## k1 and k2

$K = \frac{k_{forward}}{k_{reverse}}$

Carlos Gonzales 1H
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### k1 and k2

K can equal k1/k'1, but can it also equal k2/k'2? Is it just convention as to which step is the rate determining step?

Chem_Mod
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### Re: k1 and k2

K would apply to any step at equilibrium.

AnuPanneerselvam1H
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### Re: k1 and k2

K of reaction step 2 can equal k2/k'2. The equilibrium applies to all equations and is a ratio of forward rate reaction constant to reverse rate reaction constant.