5 mole of PCl5 and 4 mole of neon are introduced in a vessel of 110 litre and allowed to attain equilibrium at 250-C- At equilibrium- the total pressure of reaction mixture was 4-678 atm- Calculate degree of dissociation of PCl5 and equilibrium constant for the reaction-

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Byju's Answer

Standard XII

Chemistry

Introduction to Le Chatelier's Principle

5 mole of P...

Question

$5$ mole of $P C l_{5}$ and $4$ mole of neon are introduced in a vessel of $110 l i t r e$ and allowed to attain equilibrium at $250^{\circ} C$ . At equilibrium, the total pressure of reaction mixture was $4.678 a t m$ . Calculate degree of dissociation of $P C l_{5}$ and equilibrium constant for the reaction.

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Solution

$N e$
$P C l_{5}$ $P C l_{3}$ $C l_{2}$
Initial number of moles
4
5
0
0
Equilibrium number of moles
4
$5 - x$ $x$ $x$
The total number of moles is $n = \frac{P V}{R T} = \frac{4.678 \times 110}{0.0821 \times 523} = 11.98$
It is also equal to $n = 4 + 5 - x + x + x = 9 + x$
Hence, $x = 2.98 ≃ 3$
The degree of dissociation $A = \frac{2.48}{5} = 0.596$

$N e$
$P C l_{5}$ $P C l_{3}$ $C l_{2}$
Equilibrium number of moles 4
2
3
3
Equilibrium mole fraction
1/3
1/6
1/4
1/4
The equilibrium constant $B = K = \frac{(1 / 4)^{2} \times 4.678}{1 / 6} = \frac{6 \times 4.678}{16} = 1.75$
$100 (A + B) = 100 (0.596 + 1.75) = 235$

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	$N e$	$P C l_{5}$	$P C l_{3}$	$C l_{2}$
Initial number of moles	4	5	0	0
Equilibrium number of moles	4	$5 - x$	$x$	$x$

5 mole of PCl5 and 4 mole of neon are introduced in a vessel of 110 litre and allowed to attain equilibrium at 250∘C. At equilibrium, the total pressure of reaction mixture was 4.678 atm. Calculate degree of dissociation of PCl5 and equilibrium constant for the reaction.