Rate constant and Kinetic model

In summary: C is short-lived so that it is at quasi steady state, then\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.
  • #1
phyalan
22
0
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
[itex]\frac{dA}{dt}=-aA+bB[/itex]
[itex]\frac{dB}{dt}=-bB+aA[/itex]
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
[itex]\frac{1}{a}=\frac{1}{a1}+\frac{1}{a2}[/itex]?
 
Physics news on Phys.org
  • #2
phyalan said:
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
[itex]\frac{dA}{dt}=-aA+bB[/itex]
[itex]\frac{dB}{dt}=-bB+aA[/itex]
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
[itex]\frac{1}{a}=\frac{1}{a1}+\frac{1}{a2}[/itex]?

There is no functional relationship. What this system gives you is
[tex]
\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.
[/tex]

There is no way to eliminate [itex]C[/itex] from this system in order to end up with a system involving only [itex]A[/itex] and [itex]B[/itex]. The complication comes from the fact that you also have [itex]A \to C \to A[/itex] and [itex]B \to C \to B[/itex], so not all [itex]A[/itex] turns to [itex]B[/itex] before turning back to [itex]A[/itex] and vice-versa.
 
  • #3
pasmith said:
There is no functional relationship. What this system gives you is
[tex]
\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.
[/tex]

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

Maybe I change my question, is there any 'effective' rate constant between A and B in the new system
[tex]
\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.
[/tex] ?
 
  • #4
phyalan said:
Maybe I change my question, is there any 'effective' rate constant between A and B in the new system
[tex]
\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.
[/tex] ?
If C is short-lived so that it is at quasi steady state, then

[tex]C=\frac{a_1 A + b_2 B}{(a_2 + b_1)}[/tex]

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

[tex]\frac{dA}{dt}=\frac{-a_1a_2A+b_1b_2B}{(a_2 + b_1)}=-a_3A+b_3B[/tex]
[tex]\frac{dB}{dt}=\frac{a_1a_2A-b_1b_2B}{(a_2 + b_1)}=a_3A-b_3B[/tex]

Chet
 
  • Like
Likes 1 person
  • #5


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.
 

Related to Rate constant and Kinetic model

What is a rate constant?

A rate constant is a proportionality constant that relates the rate of a chemical reaction to the concentration of reactants. It is represented by the symbol k and is specific to each reaction.

How is the rate constant determined?

The rate constant can be determined experimentally by measuring the initial rates of reactions at different concentrations of reactants and plotting them on a graph. The slope of the graph is equal to the rate constant.

What factors affect the rate constant?

The rate constant is affected by temperature, concentration of reactants, and the presence of a catalyst. Generally, an increase in temperature and concentration results in an increase in the rate constant, while the presence of a catalyst can decrease the rate constant.

What is a kinetic model?

A kinetic model is a representation of the reaction mechanism that explains how reactants are converted to products. It includes the steps and intermediates involved in the reaction, as well as the rate law and rate constant.

How does the kinetic model relate to the rate constant?

The kinetic model provides a theoretical framework for understanding the relationship between the rate constant and the concentration of reactants. It allows us to predict how changes in concentration, temperature, and other factors will affect the rate constant and the overall rate of the reaction.

Similar threads

  • Quantum Physics
Replies
32
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Differential Equations
Replies
1
Views
1K
  • Quantum Interpretations and Foundations
2
Replies
38
Views
4K
  • Thermodynamics
Replies
1
Views
3K
  • Linear and Abstract Algebra
Replies
1
Views
956
  • Differential Equations
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Replies
2
Views
1K
  • Special and General Relativity
3
Replies
83
Views
4K
Back
Top