Brown and Lemay Pages

The term reaction rate refers to how fast a product is formed or a reactant is

consumed.

Consider the following generic reaction:

aA(g) + bB(g) cC(g) + dD(g)

Rateave =- 1

aΔ[A]

Δt=-

1b

Δ[B]Δt

=1c

Δ[C]Δt

=1d

Δ[D]Δt

Rates are always positive quantities.

The concentration of reactants do not change

at the same rate when coefficients are different.

Consider the following specific example:

N2(g) + 2O2(g) 2NO2(g)

Rateave =- Δ[N2]Δt

=- 12

Δ[O2]Δt

=12

Δ[NO2]Δt

=Ms

The following graph shows the decomposition

of N2 as a function of time.

There are two different rates shown in the graph.

The instantaneous rate is the rate of◦ change (Δ) at a specific instant of time, ◦ i.e. 40 s.

In a non-calculus environment, this is◦ determined by calculating the slope of a

tangent line containing that point.

The average rate is computed over a◦ time interval necessitating using two◦ different times.

Rateinst= Δ[N2]

Δt- = -

0.30 M – 0.55 M60. s – 30. s

Rateinst= 8.3 x 10-3 M s-1

Rateinst= Δ[N2]

Δt- =

This is the average rate from 20. s to 60. s.

-0.30 M – 0.67 M

60. s – 20. sRateavs = Δ[N2]Δt

- =

Rateave = 9.2 x 10-3 M s-1

V

0.00

0.20

0.40

0.60

0.80

1.00

1.20

0.0 20.0 40.0 60.0 80.0 100.0

T

C

Time (s)

[N2]

[N2] vs Time

ΔtΔ[N

2]

instantaneous rate

average rate

Chemical reactions involve the breaking of bonds (endothermic) and the forming of bonds (exothermic).

The factors affecting reaction rates:

The physical state of reactant rates.

◦ When reactants are in different phases◦ (states), the reaction is limited to their◦ area of contact.

The concentration of reactants.

◦ As the concentration of reactants◦ increase, the rate of reaction increases.

Temperature

◦ As the temperature is increased,◦ molecules have more kinetic energy◦ resulting in higher reaction rates.

Presence of a catalyst.

◦ Increases the rate of reaction without◦ being consumed.

Consider the following generic reaction:

aA(g) + bB(g) cC(g) + dD(g)

The rate law is written:

Rate = k[A]x[B]y

k is called the rate constant and is◦ temperature dependent.

The exponents x and y are called◦ reaction orders.

The overall reaction order is the sum of the orders with respect to each reactant.

The reaction orders in a rate law indicate how

the rate is affected by the concentration of the reactants.

The effect of reactant concentration can◦ not be predicted from the balanced◦ equation.

The effect of reactant concentration can◦ only be determined empirically!

The rate of a reaction will decrease as the reaction proceeds because the reactants are being consumed.

A very important distinction needs to be made at this point:

The rate of a reaction depends on concentration but k, the rate constant, does not!

The rate constant and the rate of◦ reaction are affected by temperature◦ and the presence of a catalyst.

(a) How is the rate of disappearance of N2

and O2 related to the rate of appearance of NO2 in the reaction shown below.

N2(g) + 2O2(g) 2NO2(g)Rate =- Δ[N2]

Δt=-

12

Δ[O2]Δt

=

=12

Δ[NO2]Δt

(b) If the rate of decomposition of N2 at an instant is 3.7 x 10-6 M•s-1, what is the

rate of disappearance of O2 and the rate of appearance of NO2? Rate =- Δ[N2]

Δt= 3.7 x 10-6 M•s-1

Δ[O2]Δt

= 2 × 3.7 x 10-6 M•s-1= 7.4 x 10-6 M/s

Δ[NO2]Δt

2 × 3.7 x 10-6 M•s-1= 7.4 x 10-6 M/s=

The following data was measured for the reaction of nitrogen and oxygen to form nitrogen(IV) oxide.

N2(g) + 2O2(g) 2NO2(g)

.

Trial [N2](M)

[O2](M)

Init Rate(M/s)

1 0.12 0.12 1.19×10-

3

2 0.12 0.23 2.44×10-

3

3 0.22 0.12 4.92×10-

3

(a) Determine the rate law for this reaction.

Rate 2Rate 1

= k[N2]x[O2]y

k[N2]x[O2]y

2 = 2y

y = 1

2.44 × 10-3 M/s1.19 × 10-3 M/s

=k(0.12)x(0.23)y

k(0.12)x(0.12)y

Rate 3Rate 1

= k[N2]x[O2]y

k[N2]x[O2]y

4.92 × 10-3 M/s1.19 × 10-3 M/s

=k(0.22)x(0.12 )k(0.12)x(0.12)

4=2x

x = 2

Rate = k[N2]2[O2]

(b) Calculate the rate constant.

