Net Reaction: $$\ce{6CO2(g) +6H2O(l) -> 6O2(g) + C6H12O6(aq)}$$

The net reaction shows that 12 moles become 7 moles; 6 of gas,and 1 complex molecule. This implies a decrease in entropy over the net reaction.

However, from reading further sources, it seems that the change in entropy is positive, but they don't justify it. Why is the overall change in entropy positive, over the series of reactions that make up this process? Could this be linked to the role of UV Light in this reaction?

  • 2
    $\begingroup$ The state of glucose should be solid or aqueous. $\endgroup$
    – Zhe
    Aug 5, 2020 at 16:26
  • 1
    $\begingroup$ Just looking at the net reaction, you would think the entropy decreases. Do you have any evidence that is increases? If you study photosynthesis vs glycolysis (the reverse reaction), you will learn that glycolysis is the direction toward equilibrium (, and that it is exothermic in that direction). It would be great if you could write more about your question to make sure it does not get closed. $\endgroup$ Aug 5, 2020 at 19:40
  • $\begingroup$ State of H2O should be liq, and are you including the photon absorbance? $\endgroup$
    – Andrew
    Aug 5, 2020 at 19:47
  • $\begingroup$ What do you sketch lower entropy and stores energy. It is something apparently deeply linked to life. This is a field which poses a big of puzzles and misunderstanding. Note that this is true if we look at compartments. But still quite peculiar of living things and their products. Overall photosynthesis increase the universe entropy as for hundred photons impinging not all are absorbed and not all are entering the energy balance of the reaction you have wrote. Overall efficiency is few percent, or may be 0.3%, I've forgot the precise value. $\endgroup$
    – Alchimista
    Aug 6, 2020 at 11:14
  • $\begingroup$ Light is a reactant not a catalyst here $\endgroup$
    – Andrew
    Aug 6, 2020 at 23:28

1 Answer 1


To answer this question, we need to be clear whether we are talking about the change in entropy of the system, the surroundings and the Universe.

With regards to the entropy of the Universe, the answer is direct: photosynthesis is a process we observe to happen on a macroscopic scale. Therefore, by the second law of thermodymamics, photosynthesis must cause the entropy of the Universe to increase, period.

With regards to the entropy of just the chemical reagents, if we consider those as our system, then indeed the entropy of formation of glucose is negative, equal to approximately $\mathrm{-40\ cal\ mol^{-1}\ K^{-1}}$.

The way to reconcile these facts is to note that there is a significant increase in entropy which comes from the destruction of high-energy visible photons from the Sun, whose energy is scattered into chemical bonds and multiple low-energy far infrared photons.

In a simplified analysis, if three photons of $\mathrm{700\ nm}$ light are absorbed to generate one molecule of glucose and excess energy is radiated away with black body spectrum at $\mathrm{300\ K}$, then photosynthesis is allowed by the second law of thermodynamics, so long as the photosynthetic efficiency is less than 88%. The observed efficiency of photosynthesis using red light is on the order of 59%, so this requirement is cleared with room to spare.

Depending on the level of detail you want, it gets complicated to work through all the checking and balancing of entropy changes at every individual microscopic step, but the overall results must hold as stated above.


Negative entropy and photosynthesis. Wesley Brittin and George Gamow, Proceedings of the National Academy of Sciences, 1961, 47 (5), 724-727. DOI: 10.1073/pnas.47.5.724

Entropy balance in photosynthesis. Wolfgang Yourgrau and Alwyn Van Der Merwe, Proceedings of the National Academy of Sciences, 1968, 59 (3), 734-737. DOI: 10.1073/pnas.59.3.734

Entropy production and the Second Law in photosynthesis. Robert S. Knox and William W. Parson, Biochimica et Biophysica Acta (BBA) - Bioenergetics, 2007, 1767 (10), 1189-1193. DOI: 10.1016/j.bbabio.2007.07.004


Not the answer you're looking for? Browse other questions tagged or ask your own question.