Chemistry - Is there a reason for the mathematical form of the equilibrium constant?
Solution 1:
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:
$$K=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$
Solution 2:
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 $K_{\text{equilibrium}}= \exp(-\Delta G^\circ/RT)$, where $\Delta G^\circ$ is the change in the Gibbs energy between reagents and products in their standard state, $R$ is the gas constant and $T$ is the temperature.
Solution 3:
I will make "the proof" mentioned by Silvio Levy. Let's consider a chemical reaction, written as:
$$\sum_{i} \nu_i \ce{A}_i \longrightarrow 0$$
where $\nu_i$ are the stoichiometric coefficients of the chemical species $\ce{A}_i$. 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
$$\mu_i = \mu_i^\circ + RT\ln\left(\frac{f_i}{f_i^\circ}\right)$$
where $f_i$ is the fugacity of the species $\ce{A}_i$. Then:
$$\begin{align} \sum_i \nu_i \mu_i &= \sum_ i \nu_i \left[\mu_i^\circ + RT\ln \left(\frac{f}{f^\circ}\right)\right] \\ &= \sum_i \nu_i \mu_i^\circ + \sum_i \nu_i RT \ln \left(\frac{f}{f^\circ}\right) \\ &= \sum_i \nu_i \mu_i^\circ + RT \ln \left[ \prod_i \left(\frac{f}{f^\circ}\right)^{\nu_i} \right] = 0 \end{align}$$
Then
$$\frac{\sum_{i} \nu_i\mu_i^\circ}{RT} = -\ln\left[ \prod_i \left(\frac{f}{f^\circ}\right)^{\nu_i} \right] \Longrightarrow \frac{\Delta_\mathrm{r}G^\circ}{RT}= -\ln\left[ \prod_i \left(\frac{f}{f^\circ}\right)^{\nu_i} \right]$$
By definition, activity is define as :
$$a_i=\frac{f_i}{f_i^\circ}$$
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: $a_i = [\ce{A}_i]/c^\circ$. 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.
EDIT : This is a misconsception to think that fugacities are only define for gaseous species. This is absolutely false and we can get them for liquid and solid too ! I really insist on that because this something use a lot to determine equilibrum
Solution 4:
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:
$$\ce{a X + b Y -> c Z}$$
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:
$$P(reaction)=P(\ce{a X} \in \Delta V)\times P(\ce{b Y} \in \Delta V) \times P(\ce{X\bond{->}Y} \in \Delta V)$$
Wherein the probability $P$ of an atom being in certain volume $\Delta V$ is multiplied with the final probability of a collision (here displayed as $\ce{X\bond{->}Y}$) happening in said volume. The probability of a molecule being in a set space can be determined from its concentration:
$$C=\frac nV$$
$$P(\ce{1X} \in \Delta V)= C \times 6.02 \times 10^{23} \times 10^3 \frac{L}{m^3} \times \Delta V$$
This equation assumes that $\Delta V$ is in cubic meters and concentration is in $\mathrm{mol/L}$, 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:
$$P(\ce{X} \in \Delta V)=K_S×C(\ce{X})$$
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:
$$P(reaction)=K_R K_S^{a+b} [\ce{X}]^a [\ce{Y}]^b$$
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)
$$\ce{a W + b X <=> c Y + d Z}$$
$$K_\mathrm{tot}= \frac{K_{R\ce{A}} K_S^{a+b} [\ce{W}]^a [\ce{X}]^b}{K_{R\ce{B}} K_S^{c+d} [\ce{Y}]^c [\ce{Z}]^d }$$
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.
$$1 = \frac{K_{R\ce{A}} K_S^{a+b} [\ce{W}]^a [\ce{X}]^b}{K_{R\ce{B}} K_S^{c+d} [\ce{Y}]^c [\ce{Z}]^d }$$
It can be easier however, to simplify repetitive calculations to just the concentrations, and create a new constant from all the constants shown:
$$\frac{K_{R\ce{B}} K_S^{c+d}}{K_{R\ce{A}} K_S^{a+b} }=\frac{[\ce{W}]^a [\ce{X}]^b}{[\ce{Y}]^c [\ce{Z}]^d }$$
This is the equilibrium constant, although it does change with temperature as the average kinetic energy of particles changes. The calculations 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.