Find $[\ce{C}]/[\ce{A}]$ for the following system at equilibrium:

Cyclic reversible reaction ABC

I know that at equilibrium

$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-3}}{k_3} = K_3, \tag{1}$$

but my teacher told me there was another way to express it with rate constant and gave the hint that the numerator is the sum of three terms:

$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-3}}{k_3} = \frac{\ldots + \ldots + \ldots}{\ldots}. \tag{2}$$

So, I use steady state approximation (SSA) to solve this problem:

$$ \begin{align} \frac{\mathrm d[\ce{A}]}{\mathrm dt} &= k_3[\ce{C}] + k_{-1}[\ce{B}] - (k_1 + k_{-3})[\ce{A}] = 0 \tag{3} \\ \frac{\mathrm d[\ce{B}]}{\mathrm dt} &= k_1[\ce{A}] + k_2[\ce{C}] - (k_{-1} + k_{-2})[\ce{B}] = 0 \tag{4} \\ \frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_{-2}[\ce{B}] + k_{-3}[\ce{A}] - (k_2 + k_3)[\ce{C}] = 0 \tag{5} \end{align} $$

$$[\ce{C}] = \frac{k_{-2}[\ce{B}] + k_{-3}[\ce{A}]}{k_2 + k_3} \tag{6}$$

$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}[\ce{B}]/[\ce{A}] + k_{-3}}{k_2 + k_3} \tag{7}$$


$$\frac{[\ce{B}]}{[\ce{A}]} = \frac{k_1}{k_{-1}}, \tag{8}$$

$$ \begin{align} \frac{[\ce{C}]}{[\ce{A}]} &= \frac{k_{-2}k_1/k_{-1} + k_{-3}}{k_2 + k_3} \\ &= \frac{k_{-2}k_1 + k_{-1}k_{-3}}{k_{-1}k_2 + k_{-1}k_3}. \tag{9} \end{align} $$

I don't get the correct answer term just as my teacher said. I get two or four terms in numerator, so my answer is likely wrong. Where am I mistaken?

$ % \documentclass{article} % \usepackage{chemfig} % \begin{document} % % \schemestart % A % \arrow(A--C){<=>[$k_{-3}$][$k_{3}$]}[-60,1.25,,] % C % \arrow(@A--B){<=>[${k_{-1}}$][${k_{1}}$]}[-120,1.25,,] % B % \arrow(@B--@C){<=>[$k_{2}$][$k_{-2}$]}[,1.25,,] % \schemestop % % \end{document} $

  • 2
    $\begingroup$ Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in titles. $\endgroup$
    – Poutnik
    Oct 8, 2022 at 9:49
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    $\begingroup$ with equations (3), (4) and (5), you have three equations and three unknowns. No need for equation (8) (which may not be true). $\endgroup$
    – Andrew
    Oct 8, 2022 at 14:25
  • 1
    $\begingroup$ There is missing molar amount inventory equation, like [A]+[B]+[C] = c_tot, giving 4 equations for 3 variables and 1 degree of freedom for rate constants.// For (8) there is the same objection as for (1). $\endgroup$
    – Poutnik
    Oct 8, 2022 at 14:49
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    $\begingroup$ By solving the equations without using eqn 1 or 8 I found $$\displaystyle \frac{C}{A}=\frac{(k_1+k_{-3})k_{-2} +k_{-1}k_{-3} } { (k_2+k_3)k_{-1} + k_3k_{-2}} $$ $\endgroup$
    – porphyrin
    Oct 9, 2022 at 16:22
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    $\begingroup$ Get B from any equation then substitute into another equation if you want C/A and simplify. You can only get ratios from the three rate equations in this case, you need $c_{tot}=A+B+C$ to get individual concentrations. $\endgroup$
    – porphyrin
    Oct 10, 2022 at 8:30

2 Answers 2


Using the steady state approximation (SSA), we can obtain the following:

$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}\frac{[\ce{B}]}{[\ce{A}]} + k_{-3}}{k_2 + k_3} \tag{1}$$

$$\frac{[\ce{B}]}{[\ce{A}]} = \frac{(k_1 + k_{-3})[\ce{A}] - k_3\frac{[\ce{C}]}{[\ce{A}]}}{k_{-1}} \tag{2}$$

$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}\left((k_1 + k_{-3}) - k_3\frac{[\ce{C}]}{[\ce{A}]}\right) + k_{-1}k_{-3}}{k_{-1}(k_2 + k_3)}\tag{3}$$

$$k_{-1}(k_2 + k_3)\frac{[\ce{C}]}{[\ce{A}]} = k_{-2}(k_1 + k_3) - k_{-2}k_3\frac{[\ce{C}]}{[\ce{A}]} + k_{-1}k_{-3} \tag{4}$$

$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}(k_1 + k_{-3}) + k_{-1}k_{-3}}{k_{-1}(k_2 + k_3) + k_{-2}k_3} \tag{5}$$

If you want to show three terms in the numerator (to match the teacher's hint), you can also write it this way:

$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}k_1 + k_{-2}k_{-3} + k_{-1}k_{-3}}{k_{-1}k_2 + k_{-1}k_3 + k_{-2}k_3} \tag{6}$$

  • 1
    $\begingroup$ I don't think it makes sense to use a steady-state approximation. At equilibrium, every reaction is at equilibrium. There is no net turnover from A to B that is compensated by net turnover from B via C to A, or something like that. Using the identity $$\frac{k_1 k_2 k_3}{k_{-1} k_{-2} k_{-3}} = 1$$ you can probably simplify to the expected simple fraction of two rate constants. Given that the problem statement asked for the complicated expression, it is the correct answer, I guess. $\endgroup$
    – Karsten
    Oct 15, 2022 at 13:07

[from the comments to the OP's question] Equations 8 and 1 hold for "direct equilibria". Note that you go from A to C with B as an intermediary. This means that B influences your equilibrium from A to C, so the constants pertaining to B will also appear in the expression of the equilibrium constant. Same reasoning for A to B. C influences that equilibrium just as B does to that between A and C.

The equilibrium between A and C is not influenced by the mutual reaction to B. This is the same as adding a catalyst. It changes the kinetics, but not the equilibrium constant.

The OP used a steady-state equation (bracket denotes equilibrium concentration): $$ \frac{\mathrm d[\ce{A}]}{\mathrm dt} = k_3[\ce{C}] + k_{-1}[\ce{B}] - (k_1 + k_{-3})[\ce{A}] = 0 \tag{3}$$

At equilibrium, all steps are at equilibrium, so there are two independent equations instead of just the combined one:

$$ k_3[\ce{C}] - k_{-3}[\ce{A}] = 0 \tag{3a}$$ $$ k_{-1}[\ce{B}] - k_1 [\ce{A}] = 0 \tag{3b}$$

Also, the three equilibrium constants are linked, as are the six rate constants:

$$\frac{k_1 k_2 k_3}{k_{-1} k_{-2} k_{-3}} = 1$$

or simpler

$$k_1 k_2 k_3 = k_{-1} k_{-2} k_{-3}\tag{A}$$

So if we take the solution (corrected for the swapped $k_2$ and $k_{-2}$)

$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_2 k_1 + k_2 k_{-3} + k_{-1}k_{-3}}{k_{-1}k_{-2} + k_{-1}k_3 + k_2 k_3} \tag{6}$$

we can expand the first term in the numerator with $\frac{k_{-3}}{k_{-3}}$ and use (A) to eliminate the $k_{-1}k_{-2}$ term in the denominator, we get:

$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{\frac{k_2 k_1}{k_{-3}} k_{-3} + k_2 k_{-3} + k_{-1}k_{-3}}{\frac{k_1 k_2}{k_{-3}} k_3 + k_{-1}k_3 + k_2 k_3} = \frac{k_{-3}}{k_3}$$

This reiterates that at equilibrium, there is no flux "around the triangle". Every single reaction is at equilibrium, and there is no need for a steady-state approximation.

  • $\begingroup$ To troubleshoot the answers, I set the kinetic constants to 5, 6, 7, 10, 1, 21, so that $$5 * 6* 7 = 10 * 1 * 21$$. This made it easier for me to quickly check whether expressions are correct or not, and how to factor out $$\frac{k_{-3}}{k_3}$$ in the last bit of the answer. $\endgroup$
    – Karsten
    Oct 15, 2022 at 15:30
  • 1
    $\begingroup$ The 'eq' subscripts seem excessive as square brackets already imply equilibrium concentration. Any other amount-of-substance concentration is denoted with $c$ anyways. $\endgroup$
    – andselisk
    Oct 15, 2022 at 16:19
  • $\begingroup$ @andselisk I was not aware of that convention. Do you have a source for that? I have seen brackets (always square) for defining the reaction quotient as well. Of course, this is always an approximation, using concentration divided by standard state concentration instead of activities. I have also seen brackets used to define rate laws (where concentrations are typically not a equilibrium. $\endgroup$
    – Karsten
    Oct 15, 2022 at 17:54
  • $\begingroup$ I'm not saying the usage is logical. For rate laws, the values have units, for the equilibrium constant, they don't, so the usage is a bit of a mess. $\endgroup$
    – Karsten
    Oct 15, 2022 at 17:55
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    $\begingroup$ I don't think brackets generally indicate equilibrium concentrations. My understanding is that brackets just denote concentration and that when you are solving for the equilibrium condition, you can either use the subscript (which is rare) or preface the solution with, "At equilibrium. . . ". For example, you could split equation 3 into the first equality (with d[A]/dt), which is true for all concentrations, and then explicitly state the condition of equilibrium results in the second equality (to 0), and with that condition stated, you don't need the eq subscript. $\endgroup$
    – Andrew
    Oct 15, 2022 at 21:57

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