MyQuestionIcon

1

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

Introduction to Le Chatelier's Principle

In a vessel, ...

Question

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

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ .........(i)
$N_{2} (g) + 2 H_{2} (g) ⇌ N_{2} H_{4} (g)$ ........(ii)

Initially the vessel contains $N_{2}$ and $H_{2}$ 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 :

A

K_{P_{1}} = \frac{1}{985 P_{0}^{2}}, K_{P_{2}} = \frac{3}{59 P_{0}^{2}}

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

B

K_{P_{1}} = \frac{20 P_{0}^{2}}{1}, K_{P_{2}} = \frac{3}{20 P_{0}}

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

C

K_{P_{1}} = \frac{1}{20 P_{0}^{2}}, K_{P_{2}} = \frac{20 P_{0}}{3}

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

D

K_{P_{1}} = \frac{20 P_{0}^{2}}{1}, K_{P_{2}} = \frac{20 P_{0}}{3}

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

Open in App

Solution

The correct option is A $K_{P_{1}} = \frac{1}{985 P_{0}^{2}}, K_{P_{2}} = \frac{3}{59 P_{0}^{2}}$
The molar ratio of nitrogen and hydrogen is 9 : 13.
Let their initial pressures be 9 P and 13 P respectively.
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$
The equilibrium pressures of nitrogen hydrogen and ammonia are 9P-x, 13P-3x and 2x respectivley.
$N_{2} (g) + 2 H_{2} (g) ⇌ N_{2} H_{4} (g)$
The equilibrium pressures of nitrogen hydrogen and hydrazine are 9p-x-y, 13P-3x-2y and y respectively.
The pressure of ammonia is $2 x = P_{0}$
Hence, $x = 0.5 P_{0}$
The pressure due to hydrogen is $13 P - 3 x - 2 y = 2 P_{0}$
$13 P - 2 y = 3.5 P_{0}$
$y = 6.5 P - 1.75 P_{0}$ .....(1)
Total pressure is $7 P_{0}$
$9 P - x - y + 13 P - 3 x - 2 y + 2 x + y = 7 P_{0}$
$20 P - 2 x - 2 y = 7 P_{0}$
Substitute eqn (i) in the above expression.
$20 P - P_{0} - 13 P + 3.5 P_{0} = 7 P_{0}$
$7 P = 4.5 P_{0}$
$P = 0.64 P_{0}$ ......(2)
Substitute (2) in (1)
$y = 4.2 P_{0} - 1.75 P_{0} = 2.4 P_{0}$ ....(3)
The equiibrium pressure of nitrogen is $9 P - x - y = 5.8 P_{0} - 0.5 P_{0} - 2.4 P_{0} = 3 P_{0}$
The equilibrium pressure of hydrogen is $13 P - 3 x = 8.4 P_{0} - 1.5 P_{0} = 6.9 P_{0}$
$K P_{1} = \frac{P_{N H_{3}}^{2}}{P_{N_{2}} P_{H_{2}}^{3}}$
$K P_{1} = \frac{(P_{0})^{2}}{(3 P_{0}) (6.9 P_{0})^{3}} = \frac{1}{985 P_{0}^{2}}$
$K P_{2} = \frac{P_{N_{2} H_{4}}}{P_{N_{2}} P_{H_{2}}^{2}} = \frac{2.4 P_{0}}{(3 P_{0}) (6.9 P_{0})^{2}} = \frac{1}{59.5 P_{0}^{2}}$

Suggest Corrections

thumbs-up

0

Similar questions

A (g) ⇌ 2 B (g)

, equilibrium constant at total equilibrium pressure

p_{1}

K_{P_{1}} and for C(g) ⇌ D (g) + E (g),

equilibrium constant at total equilibrium pressure

P_{2}

K_{p 2} .

If degree of dissociation of A and C are same, then the ratio

K_{P_{1}} / K_{P_{2}},

2 P_{1} = P_{2},

A ⟷ B

equilibrium constant at total pressure

p_{1}

k_{p_{1}}

C ⇌ D + E

equilibrium constant at total pressure

p_{2}

k_{p_{2}}

. If both reaction are gaseous homogeneous reations and degree of dissociation of

A

C

are same, then the ratio of

p_{1} / p_{2}

k_{p_{1}} = 2 k_{p_{2}}

Q. The relation between

K_{p}

K_{c}

for the following reactions.

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

(i i) 2 H_{2} O (g) ⇌ 2 H_{2} (g) + O_{2} (g)

For these two reactions;
(i)

K_{p_{1}} = K_{c} (R T)^{x}

K_{p_{2}} = K_{c} (R T)^{y}

Find the sum (x+y).

Q. The equilibrium constants

K_{p_{1}}

K_{p_{2}}

for the reactions

A ⇌ 2 B

P ⇌ Q + R

, respectively, are in the ratio of 2 : 3. If the degree of dissociation of

A

P

are equal, the ratio of the total pressure at equilibrium is:

Q. The values of

K_{P_{1}}

K_{P_{2}}

for the reactions

X ⇌ Y + Z . . . . . . (1)

A ⇌ 2 B . . . . . (2)

are in ratio of

9 : 1

. If degree of dissociation of

X

A

be equal, then total pressure at equilibrium

(1)

(2)

are in the ratio.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Le Chateliers Principle

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Introduction to Le Chatelier's Principle

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon