Respuesta :
Answer : The rate law for formation of NOBr based on this mechanism is, [tex]\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2[/tex]
Explanation :
The overall reaction is:
[tex]Br_2(g)+2NO(g)\rightleftharpoons 2NOBr(g)[/tex]
Rate law = [tex]k[Br_2][NO]^2[/tex]
The first step of the overall reaction is:
[tex]NO(g)+Br_2(g)\overset{k_1}{\rightarrow} NOBr_2(g)[/tex]
[tex]NO(g)+Br_2(g)\overset{k_1^-}{\leftarrow} NOBr_2(g)[/tex]
Rate law 1 = [tex]k_1[Br_2][NO][/tex]
Rate law 2 = [tex]k_1^-[NOBr_2][/tex]
The second step of the overall reaction is:
[tex]NOBr_2(g)+NO(g)\overset{k_2}{\rightarrow} 2NOBr[/tex]
Rate law 3 = [tex]k_2[NOBr_2][NO][/tex]
Now rate law of overall reaction can be obtained as follows.
We are multiplying rate law 1 and rate law 3 and dividing by rate law 2, we get:
Rate law = [tex]\frac{[k_1[Br_2][NO]]\times [k_2[NOBr_2][NO]]}{[k_1^-[NOBr_2]]}[/tex]
Rate law = [tex]\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2[/tex]
Rate law = [tex]k[Br_2][NO]^2[/tex]
The rate law for formation of NOBr based on this mechanism is, [tex]\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2[/tex]
