1.1 What is Chemistry?

http://www.ck12.org Chapter 20. Entropy and Free Energy

Q=[C]

c[D]d

[A]a[B]b(Always)

If Q >Keq, then the concentrations of the products (in the numerator) are too large compared to the concentrations of
the reactants. Therefore, the reverse reaction must be favored (reducing the products and generating more reactants)
in order for the reaction to achieve equilibrium. If Q <Keq, the situation is reversed. There are too many reactants
and not enough products, so the reaction quotient is a small number compared to Keq, and the forward reaction will
be favored until equilibrium is reached. When Q = Keq, the reaction is at equilibrium.

Free Energy at Non-Standard Concentrations

In the previous section, we looked at values for∆G°, which tells us the free energy change for a reaction being
run at standard conditions. In addition to the requirements for temperature and pressure (25°C and 1 atm), standard
conditions also specify the concentrations of each reaction component. Under standard conditions, the concentration
of each reaction component in the aqueous or gaseous states is exactly 1 M.

Plugging a value of 1 M in for every concentration in the reaction quotient expression will give us a value of Q = 1.
Unless Keqhappens to be exactly 1, reaction mixtures are not at equilibrium under standard conditions. However,
we can calculate the value of∆G for any combination of concentrations if we know the values of∆G° and Q. The
following equation (which we will not derive) shows the relationships between these quantities:

∆G =∆G° + RT ln Q

• ∆G = free energy change of the reaction under existing conditions


• ∆G° = free energy change of the reaction under standard-state conditions


• R = universal gas constant (8.314 J/K•mol)


• T = absolute temperature (in Kelvin)


• Q = reaction quotient

Although∆G° is going to be a fixed value for a given temperature,∆G will vary depending on the value of RT ln Q.

Free Energy and the Equilibrium Constant

At equilibrium, the forward and reverse reactions proceed at equal rates. The driving force in each direction is equal,
because the free energy of the reactants and products under equilibrium conditions is equivalent (∆G = 0). We also
know that, at equilibrium, Q = Keq. For a reaction that has reached equilibrium, the equation above becomes the
following:

∆G =∆G° + RT ln Q

0 =∆G° + RT ln K

∆G° = - RT ln K

This equation provides us with a way to convert between∆G° and the equilibrium constant for a given reaction.
Knowing either∆G° or Keqtells us whether the reactants or products are favored at equilibrium. This relationship is
summarized in the following table:

TABLE20.2:Relationship between equilibrium constant and∆G°

K ln K ∆G° Comments