Sulphuric Acid and the Oxoacids of Sulphur
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Sulphuric acid for JEE Main and Advanced: the contact process over V2O5 and the conditions that favour SO3, why SO3 goes into oleum and not water, the dehydrating and oxidising action of the acid, why HBr is not made with it, the structures and oxidation states of the oxoacids of sulphur, and two solved problems: counting S=O bonds and the free SO3 in 109% oleum.
Chapters
- 0:00Introduction
- 0:23Contact process: making SO₃
- 1:36From SO₃ to sulphuric acid: oleum
- 2:21Sulphuric acid: strong, dibasic, dehydrating
- 3:11Hot, concentrated acid: an oxidising agent
- 4:08Oxoacids of sulphur: read the structures
- 5:50Example 1: counting S=O bonds in three oxoacids of sulphur
- 6:53Example 2: free SO3 in 109% oleum
- 8:16Quick revision: oleum, oxidising acid and peroxo acids
Transcript
0:00Introduction
Hello everyone! Okay, a small puzzle first. Imagine a bottle of oleum with the label one hundred and nine percent. One hundred and nine percent? Nothing can be more than one hundred percent, right? By the end, you will know exactly what is hidden inside that bottle. This topic comes up often in JEE, so let's begin.
0:23Contact process: making SO₃
So, sulphuric acid is made by the contact process. It starts with S O two, sulphur dioxide, from burning sulphur or roasting sulphide ores; the small two is a subscript counting atoms. The gas is purified first, because dust and arsenic compounds would poison the catalyst. Okay? Now the key step: two S O two plus O two gives two S O three, sulphur trioxide. The two half arrows mean it is reversible. Delta H, the Greek capital delta, with a small circle for standard, is minus one hundred and ninety six point six kilojoules per mole, written k J mol to the minus one, so it is exothermic. Right? The catalyst is V two O five, vanadium pentoxide, at about seven hundred and twenty kelvin, the capital K, and two bar of pressure. Clear? Now think like Le Chatelier. The forward reaction is exothermic, and three molecules of gas become two. So which temperature, and which pressure, will favour S O three? Low temperature and high pressure, yes, exactly! And excess oxygen also helps, because adding a reactant shifts the equilibrium forward.
1:36From SO₃ to sulphuric acid: oleum
Got it? Now we have S O three, and we want sulphuric acid. So we simply pass it into water, right? No, no, careful, that is the trap! The reaction with water is so exothermic that a fog of tiny acid droplets forms, which is very hard to condense. Okay? Instead, it is absorbed in concentrated sulphuric acid. S O three plus H two S O four gives H two S two O seven, called oleum; the arrow means gives. Then the oleum is diluted: H two S two O seven plus H two O gives two H two S O four. Remember oleum; it is the key to our puzzle. Clear?
2:21Sulphuric acid: strong, dibasic, dehydrating
Now, sulphuric acid itself is a strong dibasic acid. K sub a one is very large, and K sub a two is one point two times ten to the power minus two. With me? It has low volatility, so it displaces more volatile acids from their salts. N a C l plus H two S O four gives N a HSO four plus H C l, the volatile acid. Right? Pour concentrated sulphuric acid on sugar, C twelve H twenty two O eleven, and the sugar chars black. So what has the acid taken out of it? Water, yes, exactly! It is a strong dehydrating agent. C twelve H twenty two O eleven gives twelve C plus eleven H two O, and the black solid is carbon. Clear?
3:11Hot, concentrated acid: an oxidising agent
Next, hot concentrated sulphuric acid is also an oxidising agent. With copper, C u plus two H two S O four gives C u S O four, plus S O two, plus two H two O, and the acid itself is reduced to S O two. Okay? Now a trap: we just made H C l from N a C l. So to make H B r, we simply heat N a B r with concentrated sulphuric acid, right? Oh no, careful, the H B r that forms is oxidised by the acid to bromine! Two H B r plus H two S O four gives B r two, plus S O two, plus two H two O, so the H B r is lost. Clear? H I is oxidised too. So H B r and H I are made with phosphoric acid, H three P O four, which is non-oxidising. Got it?
