Gay Lussac's law
In 1808 the Freanch chemist Gay Lussac investigating the reactions of gases
came to the conclusion that when gases combine chemically they do so in volumes
that have a simple ratio to one another, and to any gaseous product, provided
that all gases are measured at the same temperature and pressure. This became
known as Gay Lussac's law.
For example, in the reaction between hydrogen and oxygen making water:
2 volumes of hydrogen
1 volume of oxygen
hydrogen + oxygen →
water
2H2 + O2 → 2H2O
|
The volume of hydrogen needed for complete reaction is always double the
volume of oxygen, provided the temperature and pressure of the two gases are
the same.
^ top
Avogadro's hypothesis
Extending Gay Lussac's work in 1811, Avogadro suggested that "equal
volumes of all gases contain equal number of molecules (the gases being measured
at the same temperature and pressure). This became known as Avogadro's hypothesis
or Avogadro's law.
This means that in the reaction:
N2 + 3H2 → 2NH3
For a given volume of nitrogen, three times the volume of hydrogen is needed
for complete reaction. The volume of nitrogen contains a certain number of
molecules and there are three times as many molecules in the volume of hydrogen.
|
N2
|
+
|
3H2
|
→
|
2NH3
|
|
1 molecule
|
+
|
3 molecules
|
→
|
2 molecules
|
|
1 volume
|
+
|
3 volumes
|
→
|
2 volumes
|
Avogadro's hypothesis allows us to substitute the coefficients of any balanced
gaseous equation for volumes of gas, for example:
Example: Find the
volume of hydrogen required to react completely with 200 cm3
of nitrogen according to the equation:
N2 + 3H2 → 2NH3
The equation tells us that 1 volume of nitrogen reactrs completely
with 3 volumes of hydrogen:
Therefore volume of hydrogen = 3 x volume of nitrogen
Therefore volume of hydrogen = 3 x 200 cm3 = 600
cm3
Example:
Using the equation:
N2 + 3H2 →
2NH3
if 100 cm3 of nitrogen was used (this is now equivalent
to 1 'volume') it would need 3 times as much hydrogen for complete reaction
i.e. 3 volumes = 300 cm3.
The reaction would produce 2 volumes of ammonia i.e. 2 x 100 cm3
= 200 cm3
^ top
Worked examples
Q146-01 In the reaction between
sulfur dioxide and oxygen forming sulfur(VI) oxide, the gases are passed over
a hot
vanadium pentoxide catalyst acording to the equation:
2SO2 + O2 → 2SO3
Calculate the volume of oxygen needed to react fully with 200 cm
3
of sulfur dioxide.
Answer
|
The equation shows that sulfur dioxide reacts with oxygen in a 2:1
ratio
Therefore 2 volumes of sulfur dioxide react with 1 volume of oxygen
Thus for complete reaction 200 cm3 of sulfur dioxide needs
100 cm3 of oxygen
|
Q146-02 What volume of carbon
dioxide will be formed from the complete combustion of 200 cm3 methane?
Answer
|
Methane burns according to the equation:
CH4 + 2O2 →
CO2 + 2H2O
1 volume of methane produces 1 volume of carbon dioxide
Therefore 200cm3 methane produces 200cm3
of carbon dioxide
|
Q146-03 What volume of oxygen
will be required to completely oxidise 200 cm
3 of ammonia according
to the equation:
4NH3 + 5O2 → 4NO + 6H2O
Answer
|
From the equation 4 volumes of ammonia requires 5 volumes of oxygen
Therefore 200 cm3 of ammonia requires 200 x 5/4 volumes
of oxygen = 250 cm3 oxygen
|
Q146-04 What will be the
final volume of gas when 300 cm3 ethene reacts with 300 cm3
oxygen at 400ºC?
Answer
|
Equation for the reaction:
C2H4 + 3O2 →
2CO2 + 2H2O
Inspection of the equation shows that 1 volume of ethene requires
3 volumes of oxygen. In the question there is insufficient oxygen
to react with all the ethene, i.e. the oxygen is the limiting reagent
and only 100 cm3 of ethene can react.
Therefore from the equation 1 volume of ethene reacts with 3 volumes
of oxygen to produce 2 volumes of carbon dioxide and 2 volumes of
steam (at this temperature)
So 100 cm3 of ethene reacts with 3 x 100 cm3
of oxygen to produce 200 cm3 of carbon dioxide and 200cm3
of steam
Initial volume of gases = 300 cm3 + 300 cm3
Final volume of gases = 200 cm3(CO2) + 200 cm3(H2O)
+ 200 cm3 (unreacted methane) = 600
cm3 of gas mixture
|
Q146-05 What will be the
final volume of the gas mixture, when 100cm3 propane reacts with
600 cm3 oxygen at 400ºC?
Answer
|
Equation for the reaction:
C3H8 + 5O2 → 3CO2 +
4H2O
Inspection of the equation shows that 1 volume of propane requires
5 volumes of oxygen. In the question there is insufficient propane
to react with all of the oxygen, i.e. the propane is the limiting
reagent and only 500 cm3 of oxygen can react.
Therefore from the equation 1 volume of propane reacts with 5 volumes
of oxygen to produce 3 volumes of carbon dioxide and 4 volumes of
steam (at this temperature)
So 100 cm3 of propane reacts with 5 x 100cm3
of oxygen to produce 300cm3 of carbon dioxide and 400 cm3
of steam
Initial volume of gases = 100 cm3 + 600 cm3
= 700 cm3
Final volume of gases = 300cm3(CO2) + 400cm3
(H2O) + 100cm3 (unreacted oxygen) = 800
cm3 of gas mixture
|
Q146-06 When 100 cm3
of ethene is burned in excess oxygen, calculate the volume of carbon dioxide
produced (all gas volumes measured at STP)
Answer
|
Equation for the reaction:
C2H4 + 3O2 →
2CO2 + 2H2O
1 volume of ethene makes 2 volumes of CO2
100cm3 of ethene = 100/22700 moles = 4.46 x 10-3
moles
this makes 2 x 4.46 x 10-3 moles of carbon dioxide = 8.93
x 10-3 moles
This is equivalent to 8.93 x 10-3 x 22700 =
200cm3
Note: the final answer demonstrates
Gay Lussac's law of combining gas volumes - moles are proportional
to gas volumes
|
Q146-07 Calculate the volume
of oxygen required to react with excess carbon monoxide, to produce 500 cm3
of carbon dioxide gas.
Answer
|
The equation for the reaction:
2CO + O2 →
2CO2
Thus, 2 volumes of carbon monoxide react with 1 volume of oxygen
to produce 2 volumes of carbon dioxide.
If 500 cm3 of carbon dioxide gas are required then 2 volumes
= 500 cm3
1 volume of oxygen is required = 1 x 500/2 = 250cm3
of oxygen
|
Q146-08 In the combusion
of methanal, 100 cm
3 of methanal reacted with 200 cm
3
of oxygen. Calculate the
composition of the gas mixture produced, if all measurements
were taken at STP.
Answer
|
Equation for the reaction:
HCHO + O2 →
CO2 + H2O
Thus, 1 volume of methanal reacts with 1 volume of oxygen in a complete
reaction.
100 cm3 of methanal will react with only 100 cm3
of of oxygen leaving 100 cm3 of oxygen unused.
The final composition of the gas mixture = 100
cm3 of oxygen + 100 cm3 of carbon dioxide
(water is a liquid at STP)
|
Q146-09 At 200ºC carbon
disulfide vapour reacts with oxygen according to the equation:
CS2 + 3O2 → CO2 + 2SO2
Calculate the final composition of the gas mixture produced when 1.2 dm3
of carbon disulfide vapour reacts with 1.2 dm3 of oxygen gas
Answer
|
From the equation:
1 volume of carbon disulfide reacts completely with 3 volumes of
oxygen
In this case the oxygen is the limiting reagent and only 1.2/3 dm3
= 400 cm3 of the carbon disulfide can react (leaving 800
cm3 of the carbon disulfide unreacted)
400 cm3 of carbon disulfide (equivalent to 1 volume in
the equation) produces 400 cm3 of carbon dioxide and 2
x 400 cm3 = 800 cm3 of sulfur dioxide.
The final mixture = 400 cm3
of carbon dioxide, 800cm3 of sulfur dioxide and 800 cm3
of unreacted carbon disulfide
|
Q146-10 In a gunpowder mixture,
the ratio of sulfur to carbon is 3 :1 by mass. If both are fully oxidised in
the explosion to sulfur and carbon dioxide and 200cm3 of sulfur dioxide
are formed, calculate the volume of carbon dioxide that is also produced in
the explosion.
