Question

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.

Answer 1

The overall rate law is:

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

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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])`.