Sulphuric acid and the oxoacids of sulphur

The acid that every factory uses

Fertiliser, paint, the battery of a car, detergent and petrol all depend on sulphuric acid. A country's output of the acid is often used as a sign of how much industry it has. The acid is made by the contact process.

The contact process

1. Burn the sulphur. Sulphur, or the sulphur dioxide from roasting pyrites, is burned in dry air: $\text{S}\ +\ \text{O}_{2}\ \rightarrow \ \text{SO}_{2}$. The gas is purified to remove dust and substances that poison the catalyst.
2. Oxidise to sulphur trioxide. This is the key step, a reversible reaction: $2\text{SO}_{2}\ +\ \text{O}_{2}\ \rightleftharpoons \ 2\text{SO}_{3}$. The enthalpy change is about $-196.6$ kJ per mole of the reaction as written, the value in NCERT. Tables of standard values give about $-197.8$ kJ. The catalyst is vanadium(V) oxide, $\text{V}_2\text{O}_5$. The plant works at about 720 K and 1 to 2 bar.
3. Absorb the trioxide. The sulphur trioxide is not dissolved in water, because that makes a thick mist of acid that is hard to condense. It is absorbed in concentrated sulphuric acid to give oleum, $\text{H}_2\text{S}_2\text{O}_7$: $\text{SO}_{3}\ +\ \text{H}_{2}\text{SO}_{4}\ \rightarrow \ \text{H}_{2}\text{S}_{2}\text{O}_{7}$.
4. Dilute the oleum. Water is added to the oleum: $\text{H}_{2}\text{S}_{2}\text{O}_{7}\ +\ \text{H}_{2}\text{O}\ \rightarrow \ 2\text{H}_{2}\text{SO}_{4}$. The acid is about 98 percent by mass.

Apply Le Chatelier's rule to step 2. Three moles of gas become two, so a high pressure helps. The reaction gives out heat, so a low temperature helps. But at a low temperature the reaction is slow, even with the catalyst. So the plant uses 720 K. The pressure is only 1 to 2 bar, because the conversion is already about 96 to 98 percent, and a high pressure would cost a lot and need strong equipment.

The simulation uses standard enthalpies of formation and entropies at 298 K: $-296.8$ and $-395.7$ kJ/mol for $\text{SO}_2$ and $\text{SO}_3$, and 248.2, 256.8 and 205.2 J/(mol K) for $\text{SO}_2$, $\text{SO}_3$ and $\text{O}_2$. They give $\Delta H = -197.8$ kJ and $\Delta S = -188$ J/K for the reaction as written. The model treats the gases as ideal and keeps $\Delta H$ and $\Delta S$ constant.

Set 720 K and 2 bar. The conversion is about 97 percent. Raise the temperature to 900 K and the curve moves up, so the same pressure now gives a lower conversion. Now go back to 720 K and change the pressure to 1 bar and then 10 bar. The conversion changes by only a small amount, because it is already high.

Practise with new numbers

What concentrated sulphuric acid does

Concentrated sulphuric acid is a thick, oily liquid. Its density is about 1.84 g/cm$^3$. It mixes with water with a great release of heat. So the acid must always be poured into the water, a little at a time, and never the other way round. It shows four kinds of behaviour.

• A strong acid. It is a dibasic acid. The first ionisation is almost complete. The second, $\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-}$, has a constant $K_a = 1.2 \times 10^{-2}$.
• A dehydrating agent. It removes water from other substances. It chars sugar: $\text{C}_{12}\text{H}_{22}\text{O}_{11}\ \xrightarrow{\text{conc}.\ \text{H}_{2}\text{SO}_{4}}\ 12\text{C}\ +\ 11\text{H}_{2}\text{O}$. It turns blue copper sulphate crystals white: $\text{CuSO}_{4}\cdot 5\text{H}_{2}\text{O}\ \rightarrow \ \text{CuSO}_{4}\ +\ 5\text{H}_{2}\text{O}$.
• An oxidising agent when hot and concentrated. It gives sulphur dioxide: $\text{Cu}\ +\ 2\text{H}_{2}\text{SO}_{4}\ \rightarrow \ \text{CuSO}_{4}\ +\ \text{SO}_{2}\ +\ 2\text{H}_{2}\text{O}$, $\text{C}\ +\ 2\text{H}_{2}\text{SO}_{4}\ \rightarrow \ \text{CO}_{2}\ +\ 2\text{SO}_{2}\ +\ 2\text{H}_{2}\text{O}$ and $\text{S}\ +\ 2\text{H}_{2}\text{SO}_{4}\ \rightarrow \ 3\text{SO}_{2}\ +\ 2\text{H}_{2}\text{O}$.
• A non-volatile acid. It does not evaporate easily, so it drives out the more volatile acids from their salts: $\text{NaCl}\ +\ \text{H}_{2}\text{SO}_{4}\ \rightarrow \ \text{NaHSO}_{4}\ +\ \text{HCl}$ and $\text{CaF}_{2}\ +\ \text{H}_{2}\text{SO}_{4}\ \rightarrow \ \text{CaSO}_{4}\ +\ 2\text{HF}$. It also oxidises hydrogen bromide: $2\text{HBr}\ +\ \text{H}_{2}\text{SO}_{4}\ \rightarrow \ \text{Br}_{2}\ +\ \text{SO}_{2}\ +\ 2\text{H}_{2}\text{O}$.

The acid in a lead storage battery is about 38 percent by mass. Other uses include fertilisers such as superphosphate and ammonium sulphate, the refining of petroleum, and making detergents and explosives.

The oxoacids of sulphur

An oxoacid has the central atom bonded to oxygen atoms, some of them in OH groups. Sulphur forms many. Their oxidation states come from the rule that oxygen is $-2$ (but $-1$ in a peroxide link) and hydrogen is $+1$.

• Sulphurous acid, $\text{H}_{2}\text{SO}_{3}$: sulphur is $+4$. It is known only in solution.
• Sulphuric acid, $\text{H}_{2}\text{SO}_{4}$: $+6$.
• Dithionic acid, $\text{H}_{2}\text{S}_{2}\text{O}_{6}$: two $\text{SO}_3\text{H}$ groups joined by an S-S bond. Sulphur is $+5$.
• Pyrosulphuric acid (oleum), $\text{H}_{2}\text{S}_{2}\text{O}_{7}$: two $\text{SO}_3\text{H}$ groups joined by an S-O-S bridge. Sulphur is $+6$.
• Peroxomonosulphuric acid (Caro's acid), $\text{H}_{2}\text{SO}_{5}$: one O-O group. Sulphur is $+6$.
• Peroxodisulphuric acid (Marshall's acid), $\text{H}_{2}\text{S}_{2}\text{O}_{8}$: an S-O-O-S link. Sulphur is $+6$.
• Thiosulphuric acid, $\text{H}_{2}\text{S}_{2}\text{O}_{3}$: the average state of sulphur is $+2$. One sulphur atom is bonded like a sulphate sulphur, and the other, the terminal one, is $-1$.
• Dithionous acid, $\text{H}_{2}\text{S}_{2}\text{O}_{4}$ ($+3$), and disulphurous acid, $\text{H}_{2}\text{S}_{2}\text{O}_{5}$ ($+4$), are the others in the list.

The average oxidation state of sulphur in each acid is the sum that makes the molecule neutral. Take $\text{H}_{2}\text{S}_{2}\text{O}_{8}$. The six ordinary oxygen atoms give $-12$ and the two peroxide oxygen atoms give $-2$. Hydrogen gives $+2$. So the two sulphur atoms must give $+12$, and each is $+6$. The peroxide link does not change the state of sulphur. The acid is a strong oxidising agent because of that link.

Try it in Python

The program finds the equilibrium conversion of sulphur dioxide by halving an interval, as in the simulation. It prints the conversion at four temperatures.

A second program: oxidation states

The program works out the oxidation state of sulphur in six oxoacids. It treats the peroxide oxygen atoms separately.

Common mistakes

• Dissolving the trioxide in water. It makes a mist. Concentrated acid is used for the absorption.
• Thinking a low temperature is the best for the plant. The rate is too low. The plant uses 720 K with a catalyst.
• Adding water to the concentrated acid. Add the acid to the water.
• Giving sulphur $+8$ in a peroxo acid. The peroxide oxygen atoms are at $-1$, and sulphur stays at $+6$.
• Saying that the catalyst raises the yield. It only makes the reaction reach equilibrium sooner.

Remember these

• Contact process: $2\text{SO}_{2}\ +\ \text{O}_{2}\ \rightleftharpoons \ 2\text{SO}_{3}$ over $\text{V}_{2}\text{O}_{5}$, about 720 K and 1 to 2 bar. The trioxide goes into concentrated acid to give oleum, which is then diluted.
• Concentrated $\text{H}_{2}\text{SO}_{4}$ is a strong acid, a dehydrating agent, an oxidising agent when hot, and non-volatile.
• Acid into water, never water into acid.
• States of sulphur: $\text{H}_{2}\text{SO}_{3}$ $+4$, $\text{H}_{2}\text{S}_{2}\text{O}_{6}$ $+5$, $\text{H}_{2}\text{SO}_{4}$, $\text{H}_{2}\text{S}_{2}\text{O}_{7}$, $\text{H}_{2}\text{SO}_{5}$, $\text{H}_{2}\text{S}_{2}\text{O}_{8}$ all $+6$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why is sulphur trioxide absorbed in concentrated sulphuric acid and not in water?

2. A catalyst is used in the contact process. What does it do to the rate and to the equilibrium conversion?

3. What is the oxidation state of sulphur in $\text{H}_{2}\text{SO}_{5}$?

4. Write the reaction of hot concentrated sulphuric acid with carbon.

References

• NCERT, Chemistry Part I for Class XII, the chapter The p-Block Elements, on sulphuric acid and the oxoacids of sulphur.
• OpenStax, Chemistry 2e, Chapter 18, on sulphur and its compounds, and Appendix G for the standard values used in the model.