wiz-icon
MyQuestionIcon
MyQuestionIcon
8
You visited us 8 times! Enjoying our articles? Unlock Full Access!
Question

A flask containing 0.50 atm of ammonia contains some solid NH4HS which undergoes dissociation according to NH4HS(s)NH3(g)+H2S(g). Calculate the pressure of NH3 and H2S at equilibrium KP=0.11). Also calculate the total pressure.

A
PNH3 = 0.6653 atm
PH2S = 0.1653 atm
PTotal = 0.8306 atm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
PNH3 = 0.2 atm
PH2S = 0.1 atm
PTotal = 0.3 atm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
PNH3 = 0.1521 atm
PH2S = 0.1653 atm
PTotal = 0.3174 atm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
PNH3 = 0.5521 atm
PH2S = 0.0158 atm
PTotal = 0.5679 atm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A PNH3 = 0.6653 atm
PH2S = 0.1653 atm
PTotal = 0.8306 atm
NH4HS(s) NH3(g)+H2S(g)Pressure before dissociation x 0.5 0Pressure after dissociation(xP) (0.5+P) P
Let pressure equivalent to P atm is developed by NH3 and H2S on dissociation of NH4HS, when 0.5 atm of NH3 is already present.
Kp=PNH3×PH2S=(0.5+P)(P)
or 0.11=(0.5+P)(P)
P=0.1653
PNH3=0.5+0.1653=0.6653 atm
PH2S=0.1653 atm
PTotal =PNH3+PH2S
PTotal = 0.1653 atm+0.6653 atm
PTotal =0.8306 atm

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon