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Standard XII

Chemistry

Law of Mass Action

A quantity of...

Question

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:

0.02

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0.05

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0.04

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0.025

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Solution

The correct option is A 0.04
${P C l}_{5 (g)} ⇌ {P C l}_{3 (g)} + {C l}_{2 (g)}$
$0.1$ $0.2$ $0.2$
given, Volume of vessel $= 10$ litres
$K = \frac{[{P C l}_{3}] [{C l}_{2}]}{[{P C l}_{5}]}$
$= \frac{(0.20 / 10) (0.20 / 10)}{(0.1 / 10)} = 0.04$
equilibrium constant of this reaction is order.

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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 :

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. A quantity of

P C l_{5}

was heated in 10

d m^{3}

vessel

250^{o}

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}

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

10 d m^{3}

vessel

250^{0} 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}

and 0.2 mole of

C l_{2}

. The equilibrium constant of the reaction is ?

Q. Initially mole of

P {C l}_{5}

and

0.2

mole of

P {C l}_{3}

are mixed in one litre vessel. At equilibrium

0.4

mole of

P {C l}_{3}

is present. The value of

K c

for the reaction

P {C l}_{5 (g)}

\Leftrightarrow

P {C l}_{3 (g)}

is:

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A quantity of PCl5 was heated in a 10 litre vessel as 250oC;PCl5(g)⇌PCl3(g)+Cl2(g). At equilibrium the vessel contains 0.1 mole of PCl5. 0.20 mole of PCl3 and 0.2 mole of Cl2. The equilibrium constant of the reaction is:

The correct option is A 0.04PCl5(g)⇌PCl3(g)+Cl2(g) 0.1 0.2 0.2given, Volume of vessel =10 litresK=[PCl3][Cl2][PCl5]=(0.20/10)(0.20/10)(0.1/10)=0.04equilibrium constant of this reaction is order.

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:

The correct option is A 0.04
${P C l}_{5 (g)} ⇌ {P C l}_{3 (g)} + {C l}_{2 (g)}$
$0.1$ $0.2$ $0.2$
given, Volume of vessel $= 10$ litres
$K = \frac{[{P C l}_{3}] [{C l}_{2}]}{[{P C l}_{5}]}$
$= \frac{(0.20 / 10) (0.20 / 10)}{(0.1 / 10)} = 0.04$
equilibrium constant of this reaction is order.