1 L of N₂ combines with 3L of H₂ to form 2L of NH₃ under the same conditions. This illustrates the
(A)law of constant composition
(B)law of multiple proportions
(C)law of reciprocal proportions
(D)Gay Lussac's law of gaseous volumes

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Solution

This illustrates the Gay Lussac's law of gaseous volumes.

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At 298K 1 atm, 1 volume of N₂ reacts with 3 volumes of H₂ to give 2 volumes of NH₃. The ratio 1:3:2 illustrates:

N_{2} + {3 H}_{2} ⟶ {2 H}_{3}

N_{2}, H_{2}

{N H}_{3}

1 : 3 : 2

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

N_{2}, H_{2}

N H_{3}

1 : 3 : 2

At 298K 1 atm, 1 volume of N₂ reacts with 3 volumes of H₂ to give 2 volumes of NH₃. The ratio 1:3:2 illustrates:

{N H}_{3}

N_{2}

H_{2}

{2 N H}_{3} \to N_{2} + {3 H}_{2}

x % o f {N H}_{3}

{S O}_{2}

x

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