How would you use the steady state approximation to find the rate law for this reaction?

#NO + Cl_2 stackrel(k_1" ")(rightleftharpoons) NOCl_2#
#" "" "" "" """^(k_(-1))#
#NOCl_2 + NO stackrel(k_2" ")(=>) 2NOCl#

Assume #NOCl_2# is in a steady state and find the rate law for this reaction.

1 Answer
Dec 12, 2017

The overall rate law is:

#r(t) = (k_1k_2[NO]^2[Cl_2])/(k_(-1) + k_2[NO])#


Well, first you should identify the overall reaction:

#NO(g) + Cl_2(g) stackrel(k_1" ")(rightleftharpoons) color(red)(cancel(color(black)(NOCl_2(g))))#
#" "" "" "" "" "" "color(white)(.)""^(k_(-1))#
#ul(color(red)(cancel(color(black)(NOCl_2(g)))) + NO(g) stackrel(k_2" ")(=>) 2NOCl(g))#
#2NO(g) + Cl_2(g) stackrel(k_"obs"" ")(->) 2NOCl(g)#

The rate law overall is to be expressed in the form:

#r(t) = k_"obs"[NO]^m[Cl_2]^n#

Using elementary steps not at equilibrium, however, allows us to get the order directly from the coefficients:

#r(t) ~~ k_2[NO][NOCl_2]#

since #r(t)# must be an initial rate. But since #NOCl_2# is an intermediate and not a reactant, we have to get an expression for it.

From the steady state approximation,

#(d[NOCl_2])/(dt) ~~ 0#

#= k_1[NO][Cl_2] - k_(-1)[NOCl_2] - k_2[NOCl_2][NO]#

where we have added all the rate law contributions to the concentration of #NOCl_2#:

#overbrace(k_1[NO][Cl_2])^"production step 1 fwd" - overbrace(k_(-1)[NOCl_2])^"consumption step 1 rev" - overbrace(k_2[NOCl_2][NO])^"consumption step 2 fwd"#

Solving for the concentration,

#[NOCl_2] = (k_1[NO][Cl_2])/(k_(-1) + k_2[NO])#

And so, the overall rate law should be:

#color(blue)(r(t)) = k_2[NO] cdot (k_1[NO][Cl_2])/(k_(-1) + k_2[NO])#

#= color(blue)((k_1k_2[NO]^2[Cl_2])/(k_(-1) + k_2[NO]))#

If the reaction were to be known to be 2nd order with respect to #NO# and 1st order with respect to #Cl_2#, then

#k_"obs" = (k_1k_2)/(k_(-1) + k_2[NO])#.