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I have hit what seems to be a contradiction concerning entropy.

In my class, we measured the entropy change of photosynthesis which is obviously a non-spontaneous process. We used the formula, $$\Delta S_{universe}=\Delta S_{system} +\Delta S_{surroundings}.$$ For this process, we found that the entropy change of the universe was negative. That makes sense, but, I was told that for an isolated system, the entropy change of the universe is always greater than or equal to 0.

So I seem to have hit a contradiction because we can think of the universe as an isolated system, so the entropy change for the universe must always be greater than or equal to 0, but for the photosynthesis process, we found the entropy change of the universe to be negative.

My question is, when can the entropy change of the universe be negative and positive, as we consider the universe to be an isolated system.

If you want to know how we calculated the entropy change of the universe for this process, it is below:

We found the entropy change of the universe for the photosynthesis process to be negative. So here is the chemical reaction. $$6CO_2 +6 H_2O\to C_6H_{12}O_6 + 6O_2.$$ To find the entropy change of the system, we took the standard entropy changes of products minus reactants and got -262 J/mol K. Now this process is at constant pressure, so for the entropy change on the surroundings, we can use the heat of the system, which is equal to the enthalpy change of the system is also equal to the negative of the enthalpy change of the surroundings. We found $\Delta S_{surr}=-9398$ J/mol K. So the sum is -9.66 kJ/mol K

Kyle Kanos
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arch1234
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    Did you perhaps mean that you guys found that the entropy of the surroundings went down? The total entropy must always increase; photosynthesis can't do a damn thing about that. – Danu Oct 05 '14 at 01:24
  • What made you predict the entropy change of the universe based on photosynthesis? Can you elaborate how you got to that result? That one can think of the universe as an isolated system is an assumption, but it has absolutely nothing to do with the question of photosynthesis. – CuriousOne Oct 05 '14 at 01:24
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    Related: http://physics.stackexchange.com/q/95864/50583 – ACuriousMind Oct 05 '14 at 01:25
  • We found the entropy change of the universe for the photosynthesis process to be negative. So here is the chemical reaction. 6CO2 +6 H2O-->C6H12O6 + 6O2. To find the entropy change of the system, we took the standard entropy changes of products minus reactants and got -262J/molK. Now this process is at constant pressure, so for the entropy change on the surroundings, we can use the head of the system, which is equal to the enthalpy change of the system is also equal to the negative of the enthalpy change of the surroundings. We found delta_S_surr=-9398J/molk. So the sum is -9.66kj/mol*k – arch1234 Oct 05 '14 at 01:30
  • @arch1234: So where are the photons in this picture? How can you have photosynthesis without photons???? You can't, so why are you not adding them to your equation? – CuriousOne Oct 05 '14 at 01:44
  • This is how we calculated it in class. I don't understand why you I need to include the photons explicitly in the calculation. I mean, the answer is still the same. It is still negative. And my question is how could it be negative if the entropy change for an isolated system always have to be greater than or equal to 0? – arch1234 Oct 05 '14 at 01:49
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    @arch1234: If you are calculating the entropy change in the "universe", then you have to have the part of the universe that creates the light for the photosynthesis in your calculation, otherwise you are not calculating the entropy change in the whole system. Of course the entropy change in parts of the system can be negative, that's not a problem. – CuriousOne Oct 05 '14 at 02:05
  • Okay sure. So I can just replace the change in entropy of the universe with the total change in entropy, which is still an isolated system. And my argument is still there and my question is still valid. – arch1234 Oct 05 '14 at 02:21
  • @arch1234: Unless you explicitly calculate the entropy change needed to make those photons, all you have done is to make an accounting mistake. – CuriousOne Oct 05 '14 at 03:13

2 Answers2

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For this process, we found that the entropy change of the universe was negative.

You found nothing of the sort. You instead found that a tiny, tiny fraction of the photons emitted by the Sun hit the Earth, and that a tiny fraction of those photons that do hit the Earth trigger an endothermic reaction. The second law of thermodynamics does not prohibit such a reaction.

In arriving at your conclusion, you ignored all of those photons that hit the Earth and merely made the Earth get a bit warmer, you ignored all of those photons that didn't hit the Earth, and you ignored the fusion processes at the center of the Sun that ultimately created those photons. Those fusion processes are the second law of thermodynamics at work. The energy released by those fusion processes represent a very large increase in entropy. The slow process by which that energy makes its way to the surface of the Sun: That's also the second law of thermodynamics at work. This transfer diffuses energy from the Sun's core throughout the Sun. The thermal photons emitted by the Sun is yet another example of the second law of thermodynamics at work. This disperses that energy throughout the universe.

The second law of thermodynamics does not say that every process that happens must increase entropy. Energy can be harnessed to locally decrease entropy. The second law puts limits on that decrease, but it does not say that such reactions cannot happen. For example, I use my air conditioner during the summertime to decrease the entropy of my house. This does not violate the second law of thermodynamics. The entropy of the universe increases when I use my air conditioner. And as you found, plants use energy to locally decrease entropy in the plant. This does not violate the second law of thermodynamics. The entropy of the universe increases.

David Hammen
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  • This makes sense. But, is it true that the total entropy system is isolated and that for an isolated system, entropy must increase or stay constant? – arch1234 Oct 05 '14 at 03:59
  • @arch1234 - Strictly speaking, no. The second law of thermodynamics is a statistical law. Systems comprising a small number of particles can and do violate the second law of thermodynamics. On the other hand, the probability that a system comprising an Avogadro's number of particles (which isn't that large on our everyday scale) will violate the second law of thermodynamics becomes incredibly small, so small that it is essentially zero. – David Hammen Oct 05 '14 at 04:30
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ΔS universe = ΔS system +ΔS surroundings .

hotosystem I of plants has a quantum efficiency of 0.99 and an energy efficiency of >0.96. The system may be considered a canonical ensemble which consists of a Thermal bath in which the particles are suspended and monochromatic light emitted ny an incandescent lamp with a definite Tr. Please note that the lamp does not form part of the system, just as for heat exchanges in a thermodynmic system one does not consider the heating element. Now this system violates the second law. By absorption the ΔS of the light field decreases by 4/3 Uv/Tr, where Uv = hv. The absorption process for a mirror absorption/emission pigment such as chlorophyll has deltaS = 0. There are also no other entropy changes in the photosystem itself (configurationS = 0) and the small amount of heat and fluorescence emitted (quantum yield (0.99) is less than the entropy loss by the light field. So the entropy balance is negative.

robert jennings
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