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I was confused about why ATP has a lower entropy than ADP when ATP has 1 more phosphate group. According to https://chem.libretexts.org/Textbook_Maps/Biological_Chemistry/Metabolism/ATP%2F%2FADP#Why_is_ATP_hydrolysis_an_exergonic_reaction.3F It states that

The entropy, which is the level of disorder, of ADP is greater than that of ATP. Therefore, due to thermodynamics, the reaction spontaneously occurs because it wants to be at a higher entropy level. Also, the Gibbs' free energy of ATP is higher than that of ADP. Naturally, molecules want to be at a lower energy state, so equilibrium is shifted towards ADP.

Can someone please explain to me why does ATP has a lower entropy than ADP?

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    $\begingroup$ It doesn't $ $. $\endgroup$ – Mithoron Oct 4 '18 at 23:20
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If you treat the entropy as a "level of disorder" the overall entropy of ADP and phosphate seems to be higher than the entropy of ATP.

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  • $\begingroup$ Sorry for duplicating your answer! I think we were typing them at the same time. $\endgroup$ – owjburnham Oct 5 '18 at 11:04
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The key is, I suspect, in the full reaction, assuming that the reaction is the production of ADP and free inorganic phosphate:

$\ce{ATP -> ADP + P_{i} }$

When one molecule disassociates, the total entropy of the products is typically greater, simply because (in general) there are more configurations and conformations available to two molecules than there are to one.

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  • $\begingroup$ Do you mean that it is the overall entropy of ADP+Pi that is greater than ATP instead of the entropy of ADP larger than ATP? $\endgroup$ – Sharon Oct 5 '18 at 18:48
  • $\begingroup$ Yes. If anything, I would imagine that (all other things being equal) the entropy of ADP alone would be lower than that of ATP, as it has slightly fewer possible conformations. $\endgroup$ – owjburnham Oct 5 '18 at 21:31

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