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The equilibrium constant (EC) of a certain reaction is only dependent on temperature. For instance, suppose at a certain temperature, the EC of $\ce{A + 3B <=> 2C}$ is $K$. Now if I divide the whole equation by 2 to get $\ce{(1/2)A + (3/2)B <=> C}$, why will my new EC $K'$ be the square root of the previous one,namely $K$?

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    $\begingroup$ K depends on the temperature and the chemical equation. This is also true of reaction enthalpy and reaction Gibbs energy. It might be surprising because these quantities look like intensive quantities (units are joule per mole), but they double if you double the coefficients of the chemical equation. They are per amount of substance, but specifically per extent of reaction. Once you know $\Delta_r G$ changes when you double or half all coefficients, it is clear that K changes too via $\Delta_r G^\circ = R T \ln K$ $\endgroup$ – Karsten Theis Jul 19 at 3:59
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There is a simple proof behind the result. Let's take $$\ce{aA + bB \rightleftharpoons cC +dD}$$ Then the equilibrium constant of this reaction (K) will be $$\ce{K= \frac{[A]^a [B]^b}{[C]^c[D]^d}}$$ Now if you divide the whole equation by a number say n then $$\ce{\frac{a}{n}A + \frac{b}{n}B \rightleftharpoons \frac{c}{n}C +\frac{d}{n}D}$$ For this equation you have the equilibrium constant as $$\ce{ K'= \frac{[A]^{ \frac{a}{n}} [B]^{ \frac{b}{n}}}{[C]^{ \frac{c}{n}}[D]^{ \frac{d}{n}}}}$$ So $$\ce K' =(K)^\frac{1}{n}$$

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