Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1. 2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube? 1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2. Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3. Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen 2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3
Gay-Lussac’s Law explains the conditions that govern the combination of gases in a chemical reaction. This law states that gases react in volumes which are in simple ratio to each other and to the volumes of the products, provided that the temperature and pressure are constant.
For instance,
4NH3(g) + 3O2(g) = 6H2O(g) + 2N2(g)
In the reaction above, according to Gay-Lussac’s
Law, it can be said that 4cm3 of ammonia reacts with 3cm3
of oxygen to give 6cm3 of water vapour and 2cm3 of
nitrogen. This is in line with his observation in his experiment with oxygen
and hydrogen and also with hydrogen and chlorine.
2H2(g) + O2(g) = 2H2O(g)
2cm3 1cm3 2cm3
The ratios of the reactants and product are
2:1:2 and the moles are 2 moles, 1 mole, and 2 moles respectively. He also noticed the
same pattern in the case of the reaction with hydrogen and chlorine.
H2(g) + Cl2(g) = 2HCl(g)
1:
1: 2
1mole of hydrogen reacts with 1mole of chlorine
to produce 2moles of hydrogen chloride gas.
The
implication of this Law in calculation involving the volume of gases
This law can be applied in calculations involving
volumes of gases. From Gay-Lussac’s experiment, it is obvious that gases do not
just react rather they combine based on the ratios of the reacting gases and
the products are produced based on the same ratios which can be seen in a
balanced chemical equation.
Simple
Calculations Showing How Gay-Lussac’s Law is Applied
Question
1
40cm3 of carbon (II) oxide is mixed
and sparked with 200cm3 of air containing 21% of oxygen. If all the
volumes are measured at s.t.p., calculate the volume of the residual gases.
2CO + O2
= 2CO2
Solution
1st thing to do is to find the volume
of oxygen contained in 200cm3 of air.
Volume of oxygen in 200cm3 of air =
21/100 x200 = 42cm3
The volume of other gases present in 200cm3
of air = 200 – 42 = 158cm3
158cm3 of the volume of the other gases
did not take part in the reaction.
2CO + O2 = 2CO2
2cm3 1cm3 2cm3
From the equation,
2cm3 of CO = 1cm3 of O2
40cm3 of CO = ½ x40 = 20cm3 of
O2
Note:
out of the 42 cm3 of oxygen in 200 cm3 of air, only 20 cm3 reacted with the carbon (II) oxide.
Volume of oxygen which did not react = 42- 20
=22cm3
Volume of carbon (IV) oxide produced
From the equation,
2cm3 of CO = 2cm3
40cm3 of CO = 2/2 x 40 = 40cm3
So total volume of residual gases
·
Volume of other gases = 158 cm3
·
Volume of O2 which did not react = 22 cm3
·
Volume of CO2 produced =
40 cm3
The total volume of
residual gases = 220cm3
Question
2
Calculate the volume of residual gases which
would be produced, if 24cm3 of ammonia gas combines with 34cm3
of oxygen at a temperature above 100oC.
4NH3(g) + 3O2(g) = 6H2O(g)
+ 2N2(g)
Solution
Before the reaction:
Volume of ammonia = 24cm3
Volume of oxygen = 34cm3
4NH3(g) + 3O2(g) = 6H2O(g)
+ 2N2(g)
The volume of excess oxygen
From the equation,
4 cm3 of NH3 = 3cm3
of oxygen
24cm3 of NH3 = ¾ x 24 =
18cm3
Volume of O2 which did not react = 34
– 18 = 16 cm3
The volume of water vapour produced
From the equation,
4cm3 of NH3 = 6cm3
of water vapour
24cm3 of NH3 = 6/4 x 24 =
36cm3
The volume of nitrogen produced
From the equation
4cm3 of NH3 = 2cm3 of
nitrogen
24cm3 of NH3 = 2/4X 24
=12cm3
The volume of residual gases
The volume of oxygen which did not react = 16 cm3
Volume of water vapour = 36 cm3
Volume of nitrogen = 12 cm3
Total residual gases = 64 cm3
Question
3
If 12cm3 of hydrogen reacts
completely with 10cm3 of nitrogen, calculate the volume of excess
nitrogen and the volume of the residual gases.
N2(g)
+ 3H2(g) = 2NH3(g)
Solution
N2(g)
+ 3H2(g) = 2NH3(g)
1cm3 3cm3 2cm3
From the equation,
3cm3 of H2 = 1cm3
of N2
12cm3 of H2 = 1/3 X12 =
4cm3
The volume of excess nitrogen
10cm3 – 4cm3 = 6cm3
The volume of residual gases = volume of the product
(ammonia gas) + volume of excess nitrogen
The volume of ammonia gas produced
From the equation,
3cm3 of H2 = 2cm3 of
NH3
12cm3 of H2 = 2/3 X12 =
8cm3
The volume of residual gases
Volume of excess N2 = 6 cm3
Volume of NH3 produced = 8cm3
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