Kinetics The Rates and Mechanisms of Chemical Reactions

KineticsThe Rates and Mechanisms of Chemical Reactions

What does the word KINETIC imply to you?Why should we care about KINETICS?What factors affect KINETICS?

FAST SLOW

We Are Talking About Reaction RatesSpeed of any event is measured by a change

that occurs per unit of time.

The speed of reaction (i.e. reaction rate) is measured as a change in concentration (Molarity; M) of a reactant or product over a certain timescale.

◦Time is the independent variable (x-axis) and concentration is the dependent variable (y-axis)

◦Reaction rate is expressed in M/s

The 3 Fundamental Questions of Chemical Reactions1) What happens?

◦Answer given by balanced chemical equation and stoichiometry

2) To what extent does it happen?◦Answer deals with chemical equilibrium which

we will study in a later unit

3) How fast and by what mechanism?◦Chemical kinetics

Why? Importance Examples

Chemical Kinetics is very important for biological (your life), environmental (our lives) and economic (industry) processes.

◦ Biological: Large proteins (aka Enzymes) increase the rates of numerous reactions essential to life.

◦ Environmental: The maintenance or depletion of the ozone layer depends on the relative rates of reactions that produce or destroy ozone.

◦ Economic: The synthesis of ammonia (NH3) from N2 and H2 depends on rates of reactions. Fertilizer industries use catalysts to speed up these rates for economic reasons.

Schematic: ExothermicProcess where energy is released as it proceeds. Heat is given off to surroundings.

Reactants Products + Energy

Schematic: EndothermicProcess where energy is absorbed as it proceeds. Heat is consumed and surroundings become cooler.

Reactants + Energy Products

Factors that affect KINETICS

All based on COLLISION THEORY:Collision theory: For a reaction to occur, the atoms or molecules must collide with one another with enough energy (activation energy) and must collide in the right orientation.

FACTORS:1) Concentration of reactants2) Temperature3) Presence of a catalyst4) Surface area 5) Agitation6) Nature of reactants

Factor 1: Concentration


FACTORS:1) Concentration of reactants

If you increase concentration (Molarity), the rate of reaction increases.

Why? There are more molecules which increases the number of collisions altogether; however, there are better chances that molecules will collide in the right orientation.

Factor 2: Temperature


FACTORS:2) Temperature

Temperature is an averaged kinetic energy of molecules so if you increase temperature, you increase kinetic energy. This means you increase the number of collisions

Heat supplies the energy to allow the reaction to proceed (i.e. overcoming the activation energy barrier)

Think about: Why do we refrigerate milk?

Factor 3: Presence of Catalyst


FACTORS:3) Presence of a catalyst

Catalyst assist a reaction and increase the reaction rate without being consumed in the reaction.

Adding a catalyst decreases the activation energy which means more molecules will have enough energy to react.

Think about: Catalytic converter, Enzymes

Factor 3: Presence of Catalyst

Factor 4: Surface AreaAll based on COLLISION THEORY:Collision theory: For a reaction to occur, the atoms or molecules must collide with one another with enough energy (activation energy) and must collide in the right orientation.

FACTORS:4) Surface Area

Increased surface areas of molecules/particles will increase the rate of reaction. This means to break into smaller particle sizes.

More places to react give better chances for collisions in the right orientation.

How to increase surface area? Grind or crush a mixture of reactants.

Ex: A crushed aspirin will enter your blood stream faster than taking it whole.

Factor 5: Agitation


FACTORS:5) Agitation

Stirring or shaking a reaction will increase the reaction rate.

By stirring or shaking, you are introducing energy into the reaction and thus giving molecules/particles more energy to react (overcome activation energy barrier). Your mechanical energy is converted to kinetic energy.

Factor 6: Nature of Reactants


FACTORS:6) Nature of reactants

Reactants whose bonds are weaker have a lower activation energy and thus a higher rate of reaction.

All chemical reactions involve bond breaking and bond making. Bond breaking occurs on reactant side. Collisions between reactants that require less kinetic energy are needed to break weaker bonds (i.e. smaller activation energy)

Rate Laws and Reaction OrderRate Law: An equation that shows the dependence of the reaction rate on the concentration of each reactant.

a A + b B products

k is the rate constant.

∆[A]

∆trate =

rate = k[A]m[B]n

The values of the exponents (m and n) in the rate law must be determined by experiment; they cannot be deduced from the stoichiometry of the reaction.

*[ ] means molarity

Rates of Chemical Reactions

a A + b B d D + e E

rate = =

=14

–∆[O2]

∆trate = 1

2∆[N2O5]

∆t=

∆[NO2]

∆t

– 1b

∆[B]

∆t=– 1

e∆[E]

∆t=1

a∆[A]

∆t1d

∆[D]

∆t

General rate of reaction:

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

The rate of reaction can be measured based on reactants or products. The negative sign in front for reactants indicates they are consumed. Typically rates of reactions are expressed based on reactants.

Rates of Chemical Reactions

ΔT should be Δt (time, not temperature)

Rate Laws and Reaction Order

The values of the exponents in the rate law must be determined by experiment; they cannot be deduced from the stoichiometry of the reaction.

Determining a Rate Law: The Method of Initial Rates

2 NO(g) + O2(g) 2 NO2(g)

[O2]nrate = k[NO]m

Compare the initial rates to the changes in initial concentrations.


[O2]nrate = k[NO]2

m = 2

The concentration of NO doubles, the concentration of O2 remains constant, and the rate quadruples.

2m = 4

2 NO(g) + O2(g) 2 NO2(g)

[O2]rate = k[NO]2


Reaction Order with Respect to a Reactant• NO: second order• O2: first order

Overall Reaction Order• 2 + 1 = 3 (third order)

2 NO(g) + O2(g) 2 NO2(g)


=k =

M

s

(M2) (M)

1

M2 s=

rate

[NO]2 [O2]

[O2]rate = k[NO]2

Units for this third-order reaction:

2 NO(g) + O2(g) 2 NO2(g)

You can pick any experiment or trial and solve for k (should all be the same).

Example Problem #1

[HgCl2] [C2O42-] Rate (M/s)

0.164 0.15 3.2 x 10-5

0.164 0.45 2.9 x 10-4 0.082 0.45 1.4 x 10-4

a. Determine the rate law. What is the order of the reaction?

b. Determine the rate law constant (specify the units)c. What is the rate when the initial concentrations of both

reactants are 0.100 M?

Example Problem #1


0.164 0.15 3.2 x 10-5

0.164 0.45 2.9 x 10-4 0.082 0.45 1.4 x 10-4


Rate = k[HgCl2]m[C2O42-]n

= = = ,

= = = ,

Rate = k[HgCl2] [C2O42-]2

Overall third order

Example Problem #1


0.164 0.15 3.2 x 10-5

0.164 0.45 2.9 x 10-4 0.082 0.45 1.4 x 10-4


b. Determine the rate law constant (specify the units)Rate = k[HgCl2] [C2O4

2-]2

3.2x10-5M/s = k[0.164M][0.15M]2

= =

Example Problem #1


0.164 0.15 3.2 x 10-5

0.164 0.45 2.9 x 10-4 0.082 0.45 1.4 x 10-4


b. Determine the rate law constant (specify the units)c. What is the rate when the initial concentrations of both

reactants are 0.100 M?Rate = k[HgCl2] [C2O4

2-]2

)(0.100M)(0.100M)2 =

Example Problem #2[NOCl] Rate M/s3000 59802000 26601000 6654000 10640a. Determine the rate law and order.b. Determine the rate law constant.

Specify the units.c. What is the rate when the concentration

of NOCl is 8000M?

Example Problem #2[NOCl] Rate M/s3000 59802000 26601000 6654000 10640

Determine the rate law and order.Rate = k[NOCl]2; Second Order

Determine the rate law constant. Specify the units.For first trial: 6.64x10-4 (1/Ms)

For second trial: 6.65x10-4 (1/Ms)What is the rate when the concentration of NOCl is 8000M? 42560 M/s

Example #3

[A] [B] Rate (M/s)1.50 1.50 0.32

1.50 2.50 0.32 3.00 1.50 0.64 a. Determine the rate law and order.b. Determine the rate law constant.

Specify the units.

Example #3

[A] [B] Rate (M/s)1.50 1.50 0.321.50 2.50 0.32 3.00 1.50 0.64 Determine the rate law and order.

Rate = k[A]; first orderDetermine the rate law constant. Specify the units.

k = 0.21 (1/s)

Example #4[CH3COCH3] [Br2] [H+] Rate (M/s)

0.30 0.05 0.05 0.0000570.30 0.10 0.05 0.0000570.30 0.05 0.10 0.0001200.40 0.05 0.20 0.0003100.40 0.05 0.05 0.000076

a. Determine the rate law and order.b. Determine the rate law constant. Specify

the units.

Example #4[CH3COCH3] [Br2] [H+] Rate (M/s)

0.30 0.05 0.05 0.0000570.30 0.10 0.05 0.0000570.30 0.05 0.10 0.0001200.40 0.05 0.20 0.0003100.40 0.05 0.05 0.000076

Determine the rate law and order.Rate = k[CH3COCH3][H+]; second order

Determine the rate law constant. Specify the units.

k = 0.0038 (1/Ms)

Example #5[S2O8

-2] [I -] Rate (mol/L/s)

0.018 0.036 2.6 x 10-6

0.027 0.036 3.9 x 10-6

0.036 0.054 7.8 x 10-6

0.050 0.072 1.4 x 10-5

a. Determine the rate law and order.b. Determine the rate law constant.

Specify the units.

Example #5[S2O8

-2] [I -] Rate (mol/L/s)0.018 0.036 2.6 x 10-6

0.027 0.036 3.9 x 10-6

0.036 0.054 7.8 x 10-6

0.050 0.072 1.4 x 10-5

Determine the rate law and order.Rate = k[S2O8

-2]m[I-]n

= = = ,

= = = ,

Rate = k[S2O8-2][I-]

Second orderDetermine the rate law constant. Specify the units.

For first trial: 0.0040 (1/Ms)For second trial: 0.0040 (1/Ms)

Integrated Rate LawsGraphs, reaction order and concentrations.

Graphs: concentration of substance A vs. time

Scientists don’t like curves, they like straight lines. SO they have to manipulate to obtain straight line.


Rate LawOverall Reaction

Order Units for k

Rate = k Zeroth order M/s or M s–1

Rate = k[A] First order 1/s or s–1

Rate =k[A][B] Second order 1/(M • s) or M–1s–1

Rate = k[A][B]2 Third order 1/(M2 • s) or M–2s–1

*Units for reaction rates are always M/s

Zeroth-Order Reactions

A product(s)

rate = k[A]0 = k∆[A]

∆t– = k

For a zeroth-order reaction, the rate is independent of the concentration of the reactant.

Calculus can be used to derive an integrated rate law.

[A]t concentration of A at time t

[A]0 initial concentration of A

y = mx + b

[A]t = –kt + [A]0

t ½ = [A]0

2k

Zeroth-Order ReactionsA plot of [A] versus time gives a straight-line fit and the slope will be –k.

Half-Life: t1/2For zeroth order:[A]t = –kt + [A]0

[A]t1/2 = [A]0

2

= -=

• Half-life: time for A to reduce by 50%

First-Order Reactions: The Integrated Rate Law

A product(s)

rate = k[A]


∆[A]

∆t– = k[A]

x

yln = ln(x) – ln(y)Using:

[A]t[A]0

ln = –kt

ln[A]t = –kt + ln[A]0

y = mx + b


[A]0 initial concentration of A

First-Order Reactions: The Integrated Rate Law ln[A]t = –kt + ln[A]0

First-Order Reactions: Half-LifeHalf-Life: The time required for the reactant concentration to drop to one-half of its initial value.

A product(s)

rate = k[A]

[A]t[A]0

ln = –ktt = t1/2

=t1/2

[A]

2

[A]0

= –kt1/2

1

2ln t1/2 =

k

0.693or

Second-Order ReactionsA product(s)

rate = k[A]2


∆[A]

∆t– = k[A]2


[A]0 initial concentration of A= kt +

[A]0

1

[A]t

1

y = mx + b

Second-Order Reactions2 NO2(g) 2 NO(g) + O2(g)

Time (s) [NO2] ln[NO2] 1/[NO2]

0 8.00 x 10–3 –4.828 125

50 6.58 x 10–3 –5.024 152

100 5.59 x 10–3 –5.187 179

150 4.85 x 10–3 –5.329 206

200 4.29 x 10–3 –5.451 233

300 3.48 x 10–3 –5.661 287

400 2.93 x 10–3 –5.833 341

500 2.53 x 10–3 –5.980 395

Second-Order Reactions2 NO2(g) 2 NO(g) + O2(g)

= kt +[A]0

1

[A]t

1

Second-Order Reactions

A product(s)

rate = k[A]2

t = t1/2

=t1/2

[A]

2

[A]0

Half-life for a second-order reaction:

[A]0

1= kt1/2 +[A]0

2=t1/2

k[A]0

1

Second-Order Reactions=t1/2

k[A]0

1

For a second-order reaction, the half-life is dependent on the initial concentration.

Each successive half-life is twice as long as the preceding one.

Formulas Given

Example #1

A first order reaction has a half-life of 20 minutes.

a. Calculate the rate constant for this reaction.

b. How much time is required for this reaction to be 80% complete?

Example #1

A first order reaction has a half-life of 20 minutes.

