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Oxidation States

PCl 5 g⇌ PCl ...

Question

${P C l}_{5} (g) ⇌ {P C l}_{3} (g) + {C l}_{2} (g)$
In above reaction, at equilibrium condition mole fraction of ${P C l}_{5}$ is $0.4$ and mole fraction of ${C l}_{2}$ is $0.3$ . Then find out mole fraction of ${P C l}_{3}$ ?

A

0.3

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B

0.7

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C

0.4

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D

0.6

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Solution

The correct option is A $0.3$

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Similar questions

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

In above reaction at equilibrium condition mole fraction of the

P C l_{5}

is 0.4 and mole fraction of the

C l_{2}

is 0.3, then find out mole fraction of the

P C l_{3}

.

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

In above reaction, at equilibrium condition mole fraction of

P C l_{5}

0.4

and mole fraction of

C l_{2}

0.3

. Then find out mole fraction of

P C l_{3}

.

0.6

P C l_{5}

0.3

P C l_{3}

0.5

C l_{2}

1

L flask to obtain the following equilibrium:

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

If the equilibrium constant

K_{c}

for the reaction is

0.2

. Predict the direction of the reaction.

Q. Initially mole of

P {C l}_{5}

0.2

P {C l}_{3}

are mixed in one litre vessel. At equilibrium

0.4

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)}

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}

P C l_{3}

and 0.2 mole of

C l_{2}

. The equilibrium constant of the reaction is:

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