The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr : NOg+Br2g⇌NOBr2g NOBr2g+NOg→2NOBrgIf the second step is the rate determining step the order of the reaction is

Rate law= KNOBr2NuBut NO Br2 is intermediate & must not appear in th rate law eq. From 1st step Kc=NOBr2NOBr2 NOBr2=KcNOBr2∴ Rate law eq= K,KcNO2Br2Hence order of reaction with respect to NO is 2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The following mechanism has been proposed for the reaction of NO with $B r_{2}$ to form $N O B r$ :
$N O_{(g)} + B r_{2_{(g)}} ⇌ N O B r_{2 (g)}$
$N O B r_{2 (g)} + N O_{(g)} \to 2 N O B r_{(g)}$
If the second step is the rate determining step the order of the reaction is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 2.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Rate law= $K [N O B r_{2}] [N u]$
But $[N O B r_{2}]$ is intermediate & must not appear in th rate law eq.
From 1^st step $K_{c} = \frac{[N O B r_{2}]}{[N O] [B r_{2}]}$
$[N O B r_{2}] = K_{c} [N O] [B r_{2}]$