Rate = k[N2]2[O2]

1.19 × 10-3 Ms-1 = k(0.12 M)2(0.12 M)

k = 0.69 M-2s-1

(c) Calculate the rate when [N2] = 0.047 M and

[O2] = 0.16 M

Rate = 0.69 M-2s-1 × (0.047 M)2 × 0.16 M

Rate = 2.4 x 10-4 M/s

The second type of rate law is called theintegrated rate law which shows theconcentration as a function of time.

There are three possible orders for a reaction, zero, first, or second.

The integrated rate law is given by:

[A] = -kt + [A]0

Concentration vs Time

y = -0.06x + 1

0.00

0.50

1.00

0.0 5.0 10.0 15.0

Time (s)

Co

nce

ntr

atio

n (

M)

A plot of [A] vs Timeproduces a straight linewith a slope equal to thenegative rate contant, -k.

Rate vs Time

0.000.200.400.600.801.00

0.0 5.0 10.0 15.0

Time (s)

Ra

te

The reaction rate isindependent ofconcentration.

The reaction rate is independent of time.

Rate vs Concentration

0.000.200.400.600.801.00

0.00 0.50 1.00

Concentration (M)

Ra

te

For 0th order kinetics:

Rate Law: Rate = k

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

Plot needed for straight line: [A] vs T

Slope: -k

Half-Life: t1/2 =[A]0

2k

The integrated rate law for 1st order kinetics is given by:

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

ln(Concentration) vs Time

y = -0.60x

-6.00

-4.00

-2.00

0.00

0.0 5.0 10.0

Time (s)

ln(C

on

cen

tra

tiio

n)

A plot of ln[A] vs Timeproduces a straight linewith a slope equal to thenegative rate contant, -k.

The reaction rate isdirectly proportional

tothe concentration.

The reaction rate decreases but notlinearly with time.

Rate vs Time

0.000.200.400.600.801.00

0.0 5.0 10.0

Time (s)R

ate

Rate vs Concentration

y = 0.4512x - 1E-16

0.000.200.400.600.801.00

0.00 0.50 1.00 1.50

Concentration (M)

Ra

te

A first-order reaction is one in which the ratedepends on the concentration of a singlereactant to the first power.

Rate =- Δ[A]Δt

= k[A]

aA(g) Product

Δ[A]Δt= - k

[A]

ln[A]t – ln[A]0 = -kt

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

y = mx + b

The slope of the straight line gives -k,the rate constant, and ln[A]0 is the y-intercept.

For a first-order reaction, the half-life (the timerequired for one-half of a reactant todecompose) has a constant value.

t1/2 = ln 2k

= 0.693k

For a first-order reaction, the half-life is onlydependent on k, the rate constant andremains the same throughout the reaction.

The half-life is not affected by the initialconcentration of the reactant.

For 1st order kinetics:

Rate Law: Rate = k[A]

Integrated Rate Law: ln[A] = -kt + ln[A]0

Plot needed for straight line: ln[A] vs T

Slope: -k

Half-Life: t1/2 =0.693

k

The integrated rate law for 2nd order kinetics is given by:

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

A plot of 1/[A] vs Timeproduces a straight linewith a slope equal to therate contant, k.

1/Concentration vs Time

y = 0.0612x + 99.368R2 = 0.9999

0100200300400500600

0.00E+00

2.00E+03

4.00E+03

6.00E+03

8.00E+03

Time (s)

1/C

on

cen

trat

ion

(1/

M)

The reaction rate isdirectly proportional

tothe concentration.

The reaction rate decreases but notlinearly with time.

Rate vs Time

0.000.200.400.600.801.00

0.0 5.0 10.0

Time (s)R

ate

Rate vs Concentration

y = 0.4512x - 1E-16

0.000.200.400.600.801.00

0.00 0.50 1.00 1.50

Concentration (M)

Ra

te

A second-order reaction is one in which therate depends on the concentration of a

singlereactant concentration raised to the secondpower or concentrations of two differentreactants, each raised to the first power.

Rate =- Δ[A]Δt

= k[A][B]

aA(g) Product

aA(g) + bB(g) Product

- Δ[B]Δt

=

or

or

Rate =- Δ[A]2

Δt= k[A]2

The integrated rate law for a second-orderreaction is given by:

t= k1[A]t

+1

[A]0

y = mx + b

The slope of the straight line gives k, the rateconstant, and is the y-intercept.1

[A]0

For a second-order reaction, the half-life (the time required for one-half of a reactant todecompose) is double the preceding one.

t1/2 =1

k[A]0

For 2nd order kinetics:

Rate Law: Rate = k[A]2

Integrated Rate Law:

Plot needed for straight line:

Slope: k

Half-Life: t1/2 =1

1[A]

= kt +1 1

[A]0[A]1

[A]1

1[A]0

vs T

k[A]0

Particles must collide for chemical reactions to occur.

Not every collision leads to a reaction.

For a reaction to occur, an “effective collision” must take place.

An “effective collision” consists of two conditions.

The colliding particles must approach each other at the proper angle.

H2 Cl2

v1 v2

v4v3HClHCl

In addition, the colliding particles must have sufficient energy.

At the point of impact, ΔKE = ΔPE.

An elastic collision is assumed in which the law of conservation of mass-energy applies.

The KE/molecule must be sufficient to break the H-H and the CI-CI in both hydrogen and chlorine.

Keep in mind that temperature is a measure of the KE/molecule.

Some molecules will be traveling faster than others while the slower molecules will bounce off each other without reacting.

A less than ideal set of conditions for a collision.

H2 Cl2

v1 v2

v4v3Cl2H2

The bonding found in the reactants.◦ triple bonds are stronger than double bonds which are

stronger than single bonds.

The temperature of the system (reactants and products).

The initial concentration of the reactants.

The amount of surface area when reactants are in more than one phase.

The use of a catalyst to provide an alternate pathway.

In order to react, colliding molecules must have KE equal to or greater than a minimum value.

This energy is called the activation energy (Ea) which varies from reaction to reaction.◦ Ea depends on the nature of the reaction and is

independent of temperature and concentration.

The rate of a reaction depends on Ea.

At a sufficiently high temperature, a greater number of molecules have a KE > Ea.

Fraction of Molecules vs Kinetic Energy

0.00E+00

1.00E-03

2.00E-03

3.00E-03

4.00E-03

5.00E-03

6.00E-03

7.00E-03

0 100 200 300 400 500

Kinetic Energy (J)

Fra

cti

on

of

Mo

lec

ule

s

lower temperature

higher temperature

Ea

For the Maxwell-Boltzmann Distribution, only those molecules with energies in excess of Ea react.

As shown on the graph, as the temperature is increased:

The curve shifts to higher energies meaning more molecules to have energies greater than Ea.

The curve gets broader and flatter but the area remains the same under the curve.

Energy vs Reaction Pathway

Reaction Pathway

En

erg

y

Minimum Energy for Reaction

Energy Content of Reactants

Energy Content of Products

Ea

ΔH = -

Energy vs Reaction Pathway

Reaction Pathway

En

erg

y

Energy Content of Products

Energy Content of Reactants

Minimum Energy for Reaction

Ea

ΔH = +

The Arrhenius equation illustrates the dependence of the rate constant, k, on the frequency factor, A, the activation energy, Ea, the gas constant, R, and the absolute temperature, T.

k = Ae-Ea/RT

The frequency factor, A, is related to thefrequency of collisions and the probability ofthese collisions being favorably oriented.

Taking the ln of both sides gives:

ln k = ln A – Ea/RT

Rearranging the equation above shows that theln k is directly proportional to 1/T.

ln k = -Ea/R •1T

+ ln A

y = m x + b

The equation also shows that the reaction ratedecreases as the activation energy increases.

An unknown gas at 300.°C has a rate constant equal to 2.3 x 10-10 s-1 and at 350.°C the rate constant is determined to be 2.4 x 10-8 s-1.

(a) Calculate Ea for this reaction.

T1 = 300.°C = 573 K T2 = 350.°C = 623 Kk1 = 2.3 x 10-10 s-1 k2 = 2.4 x 10-8 s-1

R = 8.31 J mol-1K-1

1

T2

ln k2

k1

=Ea

R[

1

T1

- ]1

T2

.

ln2.4 × 10-8 s-1

2.3 × 10-10 s-1 =

Ea

8.31 J mol-1 K-1[

1

573 K

1

623 K- ]

Ea=2.7 × 105 J/mol

ln k2

k1

=Ea

R[

1

T1

- ]1

T2

(b) Determine the rate constant at 425°C.

ln k2

2.4 x 10-10 s-1

2.7 x 105 J mol-1

8.31 J mol-1 K-1

=

[ 1

573 K

1

698 K- ]

k2 = 6.2 x 10-6 s-1

A reaction mechanism consists of a series of elementary steps indicating how reacting particles rearrange themselves to form products.

Elementary steps are individual steps showing what molecules must collide with each other and the proper sequence of collisions.

Most reactions occur in more than one step.

A reaction mechanism provides much moreinformation than a balanced chemicalequation.

Intermediates are species that are formed and consumed between elementary steps which never appear in a balanced chemical equation.

Not all elementary steps occur at the same rate and consequently are designated as fast or slow steps.

Each elementary step has a molecularityassociated with it.

An elementary step is referred to as a unimolecular step when one molecule decomposes or rearranges to form a product as shown below:

A B

A unimolecular reaction follows a first-order rate law, i.e.

Rate = k[A]

An elementary step is referred to as a bimolecular step when two molecules collide to form a product as shown below:

A + B C

A bimolecular reaction follows a second-order rate law, i.e.

Rate = k[A][B]

An elementary step is referred to as a termolecular step when three molecules collide to form a product as shown below:

A + B + C D

A termolecular reaction follows a third-order rate law, i.e.

Rate = k[A][B][C]

Termolecular reactions are very rare.

The slowest step in a mechanism is called the rate determining step and determines the rate of the overall reaction.

Keep in mind:

The rate of the overall reaction is the same as the rate of the rate-determining step.

Write the rate expression for the slowest step.

The rate expression must only include those species that appear in the balanced equation.

Intermediates are species that are produced in one step and consumed in the next step.

Intermediates can not appear in the rate expression because their concentrations are always small and undetectable.

Determine the overall reaction and the rate expression for the suggested mechanism:

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

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

NO2(g) + NO2(g) NO3(g) + NO(g) (slow)NO3(g) + CO(g) NO2(g) + CO2(g) (fast)

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

Note that the NO3 is an intermediate becauseit is produced in one step of the reaction andconsumed in the next.

Because the rate-determining step comesfirst, the intermediate NO3 cancels out.

This is good because intermediates cannot appear in the overall reaction.

The rate expression is determined by usingthe rate-determining step and is given by:

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

The following reaction 2NO2(g) + O3(g) N2O5(g) + O2(g)

Rate = k[NO2][O3]

Which of the following mechanisms predicts a

satisfactory rate law?

Mechanism 1:NO2(g) + NO2(g) N2O2(g) + O2(g) (slow)N2O2(g) + O3(g) N2O5 (g) (fast)

Mechanism 2:NO2(g) + O3(g) NO3(g) + O2(g)

(slow)NO3(g) + NO2(g) N2O5(g) (fast)

For Mechanism 1:

NO2(g) + NO2(g) N2O2(g) + O2(g) (slow)

N2O2(g) + O3(g) N2O5 (g) (fast)

The first criteria is satisfied because after adding the elementary steps, the overall equation for the reaction is correct.

However, the predicted rate law determined from the rate-determining step is:

Rate = k[NO2]2 which does not agree with given which does not agree with given rate lawrate law

2NO2(g) + O3(g) N2O5(g) + O2(g)

For Mechanism 2:

NO2(g) + O3(g) NO3(g) + O2(g) (slow)

NO3(g) + NO2(g) N2O5(g) (fast)

2NO2(g) + O3(g) N2O5(g) + O2(g)

The first criteria is satisfied because after adding the elementary steps, the overall equation for the reaction is correct.

The predicted rate law determined from the rate-determining step is:

Rate = k[NO2][O3] which does agrees with given rate which does agrees with given rate law!law!

intermediate is not a reactant in the rate-determining step

When the intermediate is a reactant in the rate-determining step, mechanisms are more involved.

Determine the overall reaction and the rate expression for the suggested mechanism:

NO(g) + Br2(g) NOBr2(g) (fast)NOBr2(g) + NO(g) 2NOBr(g) (slow)

NO(g) + Br2(g) NOBr2(g) (fast)NOBr2(g) + NO(g) 2NOBr(g) (slow)

The predicted rate law determined from therate-determining step is:

Rate = k[NOBr2][NO]

which is incorrect because intermediates cannot appear in the rate law

2NO(g) + Br2(g) 2NOBr(g)

When this situation occurs, you must equatethe forward and reverse rates obtained fromthe fast step as shown below:

NO(g) + Br2(g) NOBr2(g) (fast)

Ratef = Rater

kf[NO][Br2] = kr[NOBr2]

[NOBr2] = kf

kr

[NO][Br2]

= k[NO][Br2][NOBr2]

Now you can use the rate law obtained originally but substitute for [NOBr2].

Rate = k[NOBr2][NO][NOBr2]= k[NO][Br2]

Rate Rate ==k[NO]k[NO]22[Br[Br22]]

(There is no need to keep track of the individual rate constants because they would be incorporated into a single constant.)

A catalyst is a substance that increases thespeed of a reaction without being consumed.

Catalysts can be either homogeneous orheterogeneous depending if they are in thesame phase as the reactants.

The function of a catalyst is to provide analternate pathway (mechanism) for a reactionwhich lowers the activation energy, Ea.

Arrhenius equation

shows that as activation energy, Ea, is made

smaller, the rate constant, k, increases.

k = Ae-Ea/RT

For the given mechanism, identify the catalyst

and the intermediate.

O3 + Br O2 + BrO

BrO + O Br + O2

Remember an intermediate is produced inone step and consumed in the other makingBrO the intermediate.

Remember a catalyst remains chemicallyunaltered during the reaction making the Brthe catalyst.