Question #31a3d

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]#