8.2. Chemical reactions[[Student version, January 17, 2003]] 265
hold the temperature fixed. Nevertheless the quantityμ 2 −μ 1 still does control the direction of the
reaction:
Your Turn 8c
a. Following Section 6.5.1 on page 186, show that in a sealed test tube, held at fixed temperature,
the Helmholtz free energyFof a+B changes byμ 2 −μ 1 when the reaction takes one step.
b. Similarly for anopentest tube, in contact with the atmosphere at pressurep,show that the
Gibbs free energyGchanges byμ 2 −μ 1 when the reaction takes one step.
T 2 Section 8.2.1′on page 294 connects the discussion to the notation used in advanced texts.
8.2.2 ∆Ggives a universal criterion for the direction of a chemical reac-
tion
Section 8.2.1 showed how the conditionμ 1 =μ 2 for the equilibrium of an isomerization reaction
recovers some ideas familiar from Chapter 6. Our present viewpoint has a number of advantages
overthe earlier one, however:
1.The analysis of Section 6.6.2 was concrete, but its applicability was limited to dilute solutions
(of buffalo). In contrast, the equilibrium conditionμ 2 =μ 1 is completely general: It’s just a
restatement of the Second Law. Ifμ 1 is bigger thanμ 2 ,then the net reaction 1→ 2 increases
the world’s entropy. Equilibrium is the situation where no such further increase is possible.
2.Interconversions between two isomers are interesting, but there’s a lot more to chemistry than
that. Our present viewpoint lets us generalize our result.
3.The analysis of Section 6.6.2 gave us a hint of a deep result when we noted that the activation
barrierE‡dropped out of the equilibrium condition. We now see that more generallyit doesn’t
matter at all what happens inside the “phone booth”mentioned at the start of Section 8.2.1.
Wemade no mention of it, apart from the fact that it ends up in the same state in which it
started.Indeed the “phone booth” may not be present at all: Our result for equilibrium holds even for
spontaneous reactions in solution, as long as they are slow enough that we have well-defined initial
concentrationsc 1 andc 2.
Burning of hydrogen Let’s follow up on point #2 above. The burning of hydrogen is a familiar
chemical reaction:
2H 2 +O 2 ⇀2H 2 O. (8.7)
Let us consider this as a reaction among three ideal gases. We take an isolated chamber at room
temperature containing twice as many moles of hydrogen as oxygen, then set off the reaction with
aspark. We’re left with a chamber with water vaporandvery tiny traces of hydrogen and oxygen.
Wenow ask, how much unreacted hydrogen remains?
Equilibrium is the situation where the world’s entropyStotis a maximum. To be at a maximum,
all the derivatives of the entropy must equal zero; in particular, there must be no change inStot
if the reaction takes one step to the left (or right). So to find the condition for equilibrium, we
compute this change and set it equal to zero.