# Question #48db7

Jul 26, 2017

$\text{Equality of forward and reverse rates of reaction.......}$

#### Explanation:

For a generalized reaction............

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

There is a $\text{rate forward} = {k}_{f} \left[A\right] \left[B\right]$....

And a $\text{rate backwards} = {k}_{r} \left[C\right] \left[D\right]$....

${k}_{f}$ and ${k}_{r}$ are simply rate constants that must be measured by experiment. The condition of chemical equilibrium DOES NOT specify cessation of chemical change BUT equality of forward and reverse rates........And thus.....

$\text{rate forward"=k_f[A][B]-="rate backwards} = {k}_{r} \left[C\right] \left[D\right]$....

And so on rearrangement.......

${k}_{f} / {k}_{r} = \text{rate forwards"/"rate backwards} = \frac{\left[C\right] \left[D\right]}{\left[A\right] \left[B\right]}$

And the quotient ${k}_{f} / {k}_{r}$ is otherwise known as the $\text{thermodynamic equilibrium constant}$, ${K}_{\text{eq}}$, which MUST be measured for a particular reaction.

And while, I admit, so far we have been very abstract we can clearly surmise that if ${K}_{\text{eq}}$ is numerically LARGE, then the products $\left[C\right]$ and $\left[D\right]$ are favoured at equilibrium. And likewise, if ${K}_{\text{eq}}$ is numerically SMALL, then the reactants $\left[A\right]$ and $\left[B\right]$ are favoured at equilibrium.

And if you can digest all that (and I don't know at which level you are) with understanding, then congratulations, you now understand $\text{chemical equilibrium}$ to undergraduate level. Anyway, you have to see how to use this knowledge in the questions you are likely to be asked. And the definition you must know is that $\text{chemical equilibrium}$ specifies equality of forward and reverse rates of reaction.