Home
Class 12
CHEMISTRY
A quantity of PCl(5) was heated in a 10 ...

A quantity of `PCl_(5)` was heated in a 10 `dm^(3)` vessel at `250^(@)C` :
`PCl_(5(g))iffPCl_(3(g))+Cl_(2(g))`
At equilibrium, the vessel contains 0.1 mole of `PCl_(5)` and 0.2 mole of `Cl_(2)`. The equilibrium constant of the reaction is

A

`0.04`

B

`0.025`

C

`0.02`

D

`0.05`

Text Solution

Verified by Experts

The correct Answer is:
A
`{:(,PCl_(5),iff,PCl_(3),+,Cl_(2)),("Moles at eqm.","0.1 mole",,"0.2 mole",,"0.2 mole"),("Molar concs.","0.1/10",,"0.2/10",,"0.2/10"):}`
`K=(0.02xx0.02)/(0.01)=0.04`
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

A quantity of PCI_(5) was heated in a 10 llitre vessel at 250^(@)C, PCI_(5)(g)hArrPCI_(3)(g) + CI_(2)(g) . At equilibrium the vessel contains 0.1 mole of PCI_(5), 0.20 mole of PCI_(3) and 0.2 mole of CI_(2) . The equilibrium constant of the reaction is

A quantity of PCI _5 was heated in a 2 litre vessel at 525 K . It dissociates as PCI_5 (g) hArr PCI _3 (g) + CI_2 (g) At equilibrium 0.2 mol each of PCI_5 , PCl_3and Cl_2 is found in the reaction mixture. The equilibrium constant K_c for the reaction is -

PCl_(5),PCl_(3)andCl_(2) are at equilibrium at 500 K with concentration 2.1 M PCl_(3) , 2.1 M Cl_(2)and"1.9 "MPCl_(5) . The equilibrium constant for the given reaction is PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))

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