2851 Views

1 mole of N2 and 3 moles of H2 produces NH3 at equilibrium calculate the concentration of NH3 at equilibrium if Kc = 4/27


Jainam shah 22nd Nov, 2018
Answers (2)
Sai 30th Apr, 2020
N2+3H2<->2NH3
Then
N2=1-X
H2=3-3X
NH3=2X
THEN K=4/27
AND
X/1-X^2=1
Vikas gehlot 22nd Nov, 2018

The Reaction can be represented as

N2+3H2=2NH3

Let x mol of N2 Got reacted till equilibrium is reached, then According to stoichiometry, Moles of the component Present at equilibrium.

N2=1-x

H2=3-3x

NH3=2x

Then Equilibrium Constant

K=(2x)^2/27(1-x)^3=4/27

therefore   x/(1-x)^2=1

Now You can calculate the X and hence the corresponding concentration at Equilibrium.

2 Comments
Comments (2)
22nd Nov, 2018
Please tell that how to calculate X value in this
Reply
22nd Nov, 2018
I am not able to solve X value so please show me further
Reply

Related Questions

UPES Integrated LLB Admission...
Apply
Ranked #28 amongst Institutions in India by NIRF | Ranked #1 in India for Academic Reputation by QS University Rankings | 16.6 LPA Highest CTC
ICFAI-LAW School BA-LLB / BBA...
Apply
Ranked 1 st among Top Law Schools of super Excellence in India - GHRDC | NAAC A+ Accredited | #36 by NIRF
Great Lakes PGPM & PGDM 2025
Apply
Admissions Open | Globally Recognized by AACSB (US) & AMBA (UK) | 17.3 LPA Avg. CTC for PGPM 2024 | Application Deadline: 1st Dec 2024
Chandigarh University Admissi...
Apply
Ranked #1 Among all Private Indian Universities In QS Asia Rankings 2025 | Scholarships worth 210 CR
Amity University, Noida Law A...
Apply
Admissions open for B.A. LL.B (Hons) , B.A. LL.B , BBA LL.B.(Hons) , B.Com.LL.B. (Hons.)
ISBR Business School PGDM Adm...
Apply
180+ Companies | Highest CTC 15 LPA | Average CTC 8 LPA | Ranked as Platinum Institute by AICTE for 6 years in a row | Awarded Best Business School...
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books