physical chemistry - Why does the equilibrium constant change on scaling the stoichiometric coefficients of a reaction?

I read in my textbook that if we multiply a chemical reaction by some factor(let's say $b$) its new equilibrium constant becomes $K^b$.But I don't understand why this happens.. What is the difference between the reaction $\ce{A} + \ce{B} = \ce{C}$ and $b \ce{A} + b \ce{B} = b \ce{C}$ ? Aren't the two reactions same? So why does the equilibrium constant change?

Answer

Have a look at this answer of mine. There you can find the derivation of the formula that defines the equilibrium constant

\begin{equation} \log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT} \end{equation}

of a general reaction

\begin{equation} \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C <=> \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F} \end{equation}

and it is also shown how this definition relates to the maybe more familiar definition

\begin{equation} K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ . \end{equation}

Now, the reason why the equilibrium constants of the general reaction above and the same reaction multiplied with a constant factor $\alpha$:

\begin{equation} \ce{\alpha\nu_{1}A + \alpha\nu_{2}B + ... + \alpha\nu_{$x$}C <=> \alpha\nu_{$x+1$}D + \alpha\nu_{$x+2$}E + ... + \alpha\nu_{$n$}F} \end{equation}

are not the same but are related to each other via:

\begin{equation} K_{\alpha} = K_{1}^{\alpha} \end{equation}

comes from the definition of $K$.

That there must be a difference is also plausible: The defining formula for $K$ relates it to the standard Gibbs free energy of reaction $\Delta G^{0}$ which in turn is defined to be the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an extend of reaction $\xi = 1 \, \text{mol}$). So, it is the amount of Gibbs free energy produced by the reaction of $\nu_{1}\, \text{mol}$ of $\text{A}$ and $\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\nu_{x+2}\, \text{mol}$ $\text{E}$ etc. But if you multiply the reaction by $\alpha$ the standard Gibbs free energy of reaction is the Gibbs free energy produced by the reaction of $\alpha\nu_{1}\, \text{mol}$ of $\text{A}$ and $\alpha\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\alpha\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\alpha\nu_{x+2}\, \text{mol}$ $\text{E}$ etc. and so there should be $\alpha$ times as much Gibbs free energy as before:

\begin{equation} \Delta G^{0}_{\alpha} = \alpha \Delta G^{0}_{1} \ . \end{equation}

And this is indeed the case when you have $K_{\alpha} = K_{1}^{\alpha}$ because according to the logarithm laws

\begin{equation} RT \log K_{1}^{\alpha} = RT \alpha \log K_{1} \ . \end{equation}