# Why is the equilbrium constant expressed the way it is?

Why are the two molarity's multiplied and not added and why is each raised to the power of the coefficient rather than multiplied by it? What is the reasoning behind this form? Was it simply determined that the probability of reaction was proportionate to both and thus they are multiplied together? (Please do not respond with 'it is the concentration quotient when the concentrations are stable' I want to know whether its form is purely empirical or has a theoretical backing)

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Equilibrium is also the state in which the forward and reverse reactions are proceeding at equal rate. Consider a hypothetical reaction

$$\ce{aA + bB} \overset{k_1}{\underset{k_{-1}}{\ce{<=>}}}\ce{ cC + dD}$$

If we assume that the forward and reverse processes are elementary reactions (a big assumption), then:

$$rate_1=k_1 [\ce{A}]^a [\ce{B}]^b$$ $$rate_{-1}=k_{-1} [\ce{C}]^c [\ce{D}]^d$$

At equilibrium: $$rate_1=rate_{-1}$$ $$k_1 [\ce{A}]^a [\ce{B}]^b=k_{-1} [\ce{C}]^c [\ce{D}]^d$$ $$\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$

The ratio of the rate constants is the equilibrium constant:

$$K=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$

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Yes, but if reactions are not elementary, why is the equilibrium constant defined the way it i's?I don't understand why the "equilibrium constant" is a constant if all reactions are not elementary. – Aditya Anand Dec 21 '15 at 10:04

The answer is both. The formula for the equilibrium constant was first observed experimentally (that is, it's empirical) for many many reactions. Then it was justified theoretically in terms of what is now called the Gibbs energy. The reason the formula is multiplicative is that (very roughly) the probability of finding a system in a certain state is proportional to the inverse exponential of that state's energy; and as you know when you take the exponential of a sum you get a product.

If you're curious to know more you should get a hold of Peter Rock's Chemical Thermodynamics. It rocks! :-) If you like math and know calculus well you can read the derivation of the famous formula $K_{\rm equilibrium}= \exp(-\Delta G^0/RT)$, where $\Delta G^0$ is the change in the Gibbs energy between reagents and products in their standards state, $R$ is the gas constant and $T$ is the temperature.

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After much research and work, I wrote a little explanation (not that the other answers weren't good, they just weren't well written for someone who didn't know about thermodynamics and other concepts.....):

The below results were determined experimentally but this explanation gives some insight into why they are the way they are. Whilst many factors affect the probability of a reaction occurring, and thus the rate, the affect of concentration can be quite easily determined. Consider the following reaction:

$\ce{aX + bY→cZ}$

The probability of it occurring can be broken down into the probability of the particles reacting in a space, and the probability of them being in a space, thus:

$prob.(reaction)=prob.(a-set-of-a-Xs-in-∆V)× prob.(a-set-of-b-Ys-in-∆V)×prob.(react in∆V)$

The probability of a molecule being in a set space can be determined from its concentration:

$C=mol/L$

$P(one X in ∆V)=mol/L×6.02×10^23×10^3×∆V$

This equation assumes that ∆V is in meters cubed and concentration is in mol/L, however this may not be the case. The main concept is that the probability of one particle being present in some area is proportional to the concentration and some scaling factor:

$P(X in ∆V)=K_S×C_x$

Based on basic probability, it is known that the probability of a particles being in the volume is equal to the probability of one particle being in there to the power of a. Thus from this, assuming the scaling constant is K_S, and that the probability of a reaction occurring at the given temperature in the given space if particles are present is equal to K_R, the probability of a reaction occurring based on concentration is as shown below:

$P(Reaction)=K_R K_S^{a+b} [X]^a [Y]^b$

If a reversible reaction is considered, the overall direction can be determined by finding the ratio between the rates of the reactions in each direction. Thus the ratio of backward reaction to forward reaction for the below reaction can be identified. (Note the constants on the top and bottoms are different)

$aW+bX↔cY+dZ$

$Overall Direction= \frac{K_{RA} K_S^{a+b} [W]^a [X]^b}{K_{RB} K_S^{c+d} [Y]^c [Z]^d }$

Dynamic Equilibrium is the state for a reversible reaction, in which the rate of both the forward and backward direction is equal and thus the overall change is zero. In this case, the ratio would equal one.

$1 = \frac{K_{RA} K_S^{a+b} [W]^a [X]^b}{K_{RB} K_S^{c+d} [Y]^c [Z]^d }$

It can be easier however, to simplify repetitive calculations to just the concentrations, and create a new constant from all the constants shown:

$\frac{K_{RB} K_S^{c+d}}{K_{RA} K_S^{a+b} }=\frac{[W]^a [X]^b}{[Y]^c [Z]^d }$

This is the equilibrium constant, although it does change with temperature as the average kinetic energy of particles changes. The calculation’s made on the concentrations come up with a value called the concentration quotient (Q) which is equal to the equilibrium constant when a dynamic equilibrium is reached.

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You can learn a lot about markup in our help center. I encourage you to use it. – Martin - マーチン Aug 4 '14 at 10:16
sorry, i couldn't find any quick reference in time. I'll try and fix this soon – J-S Aug 5 '14 at 2:38
"Based on basic probability, it is known that the probability of a particles being in the volume is equal to the probability of one particle being in there to the power of a." - does it mean "to the power ot ifs molar quantity?" O_o – CopperKettle May 3 at 12:40

I will make "the proof" mentioned by Silvio Levy. Let's consider a chemical reaction, written as:

$$\sum_{i} \nu_i \ce{A}_i \longrightarrow 0$$

At equilibrium we have

$$\Delta_\mathrm{r} G = \sum_i \nu_i \mu_i = 0$$

and we also know that the chemical potential under non-standard conditions is given by

$$\mu_i = \mu_i^\circ + RT\ln a_i$$

where $a_i$ is the activity of species $\ce{A}_i$. Then:

\begin{align} \sum_i \nu_i \mu_i &= \sum_ i \nu_i (\mu_i^\circ + RT\ln a_i) \\ &= \sum_i \nu_i \mu_i^\circ + \sum_i \nu_i RT \ln a_i \\ &= \sum_i \nu_i \mu_i^\circ + RT \ln \left( \prod_i a_i^{\nu_i} \right) = 0 \end{align}

Then

$$\frac{\sum_{i} \nu_i\mu_i^\circ}{RT} = -\ln\left( \prod_i a_i^{\nu_i} \right) \Longrightarrow \frac{\Delta_\mathrm{r}G^\circ}{RT}= -\ln\left( \prod_i a_i^{\nu_i} \right)$$

Taking the exponential we obtain

$$\exp\left(\frac{-\Delta_\mathrm{r}G^\circ}{RT}\right) = \prod_i a_i^{\nu_i}$$

and we can conclude

$$K^\circ = \exp\left(\frac{-\Delta_\mathrm{r}G^{°}}{RT}\right)=\prod_{i}a_i^{\nu_i}$$

Here it's the exact result but we can make some approximations (obviously we do it, it's better for learn) and in the case of low concentrations, the activity of a solute can be approximated to the ratio of its concentration over the standard concentration: $a_i = c_i/c^\circ$. That's why we write most of the time:

$$K^{°}=\prod_{i}[\ce{A}_i]^{\nu_i}$$

In this reaction, the stoichiometric coefficients of the reactants are negative and that's where the "fraction" comes from. It's exactly the same as you can find in all other answers.

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I hope you don't mind me touching up a little bit on the proof. It's absolutely correct but I think it's better (remains more general) to simply write $\mu_i = \mu_i^\circ + RT\ln a_i$ since fugacity only applies to gaseous species. – orthocresol May 3 at 12:58
Oh yes you're right. I have a habit of always thinking in terms of gas that's why I did it. – Shadock 2 days ago