I'm not interested in a simple telling of what the properties of the equilibrium constant are and how to use them. I'm more hoping for an explanation of where it came from/how it might be possible to derive it.

My current guess is that it is a probability equation:

The numerator in the equation is proportional to the probability reactants meet and form products in the volume and the denominator is proportional to the probability products meet and form reactants in the volume. This ratio clearly must stay constant (unless the proportionality constants are changed via temperature or such).

Would anyone help?


Equilibrium is the state at which the rates of the forward and reverse reactions are equal.


Let's take a generic reaction:

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

For simplicity's sake let's have both the forward and reverse reactions be concerts fundamental reactions. Then we can write the rate laws for the forward and reverse reactions:

$$\begin{aligned} \mathrm{rate_{forward}}&=k_1 [\ce{A}][\ce{B}]\\ \mathrm{rate_{reverse}}&=k_{-1} [\ce{C}][\ce{D}] \end{aligned}$$

If we set them equal to each other, we can rearrange the equation to get the rate constants on one side and the concentrations on the other.

$$\begin{aligned} k_1 [\ce{A}][\ce{B}]&=k_{-1} [\ce{C}][\ce{D}]\\ \dfrac{k_1}{k_{-1}}&=\dfrac{[\ce{C}][\ce{D}]}{ [\ce{A}][\ce{B}]} \end{aligned}$$

The ratio of rate constants is the equilibrium constant. Thus is derived the law of mass action.

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  • $\begingroup$ I appreciate it, but now it reduces to instead understanding the rate laws for forward and reverse reactions. It is, for instance, possible to derive PV = nRT not through experiment, but through simple logic. Nevertheless thank you. It helped. $\endgroup$ – John Smith Jun 12 '16 at 18:55
  • $\begingroup$ The rate laws arise from the frequency of molecular collisions of particular kind; that's quite obvious. This is a sound reasoning indeed, but there is more to it. Thermodynamics is kinda "more fundamental" than kinetics. Reactions may have complicated mechanisms and do not necessarily obey the simple rate laws. The equilibrium constant, though, is always the way it is, regardless of these details. $\endgroup$ – Ivan Neretin Jun 12 '16 at 18:57
  • $\begingroup$ I believe my question is solved then. I'm aware that some things in thermodynamics are too complicated to understand simply and must be instead determined experimentally, but the equilibrium constant seemed to be the kind of thing which would have a simple explanation as to why it works. Now it makes sense. $\endgroup$ – John Smith Jun 12 '16 at 19:03

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