# Why does the conversion of SO₂ to SO₃ require high pressure?

The conversion of sulphur dioxide to sulphur trioxide in the contact process for the manufacture of sulphuric acid requires the following condition:

• Temperature: 450 degrees

• Pressure: 1-2 atm, although High pressure would fasten the reaction, creating those pressures is not economical

• Vanadium Pentoxide ($\ce{V2O5}$) as catalyst
• and excess of oxygen

$\ce{2SO2 + O2 -> 2SO3}$

My textbook says that high pressure favors the rate of the reaction. The reason it states is that the volume of product is less than that of the reactants.

Can somebody explain how the volume of the product being less causes high pressure to be required?

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You might want to read something about Le Chatelier's Principle (en.wikipedia.org/wiki/Le_Chatelier%27s_principle) – Philipp Dec 27 '12 at 22:42
so according to the same principle , if he remove generating S03 as the rate it generate, it will definitely speedups the process? – sandun dhammika Jan 9 '13 at 10:46

As Philipp stated in the comments it is Le Chatelier's principle in play.

The principle in layman's language states that

If you try to bring about any physical change in a system the system will try to but not necessarily succeed in cancelling the change.

The reaction you have given is an equilibrium reaction where all the compounds are in gaseous state.

Now if you count the total number of moles on the left-hand side it will be $\ce{2SO2 + O2}$ i.e. $\ce{2 + 1}$ that is $\text{3}$. And on the right-hand side there are only $\text{2}$ moles of $\ce{SO3}$.

So ultimately if you consider the reaction to proceed in forward direction 3 moles of gas are converting into 2 moles of gas and hence the volume decreases in forward direction.

If you think according to Gas Law(or logically) an increase in pressure generally corresponds to a decrease in volume.

Now use Le Chatelier's principle to say that if I increase the pressure on the the system ,in order the system to counteract to the change will decrease it's own volume and it has no other way of achieving that but to proceed in the forward direction.

But if the query arises that high pressure should favour any and every reaction take a look at the example below:

$$\ce{PCl5 <=> PCl3 + Cl2}$$

Here the gaseous moles on the left-hand side is $\ce{1}$ and on the right-hand side $\ce{2}$. So increasing pressure makes the reaction proceed in the backwards direction as there are less number of gaseous moles on the left-hand side.

In fact, in higher level this idea is what is called $\Delta\text{n}_\text{g}$ i.e the difference($\Delta$) of gaseous($_g$) moles($\ce{n}$).

$$\Delta\text{n}_\text{g} \quad = \text{gaseous moles on right-hand side - gaseous moles on left-hand side}$$

If $\Delta\text{n}_\text{g}$ is positive, then on increasing pressure reaction proceeds backwards and vice-cersa

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so doesn't that mean high pressure would favour all reactions in terms of amount of yield? So isn't it right if I say, higher the pressure of any reaction- higher will be the yield? – Ghost Dec 28 '12 at 9:17
@Ghost I edited the answer to incorporate your query. – Aditya Sriram Dec 28 '12 at 12:13
The reaction you've provided is reversible, and hence it's able to proceed backwards. Now my question is , will high pressure favor each and every 'irreversible' reaction?(since it cant proceed backwards) – Ghost Dec 28 '12 at 13:52
Le Chatelier's principle is not applicable on irreversible reactions. If the reaction is irreversible it will proceed all the way forward and changing the pressure will have minimal to null effect on the reaction. – Aditya Sriram Dec 28 '12 at 14:01
Thanks a lot for such a detailed explanation! Greatly appreciate it. – Ghost Dec 28 '12 at 14:11