Two moles of PCl5 were heated to 327oC in a closed two litre vessel and when equilibrium was achieved, PCl5 was found to be 40% dissociated into PCl3 and Cl2. Calculate the equilibrium constant (KC) for the reaction
A
1.267molL−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.267molL−1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1.967molL−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2.967molL−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B0.267molL−1
PCl5⇌PCl3+Cl2Initially:200 Degree of dissociation: (α)=40 %=0.4 After dissociation, Moles at equilibrium: PCl5=2−(2×0.4)=1.2molPCl3=2×α=0.8molCl2=2×α=0.8mol so, PCl5⇌PCl3+Cl2 At equilibrium: 1.20.80.8 as Concentration=MolesVolume[PCl5]=1.22=0.6molL−1[PCl3]=0.82=0.4molL−1[Cl2]=0.82=0.4molL−1