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Law of Mass Action

28g of N2 a...

Question

28g of $N_{2}$ and 6g of $H_{2}$ were mixed. At equilibrium 17g of $N H_{3}$ was formed. The weight of $N_{2}$ and $H_{2}$ at equilibrium are _________ respectively.

A

11g & zero

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B

19 & 3g

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C

14g & 3g

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D

11g & 3g

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Solution

The correct option is C 14g & 3g
The reaction for the formation of ammonia is $N_{2} + 3 H_{2} \Leftrightarrow 2 N H_{3}$ .
1 mole (28 g) of nitrogen will react with 3 moles (6g) of hydrogen to form 2 moles (34g) of ammonia if the reaction is 100% completed.
But only 17 g of ammonia are actually formed.
Hence, the reaction is 50% complete.
Hence, 0.5 mole (14 g) of nitrogen and 3 moles (3g) of hydrogen will remain unreacted at equilibrium. Hence, option C is correct.

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Similar questions

28 g

N_{2}

6 g

H_{2}

400^{o} C

1

litre vessel, the equilibrium mixture contained

27.54 g

{N H}_{3}

. The appropriate value of

K_{c}

for the above reaction can be (in

{m o l e}^{- 2} {l i t r e}^{2}

)

28 g

N_{2}

6 g

H_{2}

400^{\circ} C

1

litre vessel, the equilibrium mixture contained

27.54 g

N H_{3}

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m o l e^{- 2} l i t r e^{2})

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Q. The following concentrations were obtained for the formation of

N H_{3}

N_{2}

H_{2}

at equilibrium at

500 K

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.

Q. The following concentration were obtained for the formation of

N H_{3}

N_{2}

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.

Q. The following concentration were obtained for the formation of

N H_{3}

N_{2}

H_{2}

at equilibrium for the reaction

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

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Calculate equilibrium constant.

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