Reversible monomolecular reaction with two reverse rates

Question

I think this is a simple problem but my kinetics are terrible and I was hoping for some assistance following up on the work of a previous student. I'm open to any form of assistance and I hope this question doesn't come off as too big of an ask.

I'm working on a solution to determining a rate of catalysis ($k_2$) in a reversible reaction:

$$\ce{A <=>[k1][k1 + k2] B}$$

where $\ce{A}$ and $\ce{B}$ are expressed as molfractions because it's known that $\ce{A + B = 1}$. Experimentally, it's been determined that $k_1$ is the same in both directions so I've excluded the normal +k and -k notations.

In a simpler experiment the rate of $k_1$ can be determined by the integrated rate expression derived from the reversible reaction:

$$\ce{A <=>[k1][k1] B}$$

$$ \begin{aligned} \frac{dB}{dt} &= k_1A -k_1B \\ &= k_1 (1-B) -k_1B \\ &= -k_1(2B -1) \end{aligned} $$

which leads to the integrated rate expression:

$$\int\frac{dB}{(2B -1)} = \int{-k_1 dt}$$

$$2B-1 = \exp(-2k_1 t)$$

This is a useful form of the equation because 1. it's in terms of one species only, and 2. the measurement we take originates from the square of the population difference between the species A and B and given that $A + B = 1$:

$$(A-B)^2 = (2B-1)^2$$

Experimentally, we just collect a signal over time and the decay of the signal is fit to the square of that integrated rate equation which is $(\exp(-2k_1 t))^2 = \exp(-4k_1 t)$. So in the end:

$$\text{Observed Signal} = (2B-1)^2 = \exp(-2k_1 t)$$

I am now studying the system described in the first equation where I want to introduce a species that catalyzes the back reaction, and given that I can determine the rate without the catalyst ($k_1$), I am pretty sure I can find the rate $k_2$ if I can figure out how to make this new integrated rate expression arranged in terms of $2B-1$. Essentially, can someone help me find an integrated rate expression which solves for $(2B-1)$ in terms of both $k_1$ and $k_2$ so that I can run one experiment without the catalyzed back reaction to determine $k_1$ and then run a second where I can then plug in $k_1$ as a parameter to determine $k_2$. I've been working on this for days and I feel like I'm just missing something fundamental in my understanding.

I felt this question might be relevant to others, as it seems like a good question on basic kinetics. If this is not the case, it's due to lack of understanding on my part and I do apologize. Even if there is any resources someone could point me to I would really appreciate it, as my searches have yet to turn up anything useful.

Answer 1

Just to repeat what porphyrin said in the comments:

Simple first-order reversible reactions tend towards equilibrium with a characteristic time that is the sum of the forward and backward rate constants.

So for an equilibrium $\ce{A<=>[k_f][k_r]B}$, then the system will tend toward equilibrium with a rate $r \propto e^{-(k_r + k_f)t}$.

It doesn't matter if in fact you can mechanistically "decompose" say the back reaction into two different (first-order) pathways that have distinct rate constants. If $k_r=k_{r1} + k_{r2}$, then $r \propto e^{-(k_{r1} + k_{r2} + k_f)t}$, but that doesn't help you, at least not if your measurements are just the overall kinetics of the system. Measuring kinetics will just give you the value of the net rate constant, whether you call it $(k_r + k_f)$ or of $k_{r1} + k_{r2} + k_f$ or just $k_{net}$, with no way forward to get at the individual steps.

Explicit assumptions

You will need to introduce additional assumptions to make progress. You've implicitly assumed, I think, that:

1. there exists a condition where $k_{r2}$ is zero, and in this condition, $k_r=k_{r1}$. Let's call this condition $\aleph$.
2. also that in this condition $\aleph$, the forward and reverse rate constants are the same, which is equivalent to assuming the equilibrium constant (for condition $\aleph$ ) is 1. This means that $k_f=k_r$ for condition $\aleph$.
3. that there is another condition $\beth$, where $k_{r2}$ is not zero, but that $k_{r1}$ is still equal to $k_f$. Since the total reverse rate constant $k_r$ is now higher than in condition $\aleph$, this assumption means that that the equilibrium constant in condition $\beth$ is different (lower) than in condition $\aleph$.

What you'd need to measure

In order to determine a value for $k_{r2}$ from experimental data, you would need to measure the kinetics of the system under both condition $\aleph$ and condition $\beth$.

• From measuring condition $\aleph$, you'd get the value of $k_r+k_f$, which by assumption 2. is equal to $2k_f$.
• From measuring condition $\beth$, you'd get two pieces of information:
  • First, you'd get the value of $k_{r1} + k_{r2} + k_f$ by fitting your timecourse data.
  • Second, you should be able to measure the new equilibrium constant by observing the behavior of the system at very long times. In condition $\aleph$ the system must approach $A=B$ at long times, because of assumption 2. But now in condition $\beth$, it should trend toward a different ratio between A and B, and this ratio should be $\frac{k_f}{k_{r1}+k_{r2}}$.

You should be able to estimate $k_{r1}$ and $k_{r2}$, subject to your assumptions, with either measurement on condition $\beth$.

Note that you don't even need kinetic information from condition $\beth$ to measure $k_{r2}$, you just need to measure the new equilibrium condition. Of course, if you can make both measurements on condition $\beth$, you should get the same value (within experimental uncertainty) for $k_{r2}$. If you don't, then one or more of the assumptions I've listed is violated.