# Predicting time taken to reach equilibrium

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

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

Image taken from Chemistry Libre Texts.

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?

• I wouldn't even call that a flaw. The graphs are made for kids who don't have the concept of $\infty$ yet. Sep 19 at 7:52
• Also, related: chemistry.stackexchange.com/questions/57075/… Sep 19 at 7:52
• The graphs are made by kids who don’t know about asymptotic attainment of equilibrium yet. Sep 19 at 9:46
• 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. Sep 19 at 20:28

• 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. Sep 19 at 21:25