CameraIcon

SearchIcon

MyQuestionIcon

1

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

Dalton's Law of Partial Pressures

A sample of p...

Question

A sample of pure $N O_{2}$ gas heated to $1000 K$ decomposes: $2 N O_{2} (g) ⇌ 2 N O (g) + O_{2} (g)$ . The equilibrium constant $K_{P}$ is $100 a t m$ . Analysis shows that the partial pressure of $O_{2}$ is $0.25 a t m$ at equilibrium. The partial pressure of $N O_{2}$ at equilibrium is:

A

0.03

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

B

0.25

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

C

0.025

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

D

0.05

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

Open in App

Solution

The correct option is D $0.05$
${2 N O_{2}}_{(g)} ⇋ {2 N O}_{(g)} + {O_{2}}_{(g)}$
At $t = 0$ $P_{o}$
At time $t$ $P_{o} - 2 x$ $2 x$ $x$
Given, $x = 0.25$ $a t m$
$∴ K_{p} = \frac{(2 x^{2}) \times x}{(P_{o} - 2 x)^{2}}$
$∴ 100 = \frac{4 x^{2}}{(P_{o} - 2 x)^{2}}$
$⟹ 5 = \frac{x}{P_{o} - 2 x}$
$∴ 5 P_{o} = 11 x = 11 \times 0.25$
$∴ P_{o} = 0.55$
$∴$ Pressure of $N O_{2}$ at equillibrium $= P_{o} - 2 x$
$0.55 - 2 (0.25)$
$= 0.05$ $a t m$

Suggest Corrections

thumbs-up

0

Similar questions

Q. A sample of pure

N O_{2}

gas heated to 1000k decomposes :

2 N O_{2} (g) ⇌ 2 N O (g) +)_{2} (g)

. The equilibrium constant

K_{p}

is 100 atm . Analysis shows that the partial pressure of

O_{2}

is 0.25 atm at equilibrium. The partial pressure of

N O_{2}

at equilibrium is:

1000

K, a sample of pure

N O_{2}

gas decomposes as:

2 N O_{2} (g) ⇌ 2 N O (g) + O_{2} (g)

The equilibrium constant

K_{p}

156.25

atm. Analysis shows that the partial pressure

0.25

atm at equilibrium. The partial pressure of

N O_{2}

at equilibrium is:

Q. A sample of pure

N O_{2}

gas healed lo 1000 K decomposes.

N O_{2} (g) ⇋ 2 N O (g) + O_{2} (g)

The equillibrium constant

K_{P}

is 100atm. Analysis shows that the partial pressure of

O_{2}

is 0.25 atm at equillibrium. The partial pressure of

N O_{2}

at equillibrium is:

Q. The equilibrium constant

K_{p}

for the reaction,

N_{2} (g) + O_{2} (g) ⇋ 2 N O (g)

1500^{o} C

120

N_{2}

O_{2}

at an initial pressure of

0.25 a t m

each are maintained at

1500^{o} C

till equilibrium is established, calculate the partial pressure of

N_{2}, O_{2}

N O

in the equilibrium mixture.

Q. The equilibrium constant

K_{p}

for the reaction,

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

900 {a t m}^{- 1}

800 K

. A mixture containing

{S O}_{3}

O_{2}

having initial partial pressure of

1

2 a t m

respectively is heated at constant volume to equilibriate. Calculate the partial pressure of each gas at

800 K

at equilibrium.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Dalton's Law of Partial Pressure

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Dalton's Law of Partial Pressures

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon