Recently at school, while studying equilibrium we learned that in problems where the initial concentrations of reactants and the equilibrium constant are given, there is a special formula to determine wether the change in reactants can be ignored where C is the concentration of the reactant with the smallest concentration and K is the equilibrium constant: L=C/K. We learned that if L is bigger than 500, we can consider the change in the reactants negligible. However this is where my problem lies: i do not see why the C is included, doesn't the size of K by itself tell us wether a little or a lot of reactants is transformed? Not to mention, some cases turn out flat out wrong. For example: A+B=(reversible) 2C.

Initial concentrations: A=10^-20M B=10^-20M. C=0 M


By the equation we learned, L=10^-12 and thus the variation in the reactants should not be neglected. However, after doing the math, x~10^-24 which is 4 orders of magnitude below the beginning concentration of the reactants and completely negligible, is something wrong with my logic? Or did i discover a flaw in my textbook :)


  • $\begingroup$ You can't use it that way because technically C is the smallest, being 0. So L=0. This rule of thumb was developed for systems already in equilibrium, if I remember correctly. $\endgroup$
    – IT Tsoi
    Jun 17 '16 at 7:57

Let's look at your example and set up the ICE (intial-change-equilibrium) problem. Then I'll give you more inclusive of thumb.

Let's say reaction equation is

$\ce{A + B <=> 2C}$$

The law of mass action is:

$$K = \dfrac{[\ce{C}]^2}{[\ce{A}][\ce{B}]}=10^{-8}$$

Let's set up the ICE matrix. The change in concentration $x$, will be subtracted from $[\ce{A}]$ and $[\ce{B}]$ and added to $[\ce{C}]$.

$$\begin{array}{|c|c|c|c|}\hline \ & [\ce{A}]\ (\mathrm{M}) & [\ce{B}]\ (\mathrm{M}) & [\ce{C}]\ (\mathrm{M}) \\ \hline I & 10^{-20} & 10^{-20} & 0 \\ C & -x & -x & +2x \\ E & 10^{-20} - x & 10^{-20} -x & 2x \\ \hline \end{array}$$

Now we pug in to the law of mass action:

$$10^{-8} = \dfrac{x^2}{\left(10^{-20} -x \right)^2}$$

Since all numbers have to be positive, let's take the square root of both sides:

$$10^{-4} = \dfrac{x}{10^{-20} -x }$$

Now we have a ratio that says that $x$ is four orders of magnitude smaller than $[\ce{a}]_i -x$. I could have easily used the initial concentration of $\ce{B}$ Put another way:

$$x \approx K\cdot [\ce{A}]_i$$

If this value is much less than $[\ce{A}]_i$, then we can simplify.


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