| Chem 110 |
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Fall 2003 |
| Lecture Notes::Lec 11_19 September |
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| PREVIOUS |
Last time we left off with the expression relating the standard free energy and the equilibrium constant fora chemical reaction:
We also saw that the amount of energy available from a reaction (or which must be put in to make a reaction go) is related to how far the system is from equilibrium!
In Chem 109 we noted that a chemical reaction reaches equilibrium when the rates of the forward and reverse reactions are equal. This occurs in a a reaction with, for example a large -DH, because although the % of particles with sufficient energy to go backwards is very small, there are huge numbers of them compared to the number of reactant molecules which readily go forward.
We also noted that a reaction at equilibrium has no tendency to change (appears to be stagnant) and is stable at the macroscopic level, even though it is very active at the microscopic level. From our new perspective of thermodynamics it is at its lowest free energy, DG = 0.
Let's look at why this occurs from the perspective of energy. Consider a favorable reaction, i.e. DG = (-). Again, lets look at the components of DG, and follow the theoretical progress of the reaction from 100% reactants to 100% products. Thus:
Looking at DH first we can plot the progress of the reaction (Note that this is NOT the same as a reaction progress or reaction coordinate diagram - in that case we follow a single set of molecules through the reaction looking at the instantaneous energy of the molecules. In this case we are looking at the DG of the entire reaction mixture.):
Note that the relationship is a simple linear plot beginning at a high enthalpy for the reactants and going to a lower enthalpy for the products. The negative change indicates a favorable DH contribution to the DG of the reaction.
How about S? Let's assume a favorable DS for this reaction:
Note that S will generally not be linear, rather there will be a maximum value. This is expected since the entropy of a mixture of particles will be greater than the same number of identical particles.
As result of substituting the data of these two curves into DG = DH -TDS we will get a plot something like the one below, with a minimum value of DG occurring between the reactants and the products:
Finally, let's do a calculation of DG for a reaction which is not at equilibrium from values of the equilibrium constant, and the mass action expression, Q:
Example: Calculate DG given K and Q for the reaction of glucose and ATP to give glucose-6-phosphate and ADP. (This is the first reaction of Glycolysis, the major pathway in sugar metabolism, catalyzed by the enzyme hexokinase. It is frequently held at non-equilibrium values by keeping the enzyme largely turned off.)
- K' = 5000 [The prime indicates that the standard conditions for the reaction have been modified to pH = 7.0 instead of pH = 0) and 37°C (rather than 25°C).]
- Q = 0.04
First we can find DG°':
DG°' = RT ln K = -(8.314 J/molK)(310 K) ln 5000 DG°' = - 2.195 x 104J/mol We can now find DG:
DG = DG°' + RT ln Q DG = - 2.195 x 104J/mol + (8.314 J/molK)(310 K) ln 0.04 DG = - 3.025 x 104J/mol = - 3 x 104J/mol
In Chem 109 we used le Chatelier's Principle to predict the effects of changes in temperature on reaction equilibria. Thus, for example, for an exothermic reaction, if T increases heat must have been added, therefore the reaction will use up some of the extra heat, driving the reaction to the left and lowering the temperature back toward its original value. With our new thermodynamic knowledge we can determine this relationship quantitatively.
and
Substituting we can say
dividing by - RT:
Assuming DH° and DS° are constants (a reasonable approximation if DT is not large) this is now in the form of the equation of a straight line:
Plotting:
So we can find both DH and DS (as seen in the lab).
We can also find these values non-graphically by looking at a system at two different temperature and doing a bit of algebra:
Subtracting (2) from (1) we get:
And using log rules:
Another common form is:
Reversible vs. irreversible processes. In order for a process to be completely efficient we say it must be reversible - that is we can get back to the starting point without energy loss. At the microscopic level (atoms and molecules or smaller) processes are frequently reversible. However at the macroscopic level there are inevitably heat losses due to friction etc.
For real processes a further complication arises in that the only energy available is the Free Energy, so we measure the efficiency of a process by how much of the free energy (DG) is captured.
As state above, in real processes some energy is lost to the environment as heat. Folks often lament this loss because it appears to be wasted - wouldn't it be better to capture all the DG to do work? If so why hasn't nature done so? After all life on Earth has had somewhere around 3.5 x 109 years to get it right, and glycolysis still "wastes" about 40% of the available DG. Why? Unfortunately in order to capture the maximum amount of DG we have to accomplish things in very small equilibrium steps. Very highly simplified, if you're very efficient , you're also very slow, and you get eaten! For human designed processes you have to be somewhat inefficient to get anything done during your lifetime.
So like many things we have to compromise and try to optimize the process between speed and completion versus efficiency. The challenge is to reach this optimum, which generally changes with circumstance -life's tough!
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© R A Paselk
Last modified 19 September 2003
