Lecture 5 - What Reaction Rates Can Tell Us About Reaction Mechanisms Flashcards
(14 cards)
Elementary reaction steps
For example the catalytic destruction of ozone by Cl
• A reaction mechanism can be represented as a series
of elementary reaction steps
(1) Cl + O3 —> ClO + O2
(2) ClO + O —> Cl + O2
Overall:
O3(g) + O(g) —> 2O2(g)
• The overall equation for the reaction just gives the
reaction stoichiometry (and the physical state of the
reactants and products).
• The reaction steps represent what really happens.
What is the molecularity of an elementary reaction step ?
Give an example of uni/bi/termolecular
The molecularity of an elementary reaction step is the
number of reactant molecules involved eg
A + B —> I
I —> P
Step 1 is bimolecular
Step 2 is unimolecular
O2 + O + N2 —> O3 + N2 is termolecular
Formulating rate laws
• From the rate laws for each elementary step we can formulate an overall rate law for the reaction
– this must be consistent with observation for the mechanism to be acceptable i.e. the reaction kinetics are a primary criterion for establishing a reaction mechanism. (If you want to understand how a reaction actually proceeds and you can’t see the intermediates in a useful way, then doing kinetics is the most helpful way)
– note that the mechanism must also be plausible on other criteria e.g. proposed intermediates have to be ‘sensible’ species.
Reactions approaching equilibrium
Considering reaction
A -> B where rate = ka[A]
B -> A where rate = kb[B]
Rate = d[A]/dt
At equilibrium d[A]/dt = 0
Rate of change of reactant (or products) is 0
Therefore d[A]/dt = -ka[A] + kb[B] = 0
The rates of the forward and back reaction are the same and ka[A] = kb[B] and so K = [B]/[A]
Diagram with ka and kb from slide before
ka
A ⇌ B
kb
Where at t=0, [A] = [A]0 and [B]=0
At any time [B] = [A]0 - [A]
Standard D.E. Solution
[A] = ( (kb + ka x e^-(ka + kb)t) / (ka + kb) )[A]0
At equilibrium where d[A]/dt = 0 and t = ∞
[A]eq = (kb x [A]0) / (ka + kb)
[B]eq = [A]0 - [A]eq = (ka x [A]0) / (ka + kb)
Sequencial reactions, consider:
k1 k2
A —> B —> C where each step is first order
1) Rate of decomposing of A (ie going from A to B: -d[A]/dt = k1[A] therefore [A] = [A]0 x e^-(k1t)
2) Rate of formation of B: d[B]/dt = k1[A] - k2[B]
3) Rate of formation of P: d[P]/dt = k2[B]
So 1) and 2) give (check slide 30 its sooo long to write out)
Steady state approx for:
A + B —> I (k1 is rate coefficient) Rate = k1[A][B]
I —> P (k2 is rate coefficient) Rate = k2[I]
When k1 and k2 are similar:
[A] decreases exponentially until reaches 0
[I] increases exp. (V steep at start), peaks, then depletes exp. As product forms
[P] increases exp at a slower rate than [I] to begin with
When k2 > k1:
Same thing except less [I] ie peak is at a much lower value of conc and earlier in time and [P] is much steeper
When k2»_space; k1
After a very small initial increase, [I] remains at a small and approx constant value ie d[I]/dt is roughly 0
Product is formed as soon as I forms and so very very steep curve
Steady-state approx for when k2»_space; k1
More info
• If the intermediate is highly reactive (i.e. k2 is very large)
– its concentration stays low
– it rapidly reaches a steady state, where its rate of
production balances its rate of removal i.e. assume:
d[I]/dt = 0
• This allows us to determine its concentration w.r.t. the
other reactants and so determine the overall rate law.
• This is the steady state approximation (also stationary
state principle or pseudo steady-state approximation).
• Possible reactive intermediates include free radicals,
carbocations, carbanions….
Look on slides for steady-state approx
Slides 37 - 42
Rate determining steps - how is it seen in energetics
Is it dependent on anything?
We have seen an example of a rate-determining step,
where the rate law is determined by the slowest step.
Can interpret in terms of energetics: the rate determining step has the highest activation energy.
But we have seen that a step may only be rate-determining under given conditions (ie changing conc)
Kinetic vs thermodynamic control
2 competing reactions when far away from eq:
A + B —> C k1
A + B —> D k2
Some reactions produce a variety of products, e.g. nitration of mono-substituted benzene gives p, o and m-substituted products.
Eg for the 2 parallel competing reactions:
-d[A]/dt = k1[A][B] + k2[A][B] = kobserved [A][B]
Where d[C]/dt = k1[A][B] and d[D]/dt =k2[A][B]
Hence [D]/[C] = k2/k1 and so k2/k1 is roughly e^((Ea1-Ea2) / RT) assuming the A’s are similar
The ratio of the different product concentrations depends
upon the rate coefficients, k1 and k2, and hence the activation -> kinetic control
energies, Ea1 and Ea2, for the different reaction pathways.
Kinetic vs thermodynamic control
2 competing reactions at eq:
k1
A + B ⇌ C k1
k-1
k2
A + B ⇌ D k2
k-2
K1 = k1/k-1 ≈ [C]eq / ([A]eq x [B]eq)
K2 = k2/k-2 ≈ [D]eq / ([A]eq x [B]eq)
Hence [D]eq/[C]eq = K2/K1 = e^((∆G1 - ∆G2) / RT)
The ratio of the different product concentrations depends upon the equilibrium constants, K1 and K2, and hence the standard free energies, ∆G1 and ∆G2, for the different reactions. -> Thermodynamic Control
Kinetic vs thermodynamic control
What conditions are equilibriums established?
What is the major product in kinetic and thermodynamic controlled reactions?
At long reaction times, or high T, equilibrium can be established and thermodynamic control will occur.
Kinetic control (irreversible conditions), the major product is from the fastest reaction, i.e. C, the kinetic product.
Thermodynamic control (equilibrium, reversible conditions), the major product is the more stable one, i.e. D, the thermodynamic product.
SUMMARY
• A reaction mechanism can be written as a set of
elementary reaction steps, for which rate laws can be
deduced from the molecularity of each step.
• An overall rate law for a reaction mechanism must be
consistent with the experimental rate law.
• For complex mechanisms it is often necessary to make
approximations in order to determine the rate law.
– steady state approximation.
• The overall reaction kinetics are frequently controlled
by one slow step: the rate-determining step.
• Different products can arise from thermodynamic vs.
kinetic control.
