Suppose we have chemical equation: $$\ce{ aA + bB <=> cC + dD }$$

then equilibrium constant is defined: $$K=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$

but why don't we define it as: $$K=\frac{cd[C][D]}{ab[A][B]}$$

or even as: $$K=\frac{c[C]+d[D]}{a[A]+b[B]}$$

As adding concentrations is additive and not multiplicative. And because: $$\ce{ aA + bB <=> cC + dD }$$ is the same as: $$\ce{ cC + dD <=> aA + bB }$$ then we should have: $$K=\frac{[A]^a[B]^b}{[C]^c[D]^d}$$ and specifically: $$\frac{[A]^a[B]^b}{[C]^c[D]^d}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$ but trying some examples shows that this isn't the case.

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    $\begingroup$ Mostly because of $\ln K = -{\Delta G\over RT}$. $\endgroup$ Commented Feb 7, 2016 at 21:47
  • $\begingroup$ @IvanNeretin never heard of that before, could you please give more detail? $\endgroup$
    – user153330
    Commented Feb 7, 2016 at 21:57
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    $\begingroup$ Why, en.wikipedia.org/wiki/… $\endgroup$ Commented Feb 7, 2016 at 22:02
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    $\begingroup$ chemistry.stackexchange.com/questions/4262/… $\endgroup$
    – Mithoron
    Commented Feb 8, 2016 at 21:28
  • $\begingroup$ The final pair (with reciprocals) is convention only. Had we gone with reactants divided by products, the equation given by @IvanNeretin would have no minus sign, but would remain otherwise unchanged. There's no reason that a forward and backward reaction should have the same equilibrium constant, only that their equilibria will occur under the same conditions. The one that bugs my students more is that if you look at a rxn multiplied by a constant, K changes. That is, K for A --> B is not the same K as for 2A --> 2B. $\endgroup$ Commented Feb 8, 2016 at 22:44

3 Answers 3


It can be understood in simple terms with just few intuitive keys (of course it is not very rigorous).

  1. Take in mind that $K$ is a measure of how displaced to reactive or equilibrium the reaction is. So it should be greater when products concentration is relatively high (that is why they go into the numerator) and small when reactive concentrations are relatively high.

  2. In elemental reactions (technical term) is like a balance between the speed of the reaction in the direction of products and the reverse reaction.

  3. A reaction only occurs when its reactives are in the same region of the space. So, if you have an elmental reaction of the form $a A + b B \rightarrow $, for the reaction takes place it is needed that $a$ molecules of $A$ and $b$ molecules of $B$ collide. This is the main point: How probably is this event? Well, if you only need one molecule of $A$ ($a=1$), the probability that an $A$ molecule be in this specific place (at some time) is proportional to the $A$ concentration. If $(a=2)$ we need another $A$ molecule, as the probability for a second molecule being in that space region (at some time, in particular at the same time) is independent, the probability of finding two $A$ molecules in the same space region (at the same time) is the product of the two individual probabilities (is like throwing a dice two times). In general, for the mixture the probability of the encounter is proportional to $[A]^a [B]^n$.

  4. The reaction speed in that direction is proportional to the number of encounters.

That is why you need to make products. Of course equilibrium constant $could$ be defined in other ways, for example, suppose that we define them as $K_{new}=ln(K_{old})$. In such case for the logarithm properties you'll have other expressions that includes sums. But there is not advantages of defining $K$ in that way.


At equilibrium, the forward rate of reaction is equal to the backward rate of reaction. Let us consider the equation:

$$\ce{aA +bB<=>cC +dD}$$

The forward rate of reaction is given by $$\text{rate}_\mathrm{f}=K_\mathrm{f} \cdot [\ce{A}]^a [\ce{B}]^b$$

The reverse rate of reaction is given by: $$\text{rate}_\mathrm{r}=K_\mathrm{r} \cdot [\ce{C}]^c [\ce{D}]^d$$

When the two are equal: $$K_\mathrm{f} \cdot [\ce{A}]^a [\ce{B}]^b=K_\mathrm{r} \cdot [\ce{C}]^c [\ce{D}]^d$$ $$\frac{K_\mathrm{f}}{K_\mathrm{r}}=K_\mathrm{eq}=\frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$

