At 450C the equilibrium constant K P for the reaction N 2 - 3H 2 - 2NH 3 was found to be 1-6- 10 -5 at a pressure of 200 atm- If N 2 and H 2 are taken in a ratio 1-3- What is the

The correct option is B 18 N _ 2 + 3H _ 2 ⇌ 2NH _ 3 1 nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp;3 nbsp; nbsp; nbsp; nbsp; ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Chemistry

Industrial Preparation of Ammonia

At 450C the...

Question

At $45^{0} C$ the equilibrium constant $K_{P}$ for the reaction $N_{2} + {3 H}_{2} ⇌ {2 N H}_{3}$ was found to be $1.6 \times 10^{- 5}$ at a pressure of 200 atm. If $N_{2}$ and $H_{2}$ are taken in a ratio 1:3. What is the % of ${N H}_{3}$ formed at this temperature?

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

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

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

Open in App

Solution

The correct option is B 18
$N_{2} + {3 H}_{2} ⇌ {2 N H}_{3}$
1 3 0 Moles before equation
$(1 - x)$ $3 (1 - x)$ $2 x$ Moles after equation
$K_{P} = \frac{(2 x^{2}) {(4 - 2 x)}^{2}}{(1 - x) {[3 (1 - x)]}^{3} \times P^{2}}$
$K_{P} = \frac{16 \times x^{2} \times {(2 - x)}^{2}}{27 {(1 - x)}^{4} \times P^{2}}$
$o r 1.6 \times 10^{- 5} = \frac{16 x^{2} \times {(2 - x)}^{2}}{27 ({1 - x)}^{4} \times P^{2}}$
$\frac{x^{2} {(2 - x)}^{2}}{{(1 - x)}^{4}} = \frac{1.6 \times 10^{- 5} \times 27 \times {(200)}^{2}}{16}$
$= \frac{1.6 \times 10^{- 6} \times 27 \times {(200)}^{2}}{16}$
$∴ \frac{x (2 - x)}{{(1 - x)}^{2}} = 200 \times 10^{- 3} \times \sqrt{27} = 1.039$
$∴ x = 0.301$
$∴ M o l e o f {N H}_{3} f o r m e d$ = $2 \times 0.301 = 0.602$
Total moles at equilibrium = $4 - 2 \times 0.301 = 3.398$
$∴$ % of ${N H}_{3}$ at equilibrium= $\frac{0.602}{3.398} \times 100 = 17.76$

Suggest Corrections

Similar questions

Q. At

450^{o} C

, the equilibrium constant

K_{p}

for the reaction,

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

, was found to be

1.6 \times 10^{- 5}

at a pressure of

200 a t m

. If

N_{2}

and

H_{2}

are taken in

1 : 3

ratio, what is % of

{N H}_{3}

formed at this temperature?

Q. For

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

, one mole of

N_{2}

and three mole of

H_{2}

are at pressure of 4 atm. Equilibrium pressure is found to be 3 atm. Hence,

K_{p}

is :

Q. The reaction :

N_{2} + {3 H}_{2} ⇋ {2 N H}_{3}

takes places at

450^{o} C

. 1 mole of

N_{2}

and

2

mole of

H_{2}

are mixed in a 1 litre vessel and 1 mole of

{N H}_{3}

is formed at equilibrium. Then

K_{c}

for the above reaction is:

Q. If a mixture of

N_{2}

and

H_{2}

in the ratio 1 : 3 at 50 atmosphere and

650^{o} C

is allowed to react till equilibrium is reached. Ammonia present at equilibrium was at 25 atm pressure. Calculate the equilibrium constant (

K_{P}

) for the reaction.

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

Q. For

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

, one mole of

N_{2}

and three moles of

H_{2}

are at pressure of

4

atm. Equilibrium pressure is found to be

3

atm. Hence,

K_{p}

is:

Join BYJU'S Learning Program

At 450C the equilibrium constant KP for the reaction N2+3H2⇌2NH3 was found to be 1.6×10−5 at a pressure of 200 atm. If N2 and H2 are taken in a ratio 1:3. What is the % of NH3 formed at this temperature?

At $45^{0} C$ the equilibrium constant $K_{P}$ for the reaction $N_{2} + {3 H}_{2} ⇌ {2 N H}_{3}$ was found to be $1.6 \times 10^{- 5}$ at a pressure of 200 atm. If $N_{2}$ and $H_{2}$ are taken in a ratio 1:3. What is the % of ${N H}_{3}$ formed at this temperature?