kinetics

KINETICSThe study of reaction rates.Spontaneous reactions are reactions that

will happen - but we can’t tell how fast.Diamond will spontaneously turn to

graphite – eventually.Reaction mechanism- the steps by which a

reaction takes place.

The part of chemistry that is thermodynamics

and the part that is kinetics.

FACTORS AFFECTING RATE 1. Nature of Reactants 2. Concentration of Reactants 3. Temperature 4. Catalysts 5. Surface Area of Reactants 6. Adding an Inert Gas (NO EFFECT).

COLLISION THEORYMolecules must collide to react.Only two particles collide at a time.Must have proper orientation.

COLLISION THEORYMust collide with enough

energy.Only a small number of

collisions produce reactions.

Concentration affects rates because collisions are more likely.

Temperature and rate are related.

ENERGY PLOT Know Ea, transition state, and activated complex. Know exothermic versus endothermic.

Potential Energy

Reaction Coordinate

Reactants

Products

Potential Energy

Reaction Coordinate

Reactants

Products

Activation Energy Ea

Potential Energy

Reaction Coordinate

Reactants

Products

Activated complex

Potential Energy

Reaction Coordinate

Reactants

ProductsDE}

Potential Energy

Reaction Coordinate

2BrNO

2NO + Br

Br---NO

Br---NO

2

Transition State

MAXWELL BOLTZMANN DISTRIBUTION

MAXWELL BOLTZMANN DISTRIBUTION

REACTION RATE Rate = [A] at t2 – [A] at t1

t2- t1

Rate = D[A] Dt

Change in concentration per unit time. [reactants] decreases with time. [products] increases with time

N2 + 3H2 → 2NH3

As the reaction progresses the concentration H2 goes down

Concentration

Time

[H2]

N2 + 3H2 → 2NH3

As the reaction progresses the concentration N2 goes down 1/3 as fast.

Concentration

Time

[H2]

[N2]

N2 + 3H2 → 2NH3

As the reaction progresses the concentration NH3 goes up.

Concentration

Time

[H2]

[N2]

[NH3]

CALCULATING RATES

Average rates are taken over a range of time. Worry about coefficients.

Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time

Derivative.RATES ARE ALWAYS POSITIVE!

AVERAGE SLOPE METHOD

Concentration

Time

D[H2]

Dt

INSTANTANEOUS SLOPE METHOD

Concentration

Time

D[H2] D t

INSTANTANEOUS RATE

DEFINING RATE We can define rate in terms of the disappearance

of the reactant or in terms of the rate of appearance of the product.

example N2 + 3H2 → 2NH3

-D[N2] = -3D[H2] = 2D[NH3] Dt Dt Dt Negative because [reactant] goes down.

RATE LAWSReactions are reversible.As products accumulate they can begin to turn

back into reactants. Early on the rate will depend on only the amount

of reactants present.We want to measure the reactants as soon as

they are mixed. This is called the Initial rate method.

Two key points: The concentration of the products do not appear

in the rate law because this is an initial rate. The order must be determined experimentally;

they can’t be obtained from the equation.

RATE LAWS

2 NO2 2 NO + O2

You will find that the rate will only depend on the concentration of the reactants.

Rate = k[NO2]n

This is called a rate law expression. k is called the rate constant. n is the order of the reactant -usually a positive integer.

2 NO2 2 NO + O2

Oxygen can appear only half as rapidly as the nitrogen dioxide disappears while NO appears twice as fast as oxygen appears.

Calculate the AVERAGE rate at which [NO2] changes in the first 50.0 seconds given the [NO2] at 0.0s = 0.0100M and [NO2] at 50s = 0.0079M .

RATE = −Δ [NO] = −[.0079]−[0.0100] Δt 50.0 s = −[−4.2 × 10−5 M / sec] = 4.2 × 10−5 M /sec

2 NO2 → 2 NO + O2

In terms of NO2: Rate = -Δ[NO2] = k[NO2]n

Δt

In terms of O2:

Rate’ = -Δ[O2] = k’[O2]n

Δt

So: Rate = 2 x Rate’ OR k[NO2]n = 2k’[NO2]n

RELATIVE RATES We can consider the appearance of products along with the disappearance of reactants.

The reactant’s concentration is declining, the products is increasing.

Respect the algebraic sign AND respect the stoichiometry.

Divide the rate of change in concentration of each reactant by its stoichiometric coefficient in the balanced chemical equation.

RELATIVE RATES 2 NO2 → 2 NO + O2

Thus..... Rate of reaction = - 1Δ[NO2] = 1 Δ[NO] = Δ [ O2] 2 Δtime 2 Δtime Δtime There are 2NO2 molecules consumed for every O2 molecule produced.

ANOTHER EXAMPLE Using the coefficients from the balanced equation, you should be able to give relative rates. For example: 4 PH3(g) → P4(g) + 6 H2(g)

Initial rate rxn = = + = +

PRACTICE ONE What are the relative rates of change in concentration of the products and reactant in the decomposition of nitrosyl chloride, NOCl? 2 NOCl(g) → 2 NO(g) + Cl2(g)

TYPES OF RATE LAWS

Differential Rate law - describes how rate depends on concentration.

Integrated Rate Law - Describes how concentration depends on time.

For each type of differential rate law, there is an integrated rate law and vice versa.

Rate laws can help us better understand reaction mechanisms.

TYPES OF RATE LAWSReactions are reversible.When the rate of the forward = the rate of the

reverse we have EQUILIBRIUM! To avoid this complication we will discuss reactions soon after mixing--initial reactions rates, and not worry about the buildup of products and how that starts up the reverse reaction.

So reaction rate will only depend on [reactants] and initial [reactants] right after mixing.

DETERMINING RATE LAWS

The first step is to determine the form of the rate law (especially its order).

Must be determined from experimental data. For this reaction

2 N2O5(aq) → 4NO2(aq) + O2(g)

The reverse reaction won’t play a role

DIFFERENTIAL RATE LAWDescribes how the reaction rate varies with the

concentration of various species in a system. Rates generally depend on reactant

concentrations. To find the exact relation between rate and

concentration, we must conduct experiments and collect information.

DIFFERENTIAL RATE LAW C

aA + bB → xX

Becomes Initial rxn rate = k[A]m[B]n[C]p

NOT COEFFICIENTS K = temperature dependent

DIFFERENTIAL RATE LAW

Exponents can be zero, whole numbers or fractions AND MUST BE DETERMINED BY EXPERIMENTATION!!

THE RATE CONSTANT, k◦ temperature dependent and must be evaluated by

experiment.◦ Example: rate = k[A]. If k is 0.090/hr when [A] = 0.018 mol/L◦ rate = (.0090/hr)(0.018 mol/L) = 0.00016 mol/(L• hr)

ORDER OF A REACTION

order with respect to a certain reactant is the exponent on its concentration term in the rate expression.

order of the reaction is the sum of all the exponents on all the concentration terms in the expression.

DETERMINATION OF THE RATE EXPRESSION aA + bB → xX

initial rate = k[A]om[B]o

n - the little subscript “o” means “original” or at “time zero”.

ORDER OF A REACTION

1. Zero order: The change in concentration of reactant has no effect on the rate. These are not very common. General form of rate equation: Rate = k

2. First order: Rate is directly proportional to the reactants concentration; doubling [rxt], doubles rate. These are very common! Nuclear decay reactions usually fit into this category. General form of rate equation: Rate = k [A]1 = k[A]

3. Second order: Rate is quadrupled when [rxt] is doubled and increases by a factor of 9 when [rxt] is tripled etc. These are common, particularly in gas-phase reactions. General form of rate equation: Rate = k [A]2 or Rate = k[A]1[B]1 which has an overall order of two (second order).

ORDER OF A REACTION

To determine the order of the reaction: 1. Determine skeleton rate law 2. Determine exponents using data from trials. 3. Determine k with its units. 4. Write the rate law. Adding the orders of each reactant gives the overall order of the reaction.

EXAMPLE NO(g) + Cl2(g) → NOCl(g) at 295K

Rate = k[NO]m[Cl2]n

Experiment [NO] [Cl2] Initial Rate (M/s)1 0.050 0.050 1.0 x 10-3

2 0.050 0.150 3.0 x 10-3

3 0.150 0.050 9.0 x 10-3

HOW TO SOLVE 1. Skeleton: Rate = k[NO]m[Cl2]n

2. Determine order: Look for two trials where the concentration of a reactant was held constant. Next, focus on the other reactant. Ask yourself how it’s concentration changed for the same two trials. Was it doubled? Was it tripled? Was it halved? Once you have determined the factor by which the concentration of the other reactant was changed, determine how that affected the rate for those same two trials.

HOW TO SOLVE To find the rate order of NO, find two trials where [Cl2] is constant and [NO] changes. From experiment 1 to experiment 3, [NO] triples while [Cl2] is constant. Since the initial rate increases by a factor of 9x, the rate is directly proportional to [NO]2 and the RATE ORDER FOR [NO] IS 2.

Experiment 3 9.0 x 10-3 k[0.150]n [0.050]m 9 = 3n

Experiment 1 1.0 x 10-3 k[0.050]n [0.050]m n = 2

HOW TO SOLVE To find the rate order of Cl2, find two trials where [NO] is constant and [Cl2] changes. From experiment 1 to experiment 2, [Cl2] triples while [NO] is constant. Since the initial rate triples, the rate is directly proportional to [Cl2]1 and the RATE ORDER FOR Cl2 IS 1.

Experiment 2 3.0 x 10-3 k[0.050]n [0.150]m 3 = 3m

Experiment 1 1.0 x 10-3 k[0.050]n [0.050]m m = 1

HOW TO SOLVE THE RATE LAW EXPRESSION IS RATE = k[NO]2[Cl2]1

OVERAL ORDER: 3 NOTE: The rate orders aren’t always the same as the coefficients.

HOW TO SOLVE 3. Determine k and units: Use any experimental data to find the value of k and its units. The units for K are one less that the overall order. Using experiment 1 data, plug the values into the rate law and solve for k

Rate = k[NO]2[Cl2]1 = 1.0 x 10-3M = k (0.050M)2 (0.050M)1 k = (1.0 x 10-3M/s) / [(0.050M)2 (0.050M)1] k = 8.0M-2 x s-1 or 8.0L2 / (mol2 x s) The larger the k, the faster the reaction will be. k only changes with changes in TEMPERATURE. Increasing the temperature will increase k and increase the reaction rate.

PRACTICE TWO In the following reaction, a Co-Cl bond is replaced by a Co-OH2 bond.

[Co(NH3)5Cl]+2 + H2O → [Co(NH3)5H2O]+3 + Cl

Initial rate = k{[Co(NH3)5Cl]+2}m

Using the data below, find the value of m in the rate expression and calculate the value of k.

Exp. Initial Concentration Initial rateof [Co(NH3)5Cl]+2 (M) mol/(L• min)

1 1.0 × 10-3 1.3 × 10-7

2 2.0 × 10-3 2.6 × 10-7

3 3.0 × 10-3 3.9 × 10-7

4 1.0 × 10-3 1.3 × 10-7

PRACTICE THREE The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation: BrO3

-(aq) + 5 Br –

(aq) + 6 H+(aq) → 3 Br2(l) + 3 H2O(l)

The table below gives the results of four experiments. Using these data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant. What is the value of k?

What are the units of k?

Experiment Initial [BrO3-] Initial [Br–] Initial [H+] Measured initial

rate (mol/L•s)

1 0.10 0.10 0.10 8.0 × 10-4

2 0.20 0.10 0.10 1.6 × 10-3

3 0.20 0.20 0.10 3.2 × 10-3

4 0.10 0.10 0.20 3.2 × 10-3

PRACTICE THREE The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation: BrO3

-(aq) + 5 Br –

(aq) + 6 H+(aq) → 3 Br2(l) + 3 H2O(l)

The table below gives the results of four experiments. Using these data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant. What is the value of k?

What are the units of k?

Experiment Initial [BrO3-] Initial [Br–] Initial [H+] Measured initial

rate (mol/L•s)

1 0.10 0.10 0.10 8.0 × 10-4

2 0.20 0.10 0.10 1.6 × 10-3

3 0.20 0.20 0.10 3.2 × 10-3

4 0.10 0.10 0.20 3.2 × 10-3

TYPES OF RATE LAWS

Differential Rate law – data table contains concentration and rate data Integrated Rate Law – data table contains concentration and time data. Only one reactant.

INTEGRATED RATE LAW Expresses the reaction concentration as a function of time.

Form of the equation depends on the order of the rate law (differential).

Changes Rate = D[A]n Dt

We will only work with n=0, 1, and 2

HOW TO SOLVE1. Set up your axes so that time is always on the x-axis.

2. Plot the concentration of the reactant on the y-axis of the first graph.

3. Plot the natural log of the concentration (ln [A], NOT log[A]) on the y-axis of the second graph.

4. Plot the reciprocal of the concentration on the y-axis of the third graph.

You are in search of linear data.

HOW TO SOLVE If you do the set of graphs in this order with the y-axes being “concentration”, “natural log of concentration” and “reciprocal concentration”, the alphabetical order of the y-axis variables leads to 0, 1, 2 orders respectively for that reactant.

You can now easily solve for either time or concentration once you know the order of the reactant.

Just remember y = mx + b.

Choose the set of variables that gave you the best straight line (r value closest to ±1) and insert them in place of x and y in the generalized equation for a straight line. “A” is reactant A and Ao is the initial concentration of reactant A at time zero [the y-intercept].

ZERO ORDER Slope = -k Rate is independent of concentration.

Rate = k[A]0

[A] = -kt + [A0] Remember that the rate constant is NEVER negative.

This is not a straight line.

FIRST ORDER Slope = -k Concentration is a function of time

Rate = k[A]1

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

In the form y = mx + b

This is a straight line.

SECOND ORDER Slope = k Concentration is a function of time

Rate = k[A]2

1/[A] = kt + 1/[A]0

In the form y = mx + b

This is not a straight line.

PRACTICE FOUR The decomposition of N2O5 in the gas phase was studied at constant temperature.

2 N2O5(g) → 4 NO2(g) + O2(g)

The following results were collected:

[N2O5] Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

Determine the rate law and calculate the value of k. What is the concentration of N2O5(g) at 600s? At what time is the concentration of N2O5(g) equal to 0.00150 M ?

PRACTICE FOURy = mx + b

zero order [A] = −kt + [Ao] first order ln[A] = −kt + ln[Ao] second order 1/[A] = kt + 1/[Ao]

HALF LIFE – FIRST ORDER

If the reaction is first order [A] = [A]0/2 when t = t1/2

t½ = ln2/k = 0.693/k

HALF LIFE – SECOND ORDER

If the reaction is second order [A] = [A]0/2 when t = t1/2

t½ = 1/k[A]o

PRACTICE FIVE A certain first-order reaction has a half-life of 20.0 minutes. Calculate the rate constant for this reaction. How much time is required for this reaction to be 75% complete?

PRACTICE SIX For the reaction of (CH3)3CBr with OH−,

(CH3)3CBr + OH− → (CH3)3COH + Br−

The following data were obtained in the laboratory.

TIME (s) [(CH3)3CBr]

0 0.100

30 0.074

60 0.055

90 0.041

Plot these data as ln [(CH3)3CBr] versus time. Sketch your graph.

Is the reaction first order or second order? What is the value of the rate constant?

PRACTICE SEVEN Butadiene reacts to form its dimer according to the equation

2 C4H6(g) → C8H12(g)

The following data were collected for this reaction at a given temperature:

[C4H6] Time (± 1 s)

0.01000 0

0.00625 1000

0.00476 1800

0.00370 2800

0.00313 3600

0.00270 4400

0.00241 5200

0.00208 6200

PRACTICE SEVEN What is the order of this reaction? Explain. Sketch your graph as part of your explanation. Write the rate law expression:

What is the value of the rate constant for this reaction?

What is the half-life for the reaction under the conditions of this experiment?

HALF LIFE – ZERO ORDER

Rate = k[A]0 = k (1) = k Integrated rate law is [A] = -kt + [A]o

t1/2 = [A]o

2k

Zero-order reactions are most often encountered when a substance such as a metal surface or an enzyme is required for the reaction to occur. The enzyme or catalyst may become saturated and therefore an increase in the [reactant/substrate] has no effect on the rate.

SUMMARY OF RATE LAWS

Memorize this!

REACTION MECHANISMS

The sequence of bond-making and bond-breaking steps that occurs during the conversion of reactants to products. This is using the rate determining step to relate a reaction mechanism to its rate law expression.

Kinetics can tell us something about the mechanism A balanced equation does not tell us how the reactants become products.

