power. - The reaction is 1st order for NH4
- The reaction is 1st order for NO2-
- The reaction is 2nd order overall

Units for Rate Constant (k)

- The units for the rate constant depend on the

order of the rate law.

Z overall order of reaction

- The rate law of the following reaction was

experimentally determined to be - H2(g) I2(g)? 2 HI(g)
- Rate k?H2??I2?
- What is the order of reaction for H2? I2?
- 1st order.
- What is the overall reaction order?
- 2nd order
- What are the units for the rate constant of the

reaction above?

- What are the units for the rate constant of the

reaction below? - 2N2O5(g) ? 4NO2(g) O2(g)
- Rate k N2O5

Determining Rate laws

- Rate law for any reaction must be determined

experimentally. - Zero order for a reactant means concentration

changes have no effect on reaction rate (straight

line on graph).

Example of data that would give us a 0 order of

reaction.

C

As we double B and keep A constant, rate doesnt

change.

Reaction order of B 0.

- 1st order means concentration changes give

proportional changes in reaction rate doubling

concentration doubles rate (parabola on graph

because rate decreases as reactant is used up).

- 2nd Order rate law, increasing concentration

results in a rate increase equal to the

concentration increased to the second power.

(parabola is steeper in beginning ramp at end). - Double con. 22 4 (rate increase)
- Triple con. 32 9 (rate increase)

1st

2nd

What is the reaction order of A?

C

As we double A and keep B constant, rate

increases by factor of 4.

Reaction order of A 2 because 22 4.

Once we know the reaction orders of the reactants

involved, we can write the rate law. Since the

reaction order of A was 2 and the reaction order

of B was 0, the rate law is Rate kA2 We

dont include B in the rate law because any

number raised to the 0 power 1.

- A particular reaction was found to depend on the

concentration of the hydrogen ion. The hydrogen

concentration was changed in each of the 3

reactions and the following rates of reactions

were found in each experiment. - H (M) 0.0500 0.100 0.200
- Initial rate(M/s) 1.6x10-7 3.2x10-7

6.4x10-7 - What is the order of the reaction in H?1st

order - Predict the concentration of H when initial

rate 0.80 x 10-7 M/s.0.250 M

What is the reaction order of NH4?

1

What is the reaction order of NO2-?

1

What is the rate law of the reaction?

Rate kNH4NO2-

What is the overall order of the reaction?

2

Once we determine the rate law we can apply the

experimental data to the rate law to find the

value of k.

Consider the following data for the upcoming

slides.

Calculate the value of the rate constant in the

following scenario.

- NH4 0.0200M
- NO2- 0.200M
- Rate 10.8 x 10-7 M-1s-1

Rate k NH4 NO2-

- When we know the rate law and the rate constant

for a reaction, you can calculate the rate of

reaction for any given concentrations of

reactants.

- What is the rate of reaction if the following

concentrations are present. - NH4 0.100M
- NO2- 0.100M
- k 2.7 x 10-4 M-1s-1

NH4 0.100 M NO2- 0.100 M k 2.7 x 10-4

M-1s-1

Rate k NH4 NO2-

Rate 2.7 x 10-4 M-1 s-1 (.100 M) (.100 M)

Rate 2.7 x 10-6 M / s

- Typical Problem
- Determine rate law from data.
- Determine value of k including units.
- Determine rate given concentrations not in data

table. - Common question Does k change as concentration

changes? Answer No, k only changes with

temperature.

Note The previous slides related to reactants

that were both 1st order. This is not always the

case. Handout Explanation of Determining Rate

Law(Read in Class) Practice Problem Handout

Change in Concentration with time

- Rate law tells how rate changes as concentration

changes at a particular temperature. - Using calculus, we can convert the rate laws into

equations that can give us the concentrations of

reactants or products at any time during a

reaction. - Dont worry. No calculus! The derived equations

are are given on the AP exam. You must know how

to use the equations.

Equations Listed on AP Exam

Used for First Order Reactions

Used for Second Order Reactions

First-Order Reactions

- Equation to relate beginning concentration (A0)

to concentration at any time (At). - Given any of the 3 quantities, we can solve for

the fourth (t, k, At, A0). - Make sure units of At and A0 are the same.

