|
Since we can relate the Gibbs free energy
and the reaction
quotient Q through the following equation
- DG = DG0 +
RTln(Q)
we can also obviously get the equilibrium constant as well
since Q=K and DG = 0 at equilibrium.
- 0 = DG0 + RTln(K)
- DG0 = -RTln(K)
Since the
DG here is the standard free energy change, (Gases at 1 atm,
T at 25oC), we can simply look up the values of DG in tables.
Example: What is the equilibrium constant for the following reaction?
- Fe+3(aq) + 3OH-(aq) ->
Fe(OH)3(s)
Solution: First, we need to compute DG0 the usual way through thermodynamic
data
| Compound |
DG0 (kJ/mol) |
| Fe+3(aq) |
-4.7 |
| OH-(aq) |
-157.2 |
| Fe(OH)3(s) |
-696.6 |
Next, compute DG0
- DG = (1*DGFe(OH)30) - (1*DGFe+30 + 3*DGOH-0 )
- DG = (1* -696.6) - (1*-4.7 + 3* -157.2)
- DG = -220.8 kJ/mol
(Note since
DG is negative the reaction is spontaneous: iron hydroxide
will precipitate)
Now simply plug into the relation between DG and K
- DG0 = -RTln(K)
- -220.8 kJ/mol = -(8.314*10-3 kJ/mol*K)*(298K)*ln(K)
- ln(K) = 88.9
- K = 4.1*1038
Iron(III) hydroxide is very
insoluble. Note that this number is very large: the equilibrium lies on the
right. When computing a Ksp
value, the chemical equation would be written with the solid on the left and
thus Ksp would be very small. |