Home
Class 11
CHEMISTRY
The equilibrium, P(4(g)) +6Cl(2(g)) Left...

The equilibrium, `P_(4(g)) +6Cl_(2(g)) Leftrightarrow 4PCl_(3(g))` is attained by mixing equal moles of `P_4 and Cl_2` in an evacuated vessel. Then at equilibrium

A

`[Cl_2] gt [PCl_3]`

B

`[Cl_2] gt [P_4]`

C

`[P_4] gt [Cl_2]`

D

`[PCl_3] gt [P_4]`

Text Solution

Verified by Experts

The correct Answer is:
C
`P_(4(g)) +6Cl_(2(g)) Leftrightarrow 4PCl_(3(g))`
One mole of `P_4` reacts with 6 moles of `Cl_2` i.e, at equilibrium `Cl_2` is consumed more than `P_4`. If we start the reaction with equal number of moles of `P_4 and Cl_2` then obviously at equilibrium `[P_(4)] gt [Cl_2]`
Promotional Banner

Similar Questions

Explore conceptually related problems

The equilibrium: P_(4)(g)+6Cl_(2)(g) hArr 4PCl_(3)(g) is attained by mixing equal moles of P_(4) and Cl_(2) in an evacuated vessel. Then at equilibrium:

The equilibrium, P_(4)(s) + 6 Cl_(2)(g) rarr 4PCl_(3) attained by mixing equal moles of P_(4) "and" Cl_(2) in a evacuated vessel. Then at equilibrium

The equilibrium A(g) +4B(g) hArr AB_4(g) is attained by mixing equal moles of A and B in a one litre vessel Then at equilibrium

4 mole of Sl_(2)Cl_(4)(g) is introduced innto a 10L vessel. The following equilibrium was established S_(2)Cl_(4)(g)hArr2SCl_(2)(g) at equilibrium 0.2 mol of S_(2)Cl_(4) was present in the vessel. The value of equilibrium constant is.