chem 4a: general chemistry with quantitative analysissrjcstaff.santarosa.edu/~oraola/chem1blect/ch....

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OverviewReaction Rates

Rates and Time

Initial Rates

First-Order Reactions

Half-life of First Order Reactions

Second-Order Reactions

Half-life of Second Order Reactions

Mechanisms

Activation Energy

Reactions mechanisms with a slow first step

Rections with a fast equilibrium step

Catalysis

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Slow vs. fast chemical reactions

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Slow vs. fast chemical reactions

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Microscopic View

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Ozone_Destruction.movMedia File (video/quicktime)

Factors that affect the reaction rate

1. Concentration: Molecules must collide to react.

Rate ∝ collision frequency ∝ concentration

2. Physical state: Molecules must mix to react.3. Temperature: Molecules must collide with enough energy.

Rate ∝ Collision frequency ∝temperatureRate ∝ Collision energy ∝temperature

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Rate of Reaction

Average rate =change in concentration

time interval

Average rate = −[reactant ]final − [reactant ]initial

tfinal − tinitial

Average rate =[product ]final − [product ]initial

tfinal − tinitial

Average rate = −∆[reactant ]∆t

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Rate of reactionFor the reaction

Br2(aq) + HCOOH(aq)→ H+(aq) + Br−(aq) + CO2(g)

Average ratebetween 100 and200 s

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Rate of reaction

For the reaction


Ave. rate =

− (0.00596− 0.00846) mol · L−1

(200− 100) s= 2.5× 10−5 mol · L−1 · s−1

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Rate of reaction

For the reaction


Instantaneous rate = lim∆t→0

−∆ [reactant ]∆t

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Rate of reactionFor the reaction


Instantaneousrate at t=0 and att=200- Slope of thetangent at thosepoints

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Concentrations and rateIn the case of a reaction like

2NO(g) + O2(g)→ 2NO2(g)

we need to account for the stoichiometry of the reaction when settingup the expressions for the reaction rate.

So that we have the rate expressed as a unique positive number forthe reaction, we need to write:

rate = −∆ [O2]∆t

= −12

∆ [NO]∆t

=12

∆ [NO2]∆t 14 / 80

In general terms

For the “generic” reaction

αA + βB −−→ γC + δD

the expression for the rate of reaction will be:

−1α

∆ [A]∆t

= −1β

∆ [B]∆t

=1γ

∆ [C]∆t

=1δ

∆ [D]∆t

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Rates and TimeFor the reaction

N2O5(g)→ 2NO2(g) +12

O2(g)

the followingdata wascollected at318 K

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N2O5(g)→ 2NO2(g) +12

O2(g)

Graphically, thismeans

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N2O5(g)→ 2NO2(g) +12

O2(g)

Observe that therate of change inthe concentrationchanges withtime

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Rates and TimeAgain back to the model system

2NO(g) + O2(g)→ 2NO2(g)

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Concentrations and Rate

Experiment Initial Reactant Initial rateconcentration mol·L−1

(T = 600 K) [NO] [O2] mol·L−1·s−11 1.30× 10−2 1.10× 10−2 3.21× 10−32 1.30× 10−2 2.30× 10−2 6.40× 10−33 2.60× 10−2 1.10× 10−2 12.8× 10−34 1.30× 10−2 3.30× 10−2 9.60× 10−35 3.90× 10−2 1.10× 10−2 28.8× 10−3

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(T = 600 K) [NO] [O2] mol·L−1·s−11 1.30× 10−2 1.10× 10−2 3.21× 10−32 1.30× 10−2 2.30× 10−2 6.40× 10−33 2.60× 10−2 1.10× 10−2 12.8× 10−34 1.30× 10−2 3.30× 10−2 9.60× 10−35 3.90× 10−2 1.10× 10−2 28.8× 10−3

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(T = 600 K) [NO] [O2] mol·L−1·s−11 1.30× 10−2 1.10× 10−2 3.21× 10−32 1.30× 10−2 2.30× 10−2 6.40× 10−33 2.60× 10−2 1.10× 10−2 12.8× 10−34 1.30× 10−2 3.30× 10−2 9.60× 10−35 3.90× 10−2 1.10× 10−2 28.8× 10−3

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Concentrations and RatesModel system:

2NO(g) + O2(g)→ 2NO2(g)

rate ∝ [O2]

rate ∝ [NO]2

rate ∝ [O2] [NO]2 or rate = k [O2] [NO]2

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Concentrations and RatesModel system:

2NO(g) + O2(g)→ 2NO2(g)

rate ∝ [O2]

rate ∝ [NO]2

rate ∝ [O2] [NO]2 or rate = k [O2] [NO]2

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Concentrations and Rates

Once we know the rate law

rate = k [O2][NO]2

We can calculate the value of the rate constant from the experimentaldata

k =rate

[O2][NO]2

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Concentrations and Rates

Once we know the rate law

rate = k [O2][NO]2

We can calculate the value of the rate constant from the experimentaldata

k = 12.8×10−3mol·L−1·s−1

[1.10×10−2 mol·L−1][2.60×10−2mol·L−1]2

= 1.72× 103 mol−2 · L2 · s−1

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General form of rate laws

rate = k [reactant1]x [reactant2]y ...

According to this rate law, the reaction described is:

• order x with respect to reactant 1• order y with respect to reactant 2• overall order: x + y

The order of a reaction with respect to eachreactant must be determined EXPERIMENTALLY

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Calculation of k

Units of the rate constant:

k : mol−(order−1)·L(order−1)·s−1

So that when multiplied by molorder · L−order , we get

molorder−(order−1)·L(order−1)−order ·s−1

or mol·L−1·s−1

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Examples of reaction order

rate = k[(CH3)3CBr]

H2(g) + I2(g)→ 2HI(g) rate = k[H2][I2]

CHCl3(g) + Cl2(g)→ CCl4(g) + HCl(g) rate = k [CHCl3 ] [Cl2]12

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Exercise

For each of the following reactions, use the given rate law todetermine the reaction order with respect to each reactant and theoverall order

2H2O2(aq) + 3I−(aq) + 2H+(aq)→ I−3 (aq) + 2H2O(`)rate = k [H2O2] [I−]

2NO(g) + O2(g)→ 2NO2(g) rate = k [NO]2[O2]

CH3CHO(g)→ CH4(g) + CO(g) rate = k [CH3CHO]32

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Integrated Rate Laws:First-Order Reactions

For the reaction2N2O5(g)→ 4NO2 + O2(g)

the rate can is describe by the rate law

d [N2O5]dt

= −k [N2O5]

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Determining Reaction Orders

From the data

Experiment Initial Rate Initial [A] Initial [B]( mol·L−1·s−1) ( mol·L−1) ( mol·L−1)

1 1.75×10−3 2.50×10−2 3.00×10−22 3.50×10−3 5.00×10−2 3.00×10−23 3.50×10−3 2.50×10−2 6.00×10−24 7.00×10−3 5.00×10−2 6.00×10−2

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Determining Reaction Orders

Sometimes the exponent is not easy to find by inspection. In thosecases, we solve for m with an equation of the form a = bm:

Thenm =

log alog b

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Finding m, the order with respect to A

Comparing experiments 1 and 2

Rate 2Rate 1

=�k [A]m2 �

��[B]n2�k [A]m1 �

��[B]n1=

[A]m2[A]m1

=

[[A]2[A]1

]m

then m = 1

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that can be rearranged as

d [N2O5][N2O5]

= −kdt

and integrated between the initial and final conditions

∫ [N2O5][N2O5]0

d [N2O5][N2O5]

= −∫ t

0kdt

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to give

ln(

[N2O5][N2O5]0

)= −kt

or

ln [N2O5] = −kt + ln [N2O5]0

Integrated first order rate law36 / 80

An alternative form of the integrated rate lawWe can also write

[N2O5][N2O5]0

= e−kt

or

[N2O5] = [N2O5]0 e−kt

Notice that the fraction of N2O5 remaining after a time interval t isalways e−kt

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Graphically:(integrated first order rate law)

2N2O5(g)→ 4NO2 + O2(g)

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Half-life of First Order Reactions

The half-life for a first-order reaction is independent of the initialconcentration.

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Half-life of a first-order reaction

Half-life (t 12) is the time it takes for the concentration of a reactant to

drop to one-half of its initial value.

Fast reaction =⇒ short half-life

For a first-order process:

ln(

[N2O5][N2O5]0

)= −kt

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Half-life of a first-order reaction

Half-life (t 12) is the time it takes for the concentration of a reactant to

drop to one-half of its initial value.

Fast reaction =⇒ short half-life

For a first-order process:

ln

��[N2O5]02��

�[N2O5]0

= −kt 12

t 12

=ln 2k

=0.693

k

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For the reaction2NO(g)→ 2NO(g) + O2(g)

the rate can be described by the rate law

d [NO2]dt

= −k [NO2]2

This is a second-order kinetic law

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that can be rearranged as

[NO2]−2d [NO2] = −kdt

and integrated between the initial and final conditions

∫ [NO2][NO2]0

[NO2]−2d [NO2] = −∫ t

0kdt

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to give

1[NO2]

=1

[NO2]0+ k t

the integrated second-order rate law.

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Half-life of Second Order Reactions

Notice that in this case

t 12

=1

k [NO2]0

is a function of the initial concentration of the reactant, and thereforeuseless as a characterization of the reaction!

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Graphically:(integrated second order rate law)

2NO(g)→ 2NO(g) + O2(g)

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Reaction Mechanisms

When we study a reaction like

NO2(g) + CO(g)→ NO(g) + CO2(g)

we might guess that in the process there is a collision of a moleculeof NO2 with a molecule of CO.

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Reaction Mechanisms

Click to start

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secondorder.swfMedia File (application/x-shockwave-flash)

Reaction Mechanismsbut a study of the kinetics of the reaction shows that the rate isindependent of the concentration of CO.

Furthrer experimentation has shown that the reaction occurs actuallyin two steps.

In a first step, two molecules of NO2 collide and an oxygen atom istransferred;

NO2(g) + NO2(g)→ NO3(g) + NO(g)

and then in a second step, the collision happens between NO3 andCO

NO3(g) + CO(g)→ NO2(g) + CO2(g)49 / 80

Reaction Mechanisms

Click to start

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twostepsr.swfMedia File (application/x-shockwave-flash)

Reaction Mechanisms

In summary, we have

NO2(g) + NO2(g)→ NO3(g) + NO(g) (1)

NO3(g) + CO(g)→ NO2(g) + CO2(g) (2)

NO2(g) + CO(g)→ NO(g) + CO2(g) (3)

Each of the steps is an elementary reaction. The sum of theelemntary reaction gives the overall reaction. This constitutes areaction mechanism.

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Reaction MechanismsIn summary, we have

NO2(g) + NO2(g)→��NO3(g) + NO(g) (4)

��NO3(g) + CO(g)→ NO2(g) + CO2(g) (5)

NO2(g) + CO(g)→ NO(g) + CO2(g) (6)

NO3: A species that is generated in a step and is consumed in asubsequent step is an intermediate species. It does not appear inthe overall equation.

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Rate Laws for Elementary Reactions

The rate laws for elementary reactions (as opposed to thosebelonging to overall reactions) can be written from the equation, witheach exponent in the rate law equal to the number of particles of agiven type participating in the reaction.

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Rate Laws for Elementary Reactions

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MolecularityFor an elementary reaction (step in a mechanism), the molecularityis the number of molecules (or atoms, ions) that must take part in thatspecific step.

Example:

The elemental reaction

O3 → O2 + O

is unimolecular (has a molecularity = 1)

and the elemental reaction

O3 + O3 → O2 + O2 + O2

is bimolecular (has a molecularity = 2)

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Example:


O3 → O2 + O



O3 + O3 → O2 + O2 + O2


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Example:


O3 → O2 + O



O3 + O3 → O2 + O2 + O2


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Molecularity

Termolecular and higher molecularity are uncommon, because it isvery unlikely that three (or more) molecules will collidesimultaneously under normal conditions.

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Collision Theory of Chemical ReactionsMolecular view of the reaction:

F2(g) + NO2(g)→ F (g) + FNO2(g)

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Collision Theory

rate ofreaction

=

(collisionfrequency

)fraction ofcollisions withthe requiredenergy

fraction of collisionsin which moleculeshave therequired relativeorientation

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Activation Energy

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Enthalpy Change and Activation Energy

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Activated Complex

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Collision theory

In order to react, molecules mustcollide. The rate of reactiondepends on the collisionfrequency, on the relative energyof the collisions and on theirorientation.

k = p × z × e−EaRT

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Arrhenius Equation

The product of p ana z is relativelytemperature-independent, wegather these parameters as thepre-exponential factor, A

k = A e−

EaR T

The factor e−

EaR T represents the

fraction of collisions with energyhigher than Ea

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Temperature Dependence of the Kinetic Constant

Alternatively, we can expressArrhenius equation as

ln k = ln A− EaR T

that can be rearranged in order tocompare the rate constants at twodifferent temperatures

ln(

k2k1

)= −Ea

R

(1T2− 1

T1

)

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Example ProblemThe major reaction that occurs when an egg is boiled in water is thedenaturation of the egg protein Given that Ea=42 kJ·mol−1 for thisprocess, calculate how much time is required to cook an egg at 92.2◦C [the boiling point of water in Vail, CO, at about 2400 m (8000 ft)elevation] to the same extent as an egg cooked for 3.0 min at 100.0◦C (the boiling point of pure water at sea level). What we need is toknow the ratio of the kinetic constants, for the time the reaction takesis always inversely) proportional to the value of the constant.

lnk2k1

= −EaR

(1T2− 1

T1

)

lnk2k1

= 0.7488

e0.7488 = 1.33 t = 4.0 min

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lnk2k1

= −EaR

(1T2− 1

T1

)

lnk2k1

= 0.7488

e0.7488 = 1.33 t = 4.0 min

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lnk2k1

= −EaR

(1T2− 1

T1

)

lnk2k1

= 0.7488

e0.7488 = 1.33 t = 4.0 min

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lnk2k1

= −EaR

(1T2− 1

T1

)

lnk2k1

= 0.7488

e0.7488 = 1.33 t = 4.0 min

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Rate Limiting Step

Consider the familiar mechanism

NO2(g) + NO2(g)k1−→ NO3(g) + NO(g) rate = k1[NO2]2 (7)

NO3(g) + CO(g)k2−→ NO2(g) + CO2(g) rate = k2[NO][CO] (8)

NO2(g) + CO(g)→ NO(g) + CO2(g) (9)

The first step is much slower that the second, it determines the ratelaw and the rate of the overall reaction. It is the rate-limiting step.

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Reaction mechanisms with a fast equilibrium step

On the other hand, a reaction mechanism like

2NO k1k−1N2O2(g) fast equilibrium (10)

N2O2 + O2(g)k2−→ 2NO2(g) rate = k2[N2O2][O2](slow) (11)

2NO(g) + O2(g)→ 2NO2(g) (12)

For the fast equilibrium we can write k1[NO]2 = k−1[N2O2]

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Reaction mechanisms with a fast equilibrium stepSolving for

[N2O2] =k1

k−1[NO]2

and substituting the result in the kinetic law of the slow step, we getfor the overall reaction

rate = k2

(k1

k−1

)[NO]2[O2]

which correspond to the experimental rate law

rate = k [NO]2[O2]

if we make

k = k2

(k1

k−1

)73 / 80

CatalysisHydrogen peroxide decomposes spontaneously into oxygen gas andwater

2H2O2(aq)→ O2(g) + 2H2O(`) slowbut the reaction is very slow. In the presence of iodide ions, thereaction proceeds at a much faster rate. The following mechanismhas been proposed

H2O2(aq) + I−(aq)→ IO−(aq) + H2O(`) slow (13)

H2O2(aq) + IO−(aq)→ I−(aq) + H2O(`) + O2(g) fast (14)

2H2O2(aq)→ 2H2O(`) + O2(g) (15)The mechanism is consistent with the experimentally observed

kinetic law (first order both in iodide ion and in hydrogen peroxide).Iodide ions participate, but they are not consumed by the reaction,they act as a catalyst

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Catalysis

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Properties of Catalysts

1. Provide an alternative pathway with lower activation energy2. Not consumed by the reaction3. Increase the rates of both the forward and the reverse reaction4. Do not change the relative amounts of reactants and products in

the reaction5. Bring the reaction to completion (or equilibrium) faster

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Heterogeneous catalysis

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Enzymes: Biological Catalysts

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Reaction RatesRates and TimeInitial RatesFirst-Order ReactionsHalf-life of First Order ReactionsSecond-Order ReactionsHalf-life of Second Order ReactionsMechanismsActivation EnergyReactions mechanisms with a slow first stepRections with a fast equilibrium stepCatalysis

chem 4a: general chemistry with quantitative analysissrjcstaff.santarosa.edu/~oraola/chem1blect/ch....

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