January 23, 2021, 09:19:24 PM
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Topic: activation energy in an equilibrium  (Read 399 times)

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Offline mana

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activation energy in an equilibrium
« on: December 21, 2020, 04:27:17 AM »
hi all
in the following graph, there is two equilibrium example, my question is as we can see Ea (rev) is not equal to Ea(fwd), so why we say that in equilibrium the speed of the reverse and the forward reactions are equal?

Offline mjc123

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Re: activation energy in an equilibrium
« Reply #1 on: December 21, 2020, 05:17:27 AM »
Because the speed of a reaction is not the same as the activation energy. It also depends on the concentrations of reagents. Thus for a bimolecular reaction between A and B
rate = k[A][B ] = Ae-Ea/RT[A][B ]
Equilibrium is reached when forward rate = reverse rate; not when Ea(forward) = Ea(reverse)

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