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

'a' moles of PCl5 are heated in a closed container till equilibrium is established.
PCl5(g)PCl3(g)+Cl2(g) is maintianed at a pressure of P atm. If 'b' moles of PCl5 dissociates at equilibrium, then:

A
ab=KpKp+P
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
ab=(Kp+PKp)12
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
ab=[KpKp+P]12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
ab=[KpP]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B ab=(Kp+PKp)12
PCl5PCl3+Cl2
initial moles a 0 0
at equilibrium ab b b



Total no. of moles at equilibrium,
= a - b + b + b = a + b
Let total pressure = P
Therefore,
Partial pressure of PCl5=(aba+b)P
Partial pressure of PCl3=(ba+b)P
Partial pressure of Cl2=(ba+b)P
So
Kp=PCl3×Cl2PCl5
Kp=(ba+b)P×(ba+b)P(aba+b)PKp=b2Pa2b2a2b2b2=PKpa2b21=PKpa2b2=PKp+1(ab)2=P+KpKpab=(P+KpKp)12

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Le Chateliers Principle
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon