# Question #31a3d

Apr 4, 2015

In steady-state conditions, you assume that the rate of change of an intermediate species = 0

Here's an example to illustrate this:

$A + B \to C + D$, $\text{ rate constant} = {k}_{1}$
$C + E \to F$, $\text{ rate constant} = {k}_{2}$

C is an intermediate species, so we can apply steady-state conditions.

The rate equation for formation of C (change in concentration of C over time) is:

$\frac{d \left[C\right]}{\mathrm{dt}} = {k}_{1} \left[A\right] \left[B\right] - {k}_{2} \left[C\right] \left[E\right] = 0$

Under steady-state conditions, the change in concentration over time of intermediate species is equal to zero, so we can say that this rate equation is equal to zero .

This allows rearrangement of the equation to give rate expressions for different species in the reaction.

e.g. $\left[C\right] = \frac{{k}_{1} \left[A\right] \left[B\right]}{{k}_{2} \left[E\right]}$

This is used when we want to determine a rate equation for the whole reaction that doesn't include any intermediates in the expression.

For example, if we wanted to derive an expression for the formation of product $\left[F\right]$, without the intermediate $\left[C\right]$ present in the expression.

First, you write the equation for the formation of product $\left[F\right]$:

$\frac{d \left[F\right]}{\mathrm{dt}} = {k}_{2} \left[C\right] \left[E\right]$

we can now substitute in the expression for $\left[C\right]$ (derived under steady state conditions) to get our final rate equation:

$\frac{d \left[F\right]}{\mathrm{dt}} = \frac{{k}_{1} \left[A\right] \left[B\right] \cdot \cancel{{K}_{2} \left[E\right]}}{\cancel{{k}_{2} \left[E\right]}}$

the ${k}_{2} \left[E\right]$ terms cancel to produce:

$\frac{d \left[F\right]}{\mathrm{dt}} = {k}_{1} \left[A\right] \left[B\right]$