Sulphuric Acid — Manufacture & Properties

Why sulphuric acid matters

Among the oxoacids of sulphur, $\ce{H2SO4}$ is the one that built the chemical industry. It is sometimes called the "king of chemicals" because it feeds into the manufacture of fertilisers, paints and pigments, detergents, plastics and fibres, and serves as a workhorse laboratory reagent. The acid was known to the alchemists as oil of vitriol; before the modern routes it was obtained by heating hydrated sulphates. Today almost all of it is made by the Contact Process, which displaced the older Lead Chamber process.

For NEET, the chapter rewards a clear command of three things: the four ordered steps of the Contact Process, the equilibrium reasoning behind the conditions used to oxidise $\ce{SO2}$, and the property cluster — strong dibasic acid, low volatility, dehydrating agent, oxidising agent — that distinguishes dilute behaviour from hot-concentrated behaviour.

The Contact Process at a glance

The Contact Process converts elemental sulphur (or sulphide ores) into 96–98% sulphuric acid in a continuous plant. It is named for the catalytic "contact" step in which $\ce{SO2}$ and $\ce{O2}$ react on the surface of a solid catalyst. The whole sequence can be read as four stages, summarised below before we take each one apart.

Step 1 — Producing sulphur dioxide

The feedstock gas is sulphur dioxide, generated either by burning molten sulphur in a stream of air or by roasting sulphide ores such as iron pyrites:

$$\ce{S + O2 -> SO2}$$

$$\ce{4FeS2 + 11O2 -> 2Fe2O3 + 8SO2}$$

The $\ce{SO2}$ that leaves the burner is not clean enough to pass over the catalyst. It carries dust and, critically, arsenic compounds. The gas is therefore sent through a dust precipitator, a washing-and-cooling tower, an arsenic purifier (gelatinous hydrated ferric oxide) and a drying tower. Drying matters because moisture would otherwise condense as acid mist downstream.

Step 2 — Catalytic oxidation of SO₂ to SO₃

This is the key step of the entire process — the "contact" that gives the method its name. Purified $\ce{SO2}$ is oxidised by atmospheric oxygen over a vanadium(V) oxide catalyst:

$$\ce{2SO2(g) + O2(g) <=>[\ce{V2O5}][720\,\text{K}] 2SO3(g)}\qquad \Delta_r H = -196.6\ \text{kJ mol}^{-1}$$

Three features of this equilibrium control the choice of operating conditions: it is exothermic, it is reversible, and the forward reaction proceeds with a decrease in the number of gas molecules (3 moles of gas → 2 moles).

Le Chatelier rationale for the conditions

Applying Le Chatelier's principle to each feature tells us which way the equilibrium yield of $\ce{SO3}$ shifts, but the chosen plant conditions are a compromise between yield and rate.

Note the deliberate tension on temperature. A genuinely low temperature would maximise the equilibrium amount of $\ce{SO3}$, but it would make the reaction unworkably slow. The temperature is therefore held at roughly 720 K — high enough to give a practical rate, low enough that the equilibrium still lies well towards $\ce{SO3}$. The pressure of about 2 bar is mild because the equilibrium is already favourable; pushing pressure higher gains little and costs much.

Step 3 — Absorption to oleum, then dilution

The $\ce{SO3}$ leaving the converter is not passed into water directly. Doing so produces a dense, persistent and highly corrosive mist (fog) of fine sulphuric acid droplets that is extremely difficult to condense and handle. Instead, $\ce{SO3}$ is absorbed into concentrated (98%) sulphuric acid in an absorption tower, forming oleum, also called pyrosulphuric acid:

$$\ce{SO3 + H2SO4 -> H2S2O7}\quad \text{(oleum)}$$

Oleum is then diluted with the calculated amount of water to give sulphuric acid of the desired concentration:

$$\ce{H2S2O7 + H2O -> 2H2SO4}$$

In practice the absorption and dilution are run as a continuous loop so the process is uninterrupted and economical. The acid that emerges is about 96–98% pure.

Structure of H₂SO₄

In the sulphuric acid molecule the sulphur atom sits at the centre of a roughly tetrahedral arrangement of four oxygen atoms. Two of these are joined to sulphur by $\ce{S=O}$ double bonds, and the other two are present as $\ce{S-O-H}$ hydroxyl groups. It is these two $\ce{O-H}$ protons that make the acid dibasic (diprotic).

This bonding pattern — two S–OH and two S=O — is exactly what NEET tests when it asks you to match $\ce{H2SO4}$ against the other oxoacids of sulphur. Contrast it with sulphurous acid $\ce{H2SO3}$ (two S–OH, one S=O), pyrosulphuric acid $\ce{H2S2O7}$ (an $\ce{S-O-S}$ bridge) and peroxodisulphuric acid $\ce{H2S2O8}$ (a peroxide $\ce{-O-O-}$ linkage).

Physical properties

Pure sulphuric acid is a colourless, dense, oily liquid. Its key constants from the source data are tabulated below. The high boiling point underlies its low volatility, while the enormous heat of dilution dictates the safe mixing rule.

The chemical behaviour of sulphuric acid follows from four characteristics: (a) strong acidic character, (b) low volatility, (c) strong affinity for water (dehydrating), and (d) ability to act as an oxidising agent. The next four sections take them in turn.

Strong dibasic acid

In aqueous solution sulphuric acid ionises in two steps, furnishing two protons per molecule — hence dibasic (diprotic). The first ionisation is essentially complete; the second is appreciably weaker:

$$\ce{H2SO4(aq) + H2O(l) -> H3O^+(aq) + HSO4^-(aq)}\qquad K_a\ \text{very large } (>10)$$

$$\ce{HSO4^-(aq) + H2O(l) -> H3O^+(aq) + SO4^{2-}(aq)}\qquad K_a = 1.2\times10^{-2}$$

The very large first dissociation constant means $\ce{H2SO4}$ is largely dissociated into $\ce{H+}$ and $\ce{HSO4^-}$, which is why it ranks as a strong acid (the larger the $K_a$, the stronger the acid). Because it gives two protons, it forms two series of salts:

Low volatility

With a boiling point of 611 K, sulphuric acid is far less volatile than acids such as $\ce{HCl}$, $\ce{HF}$ and $\ce{HNO3}$. This single physical fact gives it a powerful synthetic role: it can liberate a more volatile acid from its salt. On heating a metal halide or nitrate with concentrated $\ce{H2SO4}$, the volatile acid distils off while the non-volatile sulphate remains behind, pushing the reaction to completion:

$$\ce{2MX + H2SO4 -> 2HX + M2SO4}\qquad (\text{X} = \text{F, Cl, NO3})$$

This is precisely how $\ce{HCl}$, $\ce{HF}$ and $\ce{HNO3}$ are prepared in the laboratory from $\ce{NaCl}$, $\ce{CaF2}$ and a nitrate respectively. The driving force is escape of the volatile product, not any superior acid strength of $\ce{H2SO4}$ over the displaced acid.

Dehydrating agent

Concentrated sulphuric acid has a powerful affinity for water and acts as a strong dehydrating agent. Importantly, in dehydration the acid removes water (or the elements H and O in the ratio of water) without changing its own oxidation state — this distinguishes it from oxidising behaviour. Three textbook demonstrations:

The charring of sugar is the classic equation:

$$\ce{C12H22O11 ->[\ce{conc.\,H2SO4}] 12C + 11H2O}$$

And the colour change of copper sulphate makes the dehydrating action visible:

$$\ce{\underset{(blue)}{CuSO4.5H2O} ->[\ce{conc.\,H2SO4}] \underset{(white)}{CuSO4} + 5H2O}$$

Oxidising agent

Hot concentrated sulphuric acid is a moderately strong oxidising agent — in oxidising power it sits between phosphoric acid and nitric acid. When it oxidises a substrate it is itself reduced, usually to $\ce{SO2}$. It will oxidise both metals and non-metals:

$$\ce{Cu + 2H2SO4(conc.) -> CuSO4 + SO2 + 2H2O}$$

$$\ce{S + 2H2SO4(conc.) -> 3SO2 + 2H2O}$$

$$\ce{C + 2H2SO4(conc.) -> CO2 + 2SO2 + 2H2O}$$

Uses

Sulphuric acid is needed for the manufacture of hundreds of compounds and underpins many industrial processes; the bulk of output goes into fertilisers such as ammonium sulphate and superphosphate. Other major applications, drawn from the source list, are summarised below.