Physical Chemistry Third Edition

344 7 Chemical Equilibrium

Solution

a.

Kexp

⎡

⎣^2.^93 ×^10

(^4) J mol− 1
(
8 .3145 J K−^1 mol−^1
)
( 298 .15 K)
⎤
⎦
 1. 36 × 105
b.Letxceq(ATP)/c◦

1. 36 × 105 
   ( 0. 0100 −x)^2
   x
   x
   ( 0. 0100 −x)^2


2. 36 × 105
   ≈
   ( 0. 0100 ) 2


3. 36 × 105
   ≈ 7. 35 × 10 −^10
   Exercise 7.22
   Find the value of (∂G/∂ξ)T,Pat 298.15 K for the case thatc(ATP) 0 .0100 mol L−^1 and
   c(ADP)c(Pi) 0 .0200 mol L−^1 at pH 7 .00 and pMg 4 .00. Approximate all activity
   coefficients by unity.
   The hydrolysis of ATP is “coupled” to various other reactions that would otherwise
   not be spontaneous. For example, the reaction
   Pi+glucose
   glucose 6-phosphate+H 2 O (7.7-3)
   is driven by the reaction of Eq. (7.7-1). We illustrate the coupling of reactions with the
   regeneration of ATP, which is coupled to the hydrolysis of phosphoenolpyruvic acid
   (abbreviated PEP).

O

H–O–P–O–H

OO

CH 2 ==C–C–O–H

The hydrolysis of PEP is sufficiently spontaneous to produce ATP from ADP when

coupled to this reaction. The sum of the two reactions is equivalent to a spontaneous

reaction:

(A) PEP+H 2 O
Py+Pi ∆G◦′− 53 .6kJmol−^1 (7.7-4a)

(B) ADP+Pi 
AT P+H 2 O ∆G◦′+ 29 .3kJmol−^1 (7.7-4b)

(C) ADP+PEP
AT P+Py ∆G◦′− 24 .3kJmol−^1 (7.7-4c)

where Py stands for pyruvic acid and/or pyruvate ion. The hydrolysis of PEP produces

phosphoric acid, which is a reactant in the regeneration of ATP from ADP. We now