Please answer the below given query
Solution →
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Initial concentration c M 0 0
Equilibrium concentration ( c-0.2)M 0.2 M 0.2M
K = [PCl3] [Cl2] / [PCl5]
2×10-2 = (0.2 ×0.2 )/(c -0.2)
on solving,
c-0.2 = 2
c(initial concentration of PCl5) = 2.2
As vol. Of flask is 1 litre. So no. Of moles of PCl5 taken originally are 2.2
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Initial concentration c M 0 0
Equilibrium concentration ( c-0.2)M 0.2 M 0.2M
K = [PCl3] [Cl2] / [PCl5]
2×10-2 = (0.2 ×0.2 )/(c -0.2)
on solving,
c-0.2 = 2
c(initial concentration of PCl5) = 2.2
As vol. Of flask is 1 litre. So no. Of moles of PCl5 taken originally are 2.2