266 Chapter 8. Chemical forces and self-assembly[[Student version, January 17, 2003]]
Because atoms aren’t being created or destroyed, a step to the right removes two hydrogen and
one oxygen molecules from the world, and creates two water molecules. Define the symbol ∆Gby
∆G≡ 2 μH2O− 2 μH 2 −μO 2. (8.8)With this definition, Equation 8.1 says that the change in the world’s entropy for an isolated
reaction chamber is ∆Stot=−∆G/T.For equilibrium we require that ∆Stot=0. Your Turn 8c
gave another interpretation of ∆G,asthe change of free energy of an open reaction chamber. From
this equivalent point of view, setting ∆G=0amounts to requiring that the Gibbs free energy be
at a minimum.
Oxygen, hydrogen, and water vapor are all nearly ideal gases under ordinary conditions. Thus
wecan use Equation 8.3 on page 261 to simplify Equation 8.8 and put the equilibrium condition
into the form
0=∆G
kBT=
2 μ^0 H 2 O− 2 μ^0 H 2 −μ^0 O 2
kBT
+ln[(
cH 2 O
c 0) 2 (
cH 2
c 0)− 2 (
cO 2
c 0)− 1 ]
.
Wecan lump all the concentration-independent terms of this equation into one package, theequi-
librium constantof the reaction:
Keq≡e−(2μ(^0) H 2 O− 2 μ (^0) H 2 −μ (^0) O 2 )/kBT
. (8.9)
It’s also convenient to define a logarithmic measure of the equilibrium constant via
pK≡−log 10 Keq. (8.10)With this abbreviation, the condition for equilibrium becomes
(cH 2 O)^2
(cH 2 )^2 cO 2
=Keq/c 0. in equilibrium (8.11)The left side of this formula is sometimes called the “reaction quotient.”
Equation 8.11 is just a restatement of the Second Law. Nevertheless, it tells us something useful:
The condition for equilibrium is that a certain combination of the concentrations (the reaction
quotient) must equal a concentration-independent constant (the equilibrium constant divided by
the reference concentration).
In the situation under discussion (hydrogen gas reacts with oxygen gas to make water vapor),
wecan make our formulas still more explicit:
Your Turn 8d
a. Show that the results of the Example on page 261 let us write the equilibrium constant asKeq=[
e(2H^2 +O^2 −^2 H^2 O)/kBT]
×
[
c 0(
2 π^2
kBT(mH 2 O)^2
(mH 2 )^2 mO 2) 3 / 2 ]
. (8.12)
b. Check the dimensions in this formula.Equation 8.12 shows that the equilibrium constant of a reaction depends on our choice of a reference
concentrationc 0 .(Indeed it must, bcause the left side of Equation 8.11 doesnotdepend onc 0 .)
The equilibrium constant also depends on temperature. Mostly this dependence arises from
the exponential factor in Equation 8.12. Hence we get the same behavior as in isomerization
(Equation 6.24 on page 194):