MyQuestionIcon

1

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

Introduction to Alcohols

N2g and H2g...

Question

$N_{2} (g)$ and $H_{2} (g)$ are taken in a vessel in mass ratio of $7 : 1$ . The only reaction, $N_{2} + 2 H_{2} ⇌ N_{2} H_{4} (g)$ occurs. Pressure due to $N_{2} H_{4}$ at equilibrium is $0.2$ times of total pressure 'P'. Then at equilibrium.

A

Partial pressure of

N_{2} = 2 P / 15

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

B

Partial pressure of

H_{2} = 8 P / 25

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

C

2 p_{N_{2}} = p_{H_{2}}

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

D

P_{N_{2}} = 2 p_{H_{2}}

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

Open in App

Solution

The correct option is C $2 p_{N_{2}} = p_{H_{2}}$
The gases are taken in mass ratio of $7 : 1$ .
$N_{2} + 2 H_{2} ⇋ N_{2} H_{4}$
$t = 0$ $P$ $2 P$ $0$
$t = e q u i l i b r i u m$ $P - α$ $(2 P - 2 α)$ $α$
Here, $α = 0.2 P$

where,

P is Total Pressure
$P_{N_{2}} = 0.8 P$
$P_{H_{2}} = 1.6 P$
$P_{H_{2}} = 2 P_{N_{2}}$
Hence, option (C) is correct.

Suggest Corrections

thumbs-up

0

Similar questions

N_{2} (g)

H_{2} (g)

are taken in a vessel in mass ratio of

7 : 1

N_{2} + 2 H_{2} ⇌ N_{2} H_{4} (g)

occurs. Pressure due to

N_{2} H_{4}

at equilibrium is

0.2

times of total pressure '

P

'. Then at equilibrium:

N_{2}

H_{2}

1 : 3

molar ratio in a closed vessel to attain the following equilibrium

N_{2} g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

k_{p}

for reaction at total pressure of

2 P

P_{N_{2}}

at equilibrium is

\frac{P}{3}

.

N_{2}

H_{2}

are taken in 1 : 3 molar ratio in a closed vessel to attained the following equilibrium

N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}

K_{p}

for reaction, if total pressure at equilibrium is 2P and

P_{N_{2}}

Q. In a vessel, two equilibrium are simultaneously established at the same temperature as follows :

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

N_{2} (g) + 2 H_{2} (g) ⇌ N_{2} H_{4} (g)

Initially the vessel contains

N_{2}

H_{2}

in the molar ratio of

9 : 13.

The equilibrium pressure is

7 P_{0}

, in which pressure due to ammonia is

P_{0}

and due to hydrogen is

2 P_{0}

. Find the values of equilibrium constant

(K_{P}^{'} s)

for both the reactions :

Q. For the reaction

N_{2} + 3 H_{2} ⇌ 2 N H_{3},

N_{2} : H_{2}

were taken in the ratio of 1 : 3. Up to the point of equilibrium 50% each reactant has been reacted. If total pressure at equilibrium is P. The partial pressure of ammonia would be -

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Introduction to Alcohols

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Introduction to Alcohols

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon