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Standard XII

Chemistry

Derivation of Kp and Kc

A quantity of...

Question

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 :

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 C 0.04
The equilibrium reaction is $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ .

The expression for the equilibrium constant is $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}$

Substitute values in the above expression.

$K_{c} = \frac{\frac{0.2}{10} \times \frac{0.2}{10}}{\frac{0.1}{10}} = 0.04$ .

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Similar questions

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. 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 at 250∘CPCl5(g)⇌PCl3(g)+Cl2(g). At equilibrium, the vessel contains 0.1 mole of PCl5, 0.2 mole of PCl3 and 0.2 mole of Cl2. The equilibrium constant is :

The correct option is C 0.04The equilibrium reaction is PCl5(g)⇌PCl3(g)+Cl2(g).The expression for the equilibrium constant is Kc=[PCl3][Cl2][PCl5]Substitute values in the above expression.Kc=0.210×0.2100.110=0.04.

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 :