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Topic: A question about equilibrium  (Read 2385 times)

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Offline toxicgun

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A question about equilibrium
« on: May 01, 2016, 04:43:20 AM »
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
« Last Edit: May 01, 2016, 07:03:52 AM by Borek »

Offline mjc123

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Re: A question about equilibrium
« Reply #1 on: May 01, 2016, 05:42:34 AM »
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  ::equil:: 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.

Offline toxicgun

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Re: A question about equilibrium
« Reply #2 on: May 01, 2016, 06:16:55 AM »
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?

Offline mjc123

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Re: A question about equilibrium
« Reply #3 on: May 01, 2016, 05:13:31 PM »
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.

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