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In the Haber ...

Question

In the Haber process:
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$
$30 L$ of $H_{2}$ and $30 L$ of $N_{2}$ were taken for reaction which yielded only $50 %$ of expected product. What will be the composition of the gaseous mixture in the end?

20 L

N H_{3}

, 25 L

N_{2}

and 20 L

H_{2}

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10 L

N H_{3}

, 25 L

N_{2}

and 15 L

H_{2}

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20 L

N H_{3}

, 10 L

N_{2}

and 30 L

H_{2}

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20 L

N H_{3}

, 25 L

N_{2}

and 15 L

H_{2}

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Solution

The correct option is A 20 L $N H_{3}$ , 25 L $N_{2}$ and 20 L $H_{2}$
For the following reaction
$N_{2} + 3 H_{2} \to 2 N H_{3}$
We are given 30 L of $H_{2}$ and 30 L of $N_{2}$
From the equation it is clear that
for 3 volume of $H_{2}$ - 1 volume of $N_{2}$ required
thus, 30 volume of $H_{2}$ will require 10 volume of $N_{2}$
again, from 3 volume of $H_{2}$ - 2 volume of $N H_{3}$ is formed
thus, from 30 volume of $H_{2}$ ,20 volume of $N H_{3}$ is formed
so, from the given amount of $H_{2}$ and $N_{2}$ (i.e 30 L each) it is clear that
10 L of $N_{2}$ react with 30 L of $H_{2}$ to give 20 L of $N H_{3}$
but, according to question the yeild of $N H_{3}$ is 50% of the expected i.e 10 L of $N H_{3}$ is produced.

now again from the equation
2 volume of $N H_{3}$ is formed from 1 volume of $N_{2}$
thus, 10 volume of $N H_{3}$ will be formed from 5 volume of $N_{2}$
i.e 10 L of $N H_{3}$ will be formed from 5L of $N_{2}$

2 volume of $N H_{3}$ is formed from 3 volume of $H_{2}$
thus, 10 volume of $N H_{3}$ will be formed from 15 volume of $H_{2}$
i.e 10 L of $N H_{3}$ will be formed from 15 L of $H_{2}$

thus, overall 5L of $N_{2}$ and 15 L of $H_{2}$ is used to produce 10 L of $N H_{3}$
hence the remaining amount of $N_{2}$ = 30 -5 = 25 L
remaining amount of $H_{2}$ = 30-15 = 15 L
and amount of $N H_{3}$ produced is 10 L
thus option B is correct.

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Similar questions

Q. 10 L of

N_{2}

and 30 L of

H_{2}

combine to form

N H_{3}

having volume of:

Q. Equilibrium constant,

K_{c}

for the reaction

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

500 K

0.061

At a particular time, the analysis shows that composition of the reaction mixture is

3.0 m o l

L^{- 1}

N_{2}

2.0 m o l

L^{- 1}

H_{2}

and

0.5 m o l

L^{- 1}

{N H}_{3}

. Is the reaction at equilirbium? IF not in which direction does the reaction tend to proceed to reach equilibrium?

N_{2} + 3 H_{2} \to 2 N H_{3}

1 mol of

N_{2}

and 4 mol of

H_{2}

are taken in a

15 L

flask at

27^{\circ} C

. After the complete conversion of

N_{2}

into

N H_{3}

5 L

H_{2} O

is added. The pressure set up in the flask is:

Q. A mixture containing 2 moles of

N_{2}

and 3 moles of

H_{2}

gas is brought to equilibrium in a 3 L vessel.If the molar concentration of

N H_{3}

at equilibrium is 1/3 mol/L, calculate the

K_{c}

for the formation of

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

Q. The following concentration were obtained for the formation of

N H_{3}

form

N_{2}

and

H_{2}

at equilibrium for the reaction

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

[N_{2}] = 1.5 \times 10^{- 2} M

[H_{2}] = 3.0 \times 10^{- 2} M

[N H_{3}] = 1.2 \times 10^{- 2} M

Calculate equilibrium constant.

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In the Haber process: N2(g)+3H2(g)→2NH3(g)30L of H2 and 30L of N2 were taken for reaction which yielded only 50% of expected product. What will be the composition of the gaseous mixture in the end?

In the Haber process:
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$
$30 L$ of $H_{2}$ and $30 L$ of $N_{2}$ were taken for reaction which yielded only $50 %$ of expected product. What will be the composition of the gaseous mixture in the end?