Home
Class 12
CHEMISTRY
In the reaction PCl(5(g))iffPCl(3(g))+Cl...

In the reaction `PCl_(5(g))iffPCl_(3(g))+Cl_(2(g))`, the equilibrium concentrations of `PCl_(5)andPCl_(3)` are 0.4 and 0.2 mole/litre respectively. If the value of `K_(c )` is 0.5, what is the concentration of `Cl_(2)` in moles/litre?

A

`2.0`

B

`1.5`

C

`1.0`

D

`0.5`

Text Solution

Verified by Experts

The correct Answer is:
C
`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])i.e.,0.5=(0.2xx[Cl_(2)])/(0.4)`
or, `[Cl_(2)]=1.0molL^(-1)`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIA

    MTG-WBJEE|Exercise WB JEE WORKOUT (CATEGORY 2 : Single Option Correct Type )|15 Videos

  • CHEMICAL EQUILIBRIA

    MTG-WBJEE|Exercise WB JEE WORKOUT (CATEGORY 3 : One or More than One Option Correct Type )|10 Videos

  • CHEMICAL ENERGETICS

    MTG-WBJEE|Exercise WB JEE PREVIOUS YEARS QUESTIONS (ONE OR MORE THAN ONE OPTION CORRECT TYPE )|2 Videos

  • CHEMISTRY IN INDUSTRY

    MTG-WBJEE|Exercise WB JEE PREVIOUS YEARS QUESTIONS (SINGLE OPTION CORRECT TYPE)|1 Videos

Similar Questions

Explore conceptually related problems

In the reaction PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) , the equilibrium concentrations of PCl_(5) and PCl_(3) are 0.4 and 0.2 mole // litre respectively. If the value of K_(c) is 0.5 , what is the concentration of Cl_(2) in moles // litre ?

In the reaction PCI_(5)(g)hArr PCI_(3)(g)+CI_(2)(g) the equilibrium concentrations of PCI_(3) and PCI_(3) are 0.4 and 0.2 mol^(-1) , respectively. If the value of K_(c) is 0.5 , what id the concentration of CI_(2) in moles per litre ?

For the reaction, PCl_(5)(g)to PCl_(3)(g)+Cl_(2)(g)

For the reaction : PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g) :

In the reaction PCl_(5)(g) rarr PCl_(3)(g) + Cl_(2)(g) PCl_(5),PCl_(3)"and"Cl_(2) are at equilibrium at 500 K. The concentration of PCl_(3) "and" Cl_(2) is 1.59M. K_(c) = 1.79 .Calculate the concentration of PCl_(5) .

PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) , In above reaction, at equilibrium condition mole fraction of PCl_(5) is 0.4 and mole fraction of Cl_(2) is 0.3. Then find out mole fraction of PCl_(3)

For the reaction: PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) , if the initial concentration of PCl_(5)=1 mol/ l and x moles/litre of PCl_(5) are consumed at equilibrium, the correct expression of K_(p) if P is total pressure is

For PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g), write the expression of K_(c)