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

In a closed container, nitrogen and hydrogen mixture initially in a mole ratio of 1:4 reached equilibrium is found that the half hydrogen is converted to ammonia. If the original pressure was 180 atm, what will be the partial pressure of ammonia at equilibrium (there is no change in temperature)?

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

The correct option is D 36 atm
N2+3H22NH3
Original pressure P=180atm
P=PN2+PNH3
Mole fraction of nitrogen =11+4=0.2
Mole fraction of hydrogen =10.2=0.8
Partial pressure of nitrogen 0.2×180=36 atm
Partial pressure of hydrogen =18036=144 atm
Now one half of 36 atm (i.e 18 atm) of nitrogen reacts with hydrogen to produce
2×18=36 atm of ammonia

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Stoichiometric Calculations
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon