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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)

best_answer

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

t_eq 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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