Question

What is the rate law for this reaction?

The reaction of tert-butyl bromide with azide ion in aqueous solution is proposed to proceed by the following mechanism:

1. `(CH_3)_3CBr(aq) rightleftharpoons (CH_3)_3C^(+)(aq) + Br^(-)(aq)`
2. `(CH_3)_3C^(+) + N_3^(-) to (CH_3)_3CN_3(aq)`

Assuming that `(CH_3)_3C^(+)(aq)` achieves a steady-state concentration, but making no further assumptions about the relatively magnitudes of the three constants, what is the rate law for this reaction?

The answer is `r(t) = (k_1k_2[(CH_3)_3CBr][N_3^(-)])/(k_(-1)[Br^(-)]+k_2[N_3^(-)]`

However, I haven't really practiced "steady state intermediate" methodology. Thanks!

Answer 1

I think I've got it, after toying with it and following some steps online.

If we assume the intermediate is at a steady (or constant) state, we must assume,

`DeltaF = DeltaD`

where `F=` formation and `D=` disappearance. In this case, we want to model this,

`k_(1)[(CH_3)_3CBr] = k_1[(CH_3)_3C^(+)][Br^(-)]+k_2[(CH_3)_3C^(+)][N_3^(-)]`

Where the left side contains the formation of the intermediate, and the left side contains two terms. The first is the reverse, first reaction, and the second is the disappearance of the intermediate in the second reaction. Now we need to isolate the intermediate, rearranging the equation,

`k_(1)[(CH_3)_3CBr] = [(CH_3)_3C^(+)](k_1[Br^(-)]+k_2[N_3^(-)]=>`

`[(CH_3)_3C^(+)] = (k_1[(CH_3)_3CBr])/(k_1[Br^(-)]+k_2[N_3^(-)]`

Now, substituting it into the last equation, which produces the products,

`(k_1k_2[(CH_3)_3CBr][N_3^-])/(k_1[Br^(-)]+k_2[N_3^(-)]`

Voila!

Answer 2

The principle behind the steady-state approximation (SSA) is that the rate of formation of intermediate typically is assumed to occur in such a way that its change in concentration essentially zero.

The intermediate is stated to be `("CH"_3)_3"C"^(+)`, which we'll denote as `"tBu"^(+)` (tert-butyl). So first, we'll rewrite the reactions:

`"tBuBr" stackrel(k_1" ")(rightleftharpoons) "tBu"^(+) + "Br"^(-)`
`" "" "" """^(k_(-1))`

`"tBu"^(+) + "N"_3^(-) stackrel(k_2" ")(->) "tBuN"_3`

where `k_1` and `k_(-1)` are the first step's rate constants in the forward and reverse reactions, respectively, and `k_2` is the forward rate constant for step 2.

Under the SSA, we can only write a rate law for a reaction step that has not reached equilibrium, because we want it in terms of an initial rate, and equilibria have reached a final rate.

`r(t) ~~ k_2["tBu"^(+)]["N"_3^(-)]`

`"tBu"^+` is an intermediate, which we want to re-express in terms of reactants only.

In the SSA, based on that principle at the top,

`(d["tBu"^(+)])/(dt) ~~ 0`

The three elementary-step rate laws involved are:

`r_1(t) = k_1["tBuBr"]`
`r_(-1)(t) = k_(-1)["tBu"^(+)]["Br"^(-)]`
`r_2(t) = k_2["tBu"^(+)]["N"_3^(-)]`

Now, `"tBu"^(+)` is generated in step 1 (forward), consumed in step 1 (backwards), and consumed in step 2 (forwards).

The net change in concentration over time then has three contributions from the three rate laws:

`0 = (d["tBu"^(+)])/(dt)`

`= k_1["tBuBr"] - k_(-1)["tBu"^(+)]["Br"^(-)] - k_2["tBu"^(+)]["N"_3^(-)]`

We then want to find `["tBu"^(+)]`:

`k_1["tBuBr"] = k_(-1)["tBu"^(+)]["Br"^(-)] + k_2["tBu"^(+)]["N"_3^(-)]`

`["tBu"^(+)] = (k_1["tBuBr"])/(k_(-1)["Br"^(-)] + k_2["N"_3^(-)])`

As a result,

`color(blue)(r(t) = (k_1k_2["tBuBr"]["N"_3^(-)])/(k_(-1)["Br"^(-)] + k_2["N"_3^(-)]))`

If you instead use the fast-equilibrium approximation, you would get that `k_2 "<<" k_1` and

`r(t) ~~ (k_1k_2["tBuBr"]["N"_3^(-)])/(k_(-1)["Br"^(-)])`