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PCl5 decomposes as PCl5(g) hArr PCl3(g)+...

`PCl_5` decomposes as `PCl_5(g) hArr PCl_3(g)+Cl_2(g)`.If at equilibrium , total pressure is P and density of gaseous mixture is d at temperature T then degree of dissociation `(alpha)` is :
(Molecular wt. of `PCl_5=M`)

A

`alpha=1-(PM)/(dRT)`

B

`alpha=1-(dRT)/(PM)`

C

`alpha=(PM)/(dRT)-1`

D

`alpha=(dRT)/(PM)-1`

Text Solution

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The correct Answer is:
C
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PCl_(5) dissociation a closed container as : PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g)) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_(5) is alpha , the partial pressure of PCl_(3) will be:

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Knowledge Check

  • Phosphorus pentachloride dissociates as follows in a closed reaction vessel. PCl_5(g) hArr PCl_3(g) +Cl_2(g) If total pressure at equilibrium of the reactions mixture is P and degree of dissociation of PCl_5 is x, the partial pressure of PCl_3 will be:

    A
    `(x/(x+1))P`
    B
    `((2x)/(1-x))P`
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    `(x/(1-x))P`
  • Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, PCl_(5(g)) Leftrightarrow PCl_(3(g))+Cl_(2(g)) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_5 is x, the partial pressure of PCl_3 will be

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    `((x)/(x+1))P`
    B
    `((2x)/(1-x))P`
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  • Phosphorus pentachloride dissociates as follows, in a closed reaction vessel PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_(5) is x, the partial pressure of PCl_(3) wil be

    A
    `((x)/(x-1))P`
    B
    `((x)/(1-x))P`
    C
    `((x)/(1+x))P`
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    `((2x)/(1-x))P`
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