Chapter 12

Chemical Kinetics

Introduction Consider the following:

diamond graphite ΔG = -2.9 kJ Thoughts? Questions?

If this is true, why don’t we see diamonds in jewelry turning into graphite?

Hint: the reaction happens, yes, but at what pace?

This happens VERY slowly! Also ΔG is very small, so just barely spontaneous.

Introduction What have we learned about

thermochemical values and what they do/don’t tell us?

Reactions can be spontaneous but this DOES NOT mean they occur fast.

Chemical kinetics: branch of chemistry that studies reaction rates.

What Affects Rates? Brainstorm- what do you think? Remember-

in order to react, two particles must collide!

(1) Identity of reactant- large & complex = fewer chances that the collision will take place at the correct location (rate slows).

(2) Temperature- greater KE means more energetic and frequent collisions.

(3) Concentration- larger amounts mean increased likelihood of collisions.

ON

Br

ON

Br

ON

O N Br ONBr

ON

O N BrONBr No Reaction

ON

BrBr

O N

BrBr

Molecular Orientation

What Affects Rates?(4) States of matter- reactants in the same state are more able to collide & react; liquids and gases are more able to collide due to greater surface area.

(5) Adding a catalyst- adding this increases the rate.

*Is recoverable at the end of the rxn.

*Reduces activation energy.

Unit Essential Question What role do rates play in chemical

reactions?

Reaction RatesSection 12.1

Lesson Essential Question: How is the rate of a reaction determined?

Example

Using stoichiometry allows rates to be set equal to each other! Thus the rate of disappearance of a reactant can be said to equal the rate of appearance of a product.

Kinetics 2A + B C + 3D (generic reaction) Rate =[substance]

t Rate = -½[A]/tt = Ct = Dt Notice effect of coefficients! Why are the first two negative? Because reactant concentrations decrease! This gives the initial reaction rate- how quickly

the reaction occurs as soon as the reactants are mixed.

Change in [ ] per unit time

PracticeN2 + 3H2 2NH3

What would the initial rate equal in terms of each substance in the reaction?

Rate = [N2] = -1/3[H2] = ½ [NH3] t t t

Rate Laws: An introduction

Section 12.2

Rate = k[NO2]n

This is called a rate law expression. Only reactants are used! Why? k is called the rate constant.

k does not change for a given reaction UNLESS T changes or a catalyst is added.

n is the order of the reactant(s) - usually a positive integer. Value of exponent can ONLY be determined experimentally!

Does NOT com from coefficients!

2 NO2 2 NO + O2

Rate Law ExpressionRate Law Expression

When n = 1: if reactant concentration is doubled, the rate doubles; if reactant concentration is tripled, the rate triples. First order with respect to this substance.

When n = 2: if reactant concentration is doubled, the rate increases four times (22 = 4). Second order with respect to this substance.

When n = 0: reactant concentration has no effect on the rate; it simply needs to be present in some amount. Zero order with respect to this substance.

Can add orders together for all reactants to get the overall order for the entire reaction.

Understanding Rate Laws & Understanding Rate Laws & Orders: Orders: Rate = k[NO2]n

Determining Rate Order & k2NO (g) + O2 (g) 2NO2 (g)

Initial [NO] Initial [O2] Rate of NO2 formation (M/s)

0.01 0.01 0.05

0.02 0.01 0.20

0.01 0.02 0.10

Can determine rate orders by inspection or mathematically.

Second order with respect to NO and first order with respect to O2.

Determining Rate Order & k Once you know the rate orders for each

reactant you can solve for K. How? Choose any experiment and the corresponding

concentration values and rate. Plug them into the rate equation and solve for k.

k = 5 x 104 /M2s

Sections 1&2 Homework Pg. 567 #10, 13, 26, 28

AP Practice QuestionWhat is the rate law for the following:

CO (g) + Cl2 (g) COCl2 (g)

a) Rate = k[CO] c) Rate = k[CO][Cl2]

b) Rate = k[CO]2[Cl2] d) Rate = k[CO][Cl2]2

Experiment Initial [CO] M Initial [Cl2] M Initial Rate

1 0.200 0.100 3.9 x 10-25

2 0.100 0.200 3.9 x 10-25

3 0.200 0.200 7.8 x 10-25

AP Practice QuestionThe reaction (CH3)3CBr (aq) + H2O (l) (CH3)3COH (aq) + HBr (aq) has the following rate law: Rate = k[(CH3)3CBr]. What will the effect of decreasing the concentration of (CH3)3CBr?

a) Reaction rate will increase.

b) More HBr will form.

c) Reaction rate will decrease.

d) Reaction will shift to the left.

