# 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?

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}

Let's take the simplest reaction equation $$\ce {A <=> B}$$ In this reaction, your stoichiometric coefficients are simply $$1A:1B$$ Coefficients are the factor you will need to raise your ratio to when you calculate your equilibrium constant. For this reaction, the equilibrium equation would be: $$K^1_{eq} = \frac{[B]^1}{[A]^1}$$ Now, to make things easier to see, lets say your concentrations are: $$A=10 : B=5$$ This would mean that $$K^1_{eq}=\frac{10}{5} = 2$$ Now, let's multiply our equation by a coefficient say, 2: $$\ce{2A<=>2B}$$ Now when you go to calculate your equilibrium constant, your exponents for a will be 2 instead of 1: $$K^2_{eq} = \frac{[B]^2}{[A]^2}$$ Replace A and B with your concentration since they haven't changed, only their stoichiometric coefficients have: $$K^2_{eq}=\frac{10^2}{5^2} = \frac{100}{25}=4$$ Compare this new K value, call it K2 to your first K value, call it K1: $$K_2=4 , K_1=2$$ Remembering that K2 is the square of K1, prove to yourself the reaction is the same by saying: $$K_2=K_1\iff K_1^2=K_2 \iff K_2^\frac{1}{2} = K_1$$ which just shows that $$4^\frac{1}{2} = 2$$ $$2=2$$ Your rate hasn't necessarily increased, the kinetics of the reaction are just the same, it's just that an equilibrium constant serves as a way to analyze how many moles of your product you are gaining compared to how many you are losing. If you add more moles of your reactant, your product amount has to increase commensurately.

If you want to prove it to yourself, do the same thing but $$K^3_{eq} = \frac{[B]^3}{[A]^3}$$ You should still get a $$K=2$$ if you use $$B=10 , A=5$$