While heat is not a state variable, it can be written as the product of two variables which are: temperature and entropy, denoted by S:
Q = T SThe units of entropy are obviously J / K. Since it is proportional to the heat content of a system, entropy is an extrinsic state variable.
Entropy is often defined as a measure of the randomness of a system. For example, as the aroma from a cup of coffee diffuses through the room (with the help of convection), the entropy of the air-aroma system in the room increases, because the molecules in the aroma are distributed much more randomly than before. Mathematically, the number of possible positions and velocities for each molecule of 2 - Furylmethanethiol is much greater than when they were all in the coffee cup. Hence an increase in entropy involves an increase in the number of possible states the system may be in.
S = k ln (g)Suppose that we have a 2-dimensional system of four protons arranged at the corners of a square. In the absence of any external electromagnetic fields, the energy of this system is determined by the spins of the protons. When they are all aligned, either up or down, the energy of the system is at a minimum. Since there are two possible configurations with that energy, the degeneracy is 2. When all but one are aligned, the energy is higher, and there are 8 possible configurations: g = 8. When they are aligned in pairs, there are two different energy states: 4 configurations when like spins are at adjacent corners, and 2 configurations when like spins are at opposite corners:
Energy State | Configuration | Degeneracy | Entropy |
---|---|---|---|
1 | all aligned | 2 | k * ln 2 (= S_{1}) |
2 | all but one aligned | 8 | k * ln 8 = 3 S_{1} |
3 | adjacent pairs aligned | 4 | k * ln 4 = 2 S_{1} |
4 | opposite pairs aligned | 2 | k * ln 2 = S_{1} |
Therefore the total entropy of this system is 7 k ln 2 = 6.7 * 10^{-23} J/K. We will consider below the photosynthesis of 60 mols of 690 nm photons. The entropy generated by that process (assuming it takes place at 20 C) is about 24.7 kJ/K. The degeneracy associated with this entropy is e^{1.789 * 1027}. Can your calculator evaluate that number? Using the algebraic rules for exponents and logs, you could in principle write that number down as a 1 followed by 7.77 * 10^{26} zeroes. Assuming the universe is 14 billion years old and that you could write one zero each second, it would take 1.76 billion times the current age of the universe!
It is not uncommon to hear the concept of entropy used in the social sciences. Beware: unless entropy is rigorously defined in terms of degeneracies, conclusions based on statements about entropy (such as the Second Law of Thermodynamics) are meaningless.In this section, we will be concerned with biochemical processes which take place at essentially a constant temperature. Hence the heat energy created by a given reaction will be
ΔQ = T ΔS,where ΔS is the change in entropy of the reactants. We will treat each reaction as a sort of manufacturing process, where the reactants enter into the reaction with a certain energy content, called their free energy (denoted by G), and the products leave the reaction, with a different free energy. The energy difference is the entropic energy
T ΔS = G_{reactants} - G_{products}.This expression of the conservation of energy is the First Law of Thermodynamics, as applied to chemical reactions.
In an exergonic reaction, the free energies of the reactants are greater than that of the products. The entropic energy is then positive, which from the point of view of the manufacturing process represents wasted heat. Not all of this heat is actually wasted: chemical reactions require a certain amount of kinetic energy to proceed, and much of this "wasted" energy provides that kinetic energy. But the excess energy is the energy which we were so concerned with losing in the last section.
In an endergonic reaction, the free energies of the products are larger than the free energies of the reactants. This means that the entropic energy is negative, and the reaction will not proceed without an additional energy source. Endergonic processes are coupled with exergonic ones to create reactions which are as a whole exergonic. Hence all process have in the end a positive entropic energy, and result in an increase in entropy for the system. This is called the Second Law of Thermodynamics.
ΔG = 0.This situation corresponds to death, since no free energy is available to drive endergonic reactions. Life requires a steady state relatively far from equilibrium, and in all cases that we know of, this state is maintained by sustaining non-equilibrium concentration gradients across membranes. This activity requires a constant influx of energy, ultimately from the sun, and the consequent generation of entropy.
We will be interested in four primary sources of energy for endergonic reactions. These are the exergonic reactions
The negative signs on the free energy indicate that the reactants lose energy: the reactions are exergonic. These are used to drive a number of reactions critical to animals and plants.
ATP + H_{2} O -> ADP + P ΔG = - 30.6 kJ / mol NADH + H^{+} + 1/2 O_{2} -> NAD ΔG = - 219 kJ / mol FADH_{2} + 1/2 O_{2} -> FAD + H_{2} O ΔG = - 192 kJ / mol NADPH -> NADP^{+} + H^{+} ΔG = - 218 kJ / mol
Below are just a few of the most important biochemical processes. We have simplified them as much as possible for the sake of clarity, leaving in only those molecules necessary for correct energy accounting. This means that we will include energies associated with water or C O_{2} production in the waste category. Many of these processes are actually many reactions combined into one; some are naturally exergonic, while others illustrate the coupling of exergonic processes to naturally endergonic ones:
Note that Photosystem II actually takes place before Photosystem I (non-cyclic); the names reflect their order of discovery.
Photosystem II 2 γ_{680 nm} + 2 Chlorophyll -> 2 Chlorophyll Photosystem I (cyclic) 2 Chlorophyll + 2 γ_{700 nm} -> 2 Chlorophyll + ATP Photosystem I (non-cyclic) 2 Chlorophyll + 2 γ_{700 nm} -> 2 Chlorophyll + NADPH + ATP Calvin Cycle 12 NADPH + 18 ATP -> glucose Glycolysis glucose -> 2 pyruvate + 2 NADH + 2 ATP pyruvate conversion pyruvate ->AcetylCoA + NADH TriCarboxylic Acid (TCA) Cycle AcetylCoA -> 3 NADH + FADH_{2} + ATP Electron Transport NADH -> 3 ATP FADH_{2} -> 2 ATP "fat burning" palmitate -> 131 ATP pyruvate reduction to lactate pyruvate + NADH -> lactate pyruvate reduction to acetaldehyde pyruvate -> acetaldehyde acetaldehyde reduction to ethanol acetaldehyde + NADH -> ethanol Here the γ represent photons. The energy of 1 mol of 680 nm photons is 175.92 kJ; the energy of 1 mol of 700 nm photons is 170.894 kJ.
G (kJ / mol) acetaldehyde 729 AcetylCoA 923 ethanol 924 glucose 2872 lactate 1337 palmitate 9781 pyruvate 1143
By following a molecule of glucose through the biome, we can see now how the biome functions thermodynamically:
60 γ -> 1 glucose
1 glucose -> 36 ATP
The photosynthesis of glucose required 60 mols of photons, for a total energy of 10,381 kJ. Generation of one mol of glucose gives us a net entropic energy of 7509 kJ. The subsequent respiration of glucose by an animal gives us a further entropic energy of 1770 kJ. Assuming photosynthesis takes place at 20 C and animal respiration at 37 C, we have an increase in entropy of 31.3 kJ / K. We see that this apparently cyclic process entails an enormous input of external energy and the generation of a large amount of entropy.
We can quantify the efficiency of any system as the ratio of the useful energy provided by the system to the energy it requires to provide it:
ε = E_{output} / E_{input},which is of course dimensionless. This is equivalent, for our purposes, to
ε = G_{P} / G_{R},where "P" denotes products and "R" denotes reactants. Typical system-wide physiological efficiencies are about 20 %, although cyclic processes can approach 100 %. This is true because a perfectly cyclic process ends in such a state that the products are the same as the initial reactants, hence their free energies are equal. For our biomic glucose cycle, the efficiency is 10.6 %.
In this applet, we will trace the generation of entropy in detail from photosynthesis to one of three conclusions:
You will be given the number of photons which start photosynthesis, and you will be asked to compute the entropic energy, the entropy generated and the percentage efficiency for each process. Be sure to keep track of the total entropy generated because you will be asked that as well. Use the tables above as needed. Assume that photosynthesis and fermentation take place at room temperature (20 C).
Note that the coefficients in the biochemical processes listed above are always the same. Since 60 mols of photons produces 1 mol of glucose, we see that the energy content of any given constituent will be simply:
- the number of mols of photons given by the applet,
- divided by 60,
- times the coefficient in the reaction,
- times the free energy per mol,
- times the number of times the reaction occurs.
The next section begins our study of electricity.
©2012, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.
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