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Chapter 13 Zhumdahl
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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