In article <2grmib$lr at mserv1.dl.ac.uk> "" <unknown at dl.ac.uk> writes:
>Now the question: for each photon entering
>photosynthesis: is some of the energy of the photosynthetic photon lost
>as heat due to inefficiencies, or does the set of chemical reactions
>"catalyzed" by the photon require thermal energy in addition to the photon's
>energy? To put into an equation, do we have
>>photon + substrate -> product + heat
>> or
>>photon + heat -> product ?
First, photosynthesis isn't a single reaction but a sequence of
reactions with the overall effect of converting carbon dioxide and water into
glucose. There are two photosynthetic pigments (in a typical higher
plant), each with its own spectrum. Photosynthesis doesn't necessarily
even happen all at once, having so-called light and dark phases. (If I
remember correctly, the latter is somewhat misleadingly named since it
can happen in daytime too; the light phase on the other hand is the
photon-collecting phase and definitely occurs only in the presence of
light.) I think we can all agree that this muddies things slightly since
you can't write the above reactions in terms of monochromatic photons;
that would let us do some very simple calculations, but it just doesn't
work that way.
However, I think that we can agree that the first alternative is
almost certainly right. In outline, photosynthesis does exactly what
this reaction proposes: Its sole energetic input is light which it
converts into bond energy of its products; as far as I know, the amount
of energy provided by absorbed photons is greater than the energy required
to make glucose from the reactants provided so some of it must
definitely be converted into heat. Photosynthesis is amazingly
efficient, so I'm still skeptical that this could be reliably measured
however.
Note that these notes are being written from memory, so it this
matters to you, look it up at the library. There are plenty of books on
the topic, so it shouldn't be too hard.
Marc R. Roussel
mroussel at alchemy.chem.utoronto.ca
