CameraIcon

SearchIcon

MyQuestionIcon

1

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

Raoult's Law

At certain te...

Question

At certain temperature(T) for the gas phase reaction: $2 H_{2} O (g) + 2 C l_{2} (g) ⇌ 4 H C l (g) + O_{2} (g)$ , $K_{p} = 12 \times 10^{8}$ atm. If $C l_{2},$ HCI & $O_{2}$ are mixed in such a manner that the partial pressure of each is 2 atm and the mixture is brought into contact with excess of liquid water. What would be approximate partial pressure of $C l_{2}$ when equilibrium is attained at temperature(T)?

A

3.6 \times 10^{- 5}

atm

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

B

10^{- 4}

atm

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

C

3.6 \times 10^{- 3}

atm

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

D

0.01

atm

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

Open in App

Solution

The correct option is C $3.6 \times 10^{- 3}$ atm
Assume the vapour pressure of water as $2.99 \times 10^{- 2} a t m$ .

$2 H_{2} O 0.0299 0.0299 (g) + 2 C l_{2} 2 2 - 2 x (g) ⇌ 4 H C l (g) 2 2 + 4 x + O_{2} (g) 2 2 + x$ ,

The expression for the equilibrium constant is $K_{P} = \frac{P_{O_{2}} P_{H C l}^{4}}{P_{H_{2} O}^{2} P_{C l_{2}}^{2}}$ .

Substitute values in the above expression

$K_{P} = \frac{(2 + x) (2 + 4 x)^{4}}{(0.0299)^{2} (2 - 2 x)^{2}} = 12 \times 10^{8}$

Hence $x = 0.998$ and $P_{C l_{2}} = 2 - 2 x = 3.6 \times 10^{- 3}$ .

Suggest Corrections

thumbs-up

0

Similar questions

Q. At certain temperature(T) for the gas phase reaction

2 H_{2} O (g) + 2 C l_{2} (g) ⇌ 4 H C l (g) + O_{2} (g)

K_{P} = 12 \times 10^{8}

C l_{2}, H C l

O_{2}

are mixed in such a manner that the partial pressure of each is

2

atm and the mixture is brought into contact with excess of liquid water. What would be approximate partial pressure of

C l_{2}

when equilibrium is attained at temperature (T)?

[Given: Vapour pressure of water is

380

mm Hg at temperature (T)].

Q. At certain temperature

(T)

for the gas phase reaction

2 H_{2} O (g) + 2 C l_{2} (g) ⇌ 4 H C l (g) + O_{2} (g) K_{p} = 12 \times 10^{8} a t m

C l_{2}, H C l

O_{2}

are mixed in such a manner than the partial pressure of each is

2 a t m

and the mixture is brough into contact with excess of liquid water. What would be approximate partial pressure of

C l_{2}

when equilibrium is attained at temperature

(T)

?
[Given: Vapour pressure of water is

380 m m H g

(T)

]

Q. At certain temperature (

T

) for the gas phase reaction

2 H_{2} O (g) + 2 {C l}_{2} (g) ⇌ 4 H C l (g) + O_{2} (g); K_{P} = 12 \times 10^{8} a t m

{C l}_{2}

H C l

O_{2}

are mixed in such a manner that the partial pressure of each is

2 a t m

and the mixture is brough into contact with excess of liquid water. What would be approximate partial pressure of

{C l}_{2}

when equilibrium is attained at temperature (

T

)?
[Given:Vapour pressure of water is

380 m m

H g

at temperature (

T

{S O}_{2}

{C l}_{2}

are introduced into a

3 L

vessel. Partial pressure of

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

{C l}_{2}

at equilibrium are

1

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. The equilibrium constant

K_{p}

of the reaction:

2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)

is 900 atm at 800 K. A mixture containing

S O_{3}

O_{2}

having initial partial pressure of 1 and 2 atm. respectively is heated at constant volume to equilibriate. Calculate the partial pressure of

O_{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Liquids in Liquids and Raoult's Law

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon