# Equilibrium Constant Expression

#### Hunt_

One thing that always bothered me in Chemistry is how the equilibrium constant is written. It never made sense to me.

If I take a simple bimolecular reaction approaching equilibrium :

$$aA + bB \mathop{\rightleftharpoons}^{k_1}_{k_{-1}} cC + dD$$

From ART ,

$$r_1 = k_1 [A]^ \alpha ^ \beta$$

$$r_{-1} = k_{-1} [C]^ \gamma [D]^ \delta$$

Then if we consider the rate to be equal at equilibrium and the expression of the equilibrium constant ,

$$K = \frac{k_1}{k_{-1}} = \frac{[C]^ \gamma [D]^ \delta }{[A]^ \alpha ^ \beta } \neq \frac{[C]^c [D]^d }{[A]^a ^b}$$

The only way for both expressions to be equal is that the reaction is elementary , which doesnt hold for most chemical reactions.

So what does that mean ? There can be approximations here , molecularity is not equal to stoichiometry. Am I missing something here or are all the claculations I made in equilibrium chemistry just wrong ?

#### chemisttree

Homework Helper
Gold Member
One thing that always bothered me in Chemistry is how the equilibrium constant is written. It never made sense to me.

If I take a simple bimolecular reaction approaching equilibrium :

$$aA + bB \mathop{\rightleftharpoons}^{k_1}_{k_{-1}} cC + dD$$

From ART ,

$$r_1 = k_1 [A]^ \alpha ^ \beta$$

$$r_{-1} = k_{-1} [C]^ \gamma [D]^ \delta$$
Then if we consider the rate to be equal at equilibrium and the expression of the equilibrium constant ,

$$K = \frac{k_1}{k_{-1}} = \frac{[C]^ \gamma [D]^ \delta }{[A]^ \alpha ^ \beta } \neq \frac{[C]^c [D]^d }{[A]^a ^b}$$

Actually it is:
$$K = \frac{{k}_{-1}}{k_1}}$$

The only way for both expressions to be equal is that the reaction is elementary , which doesnt hold for most chemical reactions.

So what does that mean ? There can be approximations here , molecularity is not equal to stoichiometry. Am I missing something here or are all the claculations I made in equilibrium chemistry just wrong ?
They are approximations... some better than others.

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#### Gokul43201

Staff Emeritus
Gold Member
If I take a simple bimolecular reaction approaching equilibrium :

$$aA + bB \mathop{\rightleftharpoons}^{k_1}_{k_{-1}} cC + dD$$

From ART ,
What's "ART"?

$$r_1 = k_1 [A]^ \alpha ^ \beta$$

$$r_{-1} = k_{-1} [C]^ \gamma [D]^ \delta$$

Then if we consider the rate to be equal at equilibrium and the expression of the equilibrium constant ,

$$K = \frac{k_1}{k_{-1}}$$
This is not true in general. It is only true of elementary reactions. For a general reaction,

$$K = \frac{k_1}{k_{-1}}\frac{k_2}{k_{-2}}\frac{k_3~\cdot \cdot \cdot}{k_{-3}~ \cdot \cdot \cdot}$$

where 1,2,3,... are elementary steps of the overall reaction.

Also, though this is not of vital relevance, the equilibrium constant is defined exactly in terms of activities (or partial pressures) raised to stoichiometric powers. That is not an approximation.

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#### Hunt_

chemisttree,

Actually K should be k(fwd ) / k(bwd) so that the reactants appear in the numerator. What kind of approximations are we talking about here ? Anything that makes stoichiometric coefficients equal to partial orders ?

Gokul43201,

ART = absolute rate theorem

I guess its name changed , forgot to what , after it proved to be a failure in several cases and then substituted for SRT.

This is not true in general. It is only true of elementary reactions. For a general reaction,

$$K = \frac{k_1}{k_{-1}}\frac{k_2}{k_{-2}}\frac{k_3~\cdot \cdot \cdot}{k_{-3}~ \cdot \cdot \cdot}$$

where 1,2,3,... are elementary steps of the overall reaction.
But why do you want to assume it's a complex mechanism ? Not that it would make any difference , but I prefer to take a simple case here.