Another Example For the reaction: BrO3

- +

5 Br- + 6H+ 3Br2 + 3 H2O

The general form of the rate law is:

Rate = k[BrO3-]a[Br-]b[H+]c

Use experimental data to determine the values of a, b, and c.

Then determine the reaction order for each reactant and the overall reaction.

Initial concentrations (M)

Rate (M/s)

BrOBrO33-- BrBr-- HH++

0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--

44

0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--

33

0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--

33

0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--

33

Initial concentrations (M)

Rate (M/s)

BrOBrO33-- BrBr-- HH++

0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--

44

0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--

33

0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--

33

0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--

33

Now we have to see how the rates change with concentration!

Answers to Example The powers of the reactants are as

follows: a = 1; b = 1; c = 2, so:

Rate = k[BrO3-] [Br-] [H+]2

The reaction is first order with respect to BrO3

- and Br-, second order with respect to H+, and fourth order overall.

How would you calculate the value of the rate constant? k = 8.0 L3/mol3∙s

the integrated rate law

Section 12.4

Integrated Rate Law Allows us to determine information about

concentrations over time, not just at a particular instant during the reaction.

For example, with the integrated rate law we can determine how long it will take to use up a reactant, form a product, etc.

Integrated rate laws come from setting the rate law and rate expression equal to each other for each order, and integrating over time.

First Order Integrated Rate Law

ln[A]t – ln[A]0 = -kt

[A]t = concentration of reactant at a given time, t

[A]0 = initial concentration of reactant

Can be written as: ln[A]t = -kt + ln[A]0

Has the format of a line! Note: slope = -k so adjust signs accordingly

when solving for k! Graphing ln[A] vs. time gives a straight line for

first order!

Second Order Integrated Rate Law

1/[A]t – 1/[A]0 = kt

Can be written as: 1/[A]t = kt + 1/[A]0

This also has the format of a line! Note: slope = k so no need to adjust signs! Graphing 1/[A] vs. time gives a straight line

for second order!

PracticeHI decomposes in a second order reaction. The value of k = 2.40x10-21/(Ms) at 25°C. How long will it take the concentration of HI to decrease from 0.200M to 0.190M at 25°C?

1.10 x 1020 s

Half Life• Half life: time required for a reactant to

reach half of its original concentration.

• Does not depend on concentrations for first order; only depends on k: t1/2 = ln2/k = 0.693/k

• Does depend on concentrations for second order, and also k: t1/2 = 1/k[A]0

• Note: radioactive decay is first order!

AP Practice QuestionThe rate constant, k, for radioactive beryllium-11 is 0.049/s. What mass of a 0.500mg sample of beryllium-11 remains after 28 seconds?

a)0.250mg

b)0.125mg

c)0.0625mg

d)0.375mg

Find how many half lives makeup the time given. Then halve the initial amount that manytimes.

t1/2 = 14s. 14 goes into 28 twotimes, so Be goes throughtwo half lives; divide 0.500 by two twice.

Practice The rate constant for the radioactive

decay of thorium-232 is 5.0x10-11/year. What is the half life of thorium?

1.386 x 1010 years

Practice HI decomposes in a second order

process to its elements. The value of the rate constant is 2.40 x 10-21/Ms at 25°C. What is the half life for the decomposition of 0.200M HI at 25°C?

2.08 x 1021 seconds

Determining Rate Order Example If you are given data on a reaction

(concentrations of reactant and times), plot the data to determine the order.

Plot ln[A] vs. time (first order) and 1/[A] vs. time (second order). Whichever gives a straight line is the order of

the reaction! To find k, determine the slope of the line

(remember, m = –k for first order and m = k for second order).

Use appropriate half life equation to calculate t1/2.

Determining Order Example The following data was collected for the reaction:

2C4H6(g) C8H12(g)

What is: the reaction order, the rate constant value, and the half life of the reaction?

[C4H6] (mol/L) Time (±1s)

0.01000 0

0.00625 1000

0.00476 1800

0.00370 2800

0.00313 3600

0.00270 4400

0.00241 5200

0.00208 6200

Determining Order Example

CAN’T determine order by inspection because the time at which [ ]’s are measured are not the same! MUST use integrated rate laws/graphing.

