You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stoichiometry

4.0 moles of ...

Question

$4.0$ moles of $P {C l}_{5}$ dissociate at $760 K$ in a $2$ litre flask, $P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)$ at equilibrium $0.8$ mole of ${C l}_{2}$ was present in the flask. The equilibrium constant would be:

1.0 \times 10^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.0 \times 10^{- 4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.0 \times 10^{- 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.0 \times 10^{- 3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $1.0 \times 10^{- 1}$
Given,

No. of moles of $P C l_{5} = 4 m o l$

Volume of the flask $= 2 l i t r e$

No. of moles of $C l_{2}$ left in the flask $= 0.8 m o l$

Dissociation of $P C l_{5}$ can be given as

$P C l_{5} ⇌ P C l_{3} + C l_{2}$

At t=0 $4 m o l$ $0$ $0$

At equilibrium $(4 - 0.8) m o l$ $0.8 m o l$ $0.8 m o l$

To find equilibrium constant $K_{c} = \frac{[P C l_{3}] \times [C l_{2}]}{[P C l_{5}]}$ , we have to find the concentration of every component.

Concentration of the component = moles of the component/volume

Concentration of $P C l_{5}$ $= \frac{4 - 0.8 m o l e s}{2} = 1.6 m o l / L$

Concentration of $P C l_{3}$ $= \frac{0.8 m o l e s}{2} = 0.4 m o l / L$

Concentration of $C l_{2}$ $= \frac{0.8 m o l e s}{2} = 0.4 m o l / L$

Substituting these values in the formula for equilibrium constant we get,

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

$K_{c} = \frac{0.4 \times 0.4}{1.6} = 0.1$

Since, $0.1$ can also be written as $1.0 \times 10^{- 1}$ , therefore option A is correct.

Suggest Corrections

Similar questions

4.0 m o l e s

P C l_{5}

dissociate at

760 K

in a

2

litre flask.

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

at equilibrium.

0.8 m o l e

C l_{2}

was present in the flask. The equilibrium constant would be

Q. 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 :

Q. The equilibrium constant for the equilibrium

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

at a particular temperature is

2 \times 10^{- 2} m o l L^{- 1}

. The number of moles of

P C l_{5}

that must be taken in a one litre flask at the same temperature to obtain a concentration of 0.20 mole of chlorine at equilibrium is:

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

]

Join BYJU'S Learning Program

4.0 moles of PCl5 dissociate at 760K in a 2 litre flask, PCl5(g)⇌PCl3(g)+Cl2(g) at equilibrium 0.8 mole of Cl2 was present in the flask. The equilibrium constant would be:

$4.0$ moles of $P {C l}_{5}$ dissociate at $760 K$ in a $2$ litre flask, $P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)$ at equilibrium $0.8$ mole of ${C l}_{2}$ was present in the flask. The equilibrium constant would be: