What should the rate law be for the following reaction and mechanism? My book says the answer is #k[A]^(1//2)[B]# but wouldn't it be #A_2#?

SHORT REASON

Keep in mind, this is for the reaction:

#A_2 " "stackrel(k_1" ")(rightleftharpoons) " "2A# #" "##" "##" "##" "#(fast)
#" "" """^(k_(-1))#
#A + B stackrel(k_2" ")(->) AB# #" "##" "##" "#(slow)
#A + AB stackrel(k_3" ")(->) A_2B# #" "##" "#(fast)
#"-----------------------"#
#A_2 + B stackrel(k_"obs"" ")(->) A_2B#

The slow step is going to dominate the reaction time; in that so-called elementary step, the coefficients in front of the reactants are taken as the exponents in the rate law concentrations.

This means the slow step describes the reaction on a per-particle basis; it is therefore a bimolecular reaction.

As such, to a first approximation, we write the preliminary rate law as:

#r(t) = k_2[A][B]#

But #A# is an intermediate, not a reactant in the overall reaction. So we must rewrite #[A]# in terms of #[A_2]#.

• Since #2A# is in equilibrium with #A_2#, #A# is in equilibrium with #1/2A_2#.
• The rate of the forward reaction in step 1 is equal to the rate of the reverse reaction in step 2, so #[A] prop [A_2]^"1/2"#.

Therefore,

#color(blue)(r(t) = k_"obs"[A_2]^"1/2"[B])#

where #k_"obs"# may be a product and/or ratio of many rate constants. We generally don't care too much what this product is, but I will show what it is later.

So the answer shown is wrong. It must include the reactant #A_2#.

FULL REASON

Your book might or might not gloss over this, but there is the fast equilibrium approximation, which states that the equilibrium in step 1 is established quickly, so that one can write:

#K = (k_1)/(k_(-1)) = ([A]^2)/([A_2])#

where #k_1# is the forward rate constant and #k_(-1)# is the reverse rate constant for the first mechanism step. #K# is the equilibrium constant for that step.

The rate law as before could be written to begin with as:

#r(t) = k_2[A][B]#

But #A# is an intermediate, not a reactant, so we must rewrite #[A]# in terms of #[A_2]#. As mentioned, we could write out #K#, so we can rearrange #K# to get #[A]#:

#[A] = (K[A_2])^"1/2"#

#= (k_1/k_(-1))^"1/2"[A_2]^"1/2"#

As a result, we can rewrite the rate law to be:

#color(blue)(r(t) = underbrace(k_2(k_1/(k_(-1)))^"1/2")_(k_"obs")[A_2]^"1/2"[B])#