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It is possible relate the standard Gibbs free energy
(At 250C, 1 atm pressure) and the Gibbs free energy at any
temperature and pressure through the following equation:
- DG = DG0 +
RTln(Q)
where Q is the reaction
quotient that we studied during equilibrium and R is the gas constant. The
usual rules for Q apply: gases enter as partial pressures, solutions as molar
concentration and solids and liquids don't enter in at all, and all temperatures
are in Kelvin. (Since Q = K at equilibrium, we can get equilibrium data from
the free energy)
Example: What is DG0 and DG for the following reaction at 1200 K, all gases at 0.30
atm pressure?
- 2SO3(g) -> 2SO2(g) + O2(g)
Solution: First, we need to compute DG0 the usual way through thermodynamic
data
| Compound |
DHf0 (kJ/mol) |
DS0 (J/mol*K) |
| SO3(g) |
-395.7 |
+256.7 |
| SO2(g) |
-296.8 |
+248.1 |
| O2(g) |
0.0 |
+205.0 |
Next, compute DH0 and DS0 and get
DG0 at 250C and 1 atm pressure
- DH = (2*DHSO20 + 1*DHO20) - (2*DHSO30)
- DH = (2* -296.8 + 1*0) - (2*395.7)
- DH = +197.8 kJ/mol
- DS = (2*DSSO20 + 1*DSO20) - (2*DSSO30)
- DS = (2* 248.1 + 1*205) -(2*256.7)
- DS = +187.8 J/mol*K = 0.1878 kJ/mol*K
- DG0 = DH0 - T*DS0
- DG0 = 197.8 + 298*0.1878
- DG0 = 142 kJ/mol
Now that we have DG0, we can compute DG at 1200 K and 0.30 atm partial pressure for all
gases. Q has the same form as the equilibrium constant for the reaction:
- Q = (PSO2)2PO2/
(PSO3)2
- Q = (0.30)2*0.30/(0.30)2
- Q = 0.30
Finally now we can compute DG
- DG = DG0
+RTln(Q)
- DG = 142 kJ/mol + 8.314*10-3 kJ/mol*K
* 1200 K * ln(0.30)
- DG = 142 kJ/mol - 12.0 kJ/mol = 130 kJ/mol
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