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Consider the following reaction, $\ce{A <=>[k_f][k_b] B}$, $k_f$ and $k_b$ are the rate constants of forward and backward reactions respectively. Let the initial concentration of $\ce{A}$ be $a \pu{M}$ and the final equilibrium concentrations would be $(a-x_e)\pu{M}$ and $x_e\pu{M}$ respectively, where $$\frac{k_f}{k_b} = \frac{x_e}{a-x_e}$$i.e,$$x_e = \frac{k_fa}{k_f+k_b}$$ Let $x\pu{M}$ $\ce{A}$ be consumed at any time, $t$, then the net rate of formation of $\ce{B}$ is $$\frac{dx}{dt} = k_f(a-x) - k_b(x)$$ Solving for $x$, with limits from $x=0$ to $x=x$ and $t=0$ to $t=t$ $$\frac{1}{k_f + k_b} ln(\frac{k_fa}{k_fa - (k_f+k_b)x})=t$$ Substituting $x=x_e$, $t \to \infty$, but in many books, it is shown that reactions reach equilibrium after a certian time like the following

From NCERT, CHEMISTRY PART-1

Image taken from NCERT Text book for Chemistry, Class XI, Part-1.

From Chemistry Libre Texts

Image taken from Chemistry Libre Texts.

A google search result: Attainment of Chemical Equilibrium

All the above images show that equilibrium state is attained at a finite time but mathematically it takes $\infty$ time to reach equilibrium.

Is there any flaw in the equation or in the graphs?

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    $\begingroup$ I wouldn't even call that a flaw. The graphs are made for kids who don't have the concept of $\infty$ yet. $\endgroup$
    – Ivan Neretin
    Sep 19, 2021 at 7:52
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    $\begingroup$ Also, related: chemistry.stackexchange.com/questions/57075/… $\endgroup$
    – Ivan Neretin
    Sep 19, 2021 at 7:52
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    $\begingroup$ The graphs are made by kids who don’t know about asymptotic attainment of equilibrium yet. $\endgroup$
    – Karsten Theis
    Sep 19, 2021 at 9:46
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    $\begingroup$ The problem, is anything, is with the mathematics. The equations–if interpreted strictly–only reach equilibrium at infinity. But they implicitly assume that matter is continuous and not made of discrete molecules. In a real-world reaction happening quickly real reactions will reach equilibrium give or take a few molecules difference quickly: in a continuous math world this will not be true. If you wanted a better equation you would need to include a precision level for the math that defined equilibrium as being reached when the difference was a few molecules not exactly zero. $\endgroup$
    – matt_black
    Sep 19, 2021 at 20:28

1 Answer 1

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The flaw is in the graph if you take its labels in their literal sense.

enter image description here

So technically, the label "at equilibrium" is incorrect for the left panel. You can either weaken the label language (center) or change the graph (right) to have a watertight graph and labels.

Source of original graph: https://chem.libretexts.org/Courses/University_of_Kentucky/UK%3A_CHE_103_-_Chemistry_for_Allied_Health_(Soult)/Chapters/Chapter_8%3A_Properties_of_Solutions/8.2%3A_Chemical_Equilibrium

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    $\begingroup$ Don't you think your third equation has a wrong sign ? You write it : $$\frac{dx}{dt} = k_f(a-x) + k_b(x)$$ In my opinion, it should be : $$\frac{dx}{dt} = k_f(a-x) - k_b(x)$$ The same mistake is reported later on, in the next equation. $\endgroup$
    – Maurice
    Sep 19, 2021 at 21:25
  • $\begingroup$ I see that you have edited your question, with my suggestion. Nice ! Thank you ! $\endgroup$
    – Maurice
    Sep 20, 2021 at 11:51

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