N2g reacts with H2g in either of the following ways depending upon the supply of H2g- N2g-H2g - N2H2l N2g-2H2g - N2H4gIf 5L N2 g and 3L H2 g are taken initially at same temperature and pressure- then the contraction in volume after the reaction in L is-

The balanced reactions are shown below: N_2(g)+H_2(g) → N_2H_2(l) N_2(g)+2H_2(g) → N_2H_4(g) 3 moles of H_2 reacts with 2 moles of ...

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Byju's Answer

Standard XII

Chemistry

Limiting Reagent or Reactant

N2g reacts wi...

Question

$N_{2} (g)$ reacts with $H_{2} (g)$ in either of the following ways depending upon the supply of $H_{2} (g)$ .
$N_{2} (g) + H_{2} (g) \to N_{2} H_{2} (l)$
$N_{2} (g) + 2 H_{2} (g) \to N_{2} H_{4} (g)$
If $5 L N_{2} (g)$ and $3 L H_{2} (g)$ are taken initially (at same temperature and pressure), then the contraction in volume after the reaction (in L) is:

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Solution

The balanced reactions are shown below:
$N_{2} (g) + H_{2} (g) \to N_{2} H_{2} (l)$
$N_{2} (g) + 2 H_{2} (g) \to N_{2} H_{4} (g)$
$3$ moles of $H_{2}$ reacts with $2$ moles of $N_{2}$ .
Given, $5$ L of $N_{2}$ and $3$ L of $H_{2}$ .
Here, $H_{2}$ is a limiting reagent. So, the first reaction will follow.
According to the first reaction, $3$ L of $H_{2}$ will react and form $3$ L of the product. So, the final volume of solution $3 + (5 - 3) = 5$ L.
Contraction in volume is $V_{i} - V_{f} = 5 + 3 - 5 = 3$ L.

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Similar questions

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.

Q. One mole of

N_{2} (g)

is mixed with 2 moles of

H_{2} (g)

in a 4 litre vessel. If

50

% of

N_{2} (g)

is converted to

H_{2} (g)

by the following reaction :

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

What will be the value of

K_{c}

for the following equilibrium?

{N H}_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)

Q. In Haber's process of manufacturing of ammonia,

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g) : H_{25^{\circ} C}^{0} = - 92.2 k J

Molecule	$N_{2} (g)$	$H_{2} (g)$	$N H_{3} (g)$
$C_{p} (J K^{- 1} m o l e^{- 1})$	29.1	28.8	35.1

C_{p}

is independent of temperature, then, reaction at

100^{\circ} C

as compared to that of

25^{\circ} C

will be:

N_{2} (g)

and

H_{2} (g)

are taken in a vessel in mass ratio of

7 : 1

. The 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:

Q. One mole of

N_{2} (g)

is mixed with

2

mole of

H_{2} (g)

in a

4 L

vessel. If

50 %

N_{2} (g)

is converted to

N H_{3} (g)

by the following reaction:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3}

What will be the value of

K_{c}

for the following equilibrium?

N H_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)

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N2(g) reacts with H2(g) in either of the following ways depending upon the supply of H2(g).N2(g)+H2(g)→N2H2(l)N2(g)+2H2(g)→N2H4(g)If 5LN2(g) and 3LH2(g) are taken initially (at same temperature and pressure), then the contraction in volume after the reaction (in L) is: