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
53 views
in Chemistry by (73.6k points)
closed by
The value of `K_(c )` for the reaction `3O_(2)(g) hArr 2O_(3)(g)` is `2.0xx10^(-50)` at `25^(@)C`. If the equilibrium concentration of `O_(2)` in air at `25^(@)C` is `1.6xx10^(-2)`, what is the concentration of `O_(3)`?

1 Answer

0 votes
by (73.9k points)
selected by
 
Best answer
Correct Answer - `2.86 xx 10^(-28) M`
The given reaction is:
`3O_(2(g)) harr 2O_(3(g))`
Then, `K_(c) = ([O_(3(g))]^(2))/([O_(2(g))]^(3))`
It is given that `K_(c) = 2.0 xx 10^(-50)` and `[O_(2(g))] = 1.6 xx 10^(-2)`.
Then, we have
`2.0 xx 10^(-50) = ([O_(3(g))]^(2))/([1.6 xx 10^(-2)]^(3))`
`rArr [O_(3(g))]^(2) = 2.0 xx 10^(-50) xx (1.6 xx 10^(-2))^(3)`
`rArr [O_(3(g))]^(2) = 8.192 xx 10^(-56)`
`rArr [O_(3(g))] = 2.86 xx 10^(-28) M`
Hence, the concentration of `O_(3)`is `2.86 xx 10^(-28) M`.

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

...