Question

Question #31a3d

Answer 1

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 -> C + D`, `" rate constant" = k_1`
`C + E -> F`, `" 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:

`(d[C])/(dt) = k_1[A][B] - k_2[C][E] = 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. `[C] = (k_1[A][B])/(k_2[E])`

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 `[F]`, without the intermediate `[C]` present in the expression.

First, you write the equation for the formation of product `[F]`:

`(d[F])/(dt) = k_2[C][E]`

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

`(d[F])/(dt) = (k_1[A][B] * cancel(K_2[E]))/(cancel(k_2[E]))`

the `k_2[E]` terms cancel to produce:

`(d[F])/(dt) = k_1[A][B]`