Dimerisation Equilibra

This scheme is appropriate for species which are typically monomers, but under the conditions of the reaction, react together for form a dimer. For the case where one begins with a dimer, which under the conditions of the reaction de-aggregrates to form monomers, see this page.

Scheme:

`2Astackrel(K_D) harrA_2`

Equilibrium Expression:

`K_D="dimer"/"monomer"^2=A_2/A^2`

Mass Balance:

`A_0=A+2A_2`

Differential Equations:

`A'(t)=-k_1A^2+k_(-1)A_2`

`A_2'(t)=-k_(-1)A_2+k_1A^2`

Solution:

At dynamic equilibrium, the net change in concentrations is nil. We can derive the equilibrium constant, therefore, in terms of the individual rates, ie.,

`A'(t)=0=-k_1A^2+k_(-1)A_2`

Hence,

`k_1A^2=k_(-1)A_2`

and

`A_2/A^2=k_1/k_(-1)`

This illustrates an important relationship between the rates (at equilibrium) and the equilibrium constant,

`K_D=k_1/k_(-1)`

We can find the equilibrium concentrations easily using the ICE method. We construct a table of initial, change and equilibria concentrations.

	`2A`	`stackrel(K_D) harr`	`A_2`
I	`A_0`		0
C	`-x`		`+x/2`
E	`A_0-x`		`x/2`

To evaluate the equilibrium concentrations we set up the equilibrium expression and solve it for `x`.

`K_D=x/(2(A_0-x)^2)`

We obtain the quadratic equation,

`K_DA_0^2-(2K_DA_0+0.5)x+K_Dx^2=0`

For which the root can be expressed,

`x=(4K_DA_0+1-sqrt(8K_DA_0+1))/(4K_D)`

From the mass balance, we have,

`A_0-x=A` and `A_2=x/2`

Hence,

`A_2=(4K_DA_0+1-sqrt(8K_DA_0+1))/(8K_D)`

and

`A=A_0-(4K_DA_0+1-sqrt(8K_DA_0+1))/(4K_D)`