Home
Class 11
CHEMISTRY
In the dissociation of PCl(5) as PCl(5...

In the dissociation of `PCl_(5)` as
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`
If the degree of dissociation is `alpha` at equilibrium pressure P, then the equilibrium constant for the reaction is

A

`K_(p)=alpha^(2)/(1+alpha^(2)P)`

B

`K_(p)=(alpha^(2)P^(2))/(1-alpha^(2))`

C

`K_(p)=P^(2)/(1-alpha^(2))`

D

`K_(p)=(alpha^(2)P)/(1-alpha^(2))`

Text Solution

Verified by Experts

The correct Answer is:
D
`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial",1,,0,,0),("Final",1-alpha,,alpha,,alpha),("Partial pressure"=,(1-alpha)/(1+alpha).P,,alpha/(1+alpha).P,,alpha/(1+alpha).P):}`
Total "mole" `=1-alpha+alpha+alpha=1+alpha`
`K_(p)=(alpha/(1+alpha)P.alpha/(1+alpha).P)/((1-alpha)/(1+alpha)P)=(alpha^(2)P)/(1-alpha^(2))`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    CENGAGE CHEMISTRY|Exercise Exercises (Assertion-Reasoning)|15 Videos

  • CHEMICAL EQUILIBRIUM

    CENGAGE CHEMISTRY|Exercise Exercises (Integer)|7 Videos

  • CHEMICAL EQUILIBRIUM

    CENGAGE CHEMISTRY|Exercise Exercises (Multiple Correct)|26 Videos

  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    CENGAGE CHEMISTRY|Exercise Archives Subjective|15 Videos

  • CLASSIFICATION AND NOMENCLATURE OF ORGANIC COMPOUNDS

    CENGAGE CHEMISTRY|Exercise Analytical and Descriptive Type|3 Videos

Similar Questions

Explore conceptually related problems

Prove alpha=sqrt((K_(p)/(P+K_(p)))) for PCl_(5) hArr PCl_(3)+Cl_(2) where alpha is the degree of dissociation at temperature when equilibrium constant is K_(p) .

PCl_(5) dissociates into PCl_(3) and Cl_(2) , thus PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) If the total pressure of the system in equilibrium is P at a density rho and temperature T , show that the degree of dissociation alpha=(PM)/(rhoRT)-1 , where M is the relative molar mass of PCl_(5) . If the vapour density of the gas mixture at equilibrium has the value of 62 when the temperature is 230^(@)C , what is the value of P//rho ?

Knowledge Check

  • In the dissociation of PCl_5 PCl_(5(g)) Leftrightarrow PCl_(3(g)) +CL_(2(g)) if as if the degree of dissociation is alpha at equilibrium pressure P, then the equilibrium constant for the reaction is

    A
    `K_p=(alpha^2)/(1+alpha^2P)`
    B
    `K_(p)=(alpha^2 P^2)/(1-alpha^2)`
    C
    `K_(p)=(P^2)/(1-alpha^2)`
    D
    `K_(p)=(alpha^2P)/(1-alpha^2)`

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

    A
    `DeltaH = DeltaU`
    B
    `DeltaH gt DeltaU`
    C
    `Delta lt DeltaU`
    D
    None of the above

  • For the reaction PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) , the equation connecting the degree of dissociation (alpha) of PCl_(5)(g) with the equilibrium constant K_(p) is

    A
    `alpha=(K_(P)//P)/(4+K_(P)//P)`
    B
    `alpha=[(K_(p)//P)/(K_(p)//P+1)]^(1//2)`
    C
    `alpha=[(K_(P)//P)/(K_(P)//P+1)]`
    D
    `alpha=[(K_(P)//P)/(4+K_(P)//P)]^(1//2)`

    • Similar Questions

    Explore conceptually related problems

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

    PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) 'alpha' is the degree of dissociation of PCl_(5) at equilibrium pressure 'P'. Which among the following is the correct expression for degree of dissociation of 'alpha' ?

    For the reaction Pcl_(5(g))hArrPCl_(3(g))+Cl_(2(g))

    The degree of dissociation of PCl_(5)(g) for the equilibrium PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) is approximately releated to the pressure at equilibrium (P) by the relation [altlt1]

    For the reaction PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) the forward reaction at constant temeprature is favoured by

    Home
    Profile