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
238 views
in Chemistry by (73.6k points)
closed by
following equilibrium is studied by taking 1 mole of `N_(2)` and 3 moles of `H_(2)` in a 1L flask at a given temperature?
`N_(2)(g) +3H_(2) (g) hArr 2NH_(3)(g)`
`NH_(3)(g)` formed at equilibrium is neutralised by 200 mL of 1M HCl. Calculate equilibrium constant.

1 Answer

0 votes
by (73.9k points)
selected by
 
Best answer
`200 mL of 1M HCl = (1)/(1000) XX 200 = 0.2 M`
`NH_(3) ` and HCl react in equimolar Proporation to form `NH_(4)Cl`
`:. NH_(3)` evolved n the reaction = 0.2 M
`N_(2)(g) + 3H_(2) (g) hArr 2NH_(3)(g)`
No. of moles of `NH_(3)(g)` formed =0.2M
No. of moles of `N_(2)(g) " reacted " =(0.2)/(2) = 0.1M`
No. of moles of `H_(2)(g) " reacted " =0.2xx(3)/(2) =0.3M`
The molar concentration of various species at the equilibrium point is :
`{:(,N_(2)(g),+,3YH_(2)(g),hArr,2NH_(3)(g)),("Initial moles/litre",1,,3,,0),("Moles/litre at eqm. point",1-0.1,,3-0.3,,0.2),(,=0.9,,2.7,,):}`
Applying Law of chemical equilibrium :
`K_(c) = [[NH_(3)(g)]^(2))/[[N_(2)(g)][H_(2)(g)]^(3))= (0.2)^(2)/(0.9xx(2.7)^(3))=(0.04)/(17.71)=2..26xx 10^(-3)`

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

...