How Do Rate Constants Change with an Intermediate State in a Kinetic Model?

[phyalan]

Hi everyone,
I have a question on the rate constant of a kinetic model(like the one we use in describing chemical reaction). Suppose we have the following model between two states A and B:
A--(a)-->B
A<--(b)--B
where a, b are the rate constant for the transition. So the corresponding differential equation will be
$\frac{dA}{dt}=-aA+bB$
$\frac{dB}{dt}=-bB+aA$
If now I add an intermediate state C, so that if we just measure the amount of A and B, the two models gives basically the same description (i.e. the overall rate constant from A to C then to B is equivalent to a and the same for reverse direction)
A--(a1)-->C--(a2)-->B
A<--(b1)--C<--(b2)--B

What is the functional relationship between the rate constant a1,a2 and a and similarly for b1, b2 and b? I am confused on how to find out the functional form. Is it just simply
$\frac{1}{a}=\frac{1}{a1}+\frac{1}{a2}$?

 

[pasmith]

Hi everyone,
I have a question on the rate constant of a kinetic model(like the one we use in describing chemical reaction). Suppose we have the following model between two states A and B:
A--(a)-->B
A<--(b)--B
where a, b are the rate constant for the transition. So the corresponding differential equation will be
$\frac{dA}{dt}=-aA+bB$
$\frac{dB}{dt}=-bB+aA$
If now I add an intermediate state C, so that if we just measure the amount of A and B, the two models gives basically the same description (i.e. the overall rate constant from A to C then to B is equivalent to a and the same for reverse direction)
A--(a1)-->C--(a2)-->B
A<--(b1)--C<--(b2)--B

What is the functional relationship between the rate constant a1,a2 and a and similarly for b1, b2 and b? I am confused on how to find out the functional form. Is it just simply
$\frac{1}{a}=\frac{1}{a1}+\frac{1}{a2}$?

There is no functional relationship. What this system gives you is
$$\frac{dA}{dt} = -a_1A + b_1 C, \\ \frac{dB}{dt} = -b_2 B + a_2 C, \\ \frac{dC}{dt} = a_1 A + b_2 B - (a_2 + b_1) C.$$

There is no way to eliminate $C$ from this system in order to end up with a system involving only $A$ and $B$. The complication comes from the fact that you also have $A \to C \to A$ and $B \to C \to B$, so not all $A$ turns to $B$ before turning back to $A$ and vice-versa.

 

[phyalan]

There is no functional relationship. What this system gives you is
$$\frac{dA}{dt} = -a_1A + b_1 C, \\ \frac{dB}{dt} = -b_2 B + a_2 C, \\ \frac{dC}{dt} = a_1 A + b_2 B - (a_2 + b_1) C.$$

There is no way to eliminate $C$ from this system in order to end up with a system involving only $A$ and $B$. The complication comes from the fact that you also have $A \to C \to A$ and $B \to C \to B$, so not all $A$ turns to $B$ before turning back to $A$ and vice-versa.

Maybe I change my question, is there any 'effective' rate constant between A and B in the new system
$$\frac{dA}{dt} = -a_1A + b_1 C, \\ \frac{dB}{dt} = -b_2 B + a_2 C, \\ \frac{dC}{dt} = a_1 A + b_2 B - (a_2 + b_1) C.$$ ?

 

[Chestermiller]

Maybe I change my question, is there any 'effective' rate constant between A and B in the new system
$$\frac{dA}{dt} = -a_1A + b_1 C, \\ \frac{dB}{dt} = -b_2 B + a_2 C, \\ \frac{dC}{dt} = a_1 A + b_2 B - (a_2 + b_1) C.$$ ?

If C is short-lived so that it is at quasi steady state, then

$$C=\frac{a_1 A + b_2 B}{(a_2 + b_1)}$$

If we substitute this into the other differential equations, we get:

$$\frac{dA}{dt}=\frac{-a_1a_2A+b_1b_2B}{(a_2 + b_1)}=-a_3A+b_3B$$
$$\frac{dB}{dt}=\frac{a_1a_2A-b_1b_2B}{(a_2 + b_1)}=a_3A-b_3B$$

Chet

 

[blue_raver22]

I would respond with the following:

The rate constant is a fundamental parameter in kinetic models that describes the rate at which a reaction occurs. In the example provided, the rate constant a represents the rate of the forward reaction from state A to B, while b represents the rate of the reverse reaction from B to A. This differential equation can also be used to describe the overall rate of the reaction between A and B.

When an intermediate state C is added, the overall rate of the reaction can still be described by the same differential equation. However, the individual rate constants may change due to the presence of the intermediate state. The functional relationship between the original rate constants a and b, and the new rate constants a1, a2, b1, and b2, can be determined by considering the overall reaction rate and using the principles of mass balance and steady-state approximation. The relationship is not as simple as the one proposed (i.e. \frac{1}{a}=\frac{1}{a1}+\frac{1}{a2}), as it depends on the specific reaction mechanism and the concentrations of the species involved. Further analysis and experimental data may be necessary to determine the exact functional form.