In the previous article, we have studied the law of reciprocal proportions. In this article, we shall study Gay-Lussac’s Law of Combining Volumes. A French chemist Joseph L. Gay – Lussac in 1809, put forward this law.

Statement :

Whenever gases take part in a chemical reaction, either as reactants or as products, they do so in simple proportions by Volumes. Provided the volumes of gases are measured at the same temperature and pressure,

Illustration 1:

Consider following reaction

H_2(g)      +     Cl_2(g)     →     2HCl

1vol               1vol                   2 vol

Thus the simple ratio of volumes is 1 : 1 : 2

Illustration 2:

Consider following reaction

N_2(g)      +    3 H_2(g)     →     2NH_3(g)

1vol            3vol                   2 vol

Thus the simple ratio of volumes is 1 : 3 : 2

Numerical Problems:

Example – 01:

Calculate the volume of oxygen required for the complete combustion of 0.25 dm³ of methane at STP.

Solution:

CH_4(g)   +   2O_2(g)    →   CO_2(g)    +   2H₂O_(g)

1 vol            2 vol           1 vol                  2 vol

By Gay-Lussac’s law of combining volumes of gases

1 vol of methane requires 2 vol of oxygen for complete combustion.
Hence  0.25 dm³ of methane requires 2 x 0.25 = 0.50 dm³ of oxygen.

Example – 02:

Calculate the volume of hydrogen required for the complete hydrogenation of 0.25 dm³ of ethyne at STP.

Solution:

C₂H_2(g)    +    2H_2(g)  →     C₂H_6(g)

1 vol           2vol                1 vol

By Gay-Lussac’s law of combining volumes of gases

1 vol of ethyne requires 2 vol of hydrogen for complete hydrogenation.
Thus 0.25 dm³ of ethyne requires 2 x 0.25 = 0.50 dm³ of hydrogen for complete hydrogenation.

Example – 03:

Calculate the volume of hydrogen required for the complete hydrogenation of 0.25 dm³ of ethylene at STP.

Solution:

C₂H_4(g)    +    H_2(g)  →     C₂H_6(g)

1 vol           1vol                1 vol

By Gay-Lussac’s law of combining volumes of gases

1 vol of ethylene requires 1 vol of hydrogen for complete hydrogenation.

Thus 0.25 dm3 of ethylene requires 1 x 0.25 = 0.25 dm³ of hydrogen for complete hydrogenation.

Example – 04:

Calculate the volume of oxygen required for the complete combustion of 0.25 mol of methane at STP.

Solution:

CH_4(g)   +   2O_2(g)    →   CO_2(g)    +   2H₂O_(g)

1 mol            2 mol           1 mol                2 mol

By Gay-Lussac’s law of combining volumes of gases

1 mol of methane requires 2 mol of oxygen for complete combustion.

Thus 0.25 mol of methane requires 2 x 0.25 = 0.50 mol of oxygen.

One mole of any gas occupies 22.4 dm³ by volume at STP.

Volume of oxygen required = 22.4 x No. of moles = 22.4 x 0.5 = 11.2 dm³

Example – 05:

Calculate the volume of oxygen required for the complete combustion of 0.5 dm³ of H₂S at STP.

Solution:

2H₂S_(g)   +   3O_2(g)    →   2SO_2(g)    +   2H₂O_(g)

2 vol           3 vol           2 vol                  2 vol

By Gay-Lussac’s law of combining volumes of gases

2 vol of H₂S requires 3 vol of oxygen for complete combustion.

1 vol of H₂S requires 3/2 = 1.5 vol of oxygen for complete combustion.

Thus 0.5 dm³ of H₂S requires 1.5 x 0.5 = 0.75 dm³ of oxygen for complete combustion.

Example – 06:

Calculate the volume of oxygen required at STP for the complete combustion of 5.0 dm³ of ethane at 295 K and 0.993 x 10⁵ Nm^-2

Solution:

2C₂H_6(g)   +   7O_2(g)    →   4CO_2(g)    +   6H₂O_(g)

2 vol            7 vol              4 vol                6 vol

By Gay-Lussac’s Law of Combining Volumes of gases

2 vol of C₂H₆ requires 7 vol of oxygen for complete combustion.

