1.1 What is Chemistry?

20.3. Free Energy and Equilibrium http://www.ck12.org

TABLE20.2:(continued)

K ln K ∆G° Comments

<1 negative positive reactants favored over

products at equilibrium

1 0 0 equilibrium situation;

products and reactants

equally favored

>1 positive negative products favored over re-

actants at equilibrium

Example 20.4

The standard free energy of formation (∆G°f) for ammonia is -16.6 kJ/mol. Calculate the equilibrium constant for
the following reaction at 25°C (298 K):

N 2 (g) + 3H 2 (g)2NH 3 (g)

Answer:

If we know the value of∆G° for this reaction, we can calculate the equilibrium constant using the equation above.
∆G° can be calculated from∆G°fvalues using the following equation:

∆G°rxn=Σn∆G°f(products) –Σn∆G°f(reactants)

Because nitrogen and hydrogen are both in their standard elemental states, they have∆G°fvalues of zero.

∆G°rxn=Σn∆G°f(products) –Σn∆G°f(reactants)

∆G°rxn= 2(-16.6 kJ/mol) - [(0 kJ/mol) + 3(0 kJ/mol)]

∆G°rxn= -33.2 kJ/mol

Then solve∆G° = - RT ln Keqfor Keq:

∆G◦=−RT ln Keq

ln Keq=−∆G

◦

RT

Keq=e−

∆RTG◦

Before plugging values into this equation, note that we are using a value of 8.314 J/K•mol for R, but our value for
∆G° is -33.2 kJ/mol. In order for our units to cancel correctly, we need to convert -33.2 kJ/mol to -33,200 J/mol.
Then, Keqcan be calculated as follows:

Keq=e−

∆RTG◦

Keq=e−

− 33 ,200 J/mol

( 8 .314 J/K·mol)(298 K)

Keq=e^13.^4

Keq= 6. 6 × 105

Keqis very large, indicating that the products are heavily favored at 25°C.

Looking at the equation above, we see that increasing the temperature for any reaction makes the fraction in the
exponent smaller, thus moving the equilibrium constant closer to 1. If we were to run the ammonia formation
reaction above, known as the Haber process, at a higher temperature, the equilibrium constant would be smaller,