If the forward reaction causes an increase in entropy then the backwards reaction causes a decrease in entropy, so the backwards reaction is non-spontaneous. If so, then why does the backwards reaction begin to happen spontaneously and an equilibrium position is reached? I get that when you have mixing you also have a higher entropy because there are more microstates available in the given energy configuration, but why is an equilibrium position reached if the backwards reaction is supposed to be non-spontaneous? And why not just get a 50:50 mixture of that's what will increase entropy.

P.S. 1) I do A2 chemistry. 2) Graphs are more than welcome. 3) Arguments in terms of disorder are not preffered, but please use microstates and 'ways to arrange energy'.

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    $\begingroup$ This is a very broad topic, but here are several posts related to this, please have a look there first to see if they answer your question. Also read about enthalpy and entropy and Gibbs free energy on Wikipedia or an undergraduate phys. chem. text book. $\endgroup$ – porphyrin Aug 1 '16 at 6:57
  • $\begingroup$ I already had a look at other things on the website; they didn't directly address this issue. Also, I've been doing tons of research, and, again, no one directly answered my question. $\endgroup$ – Mathematician Aug 2 '16 at 7:09
  • $\begingroup$ this answer may help, chemistry.stackexchange.com/questions/42906/… $\endgroup$ – porphyrin Aug 2 '16 at 7:35
  • $\begingroup$ It becomes a broad question if you do not state in an explicit way the circumstances in which the reaction take place. Constant $p$ ? $T$ ? $E$?. If you are talking about typical $T$ and $p$ constants experiment, then maximizing the entropy of the system is not a criteria for spontaneity. Also, spontaneous/non-spontaneous do not mean anything so extreme as "it happens" or "It does not" happens at all. $\endgroup$ – user1420303 Aug 2 '16 at 12:27
  • $\begingroup$ I haven't studied it in that much depth, but let's assume constant pressure and temperature. I don't know what 'E' stands for. $\endgroup$ – Mathematician Aug 3 '16 at 5:35

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