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Why is ΔG=0 at equilibrium

  1. Jan 2, 2014 #1
    Using this graph: http://postimg.org/image/vkv0ho8yb/

    As the reaction proceeds it stops at the equilibrium as it has the lowest Gibbs energy. However, I don't understand why ΔG=0 at that point. As shown in the arrow isn't the ΔG greater in magnitude than when the reaction goes to completion? The formula given was ΔG=ΔGstd+RTlnQ so when Q=k ΔG=0. However using the graph I don't see how that is true.

    And since all reactions goes to equilibrium, how do we determine how much energy is released? Since at equilibrium, ΔG=0 how can we tell the total amount of energy released after the reaction ends at equilibrium?

    Thanks in advance for the help
  2. jcsd
  3. Jan 2, 2014 #2


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    At equilibrium (and at constant [itex]T[/itex] and [itex]P[/itex]), the Gibbs free energy is at a minimum, as you say.

    It's not that [itex]\Delta G=0[/itex] at equilibrium; it's that [itex]dG = 0[/itex] at equilibrium. At constant [itex]T[/itex] and [itex]P[/itex], the Gibbs free energy changes according to the chemical potentials [itex]\mu_{i}[/itex] and the changes in the amounts of reactants and products [itex]d N_{i}[/itex].

    [itex]dG = \sum_{i} \mu_{i} dN_{i} [/itex]

    At equilibrium, the reactants are converted in to products at a rate equal to the reverse process, so that at equilibrium, [itex]\sum_{i} \mu_{i} dN_{i} =0[/itex].

    The total amount of energy (heat?) released (at standard conditions) when the reaction goes to equilibrium I believe can be found with:

    [itex]\Delta H^{o} = \Delta G^{o} + T \Delta S^{o} [/itex]

    All of this you're going to want to look up for yourself, but that's my two cents.
  4. Jan 3, 2014 #3
    So its dG that is zero. But isn't it the formula ΔG=ΔG°+RTlnQ? What is the difference in the two things?
    Last edited: Jan 3, 2014
  5. Jan 3, 2014 #4


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    I think I see what [itex]\Delta G[/itex] is supposed to be.

    [itex]-RT ln(K_{eq}) = \Delta G^{o}[/itex]is the condition for equilibrium, and a defining equation for the standard gibbs free energy of reaction.

    Then [itex]\Delta G[/itex] is the net change (final minus initial) of Gibbs free energy from the initial concentrations of reactants/products to the equilibrium concentrations of reactants/products.

    [itex]\Delta G= \Delta G^{o}+RT ln(Q) = -RT ln(K_{eq})+RT ln(Q)[/itex]

    where [itex]Q[/itex] is the reaction quotient.

    So if [itex]Q<K_{eq}[/itex], [itex]\Delta G[/itex] is negative, and the reaction is spontaneous in the forward direction. If [itex]Q>K_{eq}[/itex], [itex]\Delta G[/itex] is positive, and the reaction is spontaneous in the opposite direction.

    I'm not a chemist, though I have taken a lot of thermodynamics; you'll still want to back up what I'm saying here by looking this up for yourself.
  6. Jan 3, 2014 #5
    The original graph is very schematic, and is provided to indicate that, at equilibrium, G passes through a minimum. The original state is not really an equilibrium state of the system because the reaction is not at equilibrium. So, calculating the G of the system in the original state (or in any subsequent state different from chemical equilibrium) is problematic. If you are willing to calculate G in the original state and in subsequent states of the system as if the system were at equilibrium throughout the progress of the reaction, then the figure would be a little more meaningful. For ideal gas behavior, G would then be equal to the number of moles of each species at the existing concentrations in the system times the ideal gas chemical potential of each of the species (including both reactants and products). Then G would vary as shown on the graph.

    Another way of analyzing things in which ΔG = 0 would actually be the condition for equilibrium would be to consider a von Hoff cell. Here, the contents of the reactor would be at equilibrium at the existing temperature and pressure. Pure reactants would be added to the cell in stoichiometric proportions through semi-permeable membranes at the same partial pressures as in the reactor, and pure products would be removed through semipermeable membranes at their partial pressures. In this mode of operation, the sum of the free energies of the products would be equal to the sum of the free energies of the reactants, irrespective of how much material was processed.

    As far as the heat effects are concerned, you have to focus on the enthalpy change (not the change in free energy). If you know how to calculate the enthalpy of a gas mixture (containing reactants and products), then getting the heat effect is no problem.
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