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Law of Mass Action

For PCl 5 g ⇌...

Question

For $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g),$ write the expression of $K_{c} .$

A

K_{c} = \frac{[P C l_{5}]}{[P C l_{3}] [C l_{2}]}

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B

K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}

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C

K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]^{5}}

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D

K_{c} = \frac{[P C l_{3}]^{3} [C l_{2}]^{2}}{[P C l_{5}]^{5}}

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Solution

The correct option is B $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}$
We know, for a reaction
$K_{c}$ = $\frac{Active mass of products}{Active mass of reactants}$
So, for the homogenous reaction, $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
$K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}$

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Similar questions

Q. An equilibrium mixture of

P {C l}_{5}, P {C l}_{3}

{C l}_{2}

at a certain temperature contians

0.5 \times 10^{- 3} M

P {C l}_{3}

1.5 \times 10^{- 2}

P {C l}_{3}

3 \times 10^{- 2} M

{C l}_{2}

. Calculate the equilibrium constant

K_{c}

for the reaction

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

Q. For the equilibrium,

P C l_{5} ⇋ P C l_{3} + C l_{2}

K_{c} = \frac{α^{2}}{1 - α} V

, temperature remaining constant. Then:

P C l_{5} ⇌ P C l_{3} + C l_{2}

Dissociation constant

(K_{c})

P C l_{5}

4

. How many moles of

P C l_{5}

should be taken in

5

L vessel to obtain

0.15

P C l_{3}

at equilibrium?

Q. Assertion :The equilibrium of

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

is not affected by changing the volume of container.
Reason:

K_{c} = \frac{[P {C l}_{3}] [{C l}_{2}]}{[P {C l}_{5}]}

Equilibrium constant,

K_{c}

does not depend on the volume of vessel.

2

P {C l}_{3}

{C l}_{2}

1

P {C l}_{3}

5

27^{o} C

. After following equilibrium is attained in gaseous phase,

P {C l}_{5} ⇌ P {C l}_{3} + {C l}_{2}

P {C l}_{5}

40

% dissociated. Calculate equilibrium constant

K_{C}

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Law of Mass Action

Standard XII Chemistry

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