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Standard XII

Chemistry

Characteristics of Equilibrium Constant

4.0 moles of ...

Question

4.0 moles of $P C l_{5}$ dissociate at 760 K in a 2 litre flask, $P C l_{5} (g) ⇌ P C l_{5} (g) + C l_{2} (g)$ at equilibrium. 0.8 mole of $C l_{2}$ was present in the flask.The equilibrium constant would be:

1.0 \times 10^{- 1}

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1.0 \times 10^{- 4}

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1.0 \times 10^{- 2}

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1.0 \times 10^{- 3}

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Solution

The correct option is A $1.0 \times 10^{- 1}$
$P C l_{5} (g)_{e q} P C l_{3} (g) + C l_{2} (g)$
At $t = 0, 4 m o l e s o f P C l_{5} (g) ⇌ 0 m o l e s o f P C l_{3} (g) 0 m o l e s o f C l_{2} (g)$
At $t i m e = t, (4 - x) m o l e s o f P C l_{5} (g) ⇌ x m o l e s o f P C l_{3} (g) x m o l e s o f C l_{2} (g)$
$4 - 0.8 m o l e s o f P C l_{5} (g) ⇌ 0.8 m o l e s o f P C l_{3} (g) 0.8 m o l e s o f C l_{2} (g)$
At equilibrium, concentration of $P C l_{5}$ left $= \frac{4 - 0.8}{2} = \frac{3.2 |}{2} = 1.6 M$
$K_{e q} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}$
$= \frac{\frac{0.8}{2} \times \frac{0.8}{2}}{1.6} = 0.1 = 1 \times 10^{- 1}$

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Similar questions

Q. A quantity of

P C l_{5}

was heated in a 10 litre vessel at

250^{\circ} C

P C l_{5 (g)} ⇌ P C l_{3 (g)} + C l_{2 (g)}

.
At equilibrium, the vessel contains 0.1 mole of

P C l_{5}

, 0.2 mole of

P C l_{3}

and 0.2 mole of

C l_{2}

. The equilibrium constant is :

4.0 m o l e s

P C l_{5}

dissociate at

760 K

in a

2

litre flask.

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

at equilibrium.

0.8 m o l e

C l_{2}

was present in the flask. The equilibrium constant would be

Q. A quantity of

P C l_{5}

was heated in a 10 litre vessel as

250^{o} C; P C l_{5 (g)} ⇌ P C l_{3 (g)} + C l_{2 (g)}

. At equilibrium the vessel contains 0.1 mole of

P C l_{5}

. 0.20 mole of

P C l_{3}

and 0.2 mole of

C l_{2}

. The equilibrium constant of the reaction is:

Q. A quantity of

P {C l}_{5}

was heated in a

10

litre vessel at

250^{o} C

P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)

At equilibrium the vessel contains

0.25

moles of

P {C l}_{5}

0.3

mole of

P {C l}_{3}

0.30

moles of

{C l}_{2}

(a) Compute the equilibrium constant for the formation of

P {C l}_{5}

250^{o} C

(b) What is

Δ G^{o}

for the dissociation of

P {C l}_{5}

? [

log 0.036 = - 1.443

]

Q. The equilibrium constant for the equilibrium

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

at a particular temperature is

2 \times 10^{- 2} m o l L^{- 1}

. The number of moles of

P C l_{5}

that must be taken in a one litre flask at the same temperature to obtain a concentration of 0.20 mole of chlorine at equilibrium is:

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4.0 moles of PCl5 dissociate at 760 K in a 2 litre flask, PCl5(g)⇌PCl5(g)+Cl2(g) at equilibrium. 0.8 mole of Cl2 was present in the flask.The equilibrium constant would be:

4.0 moles of $P C l_{5}$ dissociate at 760 K in a 2 litre flask, $P C l_{5} (g) ⇌ P C l_{5} (g) + C l_{2} (g)$ at equilibrium. 0.8 mole of $C l_{2}$ was present in the flask.The equilibrium constant would be: