Chemical Kinetics

The “Speed” of the Reaction

Or

Reaction Rates

Reaction Kinetics

Reactants ProductsRate of Change

Average Instantaneous

Rate Law

Reaction Order

Integrated Rate

Forms of Rate Law

Graphical Analysis of Rates

Initial rates Method

Half-Life

Reactants

• What happens to the reactants in a reaction?• If we measure the concentration of reactants as a

reaction proceeds, what would the graph look like? Or

• How does the concentration vary with time?– Is it linear?– Exponential?– Random?

• Do all reactions only move forward?– Assume for now there is no reverse reaction – Or the reverse reaction proceeds so slowly, we are

willing to ignore it

Map

graph

Products

• Where do the products come from?• If we measure the concentration of the

products from the first second the reaction starts,

• how would the concentration vary with time?

Map

graph

Rate of Change

• In the reaction A B• How does the [A] vary with time? Or• What is the rate of the reaction?• Rate = [A]time2 – [A]time1

Time 2 – Time 1

[A] = the molarity of A

Map

Rate of Change

• The symbol delta, , means “the change in”• So the reaction rate can be written• Rate = [A]t2 – [A]t1 or

time 2 – time 1

• Rate = [A]t

graph

Average Rate of Change• The concentration varies as time goes on.

– Because we use up reactants

• We can calculate an average rate of product consumption– over a period of time.

• Rate is like velocity of a reaction. – the rate of change of meters vs rate of change of [A] – Instead of meters per second, it is concentration per second.– To calculate speed/velocity, divide the distance traveled by the

time it took.Meters = m2 – m1 = m

t2 – t1 t sec

graph

Average Rate of Change

• What is the change in concentration? (how far did it go? If you are calculating speed)

[A] @ time2 – [A] @time1 = [A]

• How much time has elapsed?• [A] @ time2 – [A] @time1 = [A]

t2 – t1 t

graph

Instantaneous Rate

• An average rate describes what reaction rate over a time, but does not tell us the rate at any particular moment.

• The rate at any moment is the instantaneous rate

Instantaneous Rate

• If we take the average rate over a period of time and continuously make the time period smaller

• When the time period is infinitesimally small, you approach the instantaneous rate

• Graphically, it is the slope of the tangent line at the instant.– That’s why graphing programs have that tangent line

function! Rates are important in bio, physics and chem.

Map

(differential) Rate Law• Expresses how rate depends on concentration.• Rate = - [Reactant] = k [reactant]n

t• k is the rate constant

– The bigger the k value, the faster the reaction– The smaller the k value, ….?

• n = the order of the reaction and must be determined experimentally.

Map

Reaction Rates• Reaction rates are considered positive• Rateinstantaneous = kinstantaneous = - slope of tangent line• Rateaverage = k average = - ([A]2 - [A]1)

t2-t1

• So the rate constant, k, is always negative• Rate = - [Product] = k [Product]n

t– Assuming no reverse reaction!

Average Rate of Change

From 0 to 300 s =

0.01 –0.0038 = 0.000021 M/s 300 s

Product Formation

Reaction RateAverage Rate

Instantaneous Rate

Back to:Reactant/product

2NO2 2NO + O2

• Let’s consider the above reaction• How can we measure the rate?

– What data do we need?• Measure the time• Measure the concentration

– We will take advantage of color in our lab– If we are measuring light, we are doing…..

SPECTROSCOPY

Spectrometer

Source Monochrometer,LED Or

Filter

Sample Detector

The sample absorbs the light. The detector determines how much.

Many frequencies of lightOne frequencyOf light

Beer’s Law

• Beer’s law states that the amount of light absorbed depends on:– The material

• molar absorbtivity (physical property)– How much is there?

• molarity– And how big the sample holder

• The light spends more “time” in contact with a longer sample

Spectrometer

Source Monochrometer,LED Or

Filter

Sample Detector

The sample absorbs the light. The detector determines how much.

