# What does K_(eq)  << 1 mean?

Jul 9, 2016

That the equilibrium favours the reactant side.

#### Explanation:

For the reaction,

$A + B r i g h t \le f t h a r p \infty n s C + D$

There is a rate forward, ${K}_{f} \left[A\right] \left[B\right]$, and a rate backward, ${K}_{r} \left[C\right] \left[D\right]$

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

${K}_{\text{eq}} = \frac{\left[C\right] \left[D\right]}{\left[A\right] \left[B\right]}$ $=$ ${K}_{\text{f"/K_"r}}$

And thus, when ${K}_{\text{eq}}$ 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.