Page 56 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:36 9in x 6in b985-ch05 Elementary Physical Chemistry

TheFreeEnergy Functions 41

or

o
K =exp(−∆G /RT ) (5.36b)

Comment: It is important to note that the equilibrium constant is
not related to the actual free energy change (which is zero), but to
the free energy change in the standard state.

5.8. Variation of K with T
o
o
Assuming that ∆H and ∆S vary negligibly with temperature, we have:
at a constant temperature T
o
o
o
ln K = −∆G /RT = −∆H /RT +∆S /R (5.37a)
and at a different constant temperature T
o
o
o

ln K = −∆G /RT = −∆H /RT +∆S /R (5.37b)

resulting in
o

ln K =ln K +∆H /R(1/T − 1/T ) (5.38)
Example 5.2. An ideal gas, initially at T = 273 K, P =1 atm and V i =
22.4 L expands isothermally and reversibly to a final volume, V f =44.8L.
a) Calculate w, q, ∆U, ∆H, ∆S, ∆G.
The process is reversible so that

w = − P extdV = −nRT dV/V = −RT ln V f /V i
−1 −1
= −RT ln(44.8L/22.4L) = −8.3145 JK mol × 273 K × ln 2
= −1.573 kJ
∆U = 0 (ideal gas at constant temperature). Thus, q = −w and so
q =1.573 kJ
∆H =∆U +∆(PV )=0 + ∆(RT )= 0
∆S = q rev/T =1, 573.3J/273 K = 5.763 JK −1
∆G =∆H − T ∆S =0 − 273 K × 5.763 JK −1 = −1.573 kJ