Many frequencies of lightOne frequencyOf light

Spectroscopy

• Assume the concentration is directly proportional to absorbance of light

• The more stuff there is that absorbs the light– the less light that goes through …. or– More light is absorbed

Beer’s Law a = e l c = k M a = absorbtivity e = molar absorbtivity (physical property) l = length of light path c = molarity or the solution

Molarity and absorbtivityAre directly proportional

2NO2 2NO + O2

Time [NO2] [NO] [O2]

0 0.0100 0 0

50 0.0079 0.021 0.0011

100 0.0065 0.0035 0.0018

150 0.0055 0.0045 0.0023

200 0.0048 0.0052 0.0026

250 0.0043 0.0057 0.0029

300 0.0038 0.0062 0.0031

350 0.0034 0.0066 0.0033

400 0.0031 0.0069 0.0035

Compare the [NO2] in the first 50 secs and the last 50 secs

Why does the rate slow down?

Formation of Products

2NO2 2NO + O2

Rate of Consumption NO2 = Rate Formation NO

Rate = k[NO2 ] = - k[NO]Because

For every two NO2 consumed two NO formed

Formation of Products

2NO2 2NO + O2

Rate of Consumption NO2 = 2 x Rate Formation O2

Rate = k[NO2 ] = - k/2 [O2]

Because

For every two NO2 consumed

one O2 formed

Compare the Instantaneous Rates

• At any moment in time[NO2] = - [NO] = 2 - [O2]

t t t

Or

k [NO2] = - k [NO] = - k/2 [O2]

graph

Form of the Rate Law

For aA + bB cC +dD• Rate = k [A]n [B]m

– Where k is the rate constant n = order of reactant A m = order of reactant B

• n and m must be determined experimentally• n +m = order of the reaction

Experimental Order

• the order in the integrated rate lawRate = - [Reactant] = k [Reactant]n

t

n = 0, zero order

n = 1, first order

n = 2, second order

Determine order

Order of Reaction

A + B → C• Rate = k[A]n [B]m • (n + m) = order of the reaction

= 1 unimolecular

=2 bimolecular

=3 trimolecular

This means how many particles are involved in the rate determining step

Method of Initial Rates

• A series of experiments are run to determine the order of a reactant.

• The reaction rate at the beginning of the reaction and the concentration are measured

• These are evaluated to determine the order of each reactant and the overall reaction order

If you plot the concentration versus time of [N2O5], you can determine the rate at 0.90M and 0.45M.

What is the rate law for this reaction?

Rate = k [N2O5]n n = the order. It is determined experimentally.

2N2O5(soln) 4NO2(soln) + O2(g)

• At 45C, O2 bubbles out of solution, so only the forward reaction occurs.

Data

[N2O5] Rate ( mol/l • s)0.90M 5.4 x 10-4

0 45M 2.7 x 10-4

The concentration is halved, so the rate is halved

2N2O5(soln) 4NO2(soln) + O2(g)

Rate = k [N2O5]n

5.4 x 10-4 = k [0.90]n 2.7 x 10-4 = k [0 45]n after

algebra

2 = (2)n

n = 1 which is determined by the experiment

Rate = k [N2O5]1

Method of Initial Rates

• Measure the rate of reaction as close to t = 0 as you can get.

• This is the initial rate.• Vary the concentration• Compare the initial rates.

Map

NH4+ + NO2

- N2 + 2H2O

• Rate = k[NH4+1]n [NO2

-1]m

• How can we determine n and m? (order)• Run a series of reactions under identical

conditions. Varying only the concentration of one reactant.

