http://www.ck12.org Chapter 20. Entropy and Free Energy
In order to make the units agree, the value of∆G° will need to be converted to J/mol (173,400 J/mol). To solve for
Keq, the inverse of the natural logarithm, ex, will be used.
Step 2: Solve.
∆G◦=−RT ln Keq
ln Keq=−∆G◦
RT
Keq=e
−∆RTG◦
=e
8 .314 J/K−173400 J/mol·mol(298 K)
= 4. 0 × 10 −^31Step 3: Think about your result.
The large positive free energy change leads to a Keqvalue that is extremely small. Both lead to the conclusion that
the reactants are highly favored, and very few product molecules are present at equilibrium.
Sample Problem 20.3: Free Energy from Ks p
The solubility product constant (Ks p) of lead(II) iodide is 1.4× 10 −^8 at 25°C. Calculate∆G° for the dissociation of
lead(II) iodide in water.
PbI 2 (s)⇀↽Pb^2 +(aq)+2I−(aq)Step 1: List the known values and plan the problem.
Known
- Keq= Ks p= 1.4× 10 −^8
- R = 8.314 J/K•mol
- T = 25°C = 298 K
Unknown
- ∆G° =? kJ/mol
The equation relating∆G° to Keqcan be solved directly.
Step 2: Solve.
∆G° = -RT ln(Keq)
∆G° = -(8.314 J/K•mol)(298 K)ln(1.4× 10 −^8 ) = 45,000 J/mol = 45 kJ/molStep 3: Think about your result.
The large, positive∆G° indicates that the solid lead(II) iodide is nearly insoluble and is mostly present as a solid at
equilibrium.
Practice Problem- For the Haber-Bosch process at 25°C,∆H° =−92.6 kJ/mol and∆S° =−198.5 J/K•mol.
N 2 (g)+3H 2 (g)⇀↽2NH 3 (g)
a. Calculate∆G° for the reaction.
b. Calculate the value of Keq.