There was a girl
who is,
no more
for what she thought
was H2O,
was H2SO4!!!!
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
1. MAKING THE SULFUR DIOXIDE
This can be made by
burning sulfur in an
excess of air:
S8(s) + 8O2 → 8SO2(g)
. . . or by heating sulfide ores like pyrite in an excess of air:
4FeS2(s) + 11O2(g)
→ 2Fe2O3(s) + 8SO2(g)
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.
2. CONVERTING THE SULFUR DIOXIDE INTO SULFUR
TRIOXIDE
A flow scheme for this part of the process looks like this:
The reasons for all these conditions
will be explored in detail further in
this presentation.
3. CONVERTING THE SULFUR TRIOXIDE INTO
SULFURIC ACID
This could be done by simply adding
water to the sulfur trioxide – but done
this way, the reaction is so
uncontrollable that it creates a fog of
sulfuric acid.
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.
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)
Explaining the conditions
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
Explaining the conditions
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
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.
Explaining the conditions
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. Great! But
you aren't going to produce much sulfur trioxide every day. The vast majority of
what you are passing over the catalyst is oxygen which has nothing to react with.
Explaining the conditions
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.
Explaining the conditions
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. Fotrthis reason , it is called the rate-determining step.
This means that attempts to change the overall reaction rate will focus on this step.
Explaining the conditions
THE TEMPERATURE: EQUILIBRIUM
CONSIDERATIONS
http://www.chemguide.co.uk/physical/equilibria/contact.html
Explaining the conditions
THE TEMPERATURE: EQUILIBRIUM
CONSIDERATIONS
2 SO2(g)+ O2(g) D 2 SO3(g) DH = -196 kJ
From the point of view of temperature, 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.
Explaining the conditions
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!
Explaining the conditions
THE PRESSURE: EQUILIBRIUM CONSIDERATIONS
2 SO2(g)+ O2(g) D 2 SO3(g) DH = -196 kJ
We can see that 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.
Explaining the conditions
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 SO 3 to occur at an
appreciable rate.