Ch 13 III Mechanisms

Part II—Mechanisms of Chemical Reactions

First part: we determined the rate of overall reactions

This part: we examine elementary steps of reaction

[ ]22 from experimental datRate a k NO= ⇐

Recall

•Why this reaction is second-order with respect to [NO2].

•Why this reaction is zero-order with respect to [CO].

•What the reaction mechanism is, that is, the elementary steps of the reaction?

( ) ( ) ( ) ( )2 2NO g CO g NO g CO g+ → +

For ex

ample

,

Will ans

wer

Takes place in two elementary steps (or reactions):

( ) ( ) ( ) ( )2 2 3 (Step 1)NO g NO g NO g NO g+ → +

( ) ( ) ( ) ( )3 2 2 (Step 2)NO g CO g NO g CO g+ → +

Two necessary, but not sufficient, conditions for a correct reaction mechanism:

•The sum of the elementary steps = overall reaction

•Mechanism predicts correct experimental reaction order

( ) ( ) ( ) ( )2 2NO g CO g NO g CO g+ → +

CHECK: the sum of the elementary steps = overall reaction

( ) ( ) ( ) ( )2 2 3 (Step 1)NO g NO g NO g NO g+ → +

( ) ( ) ( ) ( )3 2 2 (Step 2)NO g CO g NO g CO g+ → +

( ) ( ) ( ) ( )2 2NO g CO g NO g CO g+ → +

( ) ( ) ( ) ( )2 2NO g CO g NO g CO g+ → +

NO3 is called an intermediate species or an intermediate

Intermediates are very reactive and have very small concentrations.

Digression—Elementary Steps (Reactions)

Rar

e

( ) ( ) ( ) ( )12 2 3 (Step 1)kNO g NO g NO g NO g+ ⎯⎯→ +

( ) ( ) ( ) ( )23 2 2 (Step 2)kNO g CO g NO g CO g+ ⎯⎯→ +

[ ][ ] [ ]21 1 2 2 1 2rate k NO NO k NO= =

[ ][ ]2 2 3rate =k NO CO

Circulation of Traffic

6 lane highwayAverage 60 mphMax. 18,000 cars/hr(1 car/105 ft per lane)

2 lane rough roadAverage 20 mphMax. 4,400 cars/hr(1 car/45 ft per lane)

1 lane highwayAverage 60 mphMax. 3,000 cars/hr

C

What is the maximum rate of cars passing C per hour?

Funnel stem determinesthe rate

Pouring faster doesn’t fill the flask faster

Zumdahl p 728

Some (not all) Reaction Mechanisms Have a Rate-Determining Step

( ) ( ) ( ) ( )12 2 3

kNO g NO g NO g NO g+ ⎯⎯→ +

( ) ( ) ( ) ( )23 2 2

kNO g CO g NO g CO g+ ⎯⎯→ +

[ ]21 1 2rate k NO=

[ ][ ]2 2 3rate =k NO CO

Much slower step—rate-determining step

Reactants Must Surmount an Energy Barrier to React

Collision theory of reaction rates—In order that two species react:

•They must collide. (In general, collision rate is very large!)

•The colliding particles must have sufficient KE to break bonds.

•The colliding particles must have correct orientation.

Example

( ) ( ) ( ) ( )2 2 2NO g F g FFNO g g+ → + (Elementary step)

( ) ( ) ( )2 2g NO gF FNO g+ → (F is reactive intermediate)

Collision theory of reaction rates for : A + B → C + D

z = collision

frequencyα [A][B]

Exothermic orEndothermic?Exothermic

= z0[A][B]f =e-Ea/RT

p = steric factorRange 0 to 1

/aE RTetotalblue area

area−= =

0/ [A][B]rate aE RTp ez −=

/ ]A [A [B]aE RTe−=

[A][B]k=

/A aE RTk e−=

A 1ln ln ak ETR⎛= − ⎞⎜ ⎟⎝ ⎠

y = b + m x

1. Its value is for a specific reaction, represented by a balanced equation.

2. Its units depend on the overall order of the reaction.

3. Its value does not change with concentrations of either reactants or products.

4. Its value does not change with time.

5. Its value refers to the reaction at a particular temperature and increases if the temperature increases.

6. Its value depends on whether a catalyst is present.

7. Its value must be determined experimentally for the reaction at appropriate conditions.

Details about the rate constant, k

Arrhenius Equation

FE13

( ) ( ) ( )2 5 2 22 4N O g NO g O g→ +

x y

A 1ln ln ak ETR⎛= − ⎞⎜ ⎟⎝ ⎠

y = b + m x

/A aE RTk e−=

ln(k

)A 1ln ln ak E

TR⎛= − ⎞⎜ ⎟⎝ ⎠

Temperature Dependence of Rate Constant

1/1 A aE RTk e−= 2/

2 A aE RTk e−=

( )

( )

2

1

8.3143,500 / / 31

2

0/

/ 8.3143,500 / 3 01 / 01.8

a

a

JJ mol x KE RT mol K

E RT JJ mol x Kmol K

kk

e ee

e

⎛ ⎞− ⎜ ⎟− ⋅⎝ ⎠

− ⎛ ⎞− ⎜ ⎟⋅⎝ ⎠

== =

43,500 / 1 1exp8.31 /( ) 310 300

1.8J molJ molK K K

⎛ ⎞− ⎧ ⎫− =⎨ ⎬⎜ ⎟⎩ ⎭⎝ ⎠

=

If Ea = 60 kJ/mol, then k2/k1 = 2.2

If T1 =250 and T2 = 260, then kJ/mol, then k2/k1 = 2.2

2

1

/2

11/

2

1 1expa

a

E RTa

E RT

Eee R T

kk T

−

−

⎛ ⎞⎧ ⎫−= −⎜ ⎟⎨ ⎬⎜ ⎟⎩ ⎭⎝ ⎠

=

Alternatively,

FE20

FE25

Exothermic orEndothermic?Endothermic

Another Example

If ∆Hrxn is 40 kJ/mol and Ea is 90 kJ/mol, what is the activation energy of the reverse reaction?

90-40=50 kJ/mol

=40 kJ/mol

=90 kJ/mol

FE14

TRANSITION STATE THEORY

(or activated complex)

No reaction collisions

Activated complexFE11

Determination of mechanism

Ene

rgy

Reger p 565

I and H2IWhat are intermediates?

Ea1 Ea2=Ea3

FE22

Equilibrium (dynamic)

Evaporation rate = condensation rate

If evaporation rate >> than escape rate, then system is still at equilibrium

Still at equilibriumEvaporation rate = Condensation rate

Not at equilibriumevaporation rate > condensation rate

Like forward reaction rate = reverse rate

Digression to help understand present method

rate = k3[H2I][I]

forward rate = k2 [I][H2] = k-2[H2I] = reverse rate

[H2I] = (k2/k-2) [I][H2]

forward rate = k1 [I2] = k-1[I]2 = reverse rate

[I]2 = (k1/k-1) [I2]

-

Rate = k3[H2I][I] = k3(k2/k-2) [I][H2][I] = k3(k2/k-2)[H2](k1/k-1)[I2] =k[H2][I2]

FE23