Chemistry 112

Equilibrium


Consider the elementary reaction:

A(g) B(g)

At some point in time, the concentrations of A and B will become such that the rate of the forward reaction will equal the rate of the reverse reaction.

Rate(forward) = Rate(reverse)

k1[A] = k–1[B]




How do the concentrations of A and B change with time?

In order to maintain the constant, if B increases, A must also increase. But this cannot happen because B is created from A. Chemically, if B increases (a product), the reactant A must decrease. If this were to happen, then k1/k–1 would no longer be constant, as required.

When this condition occurs, the reaction is in a state of dynamic equilibrium.

All reactions reach equilibrium eventually, even if they are not elementary.

Equilibrium is defined as the condition when the rate of the forward reaction equals the rate of the reverse reaction.

A consequence of this is that the macroscopic concentration of products and reactants is unchanging. For all practical purposes, when equilibrium is achieved, the reaction is over.

However, at the microscopic level, the reaction still continues.



For any general reaction

a A(g) + b B(g) + ... c C(g) + d D(g) + ...

The double arrow sign is used to denote a reaction that can attain equilibrium.

In principle, all reactions proceed to an equilibrium state. However, in practice, we distinguish between reactions that achieve equilibrium and those that go to completion.
For those reactions that go to completion, the amount of reactants remaining at equilibrium is negligibly small.
One of the skills that we need to be able to develop is to be able to distinguish a reaction that achieves equilibrium and one that goes to completion.

To help us make that assessment, we define a quantity known as the mass action expression:

For the above reaction:

The mass action expression is defined based on the stoichiometry of the reaction and can be written by inspection for any balanced reaction. This is also known as a reaction quotient.

If the reaction has reached equilibrium, then the concentrations of reactants and products is unchanging (at the macroscopic level) so the value of the mass action expression reaches a constant.

Kc is called the equilibrium constant defined in terms of concentration. No matter what initial concentrations of reactants are used, the equilibrium concentrations always satisfy the same Kc for a given reaction at a given temperature.



Examples

Write the equilibrium mass action expression for Kc for the following reactions:

a) SO2(g) + ½ O2(g) SO3(g)

b) H2(g) + I2(g) 2 HI(g)

c) 2 SO2(g) + O2(g) 2 SO3(g)

The equilibrium constants for a) and c) are for the same reaction balanced in a different fashion; are the values for Kc different?


For all gas phase reactions, it is often more convenient to use partial pressures rather than concentrations. The mass action expression can be written in terms of partial pressures of gases in exactly the same manner is for concentrations.

a A(g) + b B(g) + ... c C(g) + d D(g) + ...

The mass action expression written in terms of partial pressures is

To distinguish this from Kc, the subscript is changed to p – Kp is the equilibrium constant written in terms of partial pressures.



Kp - Kc relationship

Kp = Kc(RT)n

Here R is the gas constant in units of L•atm/mol•K, T is the temperature in K, and n is the change in the number of moles of gas phase products and reactants (n = total moles gas phase products – total moles gas phase reactants).

(this can be derived from the ideal gas law)