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

Byju's Answer

Standard XIII

Chemistry

Introduction to Le Chatelier's Principle

The equilibri...

Question

The equilibrium constant for the reaction $B r_{2} (l) + C l_{2} (g) ⇌ 2 B r C l (g)$ at $27^{\circ}$ C is $K_{p} = 1 atm$ . In a closed container of volume $164 L$ initially 10 moles of $C l_{2}$ are present at $27^{\circ} C$ . What minimum moles of $B r_{2} (l)$ must be introduced into this container so that the above equilibrium is maintained at a total pressure of 2.25 atm? Vapour pressure of $B r_{2} (l)$ at $27^{\circ}$ C is 0.25 atm. Assume that volume occupied by the liquid is negligible.
[ $R = 0.082 {Latmmol}^{- 1} K^{- 1}$ ,Atomic mass of bromine = 80]

Open in App

Solution

To maintain equilibrium there must be some moles of $B r_{2} (l)$ at equilibrium
$B r_{2} (l) ⇌ B r_{2} (g)$

$n = \frac{P V}{R T}$

$n = \frac{(0.25) (164)}{(0.082) (300)} = 1.667$

Now for equilibrium
$B r_{2} (l) + C l_{2} ⇌ 2 B r C l (g)$
x 10
10-x 2x

Total gaseous moles = 10+x
We have $P_{C l_{2}} + P_{B r C l} = 2 a t m$
so
$P_{C l_{2}}$ = $\frac{(10 - x)}{(10 + x)} \times 2 a t m$

$P_{B r C l}$ = $\frac{(2 x)}{(10 + x)} \times 2 a t m$

$K_{p} = \frac{p_{B r C l}^{2}}{p_{C l_{2}}}$
on putting values and solving
x = 3.333
Hence total moles of $B r_{2} (l)$ required to maintain both equilibria = 1.667 + 3.333 = 5

Suggest Corrections

Similar questions

Q. Two flasks of equal volume connected by a narrow tube (of negligible volume) are at

27^{\circ} C

and contain 0.70 mole of

H_{2}

at 0.5 atm. One of the flasks is then immersed into a hot bath, kept at

127^{\circ} C

, while the other remains at

27^{\circ} C

. The final pressure is atm.

{S O}_{2}

and

{C l}_{2}

are introduced into a

3 L

vessel. Partial pressure of

{S O}_{2} {C l}_{2}

and

{C l}_{2}

at equilibrium are

1

atm and

2

atm respectively. The value of

K_{p}

for the following reaction is

10

{S O}_{2} {C l}_{2} (g) ⇌ {S O}_{2} (g) + {C l}_{2} (g)

What will be the total pressure in atm at equilibrium?

Q. One mole of

P C l_{5} (g)

dissociates and attains equilibrium after t seconds at

27^{\circ} C

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

At equillibrium, the vapour density is found to be 100 and the volume of the system is 10.425 L at

27^{\circ} C

.
The total pressure of the system at equilibrium will be atm.

R = 0.082 L a t m m o l^{- 1} K^{- 1}

Q. What is the volume of solution

(in L)

containing

1 g mole

of sugar that will give rise to an osmotic pressure of

1 atm

27^{\circ} C

?
(Given :

R = 0.082 {L atm K}^{- 1} {mol}^{- 1}

)

Q. The value of

K_{P}

1 \times 10^{- 3} {atm}^{- 1}

25^{\circ} C

for the reaction

2 N O (g) + C l_{2} (g) ⇌ 2 N O C l (g) .

A flask contains

N O

0.02 atm

and at

25^{\circ} C

. The mole of

C l_{2}

that must be added if

1 %

of the

N O

is to be converted to

N O C l

at equilibrium is

x \times 10^{- 5}

. The volume of the flask is such that

0.2

mole of gas produces

1 atm

pressure at

25^{\circ} C

(Ignore probable association of

N O

N_{2} O_{2}

). Find the value of

x

.
(Given:

\frac{1}{(99)^{2}} \approx 1.02 \times 10^{- 4}

)

Join BYJU'S Learning Program