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Arrhenius Acid

Vapour densit...

Question

Vapour density of $P C l_{5}$ is $104.16$ but when heated at $230^{\circ} C$ its vapour density is reduced to $62$ . The percentage dissociation of $P C l_{5}$ at this temperature will be:

6.8 %

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68 %

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46 %

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64 %

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Solution

The correct option is A $68 %$
$P C l_{5} ⇋ P C l_{3} + C l_{2}$
At Initial $C$ $0$ $0$
At equilibrium $C (1 - α)$ $C α$ $C α$

Total moles at equilibrium $= C (1 - α) + C α + C α = C (1 + α)$

For a reaction at equilibrium, vapour density is inversely proportional to the number of moles.

Therefore, $\frac{T o t a l m o l e s a t e q u i l i b r i u m}{I n i t i a l t o t a l m o l e} = \frac{V a p o u r d e n s i t y i n i t i a l}{V a p o u r d e n s i t y a t e q u i l i b r i u m}$
or, $\frac{C (1 + α)}{C} = \frac{D}{d}$

$1 + α = \frac{D}{d}$

$α = \frac{D}{d} - 1$

here, $D = 104.16$

$d = 62$

Therefore, $α = \frac{104.16}{62} - 1 = 0.68 o r 68 %$

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is 104.16 but when heated to

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, its vapour density is reduced to 62. The degree of dissociation of

P C l_{5}

, at

230^{o} C

, is:

Q. The vapour density of

P C l_{5}

104.25

but when heated to

230^{o} C,

its vapour density is reduced to

62

.
The decomposition reaction of

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on heating is

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Vapour density of PCl5 is 104.16 but when heated at 230∘C its vapour density is reduced to 62. The percentage dissociation of PCl5 at this temperature will be:

Vapour density of $P C l_{5}$ is $104.16$ but when heated at $230^{\circ} C$ its vapour density is reduced to $62$ . The percentage dissociation of $P C l_{5}$ at this temperature will be: