# Why does multipying a scalar in both sides of a balanced chemical equation, change related quantities?

Stoichiometric coefficients appearing in a balanced chemical equation are unrelated to the actual concentration of reactant/products at any stage of a chemical reaction. They give information only about the ratio in which the reactants and products are consumed and produced. Multiplying by a scalar on both sides of the equation doesn't change that ratio.

Then, why does doing so result in changes in physical quantities related to the reaction, like equilibrium constant or Gibbs energy?

• It is necessary to standardise things simply to remove tremendous confusion. As pointed out by @Ivan Neretin, the effect of not reducing the stoichiometry is to have, in effect , an infinite number of equilibrium constants /$\Delta G$ etc for the same reaction. Chaos would reign! Aug 3, 2016 at 8:23

Really, what does enthalpy mean, for example? Say, a reaction $\ce{A + B -> C + D}$ has certain enthalpy. This means that running a reaction with 1 mole of A and 1 mole of B gives us certain amount of heat. Now let's multiply everything (including coefficients and enthalpy) by 2. That would be $\ce{2A + 2B -> 2C + 2D}$. In other terms, running a reaction with 2 moles of A and 2 moles of B gives us twice as much heat, just like we would expect.
Now think of the equilibrium constant. Say, we have a reaction $\ce{2NO2<->N2O4}$ and the constant just happens to be 4. This means ${\ce{[NO2]^2\over[N2O4]}}=4$. OK, and what if we write this reaction in a different way, like $\ce{4NO2<->2N2O4}$? Obviously, the equilibrium concentrations are still the same, but the expression for the equilibrium constant has changed to ${\ce{[NO2]^4\over[N2O4]^2}}$. That's a square of the previous expression, hence its numeric value would be $4^2=16$. So what? It's not like we can just up and measure this constant with some kind of "constantometer", anyway. It's the real concentrations that matter; and if we calculate the concentrations using either version of the constant, they will turn out the same.
This is also in accordance with $\ln K = {\Delta G\over RT}$. When you square the constant, its logarithm doubles, and so does $\Delta G$.