English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

2 answers

Well, suppose you have a reaction at equilibrium:
A <==> C
The forward rate constant tells you how quick to from C from A: (forward rate) = Ka[A].
The backward rate constant tells you how quick to from A from C: (backward rate) = Kc[C].
The equilibrium constant tells you when the forward rate is the same as the backward rate:
Ka[A] = Kc[C], or:
[C]/[A] = Ka/ Kc = K(eq)

2007-10-17 16:45:48 · answer #1 · answered by Hahaha 7 · 0 0

at the same time as both aspects are appropriate, that is at perfect an oblique courting. different aspects you should take under consideration are the mechanism of the reaction, the entropy benefit or turn away regarding the reaction, and the presence of better or decrease concentrations of reactants. understanding in basic terms the equilibrium consistent, you won't be able to on the on the spot derive the reaction cost or vice versa. all of it figures into the blend, yet in a decidedly non-trivial way.

2016-10-21 07:32:47 · answer #2 · answered by ? 4 · 0 0

fedest.com, questions and answers