Calculate the rate constant for this reaction. Use: = and solve for kYou get: 5.75x10-4 1/s

How much time is required for this reaction to be 80% complete? Use: and solve for t. You calculated k and [A]t is 0.2 if you assume [A]0 is 1 (percentages). You get: 2800 s

Example #2

N2O5(g) 4NO2 (g) + O2 (g) The following results were collected:[N2O5(g) ] Time(s)0.1000 00.0707 500.0500 1000.0250 2000.0125 3000.00625 400

A. Determine the rate constant from the data above.

B. Calculate the concentration of N2O5(g) after 175 seconds have passed.

Example #2

N2O5(g) 4NO2 (g) + O2 (g) The following results were collected:[N2O5(g) ] Time(s)0.1000 00.0707 500.0500 1000.0250 2000.0125 3000.00625 400

A. Determine the rate constant from the data above. First order k = 0.00693 1/s

B. Calculate the concentration of N2O5(g) after 175 seconds have passed. 0.0297M

Example #32C4H6(g) C8H18(g)

The following data were collected:[C4H6] Time (s)0.01000 00.00625 10000.00476 18000.00370 28000.00313 3600

A. What is the order of the reaction?

B. What is the rate law constant?

C. How long will it take for the [C4H6] to be 93% gone?

Example #32C4H6(g) C8H18(g)

The following data were collected:[C4H6] Time (s)0.01000 00.00625 10000.00476 18000.00370 28000.00313 3600

A. What is the order of the reaction? Second order

B. What is the rate law constant? 0.061(1/Ms)

C. How long will it take for the [C4H6] to be 93% gone? 21,780 s

Example #4

A second order reaction has a rate constant of 5.99M-1s-1. The reaction initially contains 4M [A].

A. How long will it take for the reaction to drop to 0.17M?

B. What will the concentration be after an hour?

Example #4

A second order reaction has a rate constant of 5.99M-1s-1. The reaction initially contains 4M [A].

A. How long will it take for the reaction to drop to 0.17M? 0.94s

B. What will the concentration be after an hour? 4.63x10-5 M

Example #5

The following data was collected.[A] Time(s)10 08 36 64 92 12

A. Determine the order of the reaction from the data.

B. What is the rate law constant?

C. What is the half life for the reaction?

Example #5

The following data was collected.[A] Time(s)10 08 36 64 92 12

A. Determine the order of the reaction from the data. Zeroth order

B. What is the rate law constant? 0.67 M/s (slope)

C. What is the half-life for the reaction? 7.46s

Transition State: The configuration of atoms at the maximum in the potential energy profile. This is also called the activated complex.

Reaction Rates and the Temperature: Collision Theory and the Arrhenius Equation

The Arrhenius EquationActivation Energy (Ea): The minimum energy needed for reaction. As the temperature increases, the fraction of collisions with sufficient energy to react increases.

The Arrhenius Equation

Using the Arrhenius Equation

RT

Ealn(k) = ln(A) –

y = mx + b

+ ln(A)T

1

R

–Ealn(k) =

ln(k) = ln(A) + ln(e–E /RT)a

rearrange the equation

Slope intercept formula when you plot ln(k) vs. 1/T :

R = 0.008314 kJ/K.molA: collision frequency factor


+ ln(A)T

1

R

–Ea

ln(k) =

t (°C) T (K) k (M–1 s–1) 1/T (1/K) lnk

283 556 3.52 x 10–7 0.001 80 –14.860

356 629 3.02 x 10–5 0.001 59 –10.408

393 666 2.19 x 10–4 0.001 50 –8.426

427 700 1.16 x 10–3 0.001 43 –6.759

508 781 3.95 x 10–2 0.001 28 –3.231


+ ln(A)T

1

R

–Ea

ln(k) =

Example #1 Determine the activation energy and the

collision frequency factor at 700K for the data below.

k T (K)0.011 7000.035 7300.105 7600.343 7900.789 810

Changes in Temperature will change the rate constant

Increasing temperature will increase the rate constant (thus increasing the rate).

ln k1 - ln k2 = - (Ea/R)(1/T1 - 1/T2)

Example #2

The rate constant of a first-order reaction is 0.034 at 298K. The activation energy for this reaction is 50.2 kJ/mol.

What is the rate constant at 350K?

(if the temperature is given in Celsius, add 273 to change to Kelvin)

Example #3

At 45oC the rate constant of a first order reaction is 0.8. At 135oC the rate constant is 2.4. What is the activation energy for this reaction?

Example #4The rate of a reaction increased by a factor

of 10 when the temperature goes from 25oC to 50oC. What is the activation energy of the reaction?

Reaction Mechanisms

Elementary Reaction (step): A single step in a reaction mechanism.

Reaction Mechanism: The sequence of reaction steps that describes the pathway from reactants to products.

