Equilibrium and Rates

[Higgs]

Consider an equilibrium

 

[ce]

A <=> B

[/ce]

 

Assuming both the forward and reverse reactions are elementary, the rates of reaction can be modeled by the differential equations:

 

[ce]

\frac{d}{dt} = p[A] - q

[/ce]

 

[ce]

\frac{d[A]}{dt} = q - p[A]

[/ce]

 

Where [ce]q[/ce] is the rate constant for the forward reaction and [ce]p[/ce] is the rate constant for the reverse reaction. The general solution of this system is:

 

[ce]

= \frac{p_0-q[A]_0}{p+q}e^{-pt-qt}+q\left(\frac{[A]_0+_0}{p+q}\right)

[/ce]

 

[ce]

[A] = \frac{q[A]_0-p_0}{p+q}e^{-pt-qt}+p\left(\frac{[A]_0+_0}{p+q}\right)

[/ce]

 

By taking the limit as [ce]t[/ce] goes to infinity, it can be shown that

 

[ce]

\frac{q}{p} = K

[/ce]

 

where [ce]K[/ce] is the equilibrium constant. [ce]K[/ce] can be derived from the Gibbs energy of reaction by the equation

 

[ce]

\Delta G = -RT \ln{K}

[/ce]

 

Are there any errors in this line of reasoning? If there aren't, is it possible to predict values for [ce]p[/ce] and [ce]q[/ce] theoretically (i.e. without actually performing an experiment)?

[mississippichem]

Your differential solutions are correct. So, you essentially derived the definition of the equilibrium constant.

You seem to have a good handle on things, so I might be stating the obvious, but remember:

 

[math] \frac{q}{p} = K = \frac{[A]^{a} ^{b}}{[C]^{c} [D]^{d}} [/math]

 

...So if all the equilibrium concentrations are known along with their stoichiometric coefficients, then yes. The bad news is that [math] K_{c} [/math] is determined experimentally. Looking at the situation purely kinetically:

 

[math] \frac{d}{dt} = \frac{-d[A]}{dt} = k^n [/math]

 

for [ce] A <=> B [/ce]

 

as t approaches infinity or when [ce] Q=K [/ce]

So if you want to try and get around the equilibrium experiment, you still have the individual reaction orders to deal with. In the math above, the reaction may have a molecularity of one with respect to B, but B may display psuedo-first order kinetics, or some other kinetic anomaly. You can though, determine the initial rates by evoking the steady state assumption:

 

[math] \frac{d}{dt} = 0 [/math]

 

where is the concentration of the reaction intermediate. This is assumed to remain constant once the reaction gets up and running. I can't seem to remember the derivation to arrive at initial rate though, sorry. I do remember that it gets quite hairy but involves no differential solution searching.

Just remember, whatever you do, you are always limited by the law of mas action:

 

[math](p-\xi)^{a}(q-\xi)^{b}=\frac{k'}{k} (p'+\xi)^{a'}(q'+\xi)^{b'}...[/math]

 

But don't try and derive any kinetics from the law of mass action, as it is considered obsolete. It uses concentrations instead of the modern use of activities (concentration adapted for the dielectric constant of the solvent, along with ionic radii and other wonderful variables).

 

*that Gibb's equation doesn't account for temperature dependence of entropy, and is only true at standard conditions

 

**I typed this out fairly quickly, as I was pressed for time, so read carefully and check my work. You seem to know what you are doing.

Edited October 26, 2010 by mississippichem

[Bioquímico]

I think this is the right site to place my question.

 

I am adapting the method performed by El Mernissi and Doucet on “Quantitation of [3H]ouabain binding and turnover of Na-K-ATPase along the rabbit nephron” in order to do ouabaing binding on tubule suspensions.

 

There is one useful equation that I have no idea where it came from.

 

 

 

“Assuming that the concentration of free ligand remains constant throughout incubation (…) it is possible to determine k1 and k-1 using the standard equation:

 

 

 

Log ( [EL]_eq / [EL]_eq - [EL] ) = (k₁ [L] + k_-1) t

 

 

 

Where [EL]_eq and [EL] represent the concentration of the enzyme-ligand complex at equilibrium and at time t respectively, and [L] is the concentration of free ligand.”

 

 

 

 

 

 

 

I am wondering if anyone could expand this. Thanks in advance.

 

 

 

 

PS: how did you write those equations, I mean how to write in chemical format. Thanks.