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Question

How much $P C l_{5}$ must be added to a one-litre vessel at $250^{o} C$ in order to obtain a concentration of 0.1 moles of $C l_{2}$ ?

$K_{c}$ for $P C l_{5} ⟺ P C l_{3} + C l_{2}$ is 0.0414 mol/litre.

3.415 mole

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34.15 mole

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0.03415 mole

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0.3415 mole

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Solution

The correct option is D 0.3415 mole
The equilibrium reaction is $P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g)$ .

Let a be the initial number of moles of $P C l_{5}$ .

Following table lists quantities of various species.

$P C l_{5} ⟶$ $P C l_{3}$ + $C l_{2}$
Initial number of moles $a$ 0 0

Moles at equilibrium $a - 0.1$ $0.1$ $0.1$

$∵$ The equilibrium number of moles of chlorine is 0.1.

Hence, the equilibrium concentration of $P C l_{5}, P C l_{3}$ and $C l_{2}$ are $a - 0.1, 0.1$ and $0.1$ respectively.

The expression for the equilibrium constant is $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = \frac{0.1 \times 0.1}{a - 0.1} = 0.0414$ .

Hence, $a = 0.3415$ moles.

Hence, option D is correct.

Suggest Corrections

Similar questions

Q. X moles of

P C l_{5}

must be added to one-litre vessel at

250

K to obtain a concentration of 0.1 moles of

C l_{2}

K_{c}

for,

P C l_{5} ⇌ P C l_{3} + C l_{2}

, is

0.0414

mol / litre.

The value of 1000X is ____.

Q. How much

P C l_{5}

must be added to a one litre vessel kept at

250^{o}

C in order to obtain

0.1

mole of

C l_{2}

gas?

[

K_{C}

for

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

0.0414

mol/L]

K_{c}

for

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

, is

0.04

250^{o} C

. How many moles of

P C l_{5}

must be added to a

3

litre flask to obtain a

C l_{2}

concentration of

0.15 M

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

]

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How much PCl5 must be added to a one-litre vessel at 250oC in order to obtain a concentration of 0.1 moles of Cl2? Kc for PCl5⟺PCl3+Cl2 is 0.0414 mol/litre.

How much $P C l_{5}$ must be added to a one-litre vessel at $250^{o} C$ in order to obtain a concentration of 0.1 moles of $C l_{2}$ ?

$K_{c}$ for $P C l_{5} ⟺ P C l_{3} + C l_{2}$ is 0.0414 mol/litre.