Three reactions occur simulteneously (assume ideal) at $\pu{T = 0^\circ C}$.


a) With $\Delta_r G^\circ_\text{AB} = \pu{400 J/mol}$, $\Delta_r G^\circ_\text{BC} = \pu{-100 J/mol}$, and $\Delta_r G^\circ_\text{CD} = \pu{-200 J/mol}$. If you start with 10 moles of A, calculate how many moles of B, C, and D you have when the system reaches equilibrium. Show all your work. (Hint: mass can neither be created nor destoyed)

b) Order the equilibrium concentrations from largest to smallest. Explain how you might be able to intuitively guess this ranking.

I know that given deltaG_RxN, one can calculate Keq, and use Keq to find the moles at equilibrium of reactants and products. Since the products at the end of one reaction became the reactants of the second, I fed in the mol amount of the first equation into the second equation. enter image description here enter image description here

Confused because in discussion - we went over how to do part B. and I don't think my math lines up.

  • positive delta G_RXN = more reactants than products, and vice versa
  • thus in the first rxn there should be more [A] than [B] @ equilibrium
    • in the second rxn there should be more [C] than [B]
  • and the third rxn there should be more [D] than [C].

what did i do wrong math-wise?


2 Answers 2


What you do is write $[D]=[C]K_{CD}$, $[C]=[B]K_{BC}$, and $[B]=[A]K_{AB}$. So, $$[C]=K_{BC}K_{AB}[A]$$ and $$[D]=K_{CD}K_{BC}K_{AB}[A]$$So, $$[A]+[B]+[C]+[D]=(1+K_{AB}+K_{BC}K_{AB}+K_{CD}K_{BC}K_{AB})[A]=10$$

  • $\begingroup$ The reason for using this approach is that it fully describes all the final product-next reaction molar values based on solely the initial value of A, and given and known constants (${\Delta} G^°_{RXN}$, and R, T). No information is lost. The previous notation failed to take into account reactant Keqs, so it returned incorrect estimation results. The Keq of binding of any intermediary step impacts the Keq of the subsequent step(s). $\endgroup$ Commented Apr 30, 2019 at 11:20

One thing the exercise is lacking is the physical state of A, B, C, and D. If they are all gases or all liquids, you can calculate as shown in question and in answers. Otherwise, you would need more information to complete the exercise.

For some insight regarding part b) of the exercise, you could rewrite your equations as:

$$\ce{A <=> B}$$ $$\ce{A <=> C}$$ $$\ce{A <=> D}$$

with the corresponding Gibbs free energies (by adding the relevant steps):

$$\Delta_r G^\circ_\text{AB} = \pu{400 kJ/mol}$$

$$\Delta_r G^\circ_\text{AC} = \pu{300 kJ/mol}$$

$$\Delta_r G^\circ_\text{AD} = \pu{100 kJ/mol}$$

This shows that the concentration of A will be the highest, followed by D, C, and B.


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