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

The overall reaction is given by:

The fast step reaction is given as:

The slow step reaction is given as:

(slow step )

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

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

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as takes place in this reaction.

The expression of rate of formation is:

=    (1)

Now, consider that the fast step is always is in equilibrium. Therefore,

Substitute the value of in equation (1), we get:

=

=

Thus, rate law of formation of in terms of reactants is given by .

the answer to your question is c

just did the assignment and the answer was #1 +2 n