- The logarithm of a number is the power to which

10 must be raised to equal that . - Example logarithm of 1000 3, because 10

must be raised to 3 to equal 1000. - ln is the natural logarithm. Natural logarithms

are the power to which e (2.718) must be raised

to equal that number. - Example The ln of 7.39 is equal to 2.

Therefore, e2.00 7.39.

A certain first order reaction has a rate

constant of 4.5 x 10-3 s-1. What is the

concentration of a 0.050 M sample after it reacts

for 75.0 s?

Because we need the inverse of the natural

logarithm (antilogarithm), type into calculator

ex -3.334 At 0.356 M

- m is the slope (-k)
- b is the y-intercept of the line (lnA0)

2N2O5(g) ? 4NO2(g) O2(g)

- The decomposition of dinitrogen pentoxide is a

1st order reaction with a rate constant of 5.1 x

10-4s-1 at 45ºC. - a.) If initial conc. is 0.25M, what is the

concentration after 3.2 min.? - b.) How long will it take for the concentration

of N2O5 to decrease from 0.25M to 0.15M?

A0 0.25 k 5.1 x 10-4 s-1

t 192 s

Type into calculator ex -1.485 At 0.2265 M

How long will it take for the concentration of

N2O5 to decrease from 0.25M to 0.15M?

A0 0.25 At 0.15 k 5.1 x 10-4

s-1 t ?

T 1.0 x 103 s

Pressure can be used as unit of concentration.

(remember pressure is directly related to n/V in

ideal gas equation (PV nRT).

(CH3)2O?CH4(g) H2(g) CO(g)

- The decomposition of dimethyl ether (CH3)2O, at

510ºC is a first order process with a rate

constant of 6.80 x 10-4s-1 - If the initial pressure of (CH3)2O is 135 torr,

what is its partial pressure after 1420 s?

A0 .1776 atm k 6.80 x 10-4 s-1

t 1420 s

Type into calculator ex -2.694 At 0.0676 atm

51 torr

Second-Order Reactions

- Rate of reaction depends on the reactant

concentration raised to the second power or

depends on concentrations of two different

reactants, each to the 1st power.

Here is the equation given on the AP Exam.

- Plot of 1/At versus T gives a straight line.
- Slope k
- Y-intercept 1/A0

The decomposition of NO2 is second order. The

rate constant for the reaction is k 0.543 M-1

s-1. If the initial concentration of NO2 in a

closed vessel is 0.0500 M, what is the

concentration after 0.500 hr?

Answer 1.00 x 10-3 M

Using Lab Data to Predict Order First order

reactions give straight lines when we graph

lnAt vs. t. Second order reactions give

straight lines when we graph 1/At vs. t. Once

we graph the line, the slope can be determined,

which is k. See Sample exercise 14.8 on page 540.

Homework 14.37-14.38 Should use graph paper.

Half-life (t1/2)

- Half life is the time it takes the initial

concentration of reactant to drop to one-half its

value. - Consider a reactant that is 0.120 M. At some

point in the reaction it will be 0.060 M. This

would occur after one half-life. After a second

half-life passes, the concentration would be

0.030 M. - If the half life of the reaction above is 30

seconds, what will the concentration of the

reactant be after 2 minutes? - Answer 0.0075 M

One way to deal with half-lives is to consider

the 2 in ½ to the power of the number of half

lives. Example A reaction has a half-life of

10 seconds. If its initial concentration is

10.0 M, what is the concentration after 30

seconds? This represents 3 half lives. 23 8.

1/8 .125 .125 x 10.0 M 1.25

- In 1st order reactions the concentration of the

reactant decreases by 1/2 in a series of regular

time intervals (t1/2). The decreasing rate

depends only on k and doesnt depend on A0.

Not given on exam. Can derive from equation

given earlier (see page 541 if interested.

Easier to just memorize).

Half-life is used to describe radioactive decay

and elimination of medications from the body.

CH2CH2 CH2 ? CH3CH CH2

- Conversion of cyclopropane to propene in the gas

phase is a first-order reaction with a rate

constant of 6.7 x 10-4s-1 at 500ºC. Calculate the

half-life of the reaction.

Answer 1034 s

Half-Life of 2nd Order reactions

- Unlike 1st order reactions, the half-life of

second order reactions is dependent on the

initial concentration of reactant.

Just know the information above. We wont be

doing any calculations with these to save time.

Temperature Rate

- Most reactions increase in rate with increasing

temperature. - This is due to an increase in the rate constant

with increasing temperature.

Collision Model

- Collision model for chemical reactions is based

on Kinetic Molecular Theory. - Number of collisions increase with increasing

concentration. - Collisions increases with increasing molecular

speed (temp.)

Activation Energy

Ea

- Minimum amount of energy required to initiate a

chemical reaction. - Varies from reaction to reaction
- According to the Collision Model, this energy is

the KE of colliding molecules.

Activation energy must be enough to overcome

initial resistance for the reaction to take place.

- The difference in energy between the starting

point and the highest energy along the pathway is

the activation energy (Ea). - The arrangement of the atoms at the highest

energy point is called the activated complex or

transition state.

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- Kinetic Molecular theory states that molecules in

a sample have a wide variety of energy levels. - Increasing the temperature of a sample places a

larger fraction of the molecules with energy of

Ea. - The lower the activation energy the faster the

reaction. - See figure 14.16 page 546 and sample exercise

14.10 on 547

Which energy diagram represents a highly

exothermic reaction that has a small activation

energy? Which reaction would be the slowest? Is

this reaction endothermic or exothermic?

Endothermic

Catalyst

- Catalyst substance that changes the rate of

reaction without undergoing permanent change

during the process. - Homogenous catalyst catalyst that is present in

the same phase as the reactant molecules.

- As a general rule catalysts change the rate of

reaction by the lowering of Ea (activation

energy). - Usually this is done by giving a completely

different mechanism for the reaction. - This lowers the overall Ea.

Catalyzed pathway is two steps, both with lower

Ea than the original reaction pathway.

Heterogeneous Catalysis

- Catalyst exists in a different phase from the

reactant molecules. - Usually a solid in contact with gas or liquid
- Often times metal or metal oxides.

- Adsorption usually the initial step of

heterogeneous catalysis. - Uptake of molecules into the interior of another

substance. - Increasing surface area of catalyst increases

effectiveness. - Active sites places on the surface of a

molecule where adsorption takes place.

Enzymes

- Enzymes are biological catalyst of reactions that

occur in living systems. - Most enzymes are large protein molecules.
- They are selective in the reactions that they

catalyze.

- Active sites
- Lock-and-key model

Energy Diagram Handout (Please read when I am

finished going over the rest of the slides). We

wont be practicing any of the next few slides

because of the lack of time. I will simply go

through them quickly so you are aware of the

equations and what they are related to.

Hopefully, questions related to these wont show

up on the exam, but if they do, at least seeing

the equations might help.

The fraction of the molecules that possess Ea is

given by the following equation (not given on

exam equation sheet, will likely be given in

question).

- R gas constant (8.314 J/mol)
- T Absolute temperature
- Ea the activation energy/mol
- f fraction of molecules that possess Ea

Calculate the fraction of atoms in a sample of

argon gas at 400K that have an energy of 10.0 KJ

or greater.

Because we need units to cancel and R uses

Joules, we must make sure activation energy is in

joules.

10.0 KJ 1.00 x 104 J/mol T400K

fe-3.0070 4.94 x 10-2

1/ 4.94 x 10-2 20 Roughly 1/20 atoms has

this energy.

Orientation Factor

- Collisions can occur between molecules with

sufficient Ea, and still no reaction may occur. - The collisions must occur with proper orientation

of the molecules that are reacting.

Observations of Arrhenius

- Increase in rate in relationship to temperature

increase was non-linear. - Reaction rate obeyed an equation based on 3

factors - f
- of collisions per second
- Fraction of collisions with proper orientation.

Arrhenius Equation

- k the rate constant
- R gas constant (8.314 J/mol)
- T Absolute temperature
- Ea the activation energy
- A frequency factor (constant at varied

temperatures related to frequency of collisions

and probability of proper orientation).

Taking the natural log of both sides gives a

formula in straight line form

Given on Exam

- Graph of ln k versus 1/T will be a straight line

with a slope of Ea/R and a y-intercept of ln A.

Simplifying the equation to calculate a rate

constant (k1) at a temperature (T1) when the Ea

and rate constant (k2) at a temperature (T2) are

given.

- The rate constant of a first order reaction is

3.46 x 10-2 s-1 at 298 K. What is the rate

constant at 350 K if the activation energy is

50.2 kJ/mol?

- k1 3.46 x 10-2 s-1 k2 ?
- T1 298 K T2350 K

k2 0.702 s-1

- The first-order rate constant for the reaction of

methyl chloride with water to produce methanol

and hydrochloric acid is 3.32 x 10-10 s-1 at

25ºC. Calculate the rate constant at 40ºC if the

activation energy is 116 kJ/mol.

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Reaction Mechanisms

- The process by which a reaction occurs.
- Chemical equations only show beginning and ending

substances - Can show in detail bond breaking and forming and

structural changes that occur during a reaction.

Elementary steps

- Processes that occur in a single event or

step, are elementary processes. - Particles collide with sufficient energy and

proper orientation for reaction to occur.

- The number of molecules that participate as

reactants define the molecularity of the step. - One molecule Unimolecular
- 2 reactant molecules bimolecular

Multistep Mechanisms of Rx

- Often times chemical reactions are a result of

multiple steps not show by the overall equation.

This Rx below 225ºC occurs as 2 elementary steps.

- 1st step
- 2 NO2 molecules collide
- The NO3 then collides with CO and transfers an O.
- The elementary steps must add to result in the

overall chemical equation.

- The proposed mechanism for the above reaction is

a.)Is the proposed mechanism consistent with the

overall rx? b.)Identify any intermediates.

a.) yes b.) Mo(CO)5

Rate Laws of Elementary Steps

- Every reaction is made up of a series of

elementary steps - Rate laws reflect the relative speeds of these

steps. - Rate laws must be determined experimentally

Elementary Step Reactions

- The rate law of any elementary step is based

directly on its molecularity. - Unimolecular 1st order (A?product)
- Rate kA
- Bimolecular 2nd order (AB ?prod)
- Rate kAB

Rate Laws for All elementary steps

- Consider the reaction above.
- Write the rate law for the reaction, assuming it

has a single elementary step. - Is a single step mechanism likely for this

reaction? Why or why not?

- Rate kNO2Br2
- No, termolecular rxs are rare

Rate Laws of Multistep Mechanisms

- Most reactions involve multiple steps.
- Often one step is much slower than the other.
- The Rate determining step is the slowest step

in the reaction.

- The slowest step of a multi-step reaction

determines the overall rate of reaction.

slow

fast

Step 1 is the rate determining step, because it

is the slow step.Therefore the rate of the

overall reaction is the rate of step 1.

- Rate k1NO22
- Step one is a bimolecular process
- This rate law does in fact follow the observed

rate of reaction.

Initial fast step mechanisms

- The first step in a multi-step mechanism of this

type is no longer rate limiting. - It is more difficult to derive the rate under

these circumstances.

- The experimentally determined rate law for the

reaction is - Rate kNO2Br2
- Must find a reaction mechanism that is consistent

with the rate law.

?

- An alternative that does not involve termolecular

steps - Step 1 NO(g) Br2(g) NOBr2(g) fast
- Step 2 NOBr2(g) NO 2NOBr(g) slow

k1

k-1

k2

Step 2 NOBr2(g) NO 2NOBr(g)

k2

- Step 2 is the Rate Determining step.
- The overall reaction is governed by the rate law

for this step.

Intermediates are unstable and concentrations are

unknown.

The rate of the forward and reverse rx. are equal

in dynamic equilibrium.

- k1NOBr2 k-1NOBr2
- Solve for concentration NOBr2

Substitute this relationship into rate law

2NO2 F2(g)?2NO2F(g)

- The rate law for the above reaction is Rate

kNO2F2 - Suggested mechanism is
- NO2 F2 NO2F F slow
- F NO2 NO2F fast

k1

k2

Is this mechanism acceptable? Does it satisfy the

2 requirements?

- 1st requirement is that the sum of the steps

gives the balanced equation - NO2 F2 NO2F F
- F NO2 NO2F
- 2NO2 F2 F 2NO2F F

k1

k2

Overall 2NO2 F2(g)?2NO2F(g)

- 2nd requirement is that the mechanism agree with

the experimentally determined rate law - Mechanism states that 1st step is rate limiting.
- 1st step is bimolecular
- Therefore Rate kNO2F2

- The first step of a mechanism involving the

reaction of bromine is - Br2(g) 2Br(g) (fast/equilibrium)
- What is the expression relating concentration of

Br to concentration of Br2?

k1

k-1