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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 $[\ce{A}]=2[\ce{B}]$?

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marked as duplicate by Mithoron, Todd Minehardt, Jon Custer, andselisk, airhuff Dec 27 '17 at 0:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ It's a standard result of classical thermodynamics. See any reasonable university level physical chemistry text book, and www2.mcdaniel.edu/Chemistry/ch307.notes/… covers it, but unfortunately that page doesn't display quite properly in my browser - yours may differ. $\endgroup$ – Ian Bush Dec 26 '17 at 8:41
  • $\begingroup$ "If $K_c = 1$ then $2[\ce{B}] = [\ce{A}]$" that is an intuitive guess, but unfortunately not true. Could you explain why you think that must be so? $\endgroup$ – orthocresol Dec 26 '17 at 15:55
  • $\begingroup$ Not a duplicate - the other question is based in rates for elementary reactions, this is about thermodynamics and is the general case $\endgroup$ – Ian Bush Dec 26 '17 at 16:09
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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 $\xi$). 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 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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