Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
69 views
in Chemistry by (66.0k points)
closed by
Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below:
`2NO_((g))+Br_(2(g))hArr2NOBr_((g))`.
When `0.087 "mole"` of `NO` and `0.0437 "mole"` of `Br_(2)` are mixed in a closed container at constant temperature, `0.0518 "mole"` of `NOBr` is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine.

1 Answer

0 votes
by (62.9k points)
selected by
 
Best answer
Given: `0.087 "mol of" NO "and"0.0437 "mol of" Br_(2)`
Substitution & Calculation:
`0.0518` mol of `NOBr` is formed from `0.0518 "mol of" NO "and" 0.0518//2=0.0259 "mol of" Br_(2)`.
`therefore "At equilibrium Amoun of" NO=0.087-0.0518=0.0352 "mol"`
Amount of `Br_(2)=0.0437-0.0259=0.0178 "mol"`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...