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The Contact process

The Contact process. Production of Sulfuric Acid. H 2 SO 4. There was a girl w ho is, no more f or what she thought w as H 2 O, was H 2 SO 4 !!!!. Steps of The Contact * process.

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The Contact process

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  1. The Contact process Production of Sulfuric Acid

  2. H2SO4 • There was a girl • who is, • no more • for what she thought • was H2O, • was H2SO4!!!!

  3. Steps of The Contact* process The combustion of sulfur makes sulfur dioxide S8(s) + 8O2 → 8SO2(g) The sulfur dioxide is converted into sulfur trioxide (the reversible reaction at the heart of the process); 2SO2(g) + O2(g)D 2SO3(g) 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. 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)

  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: The reasons for all these conditions will be explored in detail further in this presentation.

  7. 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.

  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. 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. 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

  10. 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.

  11. 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.

  12. 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.

  13. 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.

  14. 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.

  15. Explaining the conditions The temperature: Equilibrium considerations • http://www.chemguide.co.uk/physical/equilibria/contact.html

  16. 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.

  17. 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!

  18. 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.

  19. 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 SO3 to occur at an appreciable rate.

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