[C4H6] (mol/L) Time (±1s)

0.01000 0

0.00625 1000

0.00476 1800

0.00370 2800

0.00313 3600

0.00270 4400

0.00241 5200

0.00208 6200

Zero Order Rate Law Rate = k[A]0 = k

Rate does not change with concentration.

Integrated: [A] = -kt + [A]0

Half life: t1/2 = [A]0/2k

Typically happen on a surface because the surface area stays constant. Rate of contact to surface impacts

rate of reaction, not the concentration of the substance on the surface.

Also applies to enzyme chemistry. Rate depends on how quickly enzyme

works.

When Do Zero Order Rate Laws Occur?

AP Practice QuestionA reaction follows the rate law:

Rate = k[A]2. Which of the plots below will give a straight line?

a)1/[A] vs. 1/time

b)[A]2 vs. time

c)1/[A] vs. time

d)ln[A] vs. time

AP Practice QuestionFor the following reaction:

NO2 (g) + CO (g) NO (g) + CO2 (g)

the rate law is: Rate = k[NO2]. If a little CO is added to a mixture that had 0.10M NO2 and 0.20M CO, which statement is true?

a)Both k and the rxn. rate remain the same.

b)Both k and the rxn. rate increase.

c)Both k and the rxn. rate decrease.

d)Only k increases; rxn. rate stays the same.

AP Practice QuestionThe decomposition of ammonia to the elements is a first-order reaction with t1/2 = 200s at a certain T. How long will it take the partial pressure of NH3 to decrease from 0.100atm to 0.00625atm?

a)200s

b)400s

c)800s

d)1,000s

Remember- P is directly linked to # of moles for gases! Treat pressures like amounts.

How much pressure would one half life give? Is this the final pressurebeing asked for?

Section 4 Homework/Examples

Pg. 569 # 31, 39, 43, 46

Rate laws: a summarySection 12.5

Important Points1. Forward reaction is important- rate

laws will only contain reactants.

2. Two types of rate laws:

• Rate Law• Shows how rate depends on concentration.

• Integrated Rate Law• Shows how concentration depends on time.

Important Points cont.3. Type of law depends on type of data

collected.

• Can write the other if we have one.

4. Initial rate method used if several experiments are done.

• Order of reactant can be determined

5. Can graph experimental data to determine rate law (order).

• K can obtained from the slope

Summary of Rate Laws

Further Look at Activation Energy

Activation energy (Ea): the minimum energy required to start a reaction. Need E to break bonds! KE when molecules collide can supply

the energy needed to break bonds and start the reaction.

This is why higher T increases rates! E is then released when new bonds

form.

Activation Energy Graph

(activated complex)

The activated complex/transition state is the orientation of the reactants as bonds are breaking and forming.

Note the difference!

Arrhenius Equation Relationship between T and k can be used to

find the activation energy: k = Ae-Ea/RT

k = rate constant; A = frequency factor; Ea = activation E; R = ideal gas constant; T = temperature in K

Take ln of both sides: ln k = -(Ea/R)(1/T) + ln A

ln k vs 1/T is a straight line: slope = -Ea/R. Since R is known, Ea can be calculated.

A could also be calculated: y-intercept.

AP Practice QuestionThe energy difference between the reactants and the transition state is:

a)the free energy.

b)the heat of reaction.

c)the activation energy.

d)the kinetic energy.

AP Practice QuestionThe purpose for striking a match against the side of a box is to:

a)supply the free energy for the rxn.

b)supply the activation energy for the rxn.

c)supply the heat of reaction.

d)supply the kinetic energy for the rxn.

Reaction mechanismsSection 12.6

Lesson Essential Question:1)How do the steps of a reaction impact the rate of the reaction?

Reaction Mechanisms The reaction mechanism is a series of

steps that occur in a chemical reaction. Reactions do not typically occur in one

step. Why might this be? Think about orientations and energy! Particles need to collide in the correct

location AND with enough energy.

Reaction Mechanisms Kinetics can tell us about the mechanism; balanced

equations can’t. Ex: NO2 (g) + CO (g) NO (g) + CO2 (g) The proposed mechanism is:

NO2 (g) + NO2 (g) NO3 (g) + NO (g)

NO3 (g) + CO (g) NO2 (g) + CO2 (g) How do people know that the overall reaction doesn’t

happen in one step? Experimental data suggests the following rate law:

Rate = k[NO2]2. No CO is present!

Reaction Mechanisms

A mechanism should always end up giving the overall reaction being considered. Think of this part like Hess’ Law- cancel

out species that appear on opposite sides to arrive at the overall reaction.

