kswales1
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Hi, I need help answering this question:
I've been asked to discuss the effect on rate of reaction and yield, considering the quantity of oxygen with reference to increasing the ratio of oxygen to sulphur dioxide from that indicated in the equation to equal volumes.
The equation being used is:
SO2 (g) + ½ O2 (g) ⇌ SO3 (g) ΔH = –197 kJ mol–1

Thanks in advance for any help : )
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acestar123
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(Original post by kswales1)
Hi, I need help answering this question:
I've been asked to discuss the effect on rate of reaction and yield, considering the quantity of oxygen with reference to increasing the ratio of oxygen to sulphur dioxide from that indicated in the equation to equal volumes.
The equation being used is:
SO2 (g) + ½ O2 (g) ⇌ SO3 (g) ΔH = –197 kJ mol–1

Thanks in advance for any help : )
The proportions of sulphur dioxide and oxygen

The mixture of sulphur dioxide and oxygen going into the reactor is in equal proportions by volume.

Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. That means that the gases are going into the reactor in the ratio of 1 molecule of sulphur dioxide to 1 of oxygen.

That is an excess of oxygen relative to the proportions demanded by the equation.

SO2 (g) + ½ O2 (g) ⇌ SO3 (g) ΔH = –197 kJ mol–1

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 sulphur dioxide into sulphur trioxide.

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 sulphur dioxide.

The equilibrium is going to be tipped very strongly towards sulphur trioxide - virtually every molecule of sulphur dioxide will be converted into sulphur trioxide. Great! But you aren't going to produce much sulphur trioxide every day. The vast majority of what you are passing over the catalyst is oxygen which has nothing to react with.

By increasing the proportion of oxygen you can increase the percentage of the sulphur dioxide converted, but at the same time decrease the total amount of sulphur trioxide made each day. The 1 : 1 mixture turns out to give you the best possible overall yield of sulphur trioxide.
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