Home
Class 12
CHEMISTRY
By dissociation of 4g mol of of PCI(5) p...

By dissociation of 4g mol of of` PCI_(5)` produce 0.8 mol of `PCI_(3)` if vol of container is 1 ltr the equilibrium constant :-

A

0.2

B

0.1

C

0.4

D

1

Text Solution

Verified by Experts

The correct Answer is:
A
`{:(PCl_(5),rarr,PCl_(3)+,Cl_(2)),(4,,0,0),(4-0.8,,0.8,0.8):}`
` K_(c)= (0.8xx0.8)/(3.2)=0.2`
Promotional Banner

Topper's Solved these Questions

  • CHEMISTRY AT A GLANCE

    ALLEN|Exercise INORGANIC CHEMISTRY|300 Videos

  • CHEMISTRY AT A GLANCE

    ALLEN|Exercise ORGANIC CHEMISTRY|483 Videos

  • Chemical Equilibrium

    ALLEN|Exercise All Questions|24 Videos

  • ELECTROCHEMISTRY

    ALLEN|Exercise EXERCISE -05 [B]|38 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

In a 500mL falsk, the degree of dissociation of PCI_(5) at equilibrium is 40% and the initial amount is 5 moles. The value of equilibrium constant in mol L^(-1) for the decomposition of PCI_(5) is

The equilibrium of formation of phosgene is represented as : CO(g)+Cl_(2)(g)hArrCOCl_(2)(g) The reaction is carried out in a 500 mL flask. At equilibrium, 0.3 mol of phosgene, 0.1 mol of CO , and 0.1 mol of Cl_(2) are present. 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 -

4.0 moles of PCI_5 dissociated at 760 K in a 2 litre flask PCI_5 (g) hArr PCI_3 (g) +CI_2(g) at equilibrium . 0.8 mole of CI_2 was present in the flask .The equilibrium constant would be