CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

One mole of N2(g) is mixed with 2 moles of H2(g) in a 4 litre vessel. If 50% of N2(g) is converted to H2(g) by the following reaction :
N2(g)+3H2(g)2NH3(g)
What will be the value of Kc for the following equilibrium?
NH3(g)12N2(g)+32H2(g)

A
256
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
16
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
116
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 116
For this reaction:
N2(g)+3H2(g)2NH3(g)
t= 0 1 2 -
At eqb 0.5 0.5 1
So Kc=(1/4)2/(0.5/4)4=256
Equilibrium constant for the following equilibrium :
NH3(g)12N2(g)+32H2(g)
Kc=1/Kc=1/256=1/16.

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