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Standard XII

Chemistry

Reaction Quotient

0.6 mole of ...

Question

$0.6$ mole of $P C l_{5}$ , $0.3$ mole of $P C l_{3}$ and $0.5$ mole of $C l_{2}$ are taken in a $1$ 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.

Forward reaction

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Backward reaction

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Direction of the reaction cannot be predicted

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Reaction does not move in any direction.

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Solution

The correct option is B Backward reaction
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$

$Q_{c} = \frac{0.3 \times 0.5}{0.6} = 0.25$

$s i n c e Q_{c} > K_{c}$ , the reaction shifts towards backwards.

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Similar questions

Q. For the following reaction,

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

0.4 Mole of

P C l_{5}, 0.2

mole of

P C l_{3}

and 0.6 mole of

C l_{2}

are taken in 1 litre flask. if

K_{c}

is 0.2 then, predict the direction in which reaction proceeds.

Q. For the following reaction,

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

0.4 Mole of

P C l_{5}, 0.2

mole of

P C l_{3}

and 0.6 mole of

C l_{2}

are taken in 1 litre flask. if

K_{c}

is 0.2 then, predict the direction in which reaction proceeds.

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. In the reaction

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

the equilibrium concentration of

P C l_{5}

and

P C l_{3}

are 0.4 and 0.2 mole/litre respectively. If the value of

K_{C}

is 0.5 what is concentration of

C l_{2}

in moles/litre ?

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}

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0.6 mole of PCl5, 0.3 mole of PCl3 and 0.5 mole of Cl2 are taken in a 1L flask to obtain the following equilibrium:PCl5(g)⇌PCl3(g)+Cl(2g)If the equilibrium constant Kc for the reaction is 0.2. Predict the direction of the reaction.

The correct option is B Backward reactionPCl5(g)⇌PCl3(g)+Cl2(g)Qc=0.3×0.50.6=0.25sinceQc>Kc, the reaction shifts towards backwards.

$0.6$ mole of $P C l_{5}$ , $0.3$ mole of $P C l_{3}$ and $0.5$ mole of $C l_{2}$ are taken in a $1$ 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.

The correct option is B Backward reaction
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$

$Q_{c} = \frac{0.3 \times 0.5}{0.6} = 0.25$

$s i n c e Q_{c} > K_{c}$ , the reaction shifts towards backwards.