Also, though this is not of vital relevance, the equilibrium constant is defined exactly in terms of activities (or partial pressures) raised to stoichiometric powers. That is not an approximation.
You are right when you say that the thermodynamic equilibrium constant is in terms of activities or fugacities , but they are not raised to the stoichiometric powers or at least shouldn't be ( according to the definition of K ). That's what I dont get , they should be raised to the powers of their partial orders , which makes the value of K very different.

Thanks,

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#### Hunt_

Actually after giving this some thought , I noticed that Gokul43201 is right. Just assume the reaction consists of multiple elementary steps , K can be determined for each step from the stoichiometry . Then the overall K can be determined by multipying all the K's and its expression would be in terms of the stoichiometry indeed. It's something like this :

Reaction : $$aA \rightleftharpoons dD \ \ \ \ \ \ K = ??$$

Assume B , C are intermediates.

$$aA \rightleftharpoons bB \ \ \ \ \ \ K_1 = \frac{^b}{[A]^a}$$

$$bB \rightleftharpoons cC \ \ \ \ \ \ K_2 = \frac{[C]^c}{^b}$$

$$cC \rightleftharpoons dD \ \ \ \ \ \ K_3 = \frac{[D]^d}{[C]^c}$$

$$K = K_1 K_2 K_3 = \frac{[D]^d}{[A]^a}$$

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#### AbedeuS

Are you use the overall rate of reaction is the product of the rates combined? seems dubious imo... for a 3 step reaction:

$$A \rightarrow B \rightarrow C$$

$$\frac{d}{dt} = -\frac{d[A]}{dt}+\frac{d[C]}{dt} = k_{1}[A]^{\alpha}-k^{\beta}$$

$$\frac{d[C]}{dt} = k_{2}^{\beta}$$

But the k's are annoying to sort out depending on what bit of the equation your upto.

EDIT: Sorry, edited alot there since i messed up a calc hehe

EDIT: Another sorry, after reading closely I did realise your talking about deriving equilibrium constants from rate constants >,< doh!

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#### Hunt_

No the overall rate is not the product of all the rates of the elementary reactions. I do not even try to find the rate , although usually approximations r involved here. I just use the def of K ( since the log of K is related to a state function ) to prove that K of any equilibrium reaction can be written in terms of its stoichiometric coefficients. The pathway does not have to be 3 steps u know , nor does it have to be all reversible steps. I am satisfied for now . I think it makes sense to me.

#### AbedeuS

Ah sorry, I was confusing myself, I've personally not been taught the relationship between rate and equilibrium constants, If I had to hazard a guess, I guess it would be that since the rate forward==Rate backward than:

$$k_{1}[A]=k_{2}$$

so:

$$\frac{[A]}{} = \frac{k_{2}}{k_{1}}$$

Although I understand its um, represented differantly, when i say k im reffering to the rate backwards, and not the Rate of production of some third product like [C].

Is this the basic premise?

#### Hunt_

You're right as long as the reaction occuring is reversible.

If you wish to learn more , you can read this

#### Hunt_

$$aA \rightleftharpoons bB \ \ \ \ \ \ K_1 = \frac{^b}{[A]^a}$$

$$bB \rightleftharpoons cC \ \ \ \ \ \ K_2 = \frac{[C]^c}{^b}$$

$$cC \rightleftharpoons dD \ \ \ \ \ \ K_3 = \frac{[D]^d}{[C]^c}$$

----------------------------------------------------

$$aA \rightleftharpoons dD \ \ \ \ \ \ K = \frac{[D]^d}{[A]^a}$$

The only way I thought of it as a reasonable mechanism is that the forward rate constant of A is significantly less than the bwd rate constant of B. The fwd rate const of B should be greater than the bwd rate constant of C. Same thing holds between C and D. The forward rate constant of C should be much greater than the bwd rate const of D. This way , B and C would not accumulate and their concentrations should be small. I think it would be a good approximation because at equilibrium and [C] would tend towards zero. But this would only work as long as K of the aA <--> dD reaction is small. If K is neither large nor small but intermediate , then another argument should be used.

What do you think ?

If anyone has some knowledge or reference to the kinetics of complex reversible reactions, plz do post. I took pchem "chemical dynamics" last year but never went through such mechanisms ( althogh in organic chem / biochem , similar reactions are encountered . )

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