Answer
|
sulfur and carbon react with oxygen according to the equations:
S + O2 →
SO2
C + O2 →
CO2
1 mole of gas is produced per mole of both carbon and sulfur
Relative atomic mass of S = 32 and C = 12
moles = mass/relative mass
Therefore mole ratio of sulfur to carbon = 3/32 : 1/12
multiply through by 12 giving sulfur to carbon ratio = 36/32 : 1
Thus if 200cm3 sulfur dioxide are formed then 200 x 32
/ 36 cm3 = 177.8 cm3
of carbon dioxide are formed
|
Q146-11 Calculate the moles
of gas present at STP in a 5 dm3 flask.
Answer
|
1 moles of gas occupies a volume of 22700 cm3 at STP
Therefore 5 dm3 = 5000cm3 contains 5000/22700
moles = 0.223 moles at STP
|
Q146-12 What volume of oxygen
corresponds to 0.2 moles, measured at STP.
Answer
|
1 mole occupies a volume = 22700 cm3
Therefore 0.2 moles = 0.2 x 22700 cm3 = 4480cm3
|
Q146-13 Find the volume occupied
at STP, by a mixture containing 0.05 moles of hydrogen and 0.025 moles of oxygen
gas.
Answer
|
Total moles of gas = 0.05 + 0.025 = 0.075 moles
1 mole occupies a volume = 22700 cm3
Therefore 0.075 moles of gas occupy a volume of 0.075 x 22700 cm3
= 1680 cm3
|
Q146-14 Calculate the number
of moles of gas contained at STP in a 100 cm3 gas syringe.
Answer
|
1 mole occupies a volume = 22700 cm3 at STP
Therefore 100 cm3 is equivalent to = 100/22700 moles of
gas = 4.46 x 10-3
|
Q146-15 Find the volume occupies
at STP by a mixture containing 0.01 moles of neon and 0.03 moles of oxygen gas.
Answer
|
Total moles of gas = 0.01 + 0.03 = 0.04
1 mole of gas occupies 22700 cm3 at STP
Therefore 0.04 moles occupies 0.04 x 22700 cm3 = 896cm3
|
Q146-16 Find the amount of
oxygen corresponding to a volume at STP of 40 cm3 of oxygen gas.
Answer
|
Oxygen, O2
At STP 40cm3 = 40/22700 moles = 1.79 x 10-3
moles
Therefore 40cm3 oxygen corresponds to = 1.79
x 10-3 moles
|
Q146-17 A mixture of 400cm
3
of oxygen gas and 640 cm
3 of hydrogen gas is contained at STP, calculate
the total number of moles present.
Answer
|
Total volume of gas = 400cm3 of oxygen gas + 640 cm3
of hydrogen gas = 1040 cm3
At STP 1 mole occupies 22700 cm3
Therefore number of moles = 1040/22700 = 0.0458
moles
|
Q146-18 If 2 moles of carbon
monoxide gas are mixed with and 4 moles of hydrogen gas at STP, calculate the
volume occupied.
Answer
|
Total moles of gas = 2 + 4 = 6 moles
1 mols of gas occupies 22700 cm3 at STP
Therefore volume occupied = 6 x 22.7 dm3 = 136.2
dm3
|
Q146-19 A sample of
natural gas contains 120 moles of methane and 20 moles
of helium. Calculate the volume in m
3 occupied by the gases at STP
to three significant figures.
Answer
|
Total moles of gas = 120 + 20 = 140 moles
1 mole occupies 22.7 dm3 at STP, therefore 140 moles occupies
22.7 x 140 = 3178 dm3
1000dm3 = 1m3
Volume occupied by the natural gas mixture = 3178/1000 = 3.18
m3
|
Q146-20 A sample of
syngas contains 60 moles of carbon monoxide, 15 moles
of carbon dioxide and 60 moles of hydrogen. Calculate the volume in m
3
occupied by the gases at STP, to three significant figures.
Answer
|
Total moles of gas = 60 + 15 + 15 = 90 moles
1 mole occupies 22.7 dm3 at STP, therefore 90 moles occupies
22.7 x 90 = 2043 dm3
1000dm3 = 1m3
Therefore the volume occupied by the syngas mixture = 2043/1000 = 2.04
m3
|
^ top
Now test yourself
^ top
