Chemical kinetics for cyclic reversible reaction

Question

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

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:

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

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

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

Since

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

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

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} $

Answer 1

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:

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

Answer 2

[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.