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Dalton's Law of Partial Pressures

' a' mol of ...

Question

' $a$ ' mol of $P {C l}_{5}$ undergoes thermal dissociation as $P {C l}_{5} ⇌ P {C l}_{3} + C l_{2}$ ; the mole fraction $P {C l}_{3}$ at equilibrium is 0.25 and the total pressure is 2 atm. The partial pressure is ${C l}_{2}$ at equilibrium is:

A

2.5

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B

1

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C

0.5

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D

none of these

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Solution

The correct option is C 0.5
Mole fraction of $P C l_{3}$ and $C l_{2}$ at equilibrium are same as the number of moles of both are equal at equilibrium.
$χ_{C l_{2}} = χ_{P C l_{3}} = \frac{moles of P C l_{3}}{total number of moles at equilibrium}$

$χ_{C l_{2}} = χ_{P C l_{3}} = \frac{x}{a - x + x + x} = \frac{x}{a + x} = 0.25$

Partial pressure of $C l_{2} = p_{C l_{2}}$

$p_{C l_{2}} = χ_{C l_{2}} \times total pressure$

$p_{C l_{2}} = 0.25 \times 2 = 0.5 a t m$

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Similar questions

a

P C l_{5}

, undergoes. thermal dissociation as :

P C l_{5} ⇌ P C l_{3} + C l_{2}

, the mole fraction of

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