Equilibrium

Equilibrium by definition is the state of a chemical reaction in which its forward and reverse reactions occur at equal rates so that the concentration of the reactants and products does not change with time; however, in the living organisms, equilibrium is hardly achieved since the condition is always changing.

An example of the equilibrium would be: A D B when µA = µB

or Keq = …i.e. the concentrations of A and B are at their equilibrium values. The energy differences between A and B are compensated for by the differences in concentration.

Another example of equilibrium is equilibration between compartments:

Here µA (left) = µA (right) …that does not mean that [A] (left) =[A] (right) …as in the picture. The chemical potential energy level is the same on both sides.

Steady State:

An Example of the Steady State: the movement from A à X à B continuously; however, no changes occurs to [X] because X is formed from A as fast as it is converted to B.

thus

…no change in a concentration does not mean equilibrium. Cells typically have constant values of solute concentrations, such as pH, and transmembrane voltage but these do not represent equilibria.

Another example of steady state in cells is that the total current (I) through the membrane is equal to 0. That must be true if the transmembrane voltage is constant. For example the potential inside a cell is about -10mV and if the potential stays constant, then the current through the membrane is equal to 0…more about this later.

The Chemical Potential

The chemical potential of a substance is a great starting point. From this we can get many familiar relationships.

µA = µºA + RT ln[A] + ZAF φ + VA P

F If A D B then Keq = =

You can look at this as an application of the Boltzmann distribution. Even though B and A are different molecules they can be looked at different forms of matter at different energy levels. Thus, at equilibrium, the ratio of [B] over [A] should be given by the energy difference and the Boltzmann distribution. Alternatively, we can use the chemical potential. At equilibrium:

µA = µB è µºA - µºB = - RT ln where µºA - µºB = ∆Gº

The symbols with the superscript º contain the information of the inherent energy of A and B that has to do with the nature of the substance and the environment in which the substance lies. When the reaction is not at equilibrium, there is a tendency for the reaction to move toward equilibrium and thus there is an energy difference between reactants and products. This energy difference is the ∆G. It depends on how far the reaction is from equilibrium. [A] and [B] are the actual concentrations.

∆G = - RT ln + RTln = ∆Gº + RTln

F If A D B + C then we know that but how do we write the Boltzman distribution?

If we use the chemical potential equations it becomes clear:

µA = µB + µC ….at equilibrium the energies must equal…energies add

è µºA - µºB - µºC =

converting to the exponential form….

So we see that in the Boltzmann distribution the concentrations…or the probabilities…should be multiplied. That makes sense because the probabilities are always combined by multiplication.

Example of energy changes in a reaction and how these depend on distance from equilibrium:

In a typical cell: ATP + H2O D ADP + Pi and thus:

*H2O is not used in the equation because H2O is pure.

ATP + H2O D ADP + Pi ∆Gº = -31kJ/mol

The negative ∆Gº tells us that the reaction favors the products....can you tell why?

Hint: from ∆Gº one can determine the equilibrium concentrations.

Important: The energy available from ATP hydrolysis is NOT fixed but depends on distance from equilibrium.

Example of the use of energy by cells to promote a reaction that is not favorable…i.e. the energy of the products is greater than that of the reactants. The first reaction in glycolysis…the use of sugar by cells…needs to phosphorylate glucose:

Glucose + Pi Glucose-6-Phosphate ∆Gº = + 14kJ/mol

Note the positive ∆Gº….lets call this by its name: the standard change in free energy…but please remember what this really means.

The ratio of products to reactants at equilibrium is 3x10^-3 …not a great start…good if you are on a diet

Cell’s solution: combine this unfavorable reaction with the hydrolysis of ATP

The addition of the two reactions gives rise to a spontaneous reaction:

ð Glucose + ATP ADP + Glucose-6-Phosphate ∆Gº = - 17KJ/mol

The enzyme, hexokinase, couples these reactions together. This is only way in which energy is used to shift a process in a non spontaneous direction.

NOTE: this is not the full story because we are dealing with ∆Gº not ∆G. The real free energy change depends on the concentrations of reactants and products. The cells keeps the [ATP] about 100 times higher than the [ADP]. Therefore the energy difference for the reaction out of equilibrium is much greater.

Question: How is it that the energy available from ATP hydrolysis depends on how far the system is from equilibrium while the energy released from burning fuel (gasoline, oil, wood) is constant?

These systems are far from equilibrium! The reverse reaction is negligible.

Cells work at energies close to thermal energy and so many reactions are not far from equilibrium.

Equilibrium between solutions separated by a membrane:

Consider a bacterium swimming in almost pure water. If we add some glycerol to the medium so as to make it 1 mM, what might the concentration be within the cytosol of that cell at equilibrium?

What we need to know:

1. Water easily crosses membranes and reaches equilibrium.
2. Glycerol also permeates through membranes easily and reaches equilibrium.
3. The concentration of solutes in a cell dilute the water inside and thus when water reaches equilibrium a pressure develops inside that is higher than outside.

V’ is V bar…the volume of a mole of water

at equilibrium:

Thus a pressure difference is generated to balance the concentration difference of water. The pressure difference affects all the molecules, including the glycerol.

at equilibrium: ….V’ is the volume of a mole of glycerol.

Glycerol