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
$$\frac{dA}{dt} = -a_1A + b_1 C, \\<br /> \frac{dB}{dt} = -b_2 B + a_2 C, \\<br /> \frac{dC}{dt} = a_1 A + b_2 B - (a_2 + b_1) C.$$

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
$$\frac{dA}{dt} = -a_1A + b_1 C, \\<br /> \frac{dB}{dt} = -b_2 B + a_2 C, \\<br /> \frac{dC}{dt} = a_1 A + b_2 B - (a_2 + b_1) C.$$

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
$$\frac{dA}{dt} = -a_1A + b_1 C, \\<br /> \frac{dB}{dt} = -b_2 B + a_2 C, \\<br /> \frac{dC}{dt} = a_1 A + b_2 B - (a_2 + b_1) C.$$ ?

 

[Chestermiller]

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

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