196 Chapter 6. Entropy, temperature, and free energy[[Student version, January 17, 2003]]
is thatk+/k-=e∆E/kBT. Section 6.7 will describe how such predictions can now be confirmed
directly in the laboratory.
Notice a key feature of the equilibrium distribution: Equation 6.27 does not containE‡. All
that matters is the energy difference ∆Eof the two states. It may take a long time toarriveat
equilibrium ifE‡is big, but once the system is there the height of the barrier is immaterial. This
result is analogous to the observation at the end of Section 4.6.3 that the value of the diffusion
constantDdrops out of the Nernst relation, Equation 4.25 on page 126.
Suppose we begin withnonequilibrium populations of molecules in the two states. Then the
numberN 2 (t)inthe upper statewillchange with time, at a net rate given by the difference between
the up- and down-rates just found, and similarly for the numberN 1 (t)inthe lower state:
N ̇ 2 ≡ dN 2 /dt= −k+N 2 (t)+k-N 1 (t)
N ̇ 1 ≡ dN 1 /dt= k+N 2 (t)−k-N 1 (t). (6.28)The steps leading from the reaction scheme (6.26) to Equation 6.28 were simple but important, so
weshould pause to summarize them:
Toget rate equations from a reaction scheme like (6.26), we:- Examine the reaction diagram. Each node (state) of this diagram leads to a
differential equation for the number of molecules in the corresponding state. - Foreach node, find all the links (arrows) impinging on it. The time derivative
of the number (or concentration) of this state has a positive term for each arrow
pointing toward its node, and a negative term for each arrow pointingawayfrom
it.
(6.29)
Returning to Equation 6.28, we note that the total number of molecules is fixed:N 1 +N 2 =Ntot.
So we can eliminateN 2 (t)everywhere, replacing it byNtot−N 1 (t). When we do this, one of the
twoequations is redundant; either one gives
N ̇ 1 =k+(Ntot−N 1 )−k-N 1.This equation is already familiar to us from our study of the relaxation of concentration jumps (see
the Example on page 122). LetN 1 ,eqbethe population of the lower state in equilibrium. Since
equilibrium impliesN ̇ 1 =0,wegetN 1 ,eq=k+Ntot/(k++k-). You can readily check that the
solution of the equation is then
N 1 (t)−Neq=(N 1 (0)−N 1 ,eq)e−(k++k-)t. relaxation to chemical equilibrium (6.30)Thus,N 1 (t)approaches its equilibrium value exponentially, with the decay constantτ=(k++k-)−^1.
Weagain say that a nonequilibrium initial population “relaxes” exponentially to equilibrium. To
find how fast it relaxes, differentiate Equation 6.30: The rate at any timetis the deviation from
equilibrium at that time, or (N 1 (t)−N 1 ,eq), times 1/τ.
According to the discussion following Equation 6.26,
1 /τ=k++k-=Ce−E
‡/kBT
(1+e−∆E/kBT).Thus, unlike the equilibrium populations themselves, a reaction’s ratedoesdepend on the barrier
E‡,akeyqualitative fact already observed earlier. Indeed many important biochemical reactions