# How do you use the steady state approximation to determine the observed rate constant for an overall reaction, when knowing its mechanism and its individual rate constants?

##### 1 Answer

**The steady state approximation assumes that the change in the concentration of the intermediate is 0.**

What this means is that the rate of the first step is slow enough that *the intermediate is consumed as it is being produced, thereby making its concentration constant*. So, let's take this reaction as an example.

Overall:

#2N_2O_5 stackrel(k_"obs")(->) 4NO_2 + O_2#

Proposed Mechanism:

#N_2O_5 stackrel(k_1)(rightleftharpoons) NO_2 + NO_3#

#color(white)"a"" "" """^(k_(-1))#

#NO_2 + NO_3 stackrel(k_2)(=>) NO + NO_2 + O_2#

#NO_3 + NO stackrel(k_3)(=>) 2NO_2#

It's a bit tricky, but if you use the reactants in step 2, you can convince yourself that

**Our goal is to get #k_"obs"#, the observed rate constant for the overall reaction, in terms of none of the intermediates.**

So, using the steady state approximation, we can start by identifying the intermediates as

#(d[NO])/(dt) = 0 = k_2[NO_2][NO_3] - k_3[NO_3][NO]#

The first set of terms is positive because there,

From here, we are trying to get *the concentration of each intermediate in terms of the main reactants in the overall reaction*:

#k_2[NO_2]cancel([NO_3]) = k_3cancel([NO_3])[NO]#

#color(green)([NO] = k_2/k_3[NO_2])#

Got the one for

#(d[NO_3])/(dt) = 0 = k_1[N_2O_5] - k_2[NO_2][NO_3] - k_3[NO_3][NO] - k_(-1)[NO_2][NO_3]#

*would be* consumed in the backwards reaction in step 1. Thus, the operations end up being that you subtract

#k_1[N_2O_5] = k_2[NO_2][NO_3] + k_3[NO_3][NO] + k_(-1)[NO_2][NO_3]#

#k_1[N_2O_5] = [NO_3](k_2[NO_2] + k_3[NO] + k_(-1)[NO_2])#

#color(green)([NO_3]) = (k_1[N_2O_5])/(k_2[NO_2] + k_3[NO] + k_(-1)[NO_2])#

As you can see on this step, *we needed* *in order to figure out*

#= (k_1[N_2O_5])/(k_2[NO_2] + cancel(k_3)*k_2/cancel(k_3)[NO_2] + k_(-1)[NO_2])#

#= color(green)((k_1[N_2O_5])/(2k_2[NO_2] + k_(-1)[NO_2]))#

At this point, we don't want

*Since* *is a distinguishable product in the overall reaction*, we can use that:

#(d[O_2])/(dt) = k_2[NO_2][NO_3]#

Conveniently enough, after substitution, the

#= k_2cancel([NO_2]) ((k_1[N_2O_5])/(2k_2cancel([NO_2]) + k_(-1)cancel([NO_2])))#

#= k_2*((k_1[N_2O_5])/(2k_2 + k_(-1)))#

#= (k_1k_2)/(2k_2 + k_(-1))[N_2O_5]#

Therefore, comparing with the overall reaction:

#(d[O_2])/(dt) = k_"obs"[N_2O_5]#

#(d[O_2])/(dt) = (k_1k_2)/(2k_2 + k_(-1))[N_2O_5]#

...the observed rate constant is:

#color(blue)(k_"obs" = (k_1k_2)/(2k_2 + k_(-1)))#