REACTION MECHANISMS Two Prong Test:

Must be determined by experiment! 1. Must agree with overall stoichiometry 2. Must agree with the experimentally determined

rate law

◦ The coefficients from the rate determining step can be used to give rate orders shown in the rate law expression.

REACTION MECHANISMS

For the reaction NO2(g) + CO(g) → CO2(g) + NO(g)

#1 NO2 + NO2 → NO3 + NO SLOW

#2 NO3 + CO → CO2 + NO2 FAST

NO2 + CO → CO2 + NO

The unstable NO3, an intermediate, cancels out as does 1 NO2

REACTION MECHANISMS

The final rate expression must include ONLY those species that appear in the balanced equation for the overall reaction.

Step #1 is the rate determining step, and the rate law equation can be found using its coefficients.:

2NO2 → NO3 + NO

RATE = k[NO2]2 always written from the slow step.

REACTION MECHANISMS Each of the two reactions is called an elementary

step . The rate for a reaction can be written from its

molecularity .Molecularity is the number of pieces that must

come together.

Unimolecular step involves one molecule - Rate is rirst order.

Bimolecular step - requires two molecules - Rate is second order.

Termolecular step- requires three molecules - Rate is third order.

Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.

REACTION MECHANISMS

A → products Rate = k[A]

A+A → products Rate= k[A]2

2A → products Rate= k[A]2

A+B → products Rate= k[A][B]

A+A+B → products Rate= k[A]2[B]

2A+B → products Rate= k[A]2[B]

A+B+C → products Rate= k[A][B][C]

REACTION MECHANISMS

PRACTICE EIGHT Nitrogen oxide is reduced by hydrogen to give water and nitrogen,

2 H2(g) + 2 NO(g) → N2(g) + 2 H2O(g)

and one possible mechanism to account for this reaction is

2 NO(g) → N2O2(g)

N2O2(g) + H2(g) → N2O(g) + H2O(g)

N2O(g) + H2(g) → N2(g) + H2O(g)

What is the molecularity of each of the three steps? Show that the sum of these elementary steps is the net reaction.

REACTION MECHANISMS AND RATE EXPRESSIONS determined by experiment the rate of the overall reaction is limited by, and is exactly

equal to, the combined rates of all elementary steps up to and including the slowest step in the mechanism

the slowest step is the rate determining step reaction intermediate - produced in one step but

consumed in another. Catalyst - goes in, comes out unharmed and DOES NOT

show up in the final reaction

PRACTICE NINE The balanced equation for the reaction of the gases nitrogen dioxide and fluorine is

2 NO2(g) + F2(g) → 2 NO2F(g)

The experimentally determined rate law is Rate = k [NO2][F2]

A suggested mechanism for the reaction is

NO2 + F2 → NO2F + F Slow

F + NO2 → NO2F Fast

Is this an acceptable mechanism? That is, does it satisfy the two requirements? Justify.

CATALYSTS Speed up a reaction without being used up in

the reaction. Enzymes are biological catalysts.Homogenous Catalysts are in the same phase as

the reactants.Heterogeneous Catalysts are in a different

phase as the reactants.

HOW CATALYSTS WORK

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

New pathway has a lower activation energy, Ea.More molecules will have this activation energy.Do not change E also known as ∆H

HETEROGENEOUS CATALYSTS

different phase than reactants, usually involves gaseous reactants adsorbed on the surface of a solid catalyst◦ adsorption—refers to the collection of one substance on

the surface of another◦ absorption—refers to the penetration of one substance

into another; water is absorbed by a sponge

Pt surface

HHHH

HH

HH

Heterogenous Catalysts Hydrogen bonds to surface of metal.

Break H-H bonds

Pt surface

HH

HH

Heterogenous Catalysts

C HH CHH

Pt surface

HH

HH

Heterogenous Catalysts The double bond breaks and bonds to the catalyst.

C HH CHH

Pt surface

HH

HH

Heterogenous Catalysts The hydrogen atoms bond with the carbon

C HH CHH

Pt surface

H

Heterogenous Catalysts

C HH CHH

H HH

HOMOGENOUS CATALYSTS exists in the same phase as the reacting molecules.

Chlorofluorocarbons catalyze the decomposition of ozone.

Enzymes regulating the body processes. (Protein catalysts)

CATALYSTS AND RATECatalysts will speed up a reaction but only

to a certain point.Past a certain point adding more reactants

won’t change the rate.Zero Order