Phosphorus pentachloride dissociates as follows in a closed reaction vessel,

PCl_{5(g)}\rightleftharpoons 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

  • Option 1)

    \left ( \frac{x}{x+1} \right )P

  • Option 2)

    \left ( \frac{2x}{1-x} \right )P

  • Option 3)

    \left ( \frac{x}{x-1} \right )P

  • Option 4)

    \left ( \frac{x}{1-x} \right )P

 

Answers (1)

As we learnt in 

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

aA+bB\rightleftharpoons cC+dD


K_{c}=\frac{[C]^{c\:[D]^{d}}}{[A]^{a}\:[B]^{b}}

[A],\:[B],\:[C]\:[D]

are equilibrium concentration

 

and

 

Relation between pressure and concentration -

PV=nRT

or\:P=\frac{n}{V}RT

or\:P=CRT

R=0.0831\:bar\:inter/mol\:K

- wherein

P is pressure in Pa. C is concentration in mol / litre. T is temperature in kelvin 

Given PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)

t = 0            1                     0                    0

teq            1-x                     x                    x

Total number of moles = 1 - x + x + x = 1 + x

Thus partial pressure of PCl_{3}= \left ( \frac{x}{1+x} \right )P

 

       


Option 1)

\left ( \frac{x}{x+1} \right )P

This option is correct

Option 2)

\left ( \frac{2x}{1-x} \right )P

This option is incorrect

Option 3)

\left ( \frac{x}{x-1} \right )P

This option is incorrect

Option 4)

\left ( \frac{x}{1-x} \right )P

This option is incorrect

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