wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

At 450C the equilibrium constant KP for the reaction N2+3H22NH3 was found to be 1.6×105 at a pressure of 200 atm. If N2 and H2 are taken in a ratio 1:3. What is the % of NH3 formed at this temperature?

A
9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
18
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
36
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
24
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 18
N2+3H22NH3
1 3 0 Moles before equation
(1x) 3(1x) 2x Moles after equation
KP=(2x2)(42x)2(1x)[3(1x)]3×P2
KP=16×x2×(2x)227(1x)4×P2
or1.6×105=16x2×(2x)227(1x)4×P2
x2(2x)2(1x)4=1.6×105×27×(200)216
=1.6×106×27×(200)216
x(2x)(1x)2=200×103×27=1.039
x=0.301
MoleofNH3formed= 2×0.301=0.602
Total moles at equilibrium = 42×0.301=3.398
% of NH3 at equilibrium=0.6023.398×100=17.76

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Industrial Preparation of Ammonia
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon