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TLDR : Can someone explain the graph at 27:40 in the following lecture: Full section length is from 21:12 to 27:40

https://www.youtube.com/watch?v=Cc2l1QTTZA4&list=PLA62087102CC93765&index=15 at 21:12, the professor takes a general chemical reaction and tries to show where along its reaction stage, equilibrium lies ( I think ).

At 24:05, he begins to plot $\Delta G$ as a function of the reaction. My first question is, what does it mean to plot $\Delta G$ as a function of the reaction? I understand talking about the change in the Gibbs free energy of a particular reaction. I.e $\mu_{products}n_{products} - \mu_{reactants}n_{reactants}$, but I don't understand what the professor means in the context of the graph.

Secondly I don't understand why he has shown the $\Delta G_{products}$ to be at a lower point on the graph than $\Delta G_{products}$ at 24:40 . Is this simply because any spontaneous process acts to reduce the Gibbs free energy?

My final query is at 27:04 he says $\Delta G$ must decrease just from the entropy of mixing. Can someone explain where this is coming from?

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  • $\begingroup$ If you've ever plotted net energy as a function of the reaction progress in chemical kinetics, where the highest point is the (transition state, activation energy) the professor is doing the same thing but instead plotting $\Delta G$ as a function of the reaction progress. $\endgroup$
    – aditya_stack
    Commented Dec 17, 2019 at 16:28
  • $\begingroup$ I haven't done so before. I am trying to follow this lecture course for context in my current lecture course on thermodynamics. This is the first time the professor in this online lecture series has drawn such a graph, I am guessing this is a physical chemistry thing. If you understand the graph, could you please post an answer explaining it? $\endgroup$
    – Vishal Jain
    Commented Dec 17, 2019 at 16:35
  • $\begingroup$ I am a high school student so I am not sure if reaction progress is an actual variable, but based on my current understanding it seems that it only represents how much percentage of the reaction has been completed. A reaction is 0% complete if the medium contains only reactants and 100% complete if it only contains products. We could instead plot the energy or some other variable as a function of time, but then we would be involving the rate of the rxn, etc so instead we make up a variable called reaction progress that represents how much of the reaction has been completed. $\endgroup$
    – aditya_stack
    Commented Dec 18, 2019 at 17:34
  • $\begingroup$ Also, if we are analysing equilibrium reactions, it doesn't make sense to plot something as a function of time since we are more interested in how the energy, etc. changes with the composition of the reaction medium, not with how the reaction evolves with time. Eg. For most equilibrium reactions the reactant and product are at approximately the same energy, which is why they are interconvertible. $\endgroup$
    – aditya_stack
    Commented Dec 18, 2019 at 17:36

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It's confusing because he is using $\Delta G^0$ to represent two different things: (a) going from pure reactants to pure products at temperature T and pressure 1 bar and (b) going from pure elements at temperature T and pressure 1 bar to a mixture of reactants and products at temperature T and pressure 1 bar. I would call the latter just $G^0$ of the mixture. What he's trying to plot is $G^0$ vs extent of the reaction, showing points on the graph for a stoichiometric mixture of the reactants (no products), a corresponding stoichiometric mixture of products (no reactants), and mixtures of reactants and products at various extent of reaction in-between. The minimum point on the graph corresponds to an equilibrium mixture of reactants and products.

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  • $\begingroup$ Thanks, can you also explain why if you dont include the entropy of mixing you get a straight line and if you do, it curves to give a minimum? Also why do the products have a lower $\Delta G$ on the graph? Is this just because we are assuming the reaction to occur spontaneously ? $\endgroup$
    – Vishal Jain
    Commented Dec 18, 2019 at 9:53
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    $\begingroup$ If you don't include the entropy of mixing, then you just have products and reactants in separate containers, and $G^0=n_a\mu^0_A+n_b\mu^0_B+n_c\mu^0_C+n_d\mu^0_D$. You are just varying the amounts of the products and reactants in the containers manually subject to the stoichiometry and the extent of the reaction. This will lead to a straight line. Regarding the lower G, yes, this is because it assumes that the products have a lower G than the reactants (both in the pure state and when mixed). $\endgroup$
    – Chet Miller
    Commented Dec 18, 2019 at 12:40

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