Equilíbrio Químico e Físico

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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 167
the concentration of the S–A complex, so [A]bound = [SA]. It therefore
follows that ν/N = [SA]/[S].�is is rearranged to [SA] = [S]ν/N , which
is one of the terms needed in the expression for K′.
�e other term is [S]free which is related to [S] as follows.�e total con-
centration of S is given by [S] = [S]free + [SA], hence [S]free = [S]− [SA].
Substituting [SA] = [S]ν/N gives [S]free = [S] − [S]ν/N = (1 − ν/N)[S].
�e expression for the equilibrium constant is now developed as
K′ = [SA]c−○
[S]free[A]free
= ([S]ν/N)c−○
(1 − ν/N)[S][A]free
= νc−○
[A]free(N − ν)
where to go to the �nal expression the numerator and demoninator are
multiplied by N and [S] is cancelled. �e Scatchard equation follows by
taking the factor (N − ν) to the le�
K′N − K′ν = νc−○
[A]free
(c) �e straight line plot is ν/[A]free against ν.�e task is therefore to deter-
mine ν from the data. Note that the data given are the total concentrations
of EB in and outside the bag. It therefore follows that
[EB]total,in = [EB]bound + [EB]free = [EB]bound + [EB]out
It follows that [EB]bound = [EB]total,in − [EB]out. Recall that ν is de�ned
as ν = [EB]bound/[M], therefore
ν = [EB]total,in − [EB]out
[M]
�e given data and the derived values of ν are shown in the following table
(1 µM = 1 µmol dm−3), and plotted in Fig. 5.37.
[EB]out/µM [EB]in/µM ν (ν/[EB]out)/(µM−1)
0.042 0.292 0.250 5.95
0.092 0.590 0.50 5.41
0.204 1.204 1.00 4.90
0.526 2.531 2.01 3.81
1.150 4.150 3.00 2.61
�e data �t to quite a good straight line, the equation of which is
(ν/[EB]out)/(µM−1) = −1.17 × ν + 6.12
�e slope is −K′/c−○ where c−○ is the standard concentration, 1 mol dm−3,
but because [EB]out is used in µM, c−○ = 106 µM. It follows that the
(dimensionless) equilibrium constant is K′ = 1.17 × 106 . �e intercept
gives K′N/c−○ , hence N = 5.23 : this is the average number of binding
sites per DNAmolecule.�e graph is a good straight line, indicating that
the data �t the model quite well.