EQUILIBRIUM
7.8 FREE ENERGY AND EQUILIBRIA IN BIOCHEMICAL SYSTEMS Reactions of biochemical interest do not normally occur in the gas phase. Rather
they occur in solution, usually saline solution. Therefore the correct expression of the free energy changes of reaction and the equilibrium constant is in terms of the corresponding activities and changes in chemical potential. Determination of activities and activity coefficients over a concentration range on nonideal solutions is not a simple matter, nor is it necessary for in vitro studies of biochemical reactions.
Instead, concentrations are used with the stipulation that the “background” condi- tions must be constant over the course of the study and must be reproduced from one study to the next. For example, energy studies on dephosphorylation of adenosine
FREE ENERGY AND EQUILIBRIA IN BIOCHEMICAL SYSTEMS 103 5-triphosphate would be carried out at a specified and constant temperature, pres- sure, pH, pMg, and ionic strength where pH and pMg refer to the ion concentrations pH= −log
H+
and pMg= −log Mg2+
, and the ionic strength is constant for all dissolved salts in the reaction solution. Ionic strength is essentially the ionic charge concentration in solution calculated as the sum 12
ciz2i over all salt concentrations cithat dissolve to give ions of charge zi.
These conditions specify a unique standard state which is not the thermodynamic standard state but which is adhered to throughout the experiment and experiments with which results will be combined or compared. Under these controls, it is proper to write
GA=G◦A+RT ln [A]
for G◦Ain the specified standard state with a concentration of reactant A, with similar expressions for B, C,. . .. The change in free energy for a reaction quotientQ, now in terms of initial concentrations, is
G=G◦+RT lnQ=G◦+RT ln[C]ξ[D]ξ. . . [A]ξ[B]ξ. . .
Different stoichiometric coefficients are all expressed using the same symbolξ for simplicity. It is understood thatQis a concentration quotient, which might fortuitously be equal to the equilibrium constant but in general will not. In general,G will be different from zero but when the reaction has arrived at equilibrium, the concentrations will have arrived at KeqandG will have arrived at zero so that
G◦= −RT ln Keq
Thus the form of the free energy relation to the equilibrium constant is reproduced but only under rigorously controlled background conditions.1Change the ionic strength or the pH, for example, and you can expect to find a differentG◦and Keq.
How can G◦ change when we usually think of it as a sum of rock-firm G◦ values? By changing our background conditions, we have changed the standard state, the benchmark to which we refer all G◦values.
7.8.1 Making ATP, the Cell’s Power Supply
The difference between rG and G◦ in the metabolic degradation of glucose, a physiological energy source, to lactate and ATP can be broken down into two individual steps (Hammes, 2007), one taking up 2 mol of ATP and producing 2 mol of the diphosphate ADP
Glucose+2 ATP→2 ADP+2 glyceraldehyde-3-phosphate
1See Treptow (1996).
and the other producing 4 mol of ATP
2 Glyceraldehyde-3-phosphate+2 phosphate+4 ADP→2 lactate+4 ATP+2 H2O The summed reaction is
Glucose+2 phosphate+2 ADP→2 lactate+2 ATP+2 H2O
for a net gain of 2 mol of ATP. The first of these two reactions is unfavorable from the point of view of a positiveG◦ =2.2 kJ mol−1but at the initial concentrations chosen to be at or near physiological concentrations,
Q=(0.14)2(0.019)2
5.0 (1.85)2 =0.0196
3.61×10−4
≈4×10−71
This very small reaction quotient drives the metabolic conversion of ADP to ATP which then powers other reactions within the biological system as a whole.
PROBLEMS AND EXAMPLES Example 7.1 Solution Calorimetry
A half-liter solution calorimeter system consisted of the calorimeter itself, a tempera- ture measuring circuit, an electrical heating circuit, and calorimeter fluid. The system was prepared with a solution of magnesium ion at 0.001 M. The ionic strength was brought to 0.25 M with KCl, a neutral electrolyte, and the pH was adjusted to 7.0.
When the heating circuit was activated at a current flow of I amperes for a time t, an amount of heat energy qp =96.5 J was delivered through a heating resistor R (q =I2Rt ) producing a temperature rise of T =0.166 K.2 What was the water equivalent (calorimeter constant) of the calorimeter system?
Solution to Example 7.1 The water equivalent of a calorimeter is the heat capacity of the entire system as if it were all water even though it consists of various parts made of various materials and contains a solution of reactants and products different from pure water. The water equivalent can be found by a straightforward heat capacity calculation even though we know that the heat capacity of the system is the sum of many parts. We calculate
Cp =dqp
dT qp
T = 95.6
0.166 =576 J K−1 =0.576 kJ K−1
21 amp volt sec=1 J and E in volts=IR.
PROBLEMS AND EXAMPLES 105 Example 7.2 Adenosine 5-triphosphate ATP
The hydrolysis of adenosine 5-triphosphate ATP to adenosine 5-diphosphate ADP plus an inorganic phosphate ion can be written
ATP+H2O→ADP+phosphate
If 10 mL of ATP with concentration [ATP]=0.200 mol L−1are pipetted into a calorimeter with a heat capacity of 0.576 kJ K−1(Example 7.1) and a temperature rise of 0.107 K is found, what is the enthalpy of hydrolysis of adenosine 5-triphosphate to adenosine 5-diphosphate plus an inorganic phosphate ion under these conditions?
What is the sign ofrH ? Solution to Example 7.2
0.107 K
0.576 kJ K−1
=0.0616 kJ 10.0 mL=0.0100 L
0.0100 L(0.200 mol L−1)=0.00200 mol ATP 0.0616 kJ
0.00200 mol = −30.8 kJ mol−1
The sign ofrH is negative because the reaction is exothermic. Heat flowing out of the reacting system at constant external pressure means that its enthalpy balance must decrease.
Problem 7.1
What is the change in the entropy S for one mole of a pure substance for an infinites- imal change in T and p? Given: Entropy is an exact differential
d S(T,p)= ∂S
∂T
p
dT+ ∂S
∂p
T
d p
Problem 7.2
In an experiment on the nitrogen tetroxide reaction at 298 K N2O4(g) →← 2NO2(g)
pure N2O4(g) was introduced into a reaction vessel maintained at constant temperature and a total pressure of 2.500 bar. When equilibrium had been reached, the partial
pressure of N2O4(g) had dropped to 1.975 bar. What is Keqfor this reaction? Compare your answer to the calculated value (Section 7.2). What are the units of Keq? Problem 7.3
Supposing that a simple A(g)→←B(g) reaction has an equilibrium constant Keq= 0.100 at T =200 K and Keq =0.200 at T =300 K. What isrH◦for the reaction?
Problem 7.4