• Compare the results and determine the order of each reactant

Order

NH4+ + NO2

- N2 + 2H2O

Experiment [NH4]+

Initial

[NO2]-

Initial

Initial RateMol/L ·s

1 0.001M 0.0050 M 1.35 x 10-7

2 0.001M 0.010 M 2.70 x 10-7

3 0.002M 0.010M 5.40 x 10-7

NH4+ + NO2

- N2 + 2H2O

• Compare one reaction to the next

1.35 x 10-7 = k(.001)n(0.050)m

2.70 x 10-7 = k (0.001)n(0.010)m

Exp [NH4]+

Initial

[NO2]-

Initial

Initial RateMol/L ·s

1 0.001M 0.0050 M 1.35 x 10-7

2 0.001M 0.010 M 2.70 x 10-7

3 0.002M 0.010M 5.40 x 10-7

Form

1.35 x 10-7 = k(0.001)n(0.0050)m

2.70 x 10-7 k (0.001)n(0.010)m

In order to find n, we can do the same type of math with the second set of reactions

1.35 x 10-7 = (0.0050)m

2.70 x 10-7 (0.010)m

1/2 = (1/2)m

m = 1

NH4+ + NO2

- N2 + 2H2O

• Compare one reaction to the next

2.70 x 10-7 = k (0.001)n(0.010)m

5.40 x 10-7 = k(.002)n(0.010)m

Exp [NH4]+

Initial

[NO2]-

Initial

Initial RateMol/L ·s

1 0.001M 0.0050 M 1.35 x 10-7

2 0.001M 0.010 M 2.70 x 10-7

3 0.002M 0.010M 5.40 x 10-7

2.70 x 10-7 = k (0.001)n(0.010)m

5.40 x 10-7 k(.002)n(0.010)m

n + m = order of the reaction 1 + 1 = 2 or second order

Form

0.5 = (0.5)n

n = 1

Review

• Method of initial rate• In the form Rate = k[A]n[B]m

– Where k is the rate constant– n, m = the order of the reactant

• The order is determined experimentally• Rate law is important so we can gain an

insight into the individual steps of the reaction

The Integrated Rate Law • Expresses how concentrations depend on time• Depends on the order of the reactionRemember• Rate = k[A]n[B]m

Order = n + m• Integrated Rate law takes the form by “integrating” the

rate function. (calculus used to determine)– The value of n and m change the order of the reaction – The form of the integrated rate depends on the value of n– You get a different equation for zero, first and second order

equations.

Map

Reaction Order

• Order of the reaction determines or affects our calculations.

• Zero order indicates the use of a catalyst or enzyme. The surface area of catalyst is the rate determining factor.

• First or second order is more typical (of college problems)

Integrated Law - Zero Order

Rate = - [A] = k

tSet up the differential equation

d[A] = -kt

Integral of 1 with respect to A is [A]

Integrated Rate Law – First Order

Rate = [A] = k [A] n

t If n = 1, this is a first order reaction. If we

“integrate” this equation we get a new form.

Ln[A] = -kt + ln[A0]

where A0 is the initial concentration

Map

Why?If Rate = - [A] = k [A] 1

tThen you set up the differential equation:

d[A] = -kdt

[A]

Integral of 1/[A] with respect to [A] is the ln[A].

Integrated Rate Lawln[A] = -kt + ln[A]0

• The equation shows the [A] depends on time• If you know k and A0, you can calculate the

concentration at any time.• Is in the form y = mx +b

Y = ln[A] m = -k b = ln[A]0

Can be rewritten ln( [A]0/[A] ) = kt

• This equation is only good for first order reactions!

First Order Reaction

[N2O5] Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

Ln

[N2O

5]

Time (s)

Zero First Second

Rate Law Rate = K[A]0 Rate = K[A]1 Rate = K[A]2

Integrated Rate Law

[A] = -kt + [A]0 Ln[A] = -kt +ln[A]0 1 = kt + 1 [A] [A]0

Line [A] vs t ln[A] vs t 1 vs t [A]

Slope = - k - k k

Half-life t1/2 = [A]0

2k

t1/2 = 0.693

k

T1/2 = 1

k[A]0

Graph

Data Map

Given the Reaction

2C4H6 C8H12

[C4H6] mol/L Time (± 1 s)

0.01000 00.00625 10000.00476 18000.00370 28000.00313 36000.00270 44000.00241 52000.00208 6200

And the data

2C4H6 C8H12

Equations

Graphical AnalysisL

n [C

4H6]

__

_1__

_

[C

4H6]

DataMap

Experimental Derivation of Reaction Order

• Arrange data in the form 1/[A] or ln [A] or [A]

• Plot the data vs time • Choose the straight line

y = mx + b

• Determine the k value from the slope• Graphical rate laws

1/[A] = kt + b → 2nd

ln[A] = kt + b → 1st

[A] = kt + b → zero

Half-life

• The time it takes 1/2 of the reactant to be consumed• This can be determined

– Graphically– Calculate from the integrated rate law

Half-LifeGraphical Determination

Half-LifeAlgebraic Determination

Half-life t1/2 = [A]0

2k

t1/2 = 0.693

k

T1/2 = 1

k[A]0

Equations are derived from the Integrated Rate Laws.

Zero First Second

Map

Temperature and Rate

• Most reactions speed up as temperature increases. (EX. food spoils faster when not refrigerated.)

• As temperature increases, the rate increases.

• Typically the rate doubles for every 10ºC increase in temperature.

Since the rate law has no temperature term in it, the rate constant must depend on temperature.

Why?The temperature effect

is quite dramatic.

Temperature and Rate

• Observations: rates of reactions are affected by concentration and temperature.

• In order for a reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products.

The Collision Model

Temperature and Rate

• The more molecules present, the greater the probability of effective collisions and the faster the rate.

• The higher the temperature, the more energy available to the molecules and the faster the rate.

The Collision Model

Temperature and Rate

The Collision Model

Complication:

Not all collisions lead to products. In fact, only a small fraction of collisions lead to product.

Temperature and Rate

Activation Energy

Temperature and Rate

Activation Energy• Arrhenius: molecules must possess a

minimum amount of energy to react. –In order to form products, bonds must

be broken in the reactants.–Bond breakage requires energy.

• Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.

Temperature and Rate

Activation Energy

Temperature and Rate

Activation EnergyConsider the rearrangement of acetonitrile:

H3C N CC

NH3C H3C C N

• In H3C-NC, the C-NC bond bends until the C-N bond breaks and the NC portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state.

• The energy required for the twisting and breaking of the reactant bonds is the activation energy, Ea.

• Once the C-N bond is broken, the NC portion can continue to rotate forming a C-CN bond.

Temperature and Rate

Activation Energy

Temperature and Rate

• The change in energy, ∆E, for the reaction is the difference in energy between reactants (CH3NC) and products (CH3CN).

• The activation energy is the difference in energy between the reactants and the transition state.

• The reaction rate depends on Ea. The lower the Ea, the faster the reaction.

• Notice that if a forward reaction is exothermic (CH3NC CH3CN), then the reverse reaction is endothermic (CH3CN CH3NC).

Key Points for Activation Energy

Temperature and Rate

Activation Energy• Re-consider the reaction between Cl and NOCl:

– If the Cl effectively collides with the Cl end of NOCl, then the products are Cl2 and NO.

– If the Cl collided with the O of NOCl, then no products are formed.

• We need to quantify this effect.

Temperature and Rate

Temperature and RateAs the temperature increases, more molecules have

sufficient energy to have effective collisions and react. Fr

acti

on o

f m

olec

ules

Kinetic energy

Minimum energy needed for reaction, Ea

Lower temp

Higher temp

Temperature and RateThe Arrhenius Equation

RTaE

Aek

Arrhenius discovered most reaction rate data obeyed the equation:

• k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K.

• A is called the frequency factor. It is a measure of the probability of a favorable collision.

• Both A and Ea are specific to a given reaction.

Temperature and RateThe Arrhenius Equation• If you have a lot of data, you can determine Ea and A

graphically by rearranging the Arrhenius equation:

ART

Ek a lnln

122

1 11ln

TTR

E

kk a

• If you have only two data points, then you can use

Catalysis

• A catalyst changes the rate of a chemical reaction.

• There are two types of catalysts: homogeneous and heterogeneous.

Homogeneous Catalysis• The catalyst and reaction are in one phase.• Hydrogen peroxide decomposes very

slowly: 2H2O2(aq) 2H2O(l) + O2(g).

CatalysisHomogeneous Catalysis

2H2O2(aq) 2H2O(l) + O2(g).

• In the presence of the bromide ion, the decomposition occurs rapidly:2Br-(aq) + H2O2(aq) + 2H+(aq) Br2(aq) + 2H2O(l).

Br2(aq) + H2O2(aq) 2Br-(aq) + 2H+(aq) + O2(g).

Br- is a catalyst because it can be recovered at the end of the reaction.

• Generally, catalysts operate by lowering the activation energy for a reaction.

CatalysisHomogeneous Catalysis

CatalysisHomogeneous Catalysis• Catalysts can operate by increasing the

number of effective collisions.

From the Arrhenius equation: catalysts increase k by increasing A or decreasing Ea.

• A catalyst may add intermediates to the reaction.

Ex: When Br- is added, Br2(aq) is generated as an intermediate in the decomposition of H2O2.

Catalysis

Heterogeneous Catalysis• The catalyst is in a different phase than

the reactants and products.• First step is adsorption (the binding of

reactant molecules to the catalyst surface).

• Adsorbed species (atoms or ions) are very reactive.

CatalysisHeterogeneous Catalysis

CatalysisHeterogeneous Catalysis• Consider the hydrogenation of ethylene:

C2H4(g) + H2(g) C2H6(g), H = -136 kJ/mol.

– The reaction is slow in the absence of a catalyst.– In the presence of a metal catalyst (Ni, Pt or Pd)

the reaction occurs quickly at room temperature.

– First the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface.

– The H-H bond breaks and the H atoms migrate about the metal surface.

CatalysisHeterogeneous Catalysis• Consider the hydrogenation of ethylene:

C2H4(g) + H2(g) C2H6(g), H = -136 kJ/mol.– When an H atom collides with an ethylene

molecule on the surface, the C-C bond breaks and a C-H bond forms.

– When C2H6 forms it desorbs from the surface.

– When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds and the activation energy for the reaction is lowered.

CatalysisEnzymes• Enzymes are biological catalysts.• Most enzymes are protein molecules with

large molecular masses (10,000 to 106 amu).• Enzymes have very specific shapes.• Most enzymes catalyze very specific

reactions.• Substrates undergo reaction at the active

site of an enzyme.• A substrate locks into an enzyme and a fast

reaction occurs.

CatalysisEnzymes

Catalysis

Enzymes• The products then move away from the enzyme.• Only substrates that fit into the enzyme lock can be

involved in the reaction.• If a molecule binds tightly to an enzyme so that

another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors).

• The number of events (turnover number) catalyzed is large for enzymes (103 - 107 per second).

Nitrogen Fixation and Nitrogenase• Nitrogen gas cannot be used in the soil for plants or

animals.• Nitrogen compounds, NO3, NO2

-, and NO3- are used in

the soil.• The conversion between N2 and NH3 is a process with

a high activation energy (the NN triple bond needs to be broken).

• An enzyme, nitrogenase, in bacteria which live in root nodules of legumes, clover and alfalfa, catalyses the reduction of nitrogen to ammonia.

Catalysis

Nitrogen Fixation and Nitrogenase

Catalysis

Nitrogen Fixation and Nitrogenase• The fixed nitrogen (NO3, NO2

-, and NO3-) is consumed

by plants and then eaten by animals.• Animal waste and dead plants are attacked by

bacteria that break down the fixed nitrogen and produce N2 gas for the atmosphere.

Catalysis