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
52 views
in Chemistry by (73.9k points)
closed by
The reaction , `CO(g) + 3 H_(2) (g) hArr CH_(4) (g) + H_(2)O (g),` is at equilibrium at 1300 K in a 1 L flask. It also contains 0*30 mol of CO, 0*10 mol of `H_(2) and 0*02 " mol of " H_(2)O` and an unknown amount of `CH_(4) " in the flask. Determine the concentration of " CH_(4)` in the mixture. The equilibrium constant, `K_(c)`, for the reaction at the given temperature is` 3*90`.

1 Answer

0 votes
by (73.6k points)
selected by
 
Best answer
` K_(c) = ([ CH_(4)][H_(2)O])/([CO][H_(2)]^(3))`
` :. 3* 90 = ([CH_(4)][H_(2)O])/([CO][H_(2)]^(3)) " " ("Molar conc = No. of moles because volume of flask = 1 L")`
`[CH_(4)] = 0*0585 M = 5*85 xx 10^(-2) 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

...