270 Chapter 8. Chemical forces and self-assembly[[Student version, January 17, 2003]]
In this formulak+andk-are called therate constantsof the reaction. They are similar to the
quantities we defined for the two-state system (Section 6.6.2 on page 194), but with different units:
Equation 8.18 shows their units to bek±∼s−^1 M−^2 .Weassociate rate constants with a reaction
bywriting them next to the appropriate arrows:
X 2 +Y 2k+
k-2XY. (8.19)
Setting the rates equal,r+=r-,gives that at equilibriumcX 2 cY 2 /(cXY)^2 =k-/k+,or
cX 2 cY 2
(cXY)^2
=Keq=const. (8.20)This seems good—it’s the same conclusion we got from Equation 8.16.
Unfortunately, predictions for rates based on the logic leading to Equation 8.18 are often totally
wrong. For example, we may find that over a wide range of concentrations, doubling the concen-
tration of Y 2 has almostno effecton the forward reaction rate, while doublingcX 2 quadruplesthe
rate! We can summarize such experimental results (for our hypothetical system) by saying that
the reaction is ofzeroth orderin Y 2 andsecond orderin X 2 ;this just means the forward rate is
proportional to (cY 2 )^0 (cX 2 )^2 .Na ̈ıvely we expected it to be first order in both.
What is going on? The problem stems from our assumption that we knew the mechanism of the
reaction, that is, that an X 2 smashed into aY 2 and exchanged one atom, all in one step. Maybe
instead the reaction involves an improbable but necessary first step followed by two very rapid
steps:
X 2 +X 2 2 X+X 2 (step 1, slow)
X+Y 2 XY 2 (step 2, fast)
XY 2 +X 2XY (step 3, fast). (8.21)The slow, step of the proposed mechanism is called thebottleneck,orrate-limitingprocess.
The rate-limiting process controls the overall rate, in this case yielding the pattern of concentration
dependences (cY 2 )^0 (cX 2 )^2 .Either reaction mechanism (Reaction 8.19 or 8.21), is logically possible;
experimental rate data are needed to rule out the wrong one.
Won’t this observation destroy our satisfying kinetic interpretation of the Mass Action rule,
Equation 8.18? Luckily, no. The key insight is that in equilibrium, each elementary reaction in
Equation 8.21 mustseparatelybein equilibrium. Otherwise there would be a constant pileup of
some species, either a net reactant like X 2 or an intermediate like XY 2 .Applying the na ̈ıve rate
analysis to each step separately gives that in equilibrium
(cX)^2 cX 2
(cX 2 )^2 =Keq,^1 c^0 ,cXY 2
cXcY 2 =Keq, 2
c 0 ,(cXY)^2
cXY 2 cX=Keq,^3.Multiplying these three equations together reproduces the usual Mass Action rule for theoverall
reaction, withKeq=Keq, 1 Keq, 2 Keq, 3 :
The details of the intermediate steps in a reaction are immaterial for its overall
equilibrium.(8.22)
This slogan should sound familiar—it’s another version of the principle that “equilibrium doesn’t
care what happens inside the phone booth” (Section 8.2.2).