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Reaction Quotient

40

Question

40% of a mixture of 0.2 mole of $N_{2}$ and 0.6 mole of $H_{2}$ react to give $N H_{3}$ according to the equation, $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are :

A

4 : 5

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B

5 : 4

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C

7 : 10

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D

8 : 5

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Solution

The correct option is A $4 : 5$

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Similar questions

40

0.2

N_{2}

0.6

H_{2}

{N H}_{3}

according to the equation

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

at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is _______.

40 %

of a mixture of

0.2

N_{2}

0.6

H_{2}

N H_{3}

according to the equation,

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

at constant temperature and pressure.
Then the ratio of the final volumn to the initial volumn of gases is as-

Q. According to the equation,

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

at constant temperature and pressure, 40% of mixture of 0.2 mol of

N_{2}

H_{2}

N H_{3}

. Then the ratio of the final volume to the initial volume of gases is:

40

of a mixture of

0.2 m o l e

N_{2}

0.6 m o l e

H_{2}

N H_{3}

according to the equation,

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

at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are :

40

0.2 m o l e

N_{2}

0.6 m o l e

H_{2}

{N H}_{3}

according to the equation

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

, at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:

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