Consecutive And Parallel Reactions Flashcards
Consider A —> B —> C with rate constants k1 and k2 for A to B and B to C respectively. What is the rate of consumption of A and rates of production of B and C?
Consumption of A: -d[A]/dt = k1[A]
Production of B: d[B]/dt = k1[A] - k2[B]
Production of C: d[C]/dt = k2[B]
Yes, as when t = 0, exponential term = 0 so [B]o = 0
When t = infinity, exponential term = 0 so [B](infinity = 0
So B does behave as an intermediate, not present at the start or end of the reaction.
How do you find an integrated rate equation for C when A —> B —> C is the consecutive reaction?
Sub [A]t and [B]t into [C] = [A]o - [A]t - [B]t
Factorise [A]o out
Introduce fraction equal to 1 with same denominator
Combine fractions and expand
Simplify
If intermediate [B] is particularly valuable, what equation gives the time at which [B] is at its maximum (tmax)?
What is the use of using a pump probe when measuring reaction kinetics?
Reaction is initiated with a short fs/ns laser pulse
UV/Vis spectroscopy is used to monitor how concentration of reaction intermediate changes
High power lasers allow for generation of high concentration of intermediates
Why?
A —slow—> B —fast—> C
This is because [B]t is very low at all time due to second reaction being much faster.
Therefore the rate of reaction only depends on k1, this is the rate determining step.
For the parallel reaction scheme A—k1–>B and A—k2–>C, what is the rate of removal of [A]?
-d[A]/dt = k1[A] + k2[A] = (k1+k2) [A]
We can write k = k1+k2 so that;
-d[A]/dt = k[A]
What is the integrated rate equation for the parallel reaction scheme A—k1–>B and A—k2–>C?
Since the rate, -d[A]/dt = k[A] where k = k1+k2, is identical to a first order reaction A—>P but with a new rate constant k, the equation is effectively the same.
What is the equation for the half-life for the parallel reaction scheme A—k1–>B and A—k2–>C?
Same as the half-life for a first order reaction, but subbing in k1+k2 for k
For the parallel reaction scheme A—k1–>B and A—k2–>C, what is the rate of production of [B]?
d[B]/dt = k1[A] = k1[A]t = k1[A]o exp(-kt)
Where k = k1+k2
For the parallel reaction scheme A—k1–>B and A—k2–>C, what is the rate of production of [C]?
d[C]/dt = k2[A] = k2[A]t = k2[A]o exp(-kt)
Where k = k1+k2
What is the integral of exp(-kt)?
For the parallel reaction scheme A—k1–>B and A—k2–>C, what is the integrated rate equation for [B]t?
Since rate of production of [B] is d[B]/dt = k1[A]t = k1[A]o exp(-kt)
Rearrange and integrate LHS and RHS separately
What is the fractional yield?
It is represented by phi, which is = (moles of product formed / moles of reactant used)
What is the fractional yield of B equal to?
For A—k1—>B and A—k2—>C respectively.
[B]t = [A]o (k1 / k1+k2) (1 - exp(-kt))
When t = infinity, exp(-kt) term approaches zero.
So, [B]t = [A]o (k1 / k1+k2)
And fractional yield is [B]t / [A]o
[B]t / [A]o = k1 / k1+k2
What is the fractional yield of B + the fractional yield of C?
For A—k1—>B and A—k2—>C respectively.
Must = 1, as matter must be conserved
What is the equilibrium constant for the reversible reaction of A to B? (Assume first order kinetics)
At eqm, concentrations do not change.
So kr[B]eq = kf[A]eq
Hence K = [B]eq / [A]eq = kf/kr
This is equal to exp(-∆rG˚ / RT) as the rate constants are related to each other by the Gibbs energy of the reaction and temperature
If A and B are not in eqm, how long does it take to reach equilibrium, assuming you have x more [A] than is present at equilibrium?
[A]t = [A]eq + x(t)
[B]t = [B]eq - x(t)
How can the dynamics of relaxation to equilibrium be measured?
Using half-life, t1/2
How do you find the new eqm constant K from the forward and reverse rate constants?
Knew = [B]new / [A]new = kf/kr
So kf = Knew kr
How is half-life used to find the forwards or reverse rate constants using the newly measured eqm constant K? (First order)
t1/2 = ln2 / k = ln2 / kf+kr
Since Kf = Knew Kr
t1/2 = ln2 / Knew Kr + Kr
Therefore Kr = ln2 / t1/2 (Knew + 1)