1 vol of C₂H₆ requires 7/2 = 3.5 vol of oxygen for complete combustion.

Thus 5 dm³ of C₂H₆ requires 3.5 x 5 = 17.5 dm³ of oxygen for complete combustion.
P = 0.993 x 10⁵ Nm^-2, T = 295 K, V = 17.5 dm³
P_o = 1.013 x 10⁵ Nm^-2, T_o = 273 K, V_o = ?

Thus the volume of oxygen required at STP is 15.88 dm³.

Example – 07:

6.0 dm³ of hydrogen is reacted with 2.4 dm³ of oxygen in a closed chamber. Calculate composition of resulting mixture.

Solution:

2H_2(g)    +    O_2(g)  →    2 H₂O_(g)

2 vol           1 vol               2 vol

By Gay-Lussac’s law of combining volumes of gases.

The ratio of the volume of hydrogen to that of oxygen is 2 : 1.

In this case, oxygen is limiting reagent, and hydrogen is an excess reagent.

2.4 dm³ of oxygen can combine with 2 x 2.4 = 4.8 dm³ of hydrogen to form 2 x 2.4 = 4.8 dm³ of water vapours.

Thus unreacted hydrogen = 6.0 – 4.8 = 1. 2  dm³.

Thus resulting mixture contains 4.8 dm³ of water vapours and 1.2 dm³ of unreacted hydrogen.

Example – 08:

15 litres of nitrogen is made to react with 30 litres of hydrogen to prepare ammonia. Calculate composition of resulting mixture.

Solution:

N_2(g)      +    3 H_2(g)     →     2NH_3(g)

1vol            3vol                   2 vol

By Gay-Lussac’s law of combining volumes of gases

The ratio of the volume of nitrogen to that of hydrogen is 1 : 3.

In this case, hydrogen is limiting reagent, and nitrogen is an excess reagent.

3 x 10 = 30 litres of hydrogen can combine with 1 x 10 = 10 litres of nitrogen to form 2 x 10 = 20 litres of ammonia.

Thus unreacted nitrogen = 15.0 – 10.0 = 5 litres.

Thus resulting mixture contains 20 litres of ammonia and 5 litres of unreacted nitrogen.

Example – 09:

200 dm³ of hydrogen gas is allowed to react with 250 dm³ of chlorine gas. Calculate composition of resulting mixture.

Solution:

H_2(g)      +     Cl_2(g)     →     2HCl

1vol               1vol                   2 vol

By Gay-Lussac’s law of combining volumes of gases.

The ratio of the volume of hydrogen to that of chlorine is 1 : 1.

In this case, hydrogen is limiting reagent and chlorine is excess reagent.

200 dm³ of hydrogen can combine with 200 dm³ of chlorine to form 2 x 200 =400 dm³of hydrogen chloride.

Thus unreacted chlorine = 250 – 200 = 50 dm³.

Thus resulting mixture contains 400 dm³ of hydrogen chloride and 50 dm³ of unreacted chlorine.

Example – 10:

10 dm³ of hydrogen gas is allowed to react with 15 dm³ of chlorine gas. Calculate composition of resulting mixture.

Solution:

H_2(g)      +     Cl_2(g)     →     2HCl

1vol               1vol                   2 vol

By Gay-Lussac’s law of combining volumes of gases.

The ratio of the volume of hydrogen to that of chlorine is 1 : 1

In this case, hydrogen is limiting reagent and chlorine is excess reagent.

10 dm³ of hydrogen can combine with 10 dm³ of chlorine to form 2 x 10 = 20 dm³ of hydrogen chloride.

Thus unreacted chlorine = 15 – 10 = 5 dm³.

Thus resulting mixture contains 20 dm³ of hydrogen chloride and 5 dm3 of unreacted chlorine

In the next chaper we shall study the concept of atomic mass and equivalent mass.

Previous Topic: The Law of Reciprocal Proportions

Next Chapter: Concept of Atomic Mass and Equivalent Mass

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