Revision - Equilibrium constant

There are reactions where one can see the reverse action also to be active and see the equilibrium point. In this case

A + B → C + D and

C+D → A+B both reactions keep taking place.

At the point of equilibrium the rate of both reactions is same. Formation of A+B is equal to consumption of A+B.



Rate of forward reaction = k-f[A][B]

Rate of reverse reaction or backward reaction = k-r[C][D]

Therefore k-f[A][B] = k-r[C]{D]

This gives k-f/k-r = [C]{D]/[A][B]

The equilibrium constant is always written as products by reactants.

For the a general reaction

aA + bB ↔ cC+dD (Normal two arrows are used for reversible reaction. Only one arrow with heads on both sides is used here to tide over the inability to show two arrows.)

k-eq = [C]^c[D]^d/[A]^a[B]^b

The equilibrium constant may or may not have units.

In the case of 2A ↔ 2B +C

The units of equilibrium constant are going to be: (mol/l)^2(mol/l)/(mol/l)^2

= mol/l