Question

Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mechanism for this overall reaction is shown below. No(g) + br2(g) br2(g) (fast step; keq = k1/k−1) k2 nobr(g) + no(g) → 2 nobr(g) (slow step) what is the rate law for formation of nobr in terms of reactants based on this mechanism

Answer 1

The overall reaction is given by:

`Br_(2)(g) + 2 NO(g) \rightarrow  2 NOBr(g)`

The fast step reaction is given as:

`NO(g) + Br_(2)(g) \rightleftharpoons NOBr_(2)(g) (fast; k_(eq)= (k_(1))/(k_(-1)))`

The slow step reaction is given as:

`NOBr_(2)(g) + NO(g) \rightarrow  2 NOBr(g)` (slow step
`k_(2)`)

Now, the expression for the rate of reaction of fast reaction is:

`r_(1)=k_(1)[NO][Br_(2)]-k_(-1)[NOBr_(2)]`

The expression for the rate of reaction of slow reaction is:

`r_(2)=k_(2)[NOBr_(2)] [NO]`

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as
`[NOBr_(2)]` takes place in this reaction.

The expression of rate of formation is:

`(d(NOBr))/(dt)=r_(2)`

=
`k_(2)[NOBr_(2)][NO]` (1)

Now, consider that the fast step is always is in equilibrium. Therefore,
`r_(1)=0`

`k_(1)[NO][Br_(2)]= k_(-1)[NOBr_(2)]`

`[NOBr_(2)] = (k_(1))/(k_(-1))[NO][Br_(2)]`

Substitute the value of
`[NOBr_(2)]` in equation (1), we get:

`(d(NOBr))/(dt)=k_(2)[NOBr_(2)][NO]`

=
`k_(2) (k_(1))/(k_(-1))[NO][Br_(2)][NO]`

=
`(k_(1)k_(2))/(k_(-1))[NO]^(2)[Br_(2)]`

Thus, rate law of formation of
`NOBr` in terms of reactants is given by
`(k_(1)k_(2))/(k_(-1))[NO]^(2)[Br_(2)]`.