The rate constant for the decompoistion of a certain reaction is described by the equation:
`log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T)`
A two-step mechanism has been suggested for the reaction of nitric oxide and bromine:
`NO(g) + Br_(2)(g) overset(k_(1))rarr NOBr_(2)(g)`
`NOBr_(2)(g)+NO(g) overset(k_(2))rarr 2NOBr(g)`
The observed rate law is, rate `= k[NO]^(2)[Br_(2)]`. Hence, the rate-determining step is

Options

A. `NO(g)+Br_(2)(g) rarr NOBr_(2)(g)`
B. `NOBr_(2)(g)+NO(g) rarr 2NOBr(g)`
C. `2NO(g) + Br_(2)(g) rarr 2NOBr(g)`
D. None of these

Answer

B

Explanation

`NO(g)+Br_(2)(g) overset(k_(1))hArr NOBr_(2)(g)` …(i)
`NOBr_(2)(g)+NO(g)underset(RDA)overset(k_(2))rarr 2NOBr(g)` …(ii)
`:. Rate = k_(2)[NOBr_(2)][NO]` …(iii)
Since `NOBr_(2)` is the reactive intermediate, so its concentration is determined form step (i).
`:. k_(1) or k_(eq) = ([NOBr_(2)])/([NO][Br_(2)])`
or `[NOBr_(2)] = k_(1)[NO][Br_(2)]`
Substitute the `[NOBr_(2)]` in step (iii),
`:. Rate = k_(1)k_(2)[NO]^(2)[Br_(2)] = K[NO]^(2)[Br_(2)]`