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Spontaneity and Reaction Coupling

  1. Aug 3, 2017 #1
    Greetings everyone! I have been a long-time lurker on here and have just recently signed up. I'm unsure if this was the correct forum to post this in, so I apologize in advance. My question isn't exactly homework per se - I am currently studying for my MCAT and have been trying to wrap my head around a particular concept recently; I would love it if any of your bright individuals out there could share some insight.

    1. The problem statement, all variables and given/known data
    For a reaction that is non-spontaneous at ALL temperatures (dH is positive and TdS is negative), can this ever actually occur? (Thermodynamically speaking). Furthermore, if you were to HYPOTHETICALLY couple such a reaction in a living system (say to ATP or an even higher energy carrier such as cAMP), or in vitro to something of a similar nature, would this reaction (non-spont at all T) occur or not occur even if the combined delta G of the coupled reaction turned out to be negative? (i.e. [positive delta G of the rxn that is non-spont at all T] + [negative delta G of the highly spont. 2nd rxn] = negative overall delta G).

    2. Relevant equations
    dG = dH - TdS

    3. The attempt at a solution
    I want to say that a reaction which is non-spontaneous all temperatures will never occur (thermodynamically), as no amount of energy addition will allow this process to occur.
    But then if it is coupled to a highly exergonic reaction, in which the total delta G for the overall reaction becomes negative (thus spontaneous), the math says it should occur, but from the statement above, it shouldn't? Or does coupling completely override the non-spontaneity condition at all temperatures for the first reaction, whereby the coupled reaction is now considered a completely separate reaction with its own associated enthalpy change and entropy change?
     
  2. jcsd
  3. Aug 3, 2017 #2

    Ygggdrasil

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    Science Advisor

    The relevant equations you list is more accurately written as:

    ΔG° = ΔH° -TΔS°

    where ΔG°, ΔH°, ΔS° represent the change in free energy, enthalpy, and entropy under standard conditions (notably, all reactants and products at 1M concentration). So, yes if ΔH° > 0, and ΔS° < 0, and you have all reactants and products at 1M concentration, the reaction will proceed in the reverse direction.

    This does not, however, mean that you can never get the reaction to go in the forward direction. The relevant equation to consider here is:

    ΔG = ΔG° + RT ln(Q)

    where Q is the reaction quotient (essentially, the concentration of product and divided by the concentration of reactant). Even if ΔG° > 0 if Q is sufficiently small, the overall ΔG of the reaction can still be negative, allowing the reaction to occur. In many metabolic pathways, one can see how reactions preceeding or following a particular chemical reaction can help couple these favorable steps to allow an unfavorable reaction to occur. For example, if a highly exergonic reaction precedes an unfavorable reaction, there might be a buildup of reactants that makes Q small enough for ΔG to be negative and the reaction to proceed forward spontaneously. Similarly, if an exergonic reaction follows the unfavorable reaction, the concentration of products will be small, again making Q small.
     
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