What does $K_{e q}$ $K_(eq)$ << 1 mean?

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What does $K_{e q}$ $K_(eq)$ << 1 mean?

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Answer 1 · 2016-07-09T06:38:16

That the equilibrium favours the reactant side.

Explanation:

For the reaction,

$A + B ⇌ C + D$ $A+BrightleftharpoonsC+D$

There is a rate forward, $K_{f} [A] [B]$ $K_f[A][B]$ , and a rate backward, $K_{r} [C] [D]$ $K_r[C][D]$

When $rate forward = rate backward$ $"rate forward"="rate backward"$ , this is by definition the very condition of chemical equilibrium, which is characterized by an equilibrium constant, $K_{eq}$ $K_"eq"$ .

$K_{eq} = \frac{[C] [D]}{[A] [B]}$ $K_"eq"=([C][D])/([A][B])$ $=$ $=$ $\frac{K_{f}}{K_{r}}$ $K_"f"/K_"r"$

And thus, when $K_{eq}$ $K_"eq"$ is small, it means that at equilibrium the reverse rate is much greater than the forward rate, and that the equilibrium favours the reactant side.

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What does $K_{e q}$ $K_(eq)$ << 1 mean?

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