4:08Oxoacids of sulphur: read the structures
Now, the figure shows five oxoacids. H two S O three, sulphurous acid, has sulphur in the plus four state, with one S double bond O, the two lines; the two dots are a lone pair. H two S O four, sulphuric acid, has sulphur in the plus six state, and two S double bond O, okay? Next, H two S two O three, thiosulphuric acid. Here one sulphur is bonded to another sulphur, written S dash S on the line, and the figure draws that bond as a double bond. So it has two kinds of sulphur, with an average oxidation state of plus two. Right? Then H two S two O seven, pyrosulphuric acid, our oleum. Its two sulphur atoms are joined through an oxygen, an SOS bridge. With me? H two S two O eight, peroxodisulphuric acid, or Marshall's acid, has an SOOS bridge. H two S O five, peroxomonosulphuric acid, or Caro's acid, has one O O bond. Only these two oxoacids have an O O bond. Clear? Now, here is a trap on oxidation states. In H two S two O eight, take hydrogen as plus one and all eight oxygens as minus two. Then sulphur comes out as plus seven, right? No, no, wait, the two oxygens in the peroxide link are minus one each. So, write x for the oxidation state of sulphur. Then two, plus two times x, minus twelve, minus two, equals zero, and x equals plus six. Got it?
5:50Example 1: counting S=O bonds in three oxoacids of sulphur
Now example one. Count the S double bond O bonds in sulphurous acid, peroxodisulphuric acid and pyrosulphuric acid. Use the figure, and pause the video to try it first. Done? Okay, let's check. Sulphurous acid, H two S O three, is O S, then O H in brackets with a two outside, so it has one oxygen on sulphur and two O H groups. So just one S double bond O, okay? Next, peroxodisulphuric acid has eight oxygens. Two sit in the O O bridge, and two in O H groups. So how many S double bond O bonds are left? Four, yes, exactly! Each S O two O H end carries two of them, and there are two ends. Right? And pyrosulphuric acid also has two on each sulphur, so four again. The answer is one, four and four. Clear?
6:53Example 2: free SO3 in 109% oleum
Now example two, and it solves our puzzle. Oleum is labelled one hundred and nine percent: one hundred grams of it gives one hundred and nine grams of H two S O four on adding water. What is the percentage of free S O three? Pause the video and try it first. Done? Okay, let's check. One hundred grams went in, and one hundred and nine grams of acid came out. So where did the extra nine grams come from? From the water we added, exactly! One hundred and nine minus one hundred gives nine grams of water taken up, g for grams. Nine over eighteen, the molar mass of water, gives zero point five zero moles, mol for short. With me? Each S O three takes up one water molecule: S O three plus H two O gives H two S O four. So the free S O three is also zero point five zero moles. Okay? The molar mass of S O three is eighty, so zero point five zero times eighty gives forty grams, in one hundred grams of oleum. The answer is forty percent free S O three. So that is our label: the extra nine percent is the water taken up by the free S O three. Got it?
8:16Quick revision: oleum, oxidising acid and peroxo acids
So, quick revision. In the contact process, S O three goes into concentrated sulphuric acid as oleum, then gets diluted. Hot concentrated acid is an oxidising agent, so H B r and H I are made with phosphoric acid instead. And the peroxo acids have an O O bond, yet sulphur is still plus six. Now go and try the practice problems on this page. All the best!
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- Sulphur and Sulphur Dioxide
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- Interhalogens, Polyhalides and Pseudohalides
- Group 18: Noble Gases and Xenon Compounds
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- Mixed Problems: Set 3 (Groups 17 and 18)
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- Part A: JEE Main Level
- Part C: Concept Checks