Why do we have the concentrations powered to the coefficients in the equilibrium expression? [duplicate]








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[Why is the rate of a reaction proportional to the concentrations of reactants raised to their stoichiometric coefficients?](/questions/81019/why-is-the-rate-of-a-reaction-proportional-to-the-concentrations-of-reactants-ra)

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$$\ce{A -> 2B} \qquad K\_\mathrm{c} = \frac{[\ce{B}]^2}{[\ce{A}]}$$


1. Why do we have a power that is the same as the coefficient?
2. Assuming $K\_\mathrm{c} = 1$, shouldn't the quantities of the mixture be $2B = A$ (in other words, the amount of $\ce{A}$ in the mixture equals two times the amount of $\ce{B}$ in the mixture)!? So why do we have $[\ce{A}] = [\ce{B}]^2$ rather than $[\ce{A}]=2[\ce{B}]$?




This is due to the way the equilibrium constant is defined. 


A chemical equilibrium is a stable state at some point in the reaction (characterized by the [extent of reaction](https://en.wikipedia.org/wiki/Extent_of_reaction) $\xi$). Because it is a stable state the free enthalpy $G$ has to have a local minimum at exactly that place.  

Therefore we can say
$$
\left( \frac{\partial G}{\partial \xi} \right)\_{p,T} = 0
$$
On the other hand the following equation applies:  

$$
\Delta\_rG\_m = \left( \frac{\partial G}{\partial \xi} \right)\_{p,T} = \sum\_i \mu\_i \cdot \nu\_i
$$
With the free, molar reaction enthalpy $\Delta\_rG\_m$, the [chemical potential](https://en.wikipedia.org/wiki/Chemical_potential) of the component $i$ $\mu\_i$ and it's reaction-coefficient $\nu\_i$ (To understand this consider this general reaction equation: $\ce{\nu\_1A + \nu\_2 B <=> \nu\_3 C}$ - note that the coefficients for the educts are negative by definition). 


Therefore we can say that the following equation applies for the equilibrium state:
$$
\Delta\_rG\_m = 0
$$ 


we can rewrite above formula with the aid of the activity-dependency (basically a concentration-dependency) of the chemical potential ($\mu\_i = \mu\_i^\* - RT\cdot\ln(a\_i)$):  

$$
\begin{split}
\Delta\_rG\_m &= \sum\_i \mu\_i \cdot \nu\_i\\
&= \sum\_i \nu\_i\mu\_i^\* - RT \cdot \sum\_i \nu\_i\ln(a\_i)
\end{split}
$$
Because $a \cdot \ln(b) = \ln(b^a)$ this is the same as
$$
\Delta\_rG\_m = \sum\_i \nu\_i\mu\_i^\* - RT \cdot \sum\_i \ln(a\_i^{\nu\_i})
$$


Now we define the standard, free, reaction enthalpy $\Delta\_rG\_m^\* = \sum\_i \nu\_i\mu\_i^\*$ and because we found out (in the beginning) that for the equilibrium state $\Delta\_rG\_m$ has to be zero we can write:
$$
\Delta\_rG\_m^\* = -RT \cdot \sum\_i \ln(a\_i^{\nu\_i})
$$
Because $ \ln(a) + \ln(b) = \ln(a\cdot b)$ we can rewrite this as
$$
\Delta\_rG\_m^\* = -RT \cdot \ln\left(\prod\_ia\_i^{\nu\_i}\right)
$$ 


The last step is now to define the equilibrium constant $K$ as
$$
K = \prod\_ia\_i^{\nu\_i}
$$


And that is how the coefficients in the recation equation end up in the exponent when calculating the equilibrium constant.


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