Gibbs Free Energy

Need help preparing for the General Chemistry section of the MCAT? MedSchoolCoach expert, Ken Tao, will teach everything you need to know about Gibbs Free Energy of Chemical Thermodynamics. Watch this video to get all the mcat study tips you need to do well on this section of the exam!
Gibbs free energy is defined as the energy available to do work at a chemical reaction. The MCAT will primarily ask about the Gibbs free energy change of a reaction system, which can be calculated by the equation below, where ΔG is the change of free energy, ΔH is change in enthalpy, and TΔS is the product of system temperature times entropy change. Note that ΔH and ΔS are constant for a given reaction, but the temperature (T) of a reaction system may differ.
ΔG = ΔH - TΔS
The change in free energy of a reaction system tells us about the spontaneity of that reaction. If ΔG is positive, then the reaction is non-spontaneous, which means that, left undisturbed, the reaction will not proceed in the forward direction. If ΔG is zero, the reaction will exist at an equilibrium where reactants and products remain at equal concentrations. If ΔG is negative, the reaction is spontaneous, and as such will favor the forward direction.
Determining if a Reaction will be Endergonic or Exergonic
Consider that changes in enthalpy and entropy can be positive or negative. This implies that any chemical reaction will fall into 1 of 4 cases:
Case 1. ΔH is positive (ΔH is greater than 0) and ΔS is negative (ΔS is less than 0 and -TΔS is greater than 0) so the reaction will always be endergonic.
Case 2. ΔH is positive (ΔH is greater than 0) and ΔS is positive (ΔS is greater than 0 and -TΔS is less than 0) so the reaction will be endergonic if ΔH is greater than TΔS and the reaction will be exergonic if ΔH is less than TΔS. If ΔH = TΔS, the reaction will exist at equilibrium. In other words, at very high reaction temperatures (T) TΔS will be high, and the reaction will be exergonic. At very low reaction temperatures, TΔS will be low, so the reaction will be endergonic.
Case 3. ΔH is negative (ΔH is less than 0) and ΔS is negative (ΔS is less than 0 and -TΔS is greater than 0) so the reaction will be endergonic if ΔH is less than TΔS and the reaction will be exergonic if ΔH is greater than TΔS. If ΔH = TΔS, the reaction will exist at equilibrium. In other words, at very high reaction temperatures (T) TΔS will be high, and the reaction will be endergonic. At very low reaction temperatures, TΔS will be low, so the reaction will be exergonic.
Case 4. ΔH is negative (ΔH is less than 0) and ΔS is positive (ΔS is greater than 0 and -TΔS is less than 0) so the reaction will always be exergonic.
Reactions that fall into case 1 will always favor the reverse reaction, while reactions that fall into case 4 will always favor the forward reaction. Reactions that fall into case 2 and case 3 may favor the forward reaction, the reverse reaction, or equilibrium, depending on the temperature of the reaction system. Keeping in mind these principles (and being able to build out these 4 cases in your head from the original equation discussed above) will allow you to reason through Gibb’s free energy problems.
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[@taniyaduffystudies7694]

Thank you! Made the concept very clear to me!

[@JeanCarlos-kp8ct]

Really helped me understand it! Thank you!!!

[@shanarobinson7672]

This is soo good!

[@syednabid4086]

why did you consider h2o as gas

[@dovidleitman9583]

In combustion reactions H2O is a gaseous product

[@btr8jb]

I'm really late but it's because in combustion reactions the water created is in the form of water vapor (hence why you don't see water running when you burn something)

[@denissemedina4619]

Are we going to be provided with the states of the compounds in the equations as in whether they are liquid, gas, etc?