[toxicgun]

Let's say i have a reaction in equilibrium. A ⇌ B So the Keq is [ B]/[A] which is equal to Kforward/Kbackward right? And I multiply this reaction with 2. The new reaction is 2A ⇌ 2B and our new K'eq is [ B]² / [A]² so we have taken the square of Keq and this completely makes sense to me.But what doesn't make sense is Kforward hasn't changed and Kbackward hasn't change so how did Kforward/Kbackward change? This has been on my mind for a while and couldn't find an answer myself, i would appreciate an explanation

[mjc123]

Equilibrium is based on thermodynamics, not kinetics. It is dangerous to use kinetic arguments to determine the equilibrium constant. It is not generally true that Keq = kf/kb (though it is true that at equilibrium the rates of the forward and reverse reactions are equal). Suppose in your case the forward rate = kf [A] and the reverse rate is kb[B ]. At equilibrium
kf/kb = [B ]/[A] =Keq.
Now if [B ]/[A] is constant, obviously [B ]²/[A]² is also constant, and we can consider this the equilibrium constant for 2A  2B (though I can't see why we would want to do so). But it is not equal to kf/kb. However you write the reaction equation, the rate law doesn't change.

[toxicgun]

Thanks for your help, in school our teacher defined Keq as Kf/Kb so thats why i thought Keq always has to be equal to Kf/kb but i guess it is not really the case. So i guess Keq being equal to Kf/kb is only when the numbers are the lowest possible integers that they can be. And if i understood correctly, the reaction rate for 2A→2B is still k[A]. Right?

[mjc123]

Depending on how you define rate of reaction, it may be 2k[A]. But it need not necessarily be proportional to [A] at all, that's just a special case. Rate laws can take many forms, for most of which Keq ≠ kf/kb.