Different Types ofDifferent Types ofComposite Reactions
Simultaneous:A Y and A Z
Opposing:A B ZA Y and A Z A B Z
Consecutive: Consecutive with opposition:A X Y
ppA X Y
In order for the system to be at complete equilibrium, thetime derivative of all species concentrations must be zero!
Kinetics for a Simple Consecutive Reaction
1 2 and Xk kA X Z 1
0k tA A e with
2 1 (2)d X
k X k Adt
1
becomes
kt
dt
d Xk X k A e
1
2 1 0
2 1 0 a "standard form" o.d.e. with solutionkt
k X k A edt
d Xk X k A e
dt
1 2
2 1 0
20
2 1
kt k t
dtkX e e A
k k
and
2 11 2
0 02 1
1k t ktk e k eZ A A X Ak k
Quasi Steady State Approximation
1 2 and Xk kA X Z Consider the differential rate equations for [A], [X] and [Z].
(1)d A
k Let [A] = [A]
1 (1)
0 (2)
k Adt
d Xk X k A
Let [A]t=0 = [A]0and [X]t=0 = [Z]t=0 = 0
ktA A 11 1
022
2 1 0
or (
(2)
2.5)ktk kX
k X k Adt
A A ek k
1
0ktA A e
Make QSS approx22
2 (3)d Z
k Xdt
k k
11 0
0
tkt
k
Z k A e dt
For (2) and (2.5) to be compatible , we must have (k1/k2) << 1 1
01 kte A
Compare Exact and Steady State S l ti f k 20 kSolutions for k2 = 20 k1
k1t
Sample Problem (10.3)Given reaction with stoichiometry:
A + B Y + Zh f ll h hThe reaction follows the mechanism:
A Xk1
k
X + B Y + Z
k–1
k2
Obtain an expression for the rate using the steady state treatmenta. What is the rate determining step if the second reaction rate is
slow relative to the initial equilibrium rates?qb. What is the rate determining step if the second reaction rate is
rapid relative to the initial equilibrium rates?
Rate Determining Steps
Is the slow step ALWAYS rate determining?
ConcepTest 1p
Reaction: 2 O3 3 O2Reaction: 2 O3 3 O2
Mechanism: O3 O + O2 (fast)k1k–1
O + O3 2 O2 (slow)What is the rate equation for O3 loss?
–1k2
A. Rate = k [O3]2
at s t e ate equat o o O3 oss
B. Rate = k [O3] [O2]C. Rate = k [O3]
2 [O2]–1
3
B. Rate = k [O3 ]–1 [O2]
SolutionReaction: 2 O3 3 O2
Mechanism: O3 O + O2 (fast)
O + O 2 O (slow)
k1k–1
k2O + O3 2 O2 (slow)What is the rate equation for O3 loss?
2
1 3 1 2
1 3
1 equilibrium :Rxn k O k Ok O
O
O
1 2
2
with a slow second step, the equilibrium is maintainedk O
O
2
3 31 32 3
1 22 3
2
Rated O O
kk O
O k O kdt
Ok O O
Exam 2Exam Subject matter:
Gibbs EnergyGibbs EnergyStandard ΔG for reactions Pressure dependence of ΔG for ideal gases, liquids, solidsChemical potential
EquilibriumEquilibrium constants, ideal gas equilibriaShifts of equilibrium at constant TLe Chatelier PrincipleLe Chatelier PrincipleRelationship between standard Gibbs energy change and
the equilibrium constantKinetics
R t t l h i th d f i iti l tRates, rate laws, mechanisms, method of initial ratesArrhenius rate law – physical interpretationReaction paths, potential energy surfacesMolecular basis of reactionsMolecular bas s of react onsElementary reactions, composite reactionsFast and slow rate approximations; steady state
rate determining steps; catalyzed reactions
Exam 2Exam
~ 6 problems (weights given) – budget your timep g g g yClosed book Don’t memorize formulas/constantsYou will be given things you needYou will be given things you need
Study lecture material, examples, and homework!Exam will not be heavily numeric, but will emphasize
tsconceptsProbably more numerical than last time (bring calculators)If a problem seems lengthy, do another problem & come p g y pback later Understanding homework will be useful
Potential Energy SurfacesE gy fThe Born Oppenheimer approximation:Generally a very accurate Generally a very accurate approximation, and one of the most important concepts in all of physical science.Without it, there would be no chemistry as we know it.B-O approximation allows the construction of potential energy construction of potential energy surfaces for molecules and reactions.Still too many nuclear degrees of freedom!S l t l i t t Select only important ones
NO2 + CO NO + CO2
Expected rate = k [NO2] [CO2]Observed rate = k [NO2]2 Composite RxDeduce a possible and reasonable mechanism
NO2 + NO2 NO3 + NO slow
NO + CO NO + CO f t
2NO2 + CO NO + CO2 + NO2Overall,
NO3 + CO NO2 + CO2 fast
2 2 2
NO3 is an intermediate product that is not seen eitheras a reactant or as a product.
Let’s look at the reaction profile (ala Zeb)
as a reactant or as a product.
NO2 + CO NO + CO2
In this two-step reaction, there2 2 3NO NO NO NO
gy
pare two barriers, one for eachelementary step. The well between the two t iti t t h ld ti
3 2 2NO CO NO CO
tial
Ene
rg transition states holds a reactiveintermediate.A reactive intermediate is acompound formed transiently but
Pote
nt compound formed transiently, butabsent in the overall reaction.
Reaction Coordinate
We need to do better than just using a 1-D energy sliceWe need to do better than just using a 1 D energy slice,namely a multidimensional potential energy surfacepotential energy surface!
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