# Equilibrium constant expression after multiplying stoichiometric coefficients with a constant factor [duplicate]

We know that the equilibrium constant of a balanced chemical reaction

$$aA+bB \rightarrow cC+dD$$

is defined as $$K_\mathrm{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$. But if we multiply both the left hand side and right hand side of the reaction by $$n$$, the reaction is still the same and the reactants and products are still balanced.

$$(a \cdot n)A+(b \cdot n)B \rightarrow (c \cdot n)C+(d \cdot n)D$$

But then the equilibrium constant becomes

$$K_\mathrm{c}^\prime=\frac{[C]^{c \cdot n}[D]^{d \cdot n}}{[A]^{a \cdot n}[B]^{b \cdot n}},$$

which is different from the previous one. Which $$K_\mathrm{c}$$ is correct and how are we getting two different equilibrium constants of the same chemical reaction whereas we know equilibrium constant of a certain chemical reaction varies with temperature only?

• The numerical value of the second $K_\mathrm{c}$ is different. If you use it with the equation it belongs to, you will find that makes no difference.
– Karl
Sep 22 at 20:50
• As Karl said, you have different values but you get the same results for the equilibrium ($Q$ is also defined differently). So the model is different, but the reality is the same.
– Karsten
Sep 22 at 21:29
• Even this seems kinda miss the point that equation is hardly determined by coefficients, just by actual mechanism. You don't multiply elementary reactions! Sep 22 at 23:19
• @Mithoron what did the first answer provider mean by smallest integer in stoichiometric equations?Does that mean a balanced stoichiometric reaction can only have natural numbers as coefficients and not fractional numbers? Sep 23 at 5:47