Examine the previous mechanism:

NO2 (g) + NO2 (g) NO3 (g) + NO (g)

NO3 (g) + CO (g) NO2 (g) + CO2 (g)

NO2 (g) + CO (g) NO (g) + CO2 (g)

+

Reaction Intermediates

Example continued:

NO2 (g) + NO2 (g) NO3 (g) + NO (g)

NO3 (g) + CO (g) NO2 (g) + CO2 (g)

NO2 (g) + CO (g) NO (g) + CO2 (g) Reaction intermediates are substances that

are produced and consumed. What is/are the intermediate(s) above? Only NO3 because it is first produced, and

later consumed. NO2 is there to start with!

+

Rate-Determining Step Each step in a mechanism can be called

an elementary step. One elementary step is always slower than

the rest; it’s called the rate-determining step (RDS).

This step is essentially the only one affecting the rate of the overall reaction since the other steps, in comparison, are much faster.

Rate-Determining Step What does this mean? Rate laws CAN be written directly from the

RDS! We CANNOT do this with the overall

reaction being examined (as we saw in previous sections) because it is not the RDS!

Acceptable Reaction Mechanisms Two requirements for acceptable

mechanisms: Using a process like Hess’ Law, the

proposed mechanism must equal the overall reaction.

The RDS must equal the experimentally determined rate law.Rate cannot be faster than the

slowest step!

Example continued:

NO2 (g) + NO2 (g) NO3 (g) + NO (g) slow

NO3 (g) + CO (g) NO2 (g) + CO2 (g) fast Which is the RDS?

The first one! The rate law for the overall reaction is:

Rate = k[NO2][NO2] Rate = k[NO2]2

Rate-Determining Step

2NO2 + F2 2NO2F Proposed mechanism:

NO2 + F2 NO2F + F slow

F + NO2 NO2F fast

What is the RDS? What is the rate law for this reaction?

What is/are the intermediate(s)?

Reaction Mechanism Practice

Rate = k[NO2][F2]

first elementary step

F

Molecularity Molecularity tells you the number of molecules

colliding. Unimolecular- one molecule breaks down or

rearranges. Bimolecular- two molecules collide. Termolecular- three molecules collide. Which of these would you expect to be least

likely to occur? Termolecular; the odds of three molecules

colliding in exactly the correct orientation is highly unlikely.

Example 2NO2(g) + F2(g) 2NO2F(g)

Rate = k[NO2][F2] Suggested mechanism:

NO2 + F2 NO2F + F (slow)

F + NO2 NO2F (fast) Is this an acceptable mechanism (does it

satisfy the two requirements)? Being acceptable does not mean it is the

correct mechanism!

k1

k2

YES

Example Another possible mechanism:

NO2 + F2 NOF2 + O (slow)

NO2 + O NO3 (fast)

NOF2 + NO2 NO2F + NOF (fast)

NO3 + NOF NO2F + NO2 (fast) Perform additional reactions to decide on

the most likely mechanism. You can determine the rate law for a

reaction given a proposed mechanism. Use RDS information!

CatalystsSection 12.8

Catalysts Speeds up a reaction without being

consumed; the catalyst will still be there in the end although it may change along the way.

An iron ion will speed up the decomposition of hydrogen peroxide:

2Fe+3 + H2O2 2Fe+2 + O2 + 2H+

2Fe+2 + H2O2 + 2H+ 2Fe+3 + 2H2O Notice the iron ends up exactly how it started!

Homogeneous Catalysts Homogeneous catalysts (same state as

reactants) allow reactions to proceed by a different mechanism - a new pathway.

New pathway has a lower activation energy. More molecules will be able to overcome this

activation energy. Does not change overall E.

Starting and ending energies of reactants and products are not altered.

Homogeneous Catalysts Two additional examples:

Enzymes are biological catalysts. Chlorofluorocarbons (CFC’s) catalyze

the decomposition of ozone.

Heterogeneous Catalysts Heterogeneous catalysts (different state

from reactants) allow more surface for the reaction to occur AND allow for better orientations of reactants.

Typically the catalyst is a solid for liquid or gaseous reactants.

Ex: making ammonia (Haber process) requires an iron catalyst for nitrogen and hydrogen gases to react.

Additional Info. For reactions in equilibrium (when both

forward and reverse reactions are considered), BOTH are affected by adding a catalyst. The rates for both increase since they

share the same Ea.

Sections 7&8 Homework Pg. 571 #49, 50, 51, 53, 55, 65