The equilibrium constant of the reverse reaction:

$$\ce{cC +dD<=>aA +bB}$$

Is given as the inverse of the equilibrium constant of forward reaction. They are (usually) not equivalent: $$\frac{1}{K_\mathrm{eq}}=\frac{K_\mathrm{r}}{K_\mathrm{f}}=\frac{[\ce{A}]^a [\ce{B}]^b}{[\ce{C}]^c [\ce{D}]^d}$$

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    $\begingroup$ Looks good, but powers (a,b,c and d) in the rate equation are not necessarily the stoichiometric values in the balanced chemical equation. $\endgroup$
    – MaxW
    Commented Mar 10, 2016 at 23:34
  • $\begingroup$ @MaxW for elementary reactions, they are. $\endgroup$ Commented May 23, 2017 at 4:33

A background in differential calculus is necessary for the answer to be intuitive. The equilibrium constant, K, is related to the asymptotic behavior of the kinetics model used to describe the system.

For a chemical equilibrium system using linear differential equations the equilibrium constant, K, is often noted with a C subscript as follows:


The following model is one of the simpler examples. (In this example r is defined as reaction rate.)

Chemical Equation

$$\ce{ A <=> C }$$

In the equation for $K_C$ given above the power terms (a,b,c,d) are for higher order reactions. In this example c=1, d=0, a=1, b=0. D and B are irrelevant.

Assume the amount of A lost equals the amount of B gained.

$$ -\frac{\mathrm d[A]}{\mathrm dt} = \frac{\mathrm d[C]}{\mathrm dt} = r $$

The equilibrium constant is calculated at reaction rate equals zero. $$ r = 0 $$

Assume the loss of reactant concentration is proportional to the concentration of reactant in the system. $$ -\frac{\mathrm d[A]}{\mathrm dt} = k[A] $$

Boundary Conditions that fit this system: Assume the reaction is reversible and does not to complete. At time equals infinity $(t=\infty)$ some but not all of the initial $[A]_0$ is consumed. $$[A]_0=[A]_\infty + [C]_\infty$$ $$[C]_0=0$$ Understanding this boundary condition is the key to understanding why the equilibrium constant works only in certian situations.

Solving this system of ordinary differential equations. (This step is tough to understand without a background in differential calculus.) $$[A] = ([A]_\infty + [C]_\infty) - [C]_\infty( 1 - e^{-kt}) $$ $$[C] = [C]_\infty ( 1 - e^{-kt}) $$

As $t \to \infty$, $e^{-kt} \to 0$, $[A] \to [A]_\infty$ and $[C] \to [C]_\infty$. Because $[A]_\infty$ and $[C]_\infty$ are constants in the model, their ratio is constant.

$$K_C = \frac{[C]_\infty}{[A]_\infty}$$

Because of the boundary conditions, if the initial concentration of reactant is increased by some factor y the ratio of $K_C$ remains the same.

$$y[A]_0=[A]_\infty + [C]_\infty$$ $$[A]_0=(1/y)[A]_\infty + (1/y)[C]_\infty$$ $$K_C = \frac{y[C]_\infty}{y[A]_\infty}$$ $$K_C = \frac{[C]_\infty}{[A]_\infty}$$

If there is an initial concentration of the product in the system then the assumption comes into play that the reaction is reversible. This means the reaction can be initially reversed to a point no products exists before it proceeds forward. $$[A]_t+[C]_t = y[A]_0 $$

If one assumption in the system does not match reality, for example if a reaction is irreversible, the equilibrium constant does not work. For a system with reaction rates defined by linear differential equations, for example reversible reactions in solutions, the equilibrium constant works.

  • $\begingroup$ This doesn't seem like an answer to me. $\endgroup$ Commented Mar 10, 2016 at 0:04
  • $\begingroup$ @DGS Your background must be theoretical. $\endgroup$ Commented Mar 10, 2016 at 14:25
  • $\begingroup$ There are many, many ways to define K $\endgroup$ Commented Mar 10, 2016 at 14:26
  • $\begingroup$ I'm not sure if you didn't read the question or if you just reeeeally want to force something in that is unrelated, but you don't even give a wrong answer to this question. You just don't have any kind of answer in there at all. $\endgroup$ Commented Mar 10, 2016 at 14:33
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    $\begingroup$ Oh... The confusion makes sense. The equilibrium constant is the asymptote as time approaches infinity of the kinetics model being used. $\endgroup$ Commented Mar 10, 2016 at 19:29

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