 # Gay-Lussac's Law of Combining Gas Volumes Tutorial

## Key Concepts

• Gay-Lusac's Law of Combining Gas Volumes states that:
The volume of gases taking part in a chemical reaction show simple whole number ratios to one another when those volumes are measured at the same temperature and pressure.
• When gas A reacts with gas B to produce gas C at constant temperature and pressure, then the ratio of the gas volumes will be a simple whole number ratio:

For gaseous reaction at constant temperature and pressure
volume of Gas A : volume of Gas B : volume of Gas C
x : y : z
where x, y and z are all whole numbers

• For a reaction in which all the reactants and products are gases:

aA(g) + bB(g)cC(g)
The ratio of the volumes of gasses A, B and C is a:b:c

• For a reaction in which one of more of the reactants and/or products are NOT gases, then ONLY the volume of gases will be in a simple whole number ratio.

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## Gay-Lusac's Law: All Reactants and Products are Gases

Consider the reaction in which nitrogen gas (N2(g)) reacts with hydrogen gas (H2(g)) to produce ammonia gas (NH3(g)).

Imagine we set up a number of different experiments in which we place different volumes of nitrogen gas and hydrogen gas in a vessel at constant temperature and pressure and allowed then to react to produce ammonia gas.

When the reaction has finished, we could then measure the volumes of nitrogen gas and hydrogen gas to determine how much of each gas was consumed during the reaction, and, we could measure the volume of ammonia gas produced.

The results of this experiment might look like this:

 volume of nitrogen gas reacted (mL) volume of hydrogen gas reacted (mL) volume of ammonia gas produced (ml) 100 300 200 250 750 500 750 2250 1500

Now let's look at the ratio of nitrogen gas reacted to hydrogen gas reacted:

 volume of nitrogen gas reacted (mL) volume of hydrogen gas reacted (mL) ratio nitrogen : hydrogen 100 300 100 : 300 250 750 250 : 750 750 2250 750 : 2250

Now, let's divide throughout by the lowest number (the volume of nitrogen) to arrive at the lowest ratio:

 ratio nitrogen : hydrogen divide by nitrogen volume lowest ratio 100 : 300 100/100 : 300/100 1 : 3 250 : 750 250/250 : 750/250 1 : 3 750 : 2250 750/750 : 2250/750 1 : 3

For this reaction when the gas volumes are measured at the same temperature and pressure the ratio of the volume of nitrogen gas reacted to the volume of hydrogen gas reacted is always 1:3
In other words, for any given volume of nitrogen that reacts, the volume of hydrogen gas that reacts will be 3 times of the volume of the nitrogen gas!

volume of hydrogen gas reacted = 3 × volume of nitrogen gas reacted

V(hydrogen gas) = 3 × V(nitrogen gas)

If we perform similar calculations to find the ratio of volume of nitrogen gas reacted to the volume of hydrogen gas reacted and volume of ammonia gas produced, we find that the ratio of the volume of nitrogen gas to the volume of hydrogen gas to the volume of ammonia gas is always 1:3:2

So, 1 volume of nitrogen gas reacts with 3 volumes of hydrogen gas to produce 2 volumes of ammonia gas.

(a) V(ammonia gas) = 2 × V(nitrogen gas)

(b) V(ammonia gas) = 2/3V(hydrogen gas)

For this gaseous reaction in which all the reactants and products are gases we could write a chemical equation:

1 volume nitrogen gas + 3 volumes hydrogen gas → 2 volumes ammonia gas

1 volume N2(g) + 3 volumes H2(g) → 2 volumes NH3(g)

If we know the ratio of gas volumes at a given temperature and pressure, we can use it to calculate the volumes of any of the gases given the volume of one of the gases in the reaction:

For the reaction:

N2(g) + 3H2(g) → 2NH3(g)

the ratio of gas volumes N2(g):H2(g):NH3(g) is 1:3:2

if I know 1 L of N2(g) was consumed then I also know that:
(i) 3 × 1 L = 3 L of H2(g) was consumed
(ii) 2 × 1 L = 2 L of NH3(g) was produced.

For the general reaction in which all the reactants and products are gases:

aA(g) + bB(g)cC(g) + dD(g)

The ratio of the gas volumes of A:B:C:D when measured at the same temperature and pressure will be a:b:c:d
where a, b, c and d will be whole numbers

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## Gay-Lusac's Law Example: Not All the Reactants and Products are Gases

We need to take care when using Gay-Lusac's Law of Combining Gas volumes because it can only be used for volumes of gases.

Consider a series of constant temperature and pressure experiments in which different volumes of liquid water are electrolysed to produce hydrogen gas and oxygen gas:

volume of liquid water electrolysed (mL) volume of hydrogen gas produced (mL) volume of oxygen gas produced (mL) 100 137,222 68,611 250 343,056 171,528 750 1029,167 514,583

Now consider the ratio of the volume of liquid water to the volume of hydrogen gas produced by electrolysis:

volume of liquid water electrolysed (mL) volume of hydrogen gas produced (mL) ratio water volume:hydrogen gas volume 100 137,222 100:1327,222 250 343,056 250:343,056 750 1029,167 750:1029,167

And divide throughout by the volume of water in order to get the lowest terms:

ratio volume of liquid water:volume of hydrogen gas (mL) divided by water volume lowest terms 100:137,222 100/100:137,222/100 1:1,327.22 250:343,056 250/250:343,056/250 1:1,372.22 750:1029,167 750/750:1029,167/750 1:1,372.22

We can see that the ratio of the volume of liquid water to the volume of gaseous hydrogen is NOT a simple whole number ratio!

But let's see what happens we look at the ratio of the volume of hydrogen gas and oxygen gas produced:

volume of hydrogen gas:volume of oxygen gas divide by oxygen gas volume lowest ratio 137,222:68,611 137,222/68,611:68,611/68,611 2:1 343,056:171,528 343,056/171,528:171,528/171,528 2:1 1029,167:514,583 1029,167/514,583:514,583/514,583 2:1

The ratio of the volume of hydrogen gas produced to the volume of oxygen gas produced IS a simple whole number ratio, 2:1

When liquid water is electrolysed, 2 volumes of hydrogen gas are produced for every volume of oxygen gas produced:

V(hydrogen gas) = 2 × V(oxygen gas)

BUT, there is no simple whole number ratio relationship for the volume of liquid water that is electrolysed!

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## Worked Example

Question : Liquid ethanol combusts in oxygen gas to produce carbon dioxide gas and water vapor.

The balanced chemical equation for this reaction is:

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)

In one experiment, 0.010 L of ethanol reacted with oxygen gas to produce 11.19 L of carbon dioxide gas.

What volume of oxygen gas in litres was consumed during the reaction?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Determine the volume of oxygen gas in litres consumed during the reaction.
V(O2(g)) = ? L

2. What data (information) have you been given in the question?

Extract the data from the question:

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)

V(C2H5OH(l)) = 0.010 L

V(CO2(g)) = 11.19 L

3. What is the relationship between what you know and what you need to find out?
Gay-Lusac's Law of Combining Gas Volumes: volume of gases taking part in a chemical reaction show simple whole number ratios to one another when measured at the same temperature and pressure.

Note: ignore volume of ethanol because it is a liquid NOT a gas.

Use the volume of carbon dioxide gas produced to calculate the volume of oxygen gas consumed.

Ratio of volume of carbon dioxide produced and volume of oxygen gas consumed will be in a simple whole number ratio
V(CO2(g)) : V(O2(g)) is a simple whole number ratio

This simple whole number ratio is given by the stoichiometric ratio (mole ratio) given in the balanced chemical equation:
V(CO2(g)) : V(O2(g)) is 2:3

Divide both sides of the ratio by 2
V(CO2(g)) : V(O2(g)) is 2/2:3/2
V(CO2(g)) : V(O2(g)) is 1:3/2

So, 1 volume of CO2(g) consumes 3/2 volumes of O2(g)
V(O2(g)) = 3/2 × V(CO2(g))

4. Calculate volume of O2(g) consumed
Substitute the value for the volume of carbon dioxide gas into the equation:

V(O2(g)) = 3/2 × V(CO2(g))

V(O2(g)) = 3/2 × 11.19 = 16.785 = 16.79 L

Work backwards: use our calculated volume of O2(g) to calculate the volume of CO2(g) that would be produced by this chemical reaction.
ratio V(O2(g)) : V(CO2(g)) is 3:2
V(O2(g)) : V(CO2(g)) is 3/3:2/3
V(O2(g)) : V(CO2(g)) is 1:2/3
1 volume of O2(g) produces 2/3 volume of CO2(g)
16.79 L O2(g) produced 2/3 × 16.79 = 11.19 L
Since this volume of CO2(g) is the same as that given in the question we are reasonably confident that our answer is correct.
6. State your solution to the problem "determine the volume of oxygen gas consumed in litres":

V(O2(g)) = 16.79 L

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