Respuesta :
The overall reaction is given by:
[tex]Br_{2}(g) + 2 NO(g) \rightarrow 2 NOBr(g)[/tex]
The fast step reaction is given as:
[tex]NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})[/tex]
The slow step reaction is given as:
[tex]NOBr_{2}(g) + NO(g) \rightarrow 2 NOBr(g)[/tex] (slow step [tex]k_{2}[/tex])
Now, the expression for the rate of reaction of fast reaction is:
[tex]r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}][/tex]
The expression for the rate of reaction of slow reaction is:
[tex]r_{2}=k_{2}[NOBr_{2}] [NO][/tex]
Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [tex][NOBr_{2}][/tex] takes place in this reaction.
The expression of rate of formation is:
[tex]\frac{d(NOBr)}{dt}=r_{2}[/tex]
= [tex]k_{2}[NOBr_{2}][NO][/tex] (1)
Now, consider that the fast step is always is in equilibrium. Therefore, [tex]r_{1}=0[/tex]
[tex]k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}][/tex]
[tex][NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}][/tex]
Substitute the value of [tex][NOBr_{2}][/tex] in equation (1), we get:
[tex]\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO][/tex]
=[tex]k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO][/tex]
= [tex]\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}][/tex]
Thus, rate law of formation of [tex]NOBr[/tex] in terms of reactants is given by [tex]\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}][/tex].
