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

Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

1 mole of P...

Question

1 mole of $P C l_{5}$ is heated with 0.2 mole of chlorine in a one litre vessel till equilibrium is established.At equilibrium if 0.8 mole of chlorine is present in the vessel, the equilibrium constant for the dissociation of $P C l_{5}$ is:

0.9

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

1.8

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

1.2

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

0.5

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

Open in App

Solution

The correct option is C 1.2
The equilibrium reaction is $P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g)$ .
The following table lists the equilibrium concentration of various species.

$P C l_{5}$
$P C l_{3}$ $C l_{2}$
Initial
1
0
0.2
Change
-x
x
x
Equilibrium
1-x
x
0.2+x
But at equilibrium $[C l_{2}] = 0.2 + x = 0.8$ or $x = 0.6$ .
Thus the equilibrium concentrations of $P C l_{5}, P C l_{3}$ and $C l_{2}$ are 0.4, 0.6 and 0.8 respectively.
The expression for the equilibrium constant is $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = \frac{0.6 \times 0.2}{0.4} = 1.2$ .

Suggest Corrections

Similar questions

2

mole of

P {C l}_{5}

is heated in a one litre vessel. If

P {C l}_{5}

dissociates to the extent of

80

%, the equilibrium constant for the dissociation of

P {C l}_{5}

is:

0.1

mole of

P {C l}_{5}

is heated in a litre vessel at

533 K

. Determine the concentration of various species present at equilibrium, if the equilibrium constant for the dissociation of

P {C l}_{5}

533 K

0.414

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. 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

	$P C l_{5}$	$P C l_{3}$	$C l_{2}$
Initial	1	0	0.2
Change	-x	x	x
Equilibrium	1-x	x	0.2+x

1 mole of PCl5 is heated with 0.2 mole of chlorine in a one litre vessel till equilibrium is established.At equilibrium if 0.8 mole of chlorine is present in the vessel, the equilibrium constant for the dissociation of PCl5 is:

1 mole of $P C l_{5}$ is heated with 0.2 mole of chlorine in a one litre vessel till equilibrium is established.At equilibrium if 0.8 mole of chlorine is present in the vessel, the equilibrium constant for the dissociation of $P C l_{5}$ is: