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CHEM260 Kinetics > Pre-equilibrium And Steady State Approximations > Flashcards

Pre-equilibrium And Steady State Approximations Flashcards

1
Q

What is the rate of formation / removal of A, I and P for the following reaction?

A

-d[A]/dt = k-1[I] - k1[A]

d[I]/dt = k1[A] - k2[I] - k-1[I]

d[P]/dt = k2[I] = (k1k2/k-1) [A]

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2
Q

What would be the observed overall rate constant for this reaction?

A

First order

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3
Q

What’re the forward and reverse rate laws for this reaction?

A

Forward = k1’[A][B]

Reverse = k-1[I]

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4
Q

What is the rate for production of P in this reaction?

A

d[P]/dt = k2[I] = k2k1’/k-1 [A][B]

This is because at eqm; [I]k-1 = k1’[A][B]

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5
Q

What is the observed rate of this reaction?

A

Kobs = k1’k2/k-1 = second order observed, as k1’ is second order

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6
Q

What is the Arrhenius equation?

A

k = Aexp(-Ea/RT)

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7
Q

When is a positive Ea observed for this reaction?

A

Positive when the two forwards reaction’s Ea > the Ea of the reverse reaction.

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8
Q

When is the observed Ea negative for this reaction?

A

Negative Ea is observed when the reverse Ea > the combined Ea of the two forwards reactions.

This of course does not mean there is negative energy, but that the reaction is more complex than originally thought. This is all because rate constants are related to Gibbs energy change - Arrhenius.

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9
Q

What does diffusion limited mean?

A

Solvent acts as a barrier, limiting the number of collisions possible

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10
Q

What is a solvent cage?

A

When reactants are in close contact in solution it can be harder for molecules to separate - they are held together.

This results in a large number of collisions with reactive partner and solvent - picking up energy - increasing the probability of a reaction.

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11
Q

Is the assumption that k-1&raquo_space; k2 always justified in a dilute solution for the reaction?

A

NO!

Solvent can trap molecules together until they react

Reverse reaction is not fast compared to making product

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12
Q

What does the steady state approximation assume?

A

It assumes that after the induction period, the net rate of concentration for all intermediates is very small

This can result in a better approximation than pre-equilibrium

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13
Q

When is the steady-state approximation suitable / unsuitable?

A

Good for unstable intermediates

To apply the steady state approximation, we need to identify an intermediate that is present at low concentrations
- enzyme substrate complex ES
- activated gas-phase species C*
- radical species generated in gas phase chain reactions, .H and .Cl

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14
Q

In the steady state approximation, how can you find the concentration of the intermediate complex [AB] in the reaction shown?

A

Well at eqm, d[AB]/dt ≈ 0

The rate of production of [AB] = d[AB]/dt = k1’[A][B] - k-1[AB] - k2[AB]

This can be simplified to:
d[AB]/dt = k1’[A][B] - (k-1 + k2)[AB]

Setting this to 0 gives:

[AB] = (k1’ / k-1 + k2) [A][B]

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15
Q

How do you find the rate of production of P in the reaction shown, in terms of [A][B] and all rate constants?

A

Well, d[P]/dt = k2[AB]

[AB] = (k1’ / k-1 + k2) [A][B]

So d[P]/dt = [AB] = (k2k1’ / k-1 + k2) [A][B]

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16
Q

In the reaction shown, when is it activation controlled? What does this mean?

A

When k-1&raquo_space; k2, it is activation controlled

Here the limiting step is the actual reaction

This gives the same result as pre-equilibrium approximation

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17
Q

What does activation control mean?

A

Activation control mean the limiting step is the actual reaction, where the final step is very slow

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18
Q

In the reaction shown, when is it diffusion controlled? What does this mean?

A

When k2»k-1, the reaction is diffusion controlled

This is because the final step is fast, so the limiting step is how fast can the reactants collide and form a complex

19
Q

What is the probability for a reaction to occur before the encounter pair leave a solvent cage?

A

The probability is 1

20
Q

What does Kd mean and what does it depend on? When is it applied?

A

Kd is the bimolecular diffusion constant, it depends on the size of the molecules and viscosity of the solvent

It is used in diffusion controlled kinetics

21
Q

What does n mean in this expression?

A

n = viscosity of the solvent

22
Q

For enzymes, if [S]o is high with a low conc of enzymes, [E], what is the rate of product formation proportional to?

A

Rate of product formation is proportional to [E]o

Adding more enzymes increases the rate

23
Q

For a given [E]o and a relatively low conc of [S]o, what is the rate of product formation proportional to?

A

Rate of product formation is proportional to [S]o

Adding more substrate increases the rate

24
Q

For a given [E]o with high [S]o, what is the rate of product formation proportional to?

A

Nothing, product formation becomes independent of both [E] and [S]. Adding more enzymes or substrate has no effect on the rate of

It is said to be at its maximum velocity, Vmax

25
Q

Is pre-equilibrium approximation appropriate for the Michaelis-Menten mechanism?

A

No, if k-1»k2 then it would be a poor enzyme and hence approximation

This is because the intermediate would then not be present in very low amounts.

26
Q

Is the steady state approximation appropriate for the Michaelis-Menten mechanism?

A

Yes, assuming k-1«k2 as the intermediate will be in very low quantities

27
Q

What is the rate of [ES] formation, and hence the [ES] concentration in the reaction?

A

d[ES]/dt = 0 = k1[E][S] - (k-1 + k2)[ES]

Hence [ES] = (k1/ k-1 + k2)[E][S]

28
Q

What does [E]o equal in the reaction?

A

[E]o = [E] + [ES]

As this is the total number of enzymes present in solution, bound or un-bound

29
Q

What is the Michaelis constant equal to?

A

Km = (k-1 + k2) / k1

30
Q

How do you find [E] in the Michaelis-Menten mechanism?

A

Since [ES] = (k1/ k-1 + k2)[E][S]

We can rearrange to find [E]:
[E] = [ES]/[S] (k-1 + k2)/k1

31
Q

How can [E] = [ES]/[S] (k-1 + k2)/k1 be rearranged using the Michaelis constant Km? And how can this be adapted using [E]o?

A

[E] = Km [ES]/[S]

Since [E]o = [E] + [ES]

[E]o = Km [ES]/[S] + [ES]

This can be written as [E]o = (Km/[S] + 1)[ES]

32
Q

What is the rate of product formation in the reaction? How does the Michaelis constant Km relate to this?

A

d[P]/dt = k2[ES]

From [E]o = [ES]/[S] (k-1 + k2)/k1
[ES] = [E]o / (Km/[S] + 1)

Hence d[P]/dt = k2[ES] = k2[E]o / (Km/[S] + 1)

33
Q

What is d[P]/dt equal to when [S]o &laquo_space;Km?

A

Km/[S] +1 —> Km/[S]

Hence d[P]/dt ≈ k2/Km [E]o[S]

Therefore when more substrate is added, the rate increases

34
Q

What is d[P]/dt equal to when it becomes independent of [S]? (Adding more substrate has no further effect on the rate of production)

A

d[P]/dt ≈ k2[E]o = Vmax

35
Q

What is the equation for the Lineweaver-Burke plot?

A
36
Q

What is Vmax equal to?

A

Vmax = k2[E]o

37
Q

What is the slope and y intercept in the Lineweaver Burke plot?

A

Slope = Km / Vmax
Where Vmax = k2[E]o

Y intercept = 1/Vmax

38
Q

How do you find k2 from the y intercept of the Lineweaver Burke plot

A

Since y intercept = 1/Vmax

And Vmax = k2[E]o

If you know the conc of enzymes in solution, then you can find k2

39
Q

What does the Lindemann-Hinshelwood mechanism describe?

A

That unimolecular gas molecules collide which results in one of the molecules becoming excited and undergoing a transformation.

At high pressure reactions are 1st order, at low pressure reactions are 2nd order

40
Q

What is the rate of formation of C* for cyclopropane to propane as seen in the reaction shown, and hence the rate of formation of the product?

A

d[C]/dt = k1[C]^2 - k-1[C][C] - k2[C*]

This can be set to 0 to give [C*] (factorise)

[C*] = k1[C]^2 / (k-1[C] + k2)

Hence d[P]/dt = k2[C*] = k2k1[C]^2 / k-1[C] + k2

41
Q

How does the Lindenmann-Hinshelwood change with high pressure?

A

At high pressure [C] is large
Assume k-1[C]&raquo_space; k2

Therefore rate = (k2k1 / k-1) [C]

Hence first order

42
Q

How does the Lindenmann-Hinshelwood change with low pressure?

A

[C] is small
Assume k-1[C] &laquo_space;k2

Therefore rate = k1[C]^2

Hence second order

43
Q

What is the rate determining step in the Lindenmann-Hinshelwood mechanism when at high pressure?

A

C* —> P competes with deactivation

Deactivation occurs on collision with another cyclopropane molecule

Hence RDS is unimolecular decay - k-1

44
Q

What is the rate determining step in the Lindenmann-Hinshelwood mechanism when at low pressure?

A

Collisional activation, and hence deactivation, becomes slow as it is much less likely to collide with another molecule at low pressure.

RDS is therefore the collision rate - k1

CHEM260 Kinetics (5 decks)

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