Possible: Hydrogen reacts with oxygen to make water.

H2 + O2 H2O2

H2O2 H2O + O

H2 + O H2O

Same reaction, other approachesO2 2OH2 + O H2OH2 + O H2O

H2 2HH + O2 OH + OH + OH H2OO + H2 H2O

How do you decide which is right?The reaction mechanism must result in the

overall balance reaction

The slow step must support the rate law

Reaction MechanismsExperimental evidence suggests that the reaction between NO2 and CO takes place by a two-step mechanism:

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

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

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

elementary reaction

overall reaction

elementary reaction

An elementary reaction describes an individual molecular event.

The overall reaction describes the reaction stoichiometry and is a summation of the elementary reactions.




Reaction MechanismsExperimental evidence suggests that the reaction between NO2 and CO takes place by a two-step mechanism:

elementary reaction

overall reaction

elementary reaction

A reactive intermediate is formed in one step and consumed in a subsequent step.

Reaction MechanismsMolecularity: A classification of an elementary reaction based on the number of molecules (or atoms) on the reactant side of the chemical equation.

termolecular reaction:

unimolecular reaction:

bimolecular reaction:

O(g) + O(g) + M(g) O2(g) + M(g)

O3*(g) O2(g) + O(g)

O3(g) + O(g) 2 O2(g)

Rate Laws for Elementary Reactions

The rate law for an elementary reaction follows directly from its molecularity because an elementary reaction is an individual molecular event.

termolecular reaction:

unimolecular reaction:

bimolecular reaction:

O(g) + O(g) + M(g) O2(g) + M(g)

O3*(g) O2(g) + O(g)

rate = k[O]2[M]

rate = k[O3]

rate = k[O3][O]

O3(g) + O(g) 2 O2(g)

Rate Laws for Overall Reactions

Rate-Determining Step: The slowest step in a reaction mechanism. It acts as a bottleneck and limits the rate at which reactants can be converted to products.

1) First check to make sure the individual steps will add up to the overall reaction.

2) In elementary steps, the coefficient of the reactants become the power in the rate law.

3) Slow step will always be the basis for comparing experimental to predicted rate law.

4) If slow step is the first, then everything is easy.5) If not, then the forward rate of step above = reverse rate and some

substituting will take place.6) Intermediates cannot be in rate law.





fast step

overall reaction

slow step

Based on the slow step: rate = k1[NO2]2

k2

k1

Initial Slow Step


N2O(g) + H2(g) N2(g) + H2O(g)

2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)

slow step

overall reaction

fast step, reversible

Based on the slow step: rate = k2[N2O2][H2]

k3

k-1

Initial Fast Step

2 NO(g) N2O2(g)k1

N2O2(g) + H2(g) N2O(g) + H2O(g)k2

fast step

Rate Laws for Overall Reactionsrate = k2[N2O2][H2]

intermediate

First step: Ratereverse = k–1[N2O2]Rateforward = k1[NO]2

k1[NO]2 = k–1[N2O2]

[NO]2[N2O2] = k–1

k1

Slow step: rate = k2[N2O2][H2] rate = k2 [NO]2[H2]k–1

k1

H2O2(aq) + I–(aq) H2O(l) + IO–(aq)

H2O2(aq) + IO–(aq) H2O(l) + O2(g) + I–(aq)

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

CatalystSince the catalyst is involved in the rate-determining step, it often appears in the rate law.

rate = k[H2O2][I–]

overall reaction

rate-determiningstep

fast step

Example #1

Overall reaction: 2NO2 + CO NO3 + CO2

Proposed mechanism:NO2 + NO2 NO3 + NO (slow)NO3 + CO NO2 + CO2 (fast)

If the rate law is Rate = k[NO2], is this a possible mechanism?

Example #2

For the overall reaction:2NO + Br2 2NOBr

The proposed mechanism is:NO + NO N2O2 (fast equilibrium)

N2O2 + Br2 2NOBr (slow)

Determine the rate law.

Example #3Overall reaction:

4HBr + O2 2H2O + 2Br2

Proposed Mechanism:HBr + O2 HOOBrHOOBr + HBr 2HOBr2HOBr + 2HBr 2H2O + 2Br2

Rate law: Rate = k[HBr]2[O2]

Which is the rate determining step?

Example #4

From the following mechanism, determine the rate law.

Cl2 2Cl (fast)Cl + CHCl3 HCl + CCl3 (slow)Cl + CCl3 CCl4 (fast)

Overall reaction:Cl2 + CHCl3 HCl + CCl4

Documents

Kinetics The Rates and Mechanisms of Chemical Reactions