[NYM]

Does anybody know a relation(i.e. formula etc.) between the reaction rate and equilibirum?

Thanks.

[Yggdrasil]

Consider the reaction:

A + B <--> C + D

With forward rate constant k₁ and backward rate constant k_-1.  Write out the expression for the forward reaction rate and the backward reaction rate.  Then recall that at equilibrium the forward reaction rate is equal to the backward reaction rate.  From there you can derive an expression of the equilibrium constant (remember K = [D][C]/[A]) in terms of the forward and backward rate constants.

[mdlhvn]

Consider the reaction:

A + B <--> C + D

With forward rate constant k₁ and backward rate constant k_-1.  Write out the expression for the forward reaction rate and the backward reaction rate.  Then recall that at equilibrium the forward reaction rate is equal to the backward reaction rate.  From there you can derive an expression of the equilibrium constant (remember K = [D][C]/[A]) in terms of the forward and backward rate constants.

------In a easy way, the equilibrium constant of this reaction K_eq=k_-1/k₁

[NYM]

Then recall that at equilibrium the forward reaction rate is equal to the backward reaction rate.

Hmm, not really. Could you elaborate a bit?
I don't see why v1 = v2, where v1 = forward reaction rate and v2 = backward reaction rate.

[Yggdrasil]

Well, at equilibrium the net change of reactants and products is zero.  Since the forward rate of reaction gives the rate at which reactants are transformed to products and the backward rate gives the rate at which products are transformed to reactants, if these two rates are not equal then the net rate of change of products and reactants will be nonzero.

As an example, consider a situation where I hand you 2 boxes per minute.  If you hand me 2 boxes per minute, then the number of boxes we have stays the same and our system is in equilibrium.  However, if you hand me 3 boxes per minute, boxes will accumulate on my side.  Since the number of boxes on my side increases with time, this system is not at equilibrium.

[Borek]

if these two rates are not equal then the net rate of change of products and reactants will be nonzero

and the concentrations/amounts of substances will be changing - thus system will be not a equilibrium.

[NYM]

Thanks, everybody!

[NYM]

Wait a second -

under "Equilibrium reactions or opposed reactions" at http://en.wikipedia.org/wiki/Rate_law

Why isn't k1 = k2?
If v1 = v2, then k1 = k2, right?

Please explain

[Borek]

Why isn't k1 = k2?
If v1 = v2, then k1 = k2, right?

No. For the reaction

A + B <--> C + D

v1 = k1[A]

and

v2 = k2[C][D]

v1 & v2 depend on the concentrations, so your conclusion that if v1=v2 then k1=k2 is faulty.

[Yggdrasil]

v1 = k1[A]

and

v2 = k2[C][D]

And, at equilibrium, v1 = v2, so you can write:

k1[A] = k2[C][D]

Rearranging you can get

[C][D]/[A] = k1/k2

Since [C][D]/[A] is defined to be the equilibrium constant for the reaction that Borek poste, K_eq, we have derived the expression that:

K_eq = k1/k2

where k1 is the forward rate constant and k2 is the backward rate constant

[thienanh]

v1 = k1[A]

and

v2 = k2[C][D]

And, at equilibrium, v1 = v2, so you can write:

k1[A] = k2[C][D]

Rearranging you can get

[C][D]/[A] = k1/k2

Since [C][D]/[A] is defined to be the equilibrium constant for the reaction that Borek poste, K_eq, we have derived the expression that:

K_eq = k1/k2

where k1 is the forward rate constant and k2 is the backward rate constant

Ok, this is only true with a simple reaction. There are some reactions don’t obey this. What’s about the complicated reactions, how can we write the rate reaction if we don’t know about its mechanism?

[NYM]

All right, that's all for now, thanks!

[Yggdrasil]

Ok, this is only true with a simple reaction. There are some reactions don’t obey this. What’s about the complicated reactions, how can we write the rate reaction if we don’t know about its mechanism?

You are correct, this expression does work only for elementary reactions.  For reactions where we don't know the mechanism, I don't think there is a way to relate the equilibrium constant to the rate of reaction (especially since we wouldn't be able to get any true rate constants if we don't know the actual mechanism!).

[zeifer_roth]

errrrrr... sorrry i don't really know the exact formula.. but 1 know a general concept bout it.. "the faster the reacrion rate the lower is the equilibrium conversion"...