Industrial chemistry

3. STEPS OF THE CONTACT PROCESS
1. The combustion of sulfur makes
sulfur dioxide
S8(s) + 8O2 → 8SO2(g)
2. The sulfur dioxide is converted into
sulfur trioxide (the reversible
reaction at the heart of the
process);
2SO2(g) + O2(g) D 2SO3(g)
3. The sulfur trioxide is converted
into concentrated sulfuric acid.
H2SO4(l) + SO3(g) → H2S2O7(l)
H2S2O7(l) + H2O(l) → 2H2SO4(l)
* Called “contact” since the molecules of the gases O2
and SO2 are in contact with the surface of the solid
catalyst, V2O5

4. This can be made
by burning sulfur in an
excess of air:
S8(s) + 8O2 → 8SO2(g)
1. MAKING THE SULFUR DIOXIDE
. . . or by heating sulfide ores like pyrite in an excess of air:
4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g)

5. 2. CONVERTING THE SULFUR DIOXIDE INTO SULFUR
TRIOXIDE
2SO2(g) + O2(g) D 2SO3(g) DH = -196 kJ
This is a reversible reaction, and the formation of the sulfur trioxide is exothermic.

6. 2. CONVERTING THE SULFUR DIOXIDE INTO SULFUR
TRIOXIDE
A flow scheme for this part of the process looks like this:

7. 3. CONVERTING THE SULFUR TRIOXIDE INTO
SULFURIC ACID
This could be done by simply adding water
to the sulfur trioxide – but the reaction is
so uncontrollable that it creates a fog of
sulfuric acid.

8. 3. CONVERTING THE SULFUR TRIOXIDE INTO
SULFURIC ACID
Instead, the sulfur trioxide is first dissolved in concentrated
sulfuric acid:
H2SO4(l) + SO3(g) → H2S2O7(l)
The product is known as fuming sulfuric acid.(Olleum)
This can then be reacted safely with water to produce concentrated
sulfuric acid - twice as much as you originally used to make the
fuming sulfuric acid.
H2S2O7(l) + H2O(l) → 2H2SO4(l)

9. THE PROPORTIONS OF SULFUR DIOXIDE AND
OXYGEN
The mixture of sulfur dioxide and oxygen going
into the reactor is in equal proportions by
volume.
That is, an excess of oxygen relative to the
proportions demanded by the equation.
2SO2(g) + O2(g) D 2SO3(g) DH = -196 kJ

10. THE PROPORTIONS OF SULFUR DIOXIDE AND
OXYGEN
2SO2(g) + O2(g) D 2SO3(g) DH = -196 kJ
According to Le Chatelier's Principle, Increasing the concentration of oxygen in
the mixture causes the position of equilibrium to shift towards the right. Since
the oxygen comes from the air, this is a very cheap way of increasing the
conversion of sulfur dioxide into sulfur trioxide.
Explaining the conditions

11. THE PROPORTIONS OF SULFUR DIOXIDE AND
OXYGEN
2SO2(g) + O2(g) D 2SO3(g) DH = -196 kJ
Why not use an even higher proportion of
oxygen? This is easy to see if you take an
extreme case. Suppose you have a million
molecules of oxygen to every molecule of sulfur
dioxide.

12. THE PROPORTIONS OF SULFUR DIOXIDE AND
OXYGEN
2 SO2(g)+ O2(g) D 2 SO3(g) DH = -196 kJ
• The equilibrium is going to be tipped very
strongly towards sulfur trioxide
• virtually every molecule of sulfur dioxide
will be converted into sulfur trioxide.
• But you aren't going to produce much
sulfur trioxide every day. The vast majority of
what you are passing over the catalyst is
unreacted oxygen which has nothing to react
with.

13. THE PROPORTIONS OF SULFUR DIOXIDE AND
OXYGEN
2 SO2(g)+ O2(g) D 2 SO3(g) DH = -196 kJ
• By increasing the proportion of oxygen you
can increase the percentage of the sulfur
dioxide converted, but at the same time
decrease the total amount of sulfur trioxide
made each day.
• The 1 : 1 mixture turns out to give you the best
possible overall yield of sulfur trioxide.

14. THE REACTION MECHANISM
2 SO2(g)+ O2(g) D 2 SO3(g) DH = -196 kJ
Studies show that this step has the lowest
reaction rate, so acts as a bottleneck for
the entire process. For this reason , it is
called the rate-determining step.
This means that attempts to change the
overall reaction rate will focus on this
step.

15. THE TEMPERATURE: EQUILIBRIUM
CONSIDERATIONS
2 SO2(g)+ O2(g) D 2 SO3(g) DH = -196 kJ
• The forward (exothermic) reaction is
favored by a decrease in temperature.
Despite the fact that the forward (exothermic)
reaction is favored by a low temperature, the
rate will be too slow at very low temperatures.

16. THE TEMPERATURE: ECONOMIC CONSIDERATIONS
Extreme temperatures (in either
direction) are costly.
A temperature of 450oC is chosen…
once again a compromise which takes all
three considerations (economic,
equilibrium, and reaction rate) into
consideration!

17. THE PRESSURE: EQUILIBRIUM CONSIDERATIONS
2 SO2(g)+ O2(g) D 2 SO3(g) DH = -196 kJ
This reaction has 3 gas molecules on the
left and only 2 gas molecules on the
right.
Hence the forward reaction is favored by
high pressure!!
A pressure of 2 atm gives a sufficient
yield, so higher pressure (more costly) is
not required.

18. THE CATALYST
The catalyst chosen is V2O5 vanadium (V)
oxide,
The catalyst favors neither reaction,
because it increases the rates of both
forward and reverse reactions equally.
However, the conditions in the process are
such that the system probably never
reaches equilibrium, so the catalyst will
allow the production of SO3 to occur at an
appreciable rate.