Nitrogen and sulphur

Class 11 Chemistry, unit 11, book pages 229–244

Practise 51 MCQs Take the unit quiz Review 14 cards

Nitrogen is unreactive because its triple bond is very strong, yet its compounds take part in the nitrogen cycle, from the ammonia made by the Haber process to the nitrates that plants absorb. The unit covers the nitrogen oxides behind photochemical smog and their removal by catalytic converters. Sulphur forms stable S8 rings and shows several oxidation states, so it can act as both an oxidising and a reducing agent, and the Contact process turns it into sulphuric acid.

Learning outcomes (12)

Explain the lack of reactivity of nitrogen due to its triple bond strength and lack of polarity.
Describe the basicity of ammonia using the Brønsted-Lowry theory.
Identify the structure of the ammonium ion and explain how it is formed by an acid-base reaction.
Describe how ammonia can be displaced from ammonium salts through acid-base reactions.
Describe the natural and man-made occurrences of oxides of nitrogen and their catalytic removal from exhaust gases of internal combustion engines.
Explain the role of NO and NO2 in the formation of photochemical smog, specifically in the reaction with unburned hydrocarbons to form peroxyacetyl nitrate (PAN).
Differentiate between nitrification and denitrification.
Explain the lack of reactivity of sulphur, with reference to its boiling and the stability of its compounds.
Describe the different oxidation states of sulphur and their relative stability.
Describe the properties and uses of sulphuric acid, including its production and industrial applications.
Describe the chemical reactions and processes involving sulphur, such as combustion and oxidation.
Explain the uses of sulphur compounds in industry and everyday life, such as in fertilisers, gunpowder and rubber, and in synthetic organic chemistry, including the synthesis of dyes, drugs and fragrances.

11.1NitrogenOften examined

Nitrogen is a non-metallic element in group V (group 15) of the periodic table. It makes up about 78% of the Earth's atmosphere and exists as diatomic molecules, $N X_{2}$ . Nitrogen gas is unreactive. To understand why, look at the bonding within its molecule.

A nitrogen atom has five valence electrons: two in the 2s orbital and three unpaired electrons in the 2p orbitals. It needs three more electrons to reach the noble gas configuration of neon, so two nitrogen atoms share three pairs of electrons and form a triple covalent bond.

Orbital diagram of nitrogen: the three unpaired electrons in 2p are the ones that form the three shared pairs.

Figure 11.1: Bonding in nitrogen molecule,

N X_{2}

The triple covalent bond is very strong. Its bond energy is almost $1000 kJ mo l^{- 1}$ , so a great deal of energy is needed to break it. This is why $N X_{2}$ is chemically inert and reacts only under extreme conditions.

The two atoms are identical, so the bond is non-polar.

11.1.1Formation of ammonium ion by acid-base reactionOften examined

Ammonia gas, $NH X_{3}$ , is alkaline: it turns moist red litmus paper blue. It reacts with dilute acid to form the ammonium ion:

NH X_{3} (aq) + H X^{+} (aq) NH X_{4} X^{+} (aq)

Ammonia has a lone pair on its nitrogen atom. It donates this lone pair to a hydrogen ion and forms a coordinate covalent bond. A species that donates an electron pair is a Lewis base, which is why ammonia is basic. In Brønsted-Lowry terms, ammonia accepts an $H X^{+}$ ion, so it is a Brønsted-Lowry base.

Figure 11.2: Formation of ammonium ion

How the ammonium ion forms

Identify the lone pair. The N atom of $NH X_{3}$ has one lone pair.
Bring in the proton. $HCl$ supplies $H X^{+}$ , which the lone pair attacks. The H-Cl bond breaks and chloride keeps the bonding electrons.
$HCl H X^{+} + Cl X^{-}$
Count the bonds on nitrogen. It now has four N-H bonds, one of them coordinate, and a positive charge.
$NH X_{3} + H X^{+} [NH X_{4}] X^{+}$
Give the shape. The four identical N-H bonds around N with no lone pair make the ammonium ion tetrahedral.

Lewis base: A species that donates an electron pair to form a coordinate bond.
Brønsted-Lowry base: A species that accepts a proton ( $H X^{+}$ ).

Laboratory preparation of ammonia gasOften examined

Ammonia gas is prepared in the laboratory by heating an ammonium salt with a base. Solid ammonium chloride and solid calcium hydroxide are mixed and heated in a test tube (Figure 11.3).

2 NH X_{4} Cl + Ca (OH) X_{2} CaCl X_{2} + 2 NH X_{3} + 2 H X_{2} O

Figure 11.3: Preparation of ammonia gas

Method

Mix the two solids in the test tube and heat them.
The gas passes through the delivery tube into the U-tube, which contains calcium oxide. Calcium oxide dries the ammonia.
The dry ammonia leaves through the open tube at the right. Damp red litmus paper held there turns blue, which shows that the gas is alkaline.

Ammonia is also displaced from an ammonium salt by a base. The base takes a proton from the ammonium ion, which releases ammonia:

NH X_{4} X^{+} + OH X^{-} NH X_{3} + H X_{2} O

Haber process (manufacture of ammonia)Often examined

Ammonia is made in industry by the Haber process, in which nitrogen reacts with hydrogen. The reaction is reversible and exothermic:

N X_{2} (g) + 3 H X_{2} (g) 2 NH X_{3} (g)

Conditions of the Haber process
Condition	Value used	Reason
Temperature	$500^{\circ} C$	The forward reaction is exothermic, so a low temperature gives a higher yield. The rate would be too slow at a low temperature, so $500^{\circ} C$ is a compromise.
Pressure	$200 atm$	Four moles of gas become two moles of gas in the forward direction. High pressure shifts the equilibrium forward.
Catalyst	Iron	It speeds up the reaction, so equilibrium is reached sooner.

Worked example: how conditions change the Haber equilibrium

In the Haber process,

N X_{2} (g) + 3 H X_{2} (g) 2 NH X_{3} (g)

is exothermic. State and explain what happens to the yield of ammonia if (a) the pressure is raised, (b) the temperature is raised, (c) the iron catalyst is left out. Why is

500^{\circ} C

used?

(a) Count the gas moles. The forward reaction turns 4 mol of gas (1 + 3) into 2 mol, so high pressure shifts the equilibrium forward and the yield of $NH X_{3}$ rises.
$N X_{2} + 3 H X_{2} (4 mol) ⟶ 2 NH X_{3} (2 mol)$
(b) The forward reaction is exothermic, so a higher temperature favours the reverse reaction and the yield falls. A low temperature gives a higher yield.
(c) Without the catalyst the equilibrium position is the same, because a catalyst only speeds up the reaction. The same yield is reached much more slowly.
Why $500^{\circ} C$ : a lower temperature gives more ammonia but the rate would be too slow, so $500^{\circ} C$ is a compromise between yield and rate. A common mistake is to say that the catalyst increases the yield.

Effect of conditions on the Haber equilibrium. The accent arrow is the one that raises the yield of ammonia.

higher pressure: more NH X_{3}; higher temperature: less NH X_{3}; catalyst: same yield, faster

11.1.2Oxides of nitrogen

Nitrogen forms several oxides: $N X_{2} O$ , $NO$ and $NO X_{2}$ . Nitrogen monoxide forms naturally during thunderstorms, when lightning provides the activation energy for nitrogen and oxygen in the air to react. It also forms inside internal combustion engines, where the high temperature does the same job.

N X_{2} (g) + O X_{2} (g) 2 NO (g)

In air, the nitrogen monoxide is oxidised further to nitrogen(IV) oxide, $NO X_{2}$ :

2 NO (g) + O X_{2} (g) 2 NO X_{2} (g)

Nitrogen(IV) oxide dissolves in water droplets to form nitric acid, which falls to the ground in rain. This is part of the natural nitrogen cycle. The nitric acid is used to make fertiliser, which plants absorb to make the proteins they need for growth.

2 NO X_{2} (g) + H X_{2} O (l) + \frac{1}{2} O X_{2} (g) 2 HNO X_{3} (aq)

11.1.3Catalytic removal of oxides of nitrogen from exhaust gasesOften examined

Nitrogen monoxide, $NO$ , is a by-product of combustion in engines. To remove it, most cars use a three-way catalytic converter in the exhaust system. The catalyst contains platinum, palladium and rhodium. It helps convert pollutants into less harmful substances.

Removal of NO

Reduction of NO by CO. The catalyst makes NO react with carbon monoxide to form nitrogen and carbon dioxide.
$2 NO + 2 CO N X_{2} + 2 CO X_{2}$
Reduction of NO by unburned hydrocarbons. Unburned hydrocarbons in the exhaust also reduce NO to nitrogen.
$2 NO + 2 HC N X_{2} + 2 H X_{2} O + CO X_{2}$

In both reactions the NO is reduced to nitrogen, which is a harmless component of the atmosphere. The converter therefore cuts the emission of harmful gases from internal combustion engines.

Concept assessment exercise 11.1 (1): nitrogen, ammonium nitrate and ammonium sulphate

(a) Explain the lack of reactivity of nitrogen gas. (b) Write a balanced chemical equation, including state symbols, for the reaction of ammonia solution with dilute nitric acid. (c) Write a balanced chemical equation, including state symbols, for the reaction of solid ammonium sulphate with solid sodium hydroxide.

Part (a).

Each nitrogen atom in $N X_{2}$ shares three pairs of electrons with the other atom, so the molecule has a triple bond, $N \equiv N$ .
This triple bond is very strong (bond energy almost $1000 kJ mo l^{- 1}$ ) and non-polar, so a great deal of energy is needed to break it. Hence nitrogen is unreactive except under extreme conditions.

Part (b). Ammonia is a base. It accepts $H X^{+}$ from nitric acid to form the salt ammonium nitrate.

Write the reactants with state symbols and balance (one of each).
$NH X_{3} (aq) + HNO X_{3} (aq) NH X_{4} NO X_{3} (aq)$

Part (c). An ammonium salt heated with a strong base liberates ammonia gas, with the sodium salt and water.

Write the formulae: ammonium sulphate is $(NH X_{4}) X_{2} SO X_{4}$ , sodium hydroxide is $NaOH$ , the sodium salt is $Na X_{2} SO X_{4}$ .
Balance: two $NH X_{4} X^{+}$ need two $OH X^{-}$ , so $2 NaOH$ is used and $2 NH X_{3}$ and $2 H X_{2} O$ form.
$(NH X_{4}) X_{2} SO X_{4} (s) + 2 NaOH (s) Na X_{2} SO X_{4} (s) + 2 NH X_{3} (g) + 2 H X_{2} O (l)$

(b) NH X_{3} (aq) + HNO X_{3} (aq) NH X_{4} NO X_{3} (aq) (c) (NH X_{4}) X_{2} SO X_{4} (s) + 2 NaOH (s) Na X_{2} SO X_{4} (s) + 2 NH X_{3} (g) + 2 H X_{2} O (l)

Concept assessment exercise 11.1 (2): Haber process and ammonium sulphate

(a) Ammonia is made in industry using the Haber process. Write the balanced equation to show the synthesis of ammonia from its elements, including the catalyst used. (b) Write a balanced equation, including state symbols, for the formation of ammonium sulphate fertilizer from ammonia solution.

Part (a). Nitrogen and hydrogen combine reversibly over an iron catalyst.

Write the elements as $N X_{2}$ and $H X_{2}$ and balance the atoms: one $N X_{2}$ needs three $H X_{2}$ to give two $NH X_{3}$ .
$N X_{2} (g) + 3 H X_{2} (g) Fe 2 NH X_{3} (g)$

Part (b). Ammonia solution neutralises sulphuric acid to give ammonium sulphate.

Each $H X_{2} SO X_{4}$ can give two $H X^{+}$ , so two $NH X_{3}$ are needed to form $(NH X_{4}) X_{2} SO X_{4}$ .
$2 NH X_{3} (aq) + H X_{2} SO X_{4} (aq) (NH X_{4}) X_{2} SO X_{4} (aq)$

(a) N X_{2} (g) + 3 H X_{2} (g) Fe 2 NH X_{3} (g) (b) 2 NH X_{3} (aq) + H X_{2} SO X_{4} (aq) (NH X_{4}) X_{2} SO X_{4} (aq)

11.1.4Photochemical smog

Photochemical smog is a mixture of pollutants formed when oxides of nitrogen and volatile organic compounds (VOCs) react with sunlight, creating a brown haze. Photo means light, and smog is a mixture of smoke and fog. It is more common in summer, when there is the most sunlight.

Nitrogen oxides, mainly \ce{NO} and \ce{NO2}, play a major part. One of the key reactions involves NO and \ce{NO2} reacting with unburned hydrocarbons to form peroxyacetyl nitrate (PAN). In the morning, the NO and hydrocarbon levels rise first, followed by an increase in nitrogen dioxide.

\text{nitrogen oxides} + \text{hydrocarbons} \xrightarrow{\text{UV light}} \text{PAN} + \ce{O3}

Peroxyacetyl nitrate (PAN) is a secondary pollutant. It is not emitted directly but forms in the atmosphere, and it is an eye irritant and a respiratory irritant.

11.1.5Mechanism of the formation of PAN

PAN forms in three stages. Each stage is shown below.

Formation of PAN

Photodissociation of NO2. In sunlight, \ce{NO2} splits into NO and an oxygen atom.
\ce{NO2 ->[sunlight] NO + O}
Formation of peroxyacyl radicals. The oxygen atom reacts with unburned hydrocarbons, such as VOCs, to form peroxyacyl radicals, \ce{RCO3}. A radical is a species with an unpaired electron, shown by the dot.
\ce{O + hydrocarbon -> RCO3^{.}}
Reaction with NO. The peroxyacyl radical reacts with NO to form PAN.
\ce{RCO3^{.} + NO -> R-CO2ONO2}

Peroxyacetyl nitrate (PAN) PAN: an acetyl group (R = \ce{CH3}) joined through a peroxy link to \ce{NO2}. The book writes the general form \ce{R-CO2ONO2}.

PAN is a major constituent of photochemical smog and a potent respiratory irritant. It is stable and can be carried long distances. It acts as a reservoir for NOx: it holds the nitrogen oxides, so they are not converted back to NO and \ce{NO2} straight away. In the lower atmosphere PAN decomposes thermally, releasing \ce{NO2} and reactive organic radicals, which add to smog and to photochemical reactions.

11.2.1Nitrification

Nitrification is an oxidation process. It is the natural process in which nitrogen compounds, such as ammonium ions, are oxidised. Ammonium ions are first oxidised to nitrite ions, which are then oxidised to nitrate ions. Bacteria are needed; the book names Nitrosomonas, a group of specialised bacteria that convert ammonia into nitrites. Nitrification is the first and essential stage of the nitrogen cycle.

Nitrification: Oxidation of ammonium ions to nitrite ions and then to nitrate ions, carried out by bacteria such as Nitrosomonas.

The overall reactions are:

\ce{2NH4+ + 3O2 -> 2NO2- + 4H+ + 2H2O}

\ce{2NO2- + O2 -> 2NO3-}

Worked example: electrons and charge in nitrification

Show that \ce{2NH4+ + 3O2 -> 2NO2- + 4H+ + 2H2O} is balanced in atoms and charge, and that the electrons lost by nitrogen equal the electrons gained by oxygen.

Count atoms. Left: N 2, H 8, O 6. Right: N 2, H 4+4=8, O 4+2=6. Atoms balance.
Count charge. Left: 2(+1)=+2. Right: 2(-1)+4(+1)=+2. Charge balances.
+2=+2
Oxidation number of N. In \ce{NH4+}: x+4(+1)=+1, so x=-3. In \ce{NO2-}: x+2(-2)=-1, so x=+3. Each N goes from -3 to +3 and loses 6 electrons, so two N atoms lose 12 electrons.
2\times 6=12\ e^-\ \text{lost}
Oxygen goes from 0 in \ce{O2} to -2. Each \ce{O2} gains 4 electrons, so three \ce{O2} gain 12 electrons, equal to the number lost.
3\times 4=12\ e^-\ \text{gained}
The second step, \ce{2NO2- + O2 -> 2NO3-}, follows the same check. Each N goes from +3 to +5 (2 electrons, so 4 for two N) and one \ce{O2} gains 4 electrons. Do not take N in \ce{NH4+} as +1: that is the charge on the ion, not the oxidation number of N.

12\ e^-\ \text{lost by N}=12\ e^-\ \text{gained by O}

Oxidation states of nitrogen in nitrification
Species	Oxidation state of N	Change
\ce{NH4+}	-3	starting point
\ce{NO2-}	+3	oxidised (N loses electrons)
\ce{NO3-}	+5	oxidised again

11.2.2Denitrification

Denitrification is a biological process in which nitrate ions are reduced. The nitrogen is released back into the atmosphere as nitrogen gas, \ce{N2}. It is the final stage of the nitrogen cycle and takes place in the absence of oxygen. It is carried out by denitrifying bacteria such as Clostridium and Pseudomonas.

Denitrification: Biological reduction of nitrate ions to nitrogen gas, carried out in the absence of oxygen by bacteria such as Clostridium and Pseudomonas.

The overall reaction is:

\ce{2NO3- + 10e- + 12H+ -> N2 + 6H2O}

Checking the balance

Count the atoms. Left: 2 N, 6 O and 12 H. Right: 2 N, 6 O (from 6\,\ce{H2O}) and 12 H.
2 = 2\ (\mathrm{N}),\quad 6 = 6\ (\mathrm{O}),\quad 12 = 12\ (\mathrm{H})
Count the charge. Left: 2(-1) + 10(-1) + 12(+1) = 0. Right: 0.
-2 - 10 + 12 = 0

Figure 11.4: Nitrification and denitrification. The oxidation state of nitrogen rises from -3 to +5 in nitrification and falls back to 0 in \ce{N2} during denitrification.

Concept assessment exercise 11.2 (1): CO and NO from car engines

In many countries, new cars have to comply with regulations which are intended to reduce the pollutants coming from their internal combustion engines. Two pollutants that may be formed in an internal combustion engine are carbon monoxide, CO, and nitrogen monoxide, NO. (a)(i) Outline how each of these pollutants may be formed in an internal combustion engine. (ii) State the main hazard associated with each of these pollutants. (b) Pollutants such as CO and NO are removed from the exhaust gases of internal combustion engines by catalytic converters which are placed in the exhaust system of a car. (i) What metal is most commonly used as the catalyst in a catalytic converter? (ii) Construct one balanced equation for the reaction in which both CO and NO are removed from the exhaust gases by a catalytic converter.

Part (a)(i). How they form.

Carbon monoxide forms by incomplete combustion of the fuel (hydrocarbons) when there is not enough oxygen in the cylinder.
\ce{2C8H18 + 17O2 -> 16CO + 18H2O}
Nitrogen monoxide forms because the temperature in the cylinder is high enough for nitrogen and oxygen from the air to combine.
\ce{N2 + O2 -> 2NO}

Part (a)(ii). Main hazards.

CO is poisonous. It combines with haemoglobin in the blood and stops it carrying oxygen.
NO is oxidised to \ce{NO2} in air. Nitrogen oxides cause acid rain and, with hydrocarbons and sunlight, photochemical smog, and they irritate the lungs.

Part (b).

The metal most commonly used is platinum (the book's converter uses platinum, palladium and rhodium).
In the converter, CO reduces NO to nitrogen and is itself oxidised to carbon dioxide. Balance: two CO and two NO give two \ce{CO2} and one \ce{N2}.
\ce{2CO + 2NO -> 2CO2 + N2}

\text{(b)(ii) }\ce{2CO + 2NO -> 2CO2 + N2}\quad(\text{catalyst: platinum})

Concept assessment exercise 11.2 (2): nitrification and denitrification

Differentiate between nitrification and denitrification.

Nitrification is the oxidation of ammonia or ammonium ions first to nitrite and then to nitrate, carried out by nitrifying bacteria in the presence of oxygen. The oxidation state of nitrogen rises from -3 to +3 to +5.
\ce{NH4+ -> NO2- -> NO3-}
Denitrification is the reduction of nitrates, by denitrifying bacteria in the absence of oxygen, through nitrite, \ce{NO} and \ce{N2O} to nitrogen gas, which returns to the atmosphere. The oxidation state falls from +5 to 0.
\ce{NO3- -> NO2- -> NO -> N2O -> N2}
So nitrification converts ammonia into nitrites and nitrates (usable by plants), while denitrification converts nitrates back into \ce{N2}.

\text{Nitrification: }\ce{NH4+ -> NO2- -> NO3-}\qquad \text{Denitrification: }\ce{NO3- -> N2}

11.3Sulphur

Sulphur atoms form cyclic octa-atomic molecules, \ce{S8}, in which eight atoms are joined in a ring. Elemental sulphur is a bright yellow, crystalline solid at room temperature (Figure 11.5).

Figure 11.5: Crystal structure of sulphur, the \ce{S8} ring

A sulphur atom has six valence electrons. The possible oxidation states of sulphur are -2, +2, +4 and +6. The +6 state is the most stable: it is found in sulphuric acid and sulphates, and it is the highest oxidation state of sulphur observed under normal conditions. The +6 state gives sulphur a complete octet by gaining or sharing electrons with other atoms.

Oxidation states of sulphur
Oxidation state	Example from the book	Note
-2	\ce{H2S} (hydrogen sulphide)	Formed when sulphur reacts with hydrogen
+2	Not given in the book	Listed as possible
+4	\ce{SO2} (sulphur dioxide)	Formed when sulphur burns
+6	\ce{H2SO4}, sulphates	The most stable state

Sulphur forms a stable structure. Its atoms form single bonds with each other to make \ce{S8} rings, and each sulphur atom carries two lone pairs. A great deal of energy is needed to break these stable single bonds, which makes sulphur relatively inert under normal conditions. Compounds such as hydrogen sulphide and sulphur dioxide are also stable, because sulphur shares electrons in a way that satisfies the octet rule.

The chemical environment also affects stability. Sulphur atoms in a complex or a coordination compound can have a different stability from free elemental sulphur.

11.3.1Properties of sulphur

Sulphur is a non-metal in group VI of the periodic table, below oxygen. It can act as both an oxidising agent and a reducing agent. Which role it plays depends on its oxidation state and on whether it gains or loses electrons in the reaction.

Oxidising agent: A substance that accepts electrons from other substances, so it is reduced itself.
Reducing agent: A substance that gives electrons to other substances, so it is oxidised itself.

Sulphur can gain two electrons to reach the electron configuration of a noble gas, with eight valence electrons. In this process sulphur is an oxidising agent, because it accepts electrons from other substances. Sulphur can also lose electrons and exist in the +4 and +6 oxidation states. In this case it acts as a reducing agent.

11.3.2Sulphur as an oxidising agentOften examined

In the following reactions sulphur is reduced from 0 to -2, so it acts as an oxidising agent.

1. Reaction with hydrogen

Sulphur combines directly with hydrogen to form hydrogen sulphide.

\ce{S + H2 -> H2S}

2. Reaction with metals

Sulphur reacts with metals such as zinc and iron on heating to form metal sulphides.

\ce{Zn + S -> ZnS}

\ce{Fe + S -> FeS}

Oxidation states in the metal reactions

In \ce{ZnS}, the oxidation state of S falls from 0 to -2 (S gains two electrons, so it is reduced). Zinc rises from 0 to +2 (Zn is oxidised).
\ce{FeS} follows the same pattern: S goes from 0 to -2 and Fe from 0 to +2.

11.3.3Sulphur as a reducing agentOften examined

In the following reactions sulphur acts as a reducing agent, because its oxidation state increases from 0 to +4 or +6.

1. Reaction with oxygen (combustion, or oxidation of sulphur)

Sulphur burns in air with a blue flame to form sulphur dioxide.

\ce{S + O2 -> SO2}

If sulphur is exposed to moist air, it may be oxidised to sulphuric acid.

\ce{2S + 2H2O + 3O2 -> 2H2SO4}

2. Reaction with oxidising agents

When sulphur is treated with oxidising agents such as \ce{HNO3} or \ce{H2SO4}, it oxidises itself and reduces the acid.

\ce{S + 6HNO3 -> H2SO4 + 6NO2 + 2H2O}

Oxidation states in the nitric acid reaction

Sulphur goes from 0 to +6 in \ce{H2SO4}: it is oxidised, so it is the reducing agent.
Nitrogen goes from +5 in \ce{HNO3} to +4 in \ce{NO2}: it is reduced.

\ce{S + 2H2SO4 -> 3SO2 + 2H2O}

Oxidation states in the sulphuric acid reaction

Sulphur from \ce{H2SO4} goes from +6 to +4 in \ce{SO2}: it is reduced.
Elemental sulphur goes from 0 to +4 in \ce{SO2}: it is oxidised, so it is the reducing agent.

Worked example: electron balance in sulphur and concentrated sulphuric acid

In \ce{S + 2H2SO4 -> 3SO2 + 2H2O}, show that the electrons lost equal the electrons gained, and name the oxidising and reducing agents.

Assign oxidation numbers to sulphur. Free sulphur is 0. In \ce{H2SO4}: 2(+1)+x+4(-2)=0, so x=+6. In \ce{SO2}: x+2(-2)=0, so x=+4.
\ce{S}: 0,\quad \text{S in }\ce{H2SO4}: +6,\quad \text{S in }\ce{SO2}: +4
Elemental S goes from 0 to +4: it loses 4 electrons, so it is oxidised and is the reducing agent.
S in \ce{H2SO4} goes from +6 to +4: each gains 2 electrons. Two \ce{H2SO4} gain 4 electrons, so the acid is reduced and is the oxidising agent.
2\times 2=4\ e^-
Electrons lost (4) equal electrons gained (4), which is why the ratio is 1 S to 2 \ce{H2SO4}. Of the three \ce{SO2}, one comes from S and two from the acid.

Oxidation number changes in the reaction. Sulphur loses 4 electrons, and two sulphuric acid molecules gain 4 between them.

\text{S: reducing agent;}\quad \ce{H2SO4}\text{: oxidising agent}

11.3.4Uses of sulphur and its compounds

Sulphur and its compounds have a wide range of applications in industry and everyday life.

Use	How sulphur is used
Fertiliser	Sulphuric acid and ammonium sulphate supply sulphur to crops. Sulphur helps plants make amino acids and vitamins, which promotes their growth and health.
Gunpowder	Gunpowder contains sulphur, charcoal and potassium nitrate. Sulphur plays an important role in the combustion and the explosive properties of gunpowder.
Rubber industry	Sulphur is used in the vulcanisation of rubber.
Synthetic organic chemistry	Thiol (-SH) compounds are used to make organic molecules and pharmaceuticals. Sulphonyl chlorides are important intermediates in making sulphonamides, a class of antibiotics.
Dyes and pharmaceuticals	Sulphur compounds are used in the production of dyes, pigments and pharmaceuticals.
Food preservatives	Sulphur dioxide and sulphites, such as \ce{Na2SO3}, are used as food preservatives. They are reducing agents, so they inhibit the oxidation of food.

11.4Sulphuric acid

Sulphuric acid, \ce{H2SO4}, is a colourless, oily liquid. Its density is 1.83\,\mathrm{g\,cm^{-3}}, which is almost as heavy as water. It solidifies at 10.37\,^\circ\mathrm{C} to colourless crystals.

11.4.1Contact process (manufacture of sulphuric acid)Often examined

The chemical industry makes sulphuric acid by the Contact process. The raw materials are sulphur, air and water. Sulphur extracted from the impurities in fossil fuels can also be used in the Contact process.

Stages of the Contact process

Oxidation (burning) of sulphur. Sulphur burns in air to form sulphur dioxide.
\ce{S + O2 -> SO2}
Purification of sulphur dioxide. Impurities such as arsenic oxide poison the catalyst used in the next step, so the gas is purified first.
Oxidation of sulphur dioxide. Pure sulphur dioxide is heated with oxygen over a vanadium(V) oxide catalyst to form sulphur trioxide. The reaction is reversible and exothermic. Conditions: 450 to 500\,^\circ\mathrm{C}, 2 to 3 atmospheric pressure.
\ce{2SO2(g) + O2(g) <=> 2SO3(g)}\qquad \Delta H^\circ = -198\,\mathrm{kJ}
Production of oleum. Sulphur trioxide is not dissolved directly in water, because the reaction is so exothermic that the mixture boils violently and splatters. It is dissolved in concentrated sulphuric acid instead, forming fuming sulphuric acid (oleum). Oleum is then diluted with water in a controlled manner to give concentrated sulphuric acid.
\ce{SO3 + H2SO4 -> H2S2O7}

Figure 11.6: Contact process manufacture of sulphuric acid. The converter (\ce{V2O5} catalyst) oxidises \ce{SO2} to \ce{SO3}.

Worked example: the acid from 64 kg of sulphur by the Contact process

(a) Write the equation for each stage of the Contact process, from sulphur to sulphuric acid. (b) What mass of \ce{H2SO4} can be made from 64\,\mathrm{kg} of sulphur if every stage is complete? (\ce{S}=32, \ce{H}=1, \ce{O}=16)

Given

m(\ce{S}) = 64\,\mathrm{kg}

To find

m(\ce{H2SO4})

(a) Stage 1 burns sulphur and stage 2 oxidises \ce{SO2} over the vanadium(V) oxide catalyst. Stage 3 dissolves \ce{SO3} in concentrated acid to give oleum, since the reaction with water is too exothermic. Diluting the oleum with water gives the acid, as in Figure 11.6.
\ce{S + O2 -> SO2},\quad \ce{2SO2 + O2 <=> 2SO3},\quad \ce{SO3 + H2SO4 -> H2S2O7}
(b) Follow the amounts of sulphur. One S atom gives one \ce{SO2}, one \ce{SO3} and finally one \ce{H2SO4} (diluting the oleum with water gives back the acid that absorbed the \ce{SO3}, plus one new \ce{H2SO4} per \ce{SO3}). The mole ratio \ce{S}:\ce{H2SO4} is 1 : 1.
Find the moles of sulphur.
n(\ce{S})=\dfrac{64\,000\,\mathrm{g}}{32\,\mathrm{g\,mol^{-1}}}=2000\,\mathrm{mol}
Find the molar mass of the acid, then the mass.
M(\ce{H2SO4})=2+32+64=98\,\mathrm{g\,mol^{-1}},\quad m=2000\times98=196\,000\,\mathrm{g}

m(\ce{H2SO4})=196\,\mathrm{kg}

11.4.2Properties of sulphuric acidOften examined

Sulphuric acid behaves as a strong acid in dilute solution. It also acts as a dehydrating agent and, when hot and concentrated, as an oxidising agent.

1. Ionisation

Sulphuric acid ionises in water to give hydrogen ions and sulphate ions.

\ce{H2SO4 -> 2H+ + SO4^2-}

2 to 4. Reactions of dilute acid

Reaction	Equation	Products
Reaction with a base (neutralisation)	\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}	Salt and water
Reaction with an active metal	\ce{Zn + H2SO4 -> ZnSO4 + H2}	Salt and hydrogen gas
Reaction with a metal carbonate	\ce{H2SO4 + Na2CO3 -> Na2SO4 + H2O + CO2}	Salt, water and carbon dioxide

5. Sulphuric acid as a dehydrating agent

Sulphuric acid has a great affinity for water, so it removes water from other compounds.

\ce{HCOOH ->[H2SO4] H2O + CO}

\ce{C2H5OH ->[H2SO4] C2H4 + H2O}

6. Sulphuric acid as an oxidising agent

Hot concentrated sulphuric acid is an oxidising agent, because it can lose oxygen and be reduced to sulphur dioxide. It reacts with metals such as copper and tin to liberate sulphur dioxide.

\ce{Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O}

Oxidation states in the copper reaction

Copper goes from 0 to +2 in \ce{CuSO4}: it is oxidised.
Sulphur goes from +6 in \ce{H2SO4} to +4 in \ce{SO2}: \ce{H2SO4} is reduced and is the oxidising agent.

\ce{2HI + H2SO4 -> I2 + SO2 + 2H2O}

Oxidation states in the hydrogen iodide reaction

Iodine goes from -1 in \ce{HI} to 0 in \ce{I2}: it is oxidised.
Sulphur goes from +6 to +4: \ce{H2SO4} is reduced and is the oxidising agent.

Worked example: dehydrating agent or oxidising agent

Concentrated sulphuric acid takes part in three reactions: with ethanol (to give ethene), with methanoic acid (to give carbon monoxide) and with hot copper (to give sulphur dioxide). In which is it a dehydrating agent and in which an oxidising agent?

A dehydrating agent removes the elements of water from a compound. It does not change any oxidation number. Check ethanol: in \ce{C2H5OH}, 2x+6(+1)+(-2)=0 gives x=-2 for each C; in \ce{C2H4}, 2x+4(+1)=0 gives x=-2. No change, so this is only loss of water.
Methanoic acid is the same. In \ce{HCOOH}: 2+x-4=0, so C is +2. In \ce{CO}: x-2=0, so C is +2. Water is removed, and nothing is oxidised or reduced.
An oxidising agent is reduced. In the copper reaction Cu goes from 0 to +2 (oxidised) and S from +6 to +4 (reduced), so the acid is the oxidising agent.
Summary: a change in oxidation number means oxidising agent; removal of \ce{H2O} with no change means dehydrating agent.

The role of sulphuric acid decided by oxidation numbers
Reaction	Change of oxidation number	Role of \ce{H2SO4}
\ce{C2H5OH ->[H2SO4] C2H4 + H2O}	C stays at -2; S unchanged at +6	Dehydrating agent
\ce{HCOOH ->[H2SO4] H2O + CO}	C stays at +2; S unchanged at +6	Dehydrating agent
\ce{Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O}	Cu: 0\to+2; S: +6\to+4	Oxidising agent

\text{ethanol, methanoic acid: dehydrating;}\quad \text{copper: oxidising}

11.4.3Uses of sulphuric acid

Sulphuric acid is an extremely important industrial chemical. It is one of the most widely produced compounds in the world and is highly stable. It is used to make fertilisers, detergents, paints, pigments, dyes, synthetic fibres and plastics. It is also used in car batteries, in tanning leather and in cleaning metal surfaces (pickling).

Concept assessment exercise 11.3: sulphuric acid manufacture

Sulphuric acid is used in many industrial processes of major importance. The first stage in the manufacture of sulphuric acid is to pass air over burning sulphur. (a) Write an equation for sulphur burning in air. (b) The emerging gas is passed over a catalyst maintained at 450 to 550\,^\circ\mathrm{C} in the reaction chamber. Name the catalyst used in the Contact process. (c) Write an equation which shows sulphuric acid as a dehydrating agent.

Part (a).

Sulphur burns in oxygen of the air to give sulphur dioxide.
\ce{S + O2 -> SO2}

Part (b).

The \ce{SO2} is oxidised to \ce{SO3} over a catalyst of vanadium(V) oxide, \ce{V2O5}.
\ce{2SO2(g) + O2(g) <=>[V2O5] 2SO3(g)}

Part (c). Concentrated sulphuric acid has a great affinity for water and removes the elements of water from other compounds, as the book shows with ethanol.

Concentrated \ce{H2SO4} removes water from ethanol, leaving ethene.
\ce{C2H5OH ->[{conc.} H2SO4] C2H4 + H2O}

\text{(a) }\ce{S + O2 -> SO2}\quad \text{(b) }\ce{V2O5}\quad \text{(c) }\ce{C2H5OH ->[{conc.} H2SO4] C2H4 + H2O}

Key terms

Ammonium ion: The ion \ce{NH4+}, formed when ammonia accepts a proton (\ce{H+}) from an acid.
Lewis base: A species that donates an electron pair to form a coordinate covalent bond.
Brønsted-Lowry base: A species that accepts a proton (\ce{H+}).
Haber process: The industrial manufacture of ammonia from nitrogen and hydrogen, carried out at about 500\,^\circ\mathrm{C} and 200\,\mathrm{atm} with an iron catalyst.
Photochemical smog: A mixture of pollutants formed when oxides of nitrogen and volatile organic compounds (VOCs) react with sunlight, creating a brown haze.
Peroxyacetyl nitrate (PAN): A secondary pollutant and eye irritant formed in photochemical smog from NO, NO2 and unburned hydrocarbons.
Catalytic converter: A device in a vehicle's exhaust system, containing platinum, palladium and rhodium, that reduces NO using CO and unburned hydrocarbons.
Nitrification: Oxidation of ammonium ions to nitrite and then nitrate ions, carried out by bacteria such as Nitrosomonas.
Denitrification: Biological reduction of nitrate ions to nitrogen gas in the absence of oxygen, carried out by bacteria such as Clostridium and Pseudomonas.
Oxidising agent: A substance that accepts electrons from other substances, so it is reduced itself.
Reducing agent: A substance that gives electrons to other substances, so it is oxidised itself.
Dehydrating agent: A substance with a great affinity for water, which removes water from other compounds; sulphuric acid is one.
Contact process: The industrial manufacture of sulphuric acid from sulphur, air and water through the steps of burning sulphur, oxidising \ce{SO2} to \ce{SO3} over a vanadium(V) oxide catalyst, and absorbing \ce{SO3} in sulphuric acid.
Oleum: Fuming sulphuric acid, \ce{H2S2O7}, formed when \ce{SO3} dissolves in concentrated sulphuric acid.

Multiple choice questionsp. 241–242

Which equations represent stages in the Contact process for manufacturing sulphuric acid?

A\ce{S + O2 -> SO2}Answer
B\ce{H2O + SO2 -> H2SO3}
C\ce{H2SO3 + O2 -> H2SO4}
D\ce{SO3 + H2O -> H2SO4}

Why

The Contact process has three stages: sulphur is burned to \ce{SO2}, the \ce{SO2} is oxidised to \ce{SO3} over vanadium(V) oxide, and the \ce{SO3} is absorbed to give sulphuric acid. Option A is the first stage.

Option B gives sulphurous acid, which is not a stage of the process. Option C is not a stage either: the oxidation stage is \ce{2SO2 + O2 <=> 2SO3} over \ce{V2O5}, and the process never converts \ce{SO2} into \ce{H2SO3} first.

Option D is a correct overall equation for absorption, but the book does not carry out that step directly. \ce{SO3} is dissolved in concentrated \ce{H2SO4} to form oleum, which is then diluted with water, so option D is not the stage the book describes.

Note. Options A and D are both correct equations for stages, so the question has two readings. The key is A because the book's absorber dissolves SO3 in concentrated sulphuric acid first, not in water. If the printed key lists both, this is the reason.

Nitrogen is absorbed by the plants in the form of

AAmmonium
BNitrites
CNitratesAnswer
DAll

Why

The book's nitrogen cycle ends with nitric acid, which is used to make fertiliser and is absorbed by plants as nitrates. Nitrites are only an intermediate in nitrification, so they are not the form the book names.

Option D is wrong because it includes nitrites, which are not the form taken up. Ammonium is the starting point of nitrification and is not the form the book names either.

Note. Some plants also take up ammonium, so option A is not strictly wrong. C is the form the book names (page 232).

iii

Nitrogen fixation is the conversion of

A\ce{N2} to \ce{N}
B\ce{N2} to \ce{NH3}Answer
C\ce{N2} to \ce{NO3-}
D\ce{N2} to urea

Why

Nitrogen fixation takes unreactive \ce{N2} and converts it into a combined form that living things can use. In the book's nitrogen cycle (Figure 11.4) fixation brings atmospheric \ce{N2} into the soil as ammonia or ammonium, so option B, \ce{N2 -> NH3}, is correct.

Option A would break the triple bond into single atoms, which is not what fixation does. Option D is not a product of fixation. Option C describes oxidation of nitrogen to nitrate, which is a later step in the cycle rather than fixation itself.

Sulphur dioxide is an important food preservative. Which property makes sulphur dioxide useful in this role?

AIt is a gas.
BIt is a reducing agent.Answer
CIt reacts with oxygen to form sulphur trioxide.
DIt reacts with water to form an acidic solution.

Why

The book says that sulphur dioxide and sulphites are reducing agents and that this inhibits the oxidation of food. A reducing agent takes up oxygen, so the food is protected from being oxidised.

Option A is not the property that matters, since many gases do not preserve food. Option C describes \ce{SO2} being oxidised, which is the same oxygen-consuming change, but the book names the property as the reducing action. Option D describes acidity, which does not stop the oxidation of food.

Most modern cars are fitted with three-way catalytic converters in the exhaust system. Which three gases are removed by such a catalytic converter?

Acarbon monoxide, hydrocarbons, nitrogen oxidesAnswer
Bcarbon monoxide, carbon dioxide, nitrogen oxides
Ccarbon monoxide, nitrogen oxides, sulphur dioxide
Dhydrocarbons, nitrogen oxides, sulphur dioxide

Why

The three-way converter reduces carbon monoxide, unburned hydrocarbons and nitrogen monoxide, converting them into \ce{N2}, \ce{CO2} and water. The three pollutants removed are therefore carbon monoxide, hydrocarbons and nitrogen oxides.

Carbon dioxide (option B) is a product of the converter, not a pollutant it removes. Sulphur dioxide (options C and D) is not one of the gases the converter targets.

Conversion of nitrates to nitrogen gas is called

AAmmonification
BNitrification
CNitrogen fixation
DDenitrificationAnswer

Why

Denitrification is the biological reduction of nitrate ions to nitrogen gas, \ce{2NO3- + 10e- + 12H+ -> N2 + 6H2O}, which returns nitrogen to the atmosphere in the absence of oxygen.

Nitrification (option B) is the opposite process, an oxidation that changes ammonium into nitrite and then nitrate. Nitrogen fixation (option C) takes nitrogen from the air into the soil, and ammonification (option A) is not a process in this unit.

vii

In the Contact process, what is the nature of the gaseous product and what is the identity of the catalyst?

AAcidic, catalyst \ce{Fe}
BAcidic, catalyst \ce{V2O5}Answer
CBasic, catalyst \ce{Fe}
DBasic, catalyst \ce{V2O5}

Why

In the converter \ce{SO2} is oxidised to \ce{SO3}, \ce{2SO2 + O2 <=> 2SO3}, over vanadium(V) oxide, \ce{V2O5}. Sulphur trioxide is an acidic oxide, so the gaseous product is acidic and the catalyst is \ce{V2O5}.

Iron is the catalyst of the Haber process, not the Contact process, so options A and C are wrong. The gas in the Contact process is acidic, not basic, so option D is wrong too.

viii

Conversion of nitrites to nitrates is called

ANitrosococcus
BClostridium
CNitrobacterAnswer
DNitrosomonas

Why

The conversion of nitrite to nitrate is the second step of nitrification, \ce{2NO2- + O2 -> 2NO3-}, and it is carried out by Nitrobacter bacteria.

Nitrosomonas (and the similar Nitrosococcus) carry out the first step, ammonium to nitrite. Clostridium is a denitrifying bacterium, which reduces nitrate, so it is not the answer.

Note. Two printing errors. The stem reads 'is called', but the options are organisms, so it should read 'is carried out by'. The book's text (page 234) names Nitrosomonas for both steps of nitrification, which is wrong for the nitrite to nitrate step; the key is Nitrobacter.

Which oxide of nitrogen is released from car engines?

ANitrogen dioxide
BNitric oxideAnswer
CNitrous oxide
DNitrogen tetroxide

Why

In an internal combustion engine nitrogen and oxygen from the air react to form nitrogen monoxide, \ce{N2 + O2 -> 2NO}, which is the oxide released from the exhaust.

Nitrogen dioxide is formed later when that nitric oxide is oxidised by the air, so it is not the oxide released directly. Nitrous oxide (\ce{N2O}) and nitrogen tetroxide (\ce{N2O4}) are not formed in the engine described in this unit.

Which compound of sulphur is used to bleach wood pulp?

ASulphuric acid
BSodium sulphate
CSulphur dioxideAnswer
DSodium sulphite

Why

Sulphur dioxide is a reducing agent and is the sulphur compound used to bleach wood pulp. The unit names it as a reducing agent and food preservative, and bleaching is the standard use of the same property.

Sulphuric acid is a strong acid, not a bleach, and sodium sulphate is a salt that does not bleach. Sodium sulphite is a related reducing compound, but sulphur dioxide is the answer the standard use points to.

Note. The stem prints 'copmpound', corrected here. The book prints no key and its bleaching use is not in the unit text, so this key is worked out from the standard use of sulphur dioxide. Sodium sulphite is the closest rival.

Short questionsp. 243–244

Ammonia is manufactured from nitrogen and hydrogen by the Haber process. Hydrogen is usually obtained by reacting methane and steam; the by-product is carbon monoxide. The reaction between nitrogen and hydrogen is exothermic and incomplete.

Solution

(a)(i) Write an equation for the Haber process.

The Haber process combines nitrogen and hydrogen reversibly. Because the reaction is incomplete, the equation uses a double arrow.

\ce{N2(g) + 3H2(g) <=> 2NH3(g)}

The balance: one \ce{N2} gives two N atoms, which make two \ce{NH3}. Those need six H atoms, so three \ce{H2} molecules are needed.

\ce{N2(g) + 3H2(g) <=> 2NH3(g)}

(a)(ii) State the three conditions necessary for the efficient working of a Haber process plant.

Each condition balances the rate of reaction against the yield of ammonia.

Temperature of about 500\,^\circ\mathrm{C}. The forward reaction is exothermic, so a low temperature gives a higher yield, but the rate becomes too slow. 500\,^\circ\mathrm{C} is a compromise that gives a reasonable rate.
Pressure of about 200\,\mathrm{atm}. Four moles of gas (one of \ce{N2} and three of \ce{H2}) form two moles of ammonia, so high pressure shifts the equilibrium to the right.
Iron as the catalyst. It speeds up the reaction so that equilibrium is reached quickly, without changing the yield.

500\,^\circ\mathrm{C},\quad 200\,\mathrm{atm},\quad \text{iron catalyst}

Write the equations that are evident for the statement that sulphur acts both as a reducing as well as an oxidising agent.

Solution

Sulphur is an oxidising agent when its oxidation state falls from 0 to -2, which happens when it gains electrons from a metal or from hydrogen.

\ce{S + H2 -> H2S}\qquad\ce{Zn + S -> ZnS}\qquad\ce{Fe + S -> FeS}

Sulphur is a reducing agent when its oxidation state rises from 0 to +4 or +6, as in burning and in reaction with oxidising acids.

\begin{gathered}\ce{S + O2 -> SO2}\\ \ce{S + 6HNO3 -> H2SO4 + 6NO2 + 2H2O}\\ \ce{S + 2H2SO4 -> 3SO2 + 2H2O}\end{gathered}

Long questionsp. 243–244

The composition of the exhaust gases (fumes) from a petrol (gasoline) engine is given below.

Gas	Percentage by volume
Carbon dioxide	9
Water vapour	8
Oxygen	4
Hydrogen	2
Carbon monoxide	3–6
Hydrocarbons	0.2
Nitrogen oxides	0.4–0.05

Solution

(a)(i) What gas, omitted from the table, makes up most of the remainder of the exhaust fumes?

Air is about 78\% nitrogen. Nitrogen passes through the engine unchanged, and the table lists the other gases, so the gas omitted from the table is nitrogen, \ce{N2}.

\ce{N2}

(a)(ii) Explain why there is always water vapour in the exhaust fumes.

Petrol is a mixture of hydrocarbons, which contain hydrogen. When the fuel burns, its hydrogen is oxidised to water, and that water leaves the engine as water vapour.

\ce{2H2 + O2 -> 2H2O}

Any petrol engine that burns hydrocarbons therefore gives off water vapour, so it is always present in the fumes.

(a)(iii) The two gases given in the table can be removed by reaction in a platinised catalytic converter. The products of the reaction in the converter are not hazardous. Write an equation which shows how it is removed to form less harmful substances.

The two gases that the converter removes are carbon monoxide and nitrogen monoxide (the nitrogen oxides row). On the platinum, palladium and rhodium catalyst they react to give harmless \ce{N2} and \ce{CO2}.

\ce{2NO + 2CO -> N2 + 2CO2}

Both sides have two N, two C and four O atoms, so the equation is balanced.

\ce{2NO + 2CO -> N2 + 2CO2}

(a)(iv) Suggest one reason why the exhaust fumes from a car fitted with a catalytic converter are still hazardous to human health to some degree.

The converter works only under specific conditions, and it needs to be hot before it converts the pollutants well. Until it reaches its working temperature, for example just after the engine starts, carbon monoxide, nitrogen oxides and unburned hydrocarbons pass through unchanged.

Even when it is working, the conversion is not complete, so some of the harmful gases listed in the table still escape into the air.

Note. The printed nitrogen oxides range reads '0.4–0.05', with the larger number first. It is reproduced as printed. The question does not say which two gases of the table the converter removes; the answer takes them as carbon monoxide and nitrogen oxides, the pair the book's converter equations use.

Nitrogen, which makes up about 80\% of the Earth's atmosphere, is very unreactive.

Solution

(a)(i) Explain the lack of reactivity of nitrogen.

Triple bond

Two nitrogen atoms share three pairs of electrons, \ce{N#N}, so the molecule has a very strong triple bond. Its bond energy is almost 1000\,\mathrm{kJ\,mol^{-1}}, so a great deal of energy is needed to break it.

No polarity

Both atoms are the same, so the molecule is non-polar. No part of it carries a partial positive or negative charge, so reagents have no polar site to attract and attack. Nitrogen therefore reacts only under extreme conditions.

(b)(i) State one large-scale use of ammonium nitrate.

Ammonium nitrate is used as a nitrogen fertiliser. Plants take up its nitrogen to make proteins, which they need for growth.

Nitrogen fertiliser

(c)(i) When solid ammonium nitrate is heated with solid sodium hydroxide in a test-tube, three products are formed. A colourless alkaline gas, Y, is given off, and a colourless liquid can be seen on the cooler parts of the test-tube. A white solid remains in the tube. Identify gas Y.

Gas Y is ammonia, \ce{NH3}. It is colourless and alkaline, so it turns moist red litmus paper blue, and it is released when an ammonium salt reacts with a base.

\ce{NH3}

(c)(ii) Write an equation, with state symbols, for the reaction of ammonium nitrate with sodium hydroxide.

Ammonium nitrate is the ammonium salt and sodium hydroxide is the base. The products are ammonia gas, water (the liquid on the cooler parts) and sodium nitrate (the white solid).

\ce{NH4NO3(s) + NaOH(s) -> NH3(g) + H2O(l) + NaNO3(s)}

Check: N is 2 on each side, H is 5 on each side and O is 4 on each side.

\ce{NH4NO3(s) + NaOH(s) -> NH3(g) + H2O(l) + NaNO3(s)}

(c)(iii) In order to produce gas Y in a pure state in the laboratory, it must be passed through a drying agent. Why is concentrated sulphuric acid not suitable for drying gas Y?

Concentrated sulphuric acid is an acid, and ammonia is a base. The acid would neutralise the ammonia, forming ammonium sulphate, so the ammonia would be lost rather than dried.

\ce{2NH3 + H2SO4 -> (NH4)2SO4}

A basic drying agent is used instead, such as calcium oxide, which is the drying agent in the laboratory set-up of the book (Figure 11.3).

Note. The question says about 80% of the atmosphere is nitrogen, while the book's text (page 230) gives 78%. The 80% is reproduced as printed.

Sulphur and its compounds are found in volcanoes, in organic matter and in minerals. Sulphuric acid, an important industrial chemical, is manufactured from sulphur by the Contact process. The Contact process may be considered to be a three-stage process in which sulphur is converted into sulphuric acid. Each stage consists of a single chemical reaction.

Solution

(a) Write a balanced equation for each of these reactions in the correct sequence.

The three stages, in order, are burning sulphur, oxidising sulphur dioxide to sulphur trioxide over a catalyst, and absorbing sulphur trioxide to form sulphuric acid.

The three stages of the Contact process, each a single reaction, from sulphur to sulphuric acid.

Stage 1: sulphur burns in air to form sulphur dioxide.
\ce{S + O2 -> SO2}
Stage 2: sulphur dioxide is oxidised by oxygen over vanadium(V) oxide. The reaction is reversible, so the double arrow is used.
\ce{2SO2 + O2 <=> 2SO3}
Stage 3: sulphur trioxide is absorbed. The book dissolves it in concentrated sulphuric acid to form oleum, which is then diluted with water. Taken as a single reaction, the stage is:
\ce{SO3 + H2O -> H2SO4}

The oleum route the book describes is \ce{SO3 + H2SO4 -> H2S2O7} followed by \ce{H2S2O7 + H2O -> 2H2SO4}, which gives the same overall change.

\ce{S + O2 -> SO2};\quad \ce{2SO2 + O2 <=> 2SO3};\quad \ce{SO3 + H2O -> H2SO4}

(b) Give three different operating conditions that are used in the second stage.

The second stage is the oxidation of sulphur dioxide to sulphur trioxide. Three conditions are used:

Temperature of 450–500\,^\circ\mathrm{C}.
Pressure of 2 to 3 atmospheres, about atmospheric pressure.
Catalyst: vanadium(V) oxide, \ce{V2O5}.

450\text{-}500\,^\circ\mathrm{C},\quad 2\text{-}3\,\mathrm{atm},\quad \ce{V2O5}

Sulphuric acid is used to manufacture fertilisers. It is also used in making detergents, paints, pigments, dyes and synthetic fibres.

Manufacture of fertilisers

(d) Sulphur dioxide is used to preserve dried fruits and vegetables. What chemical property of \ce{SO2} enables it to be used as a food preservative?

Sulphur dioxide is a reducing agent. It takes up oxygen and so inhibits the oxidation of the food, which is why it works as a preservative.

Reducing agent

Note. The book's text gives the second-stage temperature as 450–500 °C (page 238). A different concept question on page 241 quotes 450–550 °C; the exercise answer uses the 450–500 °C of the teaching text.

FBISE 2025 questions

Questions from the FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 91 mark

Peroxyacetyl nitrate (PAN) is formed by the reaction of oxides of nitrogen with:

AOxides of sulphur
BHydrogen sulphide
COxides of phosphorus
DUnburned hydrocarbonsAnswer

Why

Peroxyacetyl nitrate (PAN) is a secondary pollutant of photochemical smog. In sunlight \ce{NO2} splits into NO and an oxygen atom, the oxygen atom reacts with unburned hydrocarbons (volatile organic compounds) to give peroxyacyl radicals \ce{RCO3}, and these react with NO to give PAN, \ce{R-CO2ONO2}. PAN contains carbon, which comes from the hydrocarbons.

Oxides of sulphur (option A) are linked with acid rain, not PAN. Hydrogen sulphide and oxides of phosphorus (options B and C) cannot supply the carbon-containing acyl group of PAN.

2025 Annual I, Set B, MCQ 101 mark

Which of the following halides can reduce \ce{H2SO4} to \ce{H2S}?

A\ce{NaF}
B\ce{NaCl}
C\ce{NaBr}
D\ce{NaI}Answer

Why

Concentrated sulphuric acid is an oxidising agent, and it is reduced further when it reacts with a more powerful reducing agent. The reducing power of the halide ions increases down the group, \ce{F- < Cl- < Br- < I-}. The iodide ion is the strongest reducing agent, so it reduces the sulphur from +6 in \ce{H2SO4} all the way to -2 in \ce{H2S}.

\ce{8HI + H2SO4 -> 4I2 + H2S + 4H2O}

Bromide (option C) is the tempting choice, but it is a weaker reducing agent and only reduces the acid to \ce{SO2} (\ce{2HBr + H2SO4 -> Br2 + SO2 + 2H2O}, similar to the book's reaction of HI and \ce{H2SO4} to give \ce{SO2}). Fluoride and chloride (options A and B) are not oxidised by the acid at all, so they cannot reduce it.

2025 Annual I, Set B, Q.2 (vi) OR1+2 = 3 marks

What is acid rain? Describe its formation with the help of chemical reactions involved.

Solution

Acid rain (1 mark)

Acid rain is rain that is more acidic than normal rain because acids formed from the pollutant oxides of sulphur and nitrogen dissolve in rain water. Sulphur dioxide is the main cause, and the oxides of nitrogen add to it.

Formation (2 marks)

Sulphur dioxide is formed when fuels containing sulphur are burned. It is oxidised in the air to sulphur trioxide (the oxides of nitrogen act as catalysts) and the trioxide dissolves in water to form sulphuric acid.

\ce{S + O2 -> SO2}

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

\ce{SO3 + H2O -> H2SO4}

Nitrogen monoxide is formed from nitrogen and oxygen at the high temperature of engines (or by lightning). It is oxidised to nitrogen dioxide, which dissolves in water droplets to form nitric acid.

\ce{N2(g) + O2(g) -> 2NO(g)}

\ce{2NO(g) + O2(g) -> 2NO2(g)}

\ce{2NO2(g) + H2O(l) + 1/2 O2(g) -> 2HNO3(aq)}

The sulphuric acid and nitric acid fall to the ground in the rain.

2025 Annual I, Set B, Q.2 (x)3 marks

Describe the use of compounds of sulphur in any three chemical industries.

Solution

Compounds of sulphur are used in many industries. Three of them (1 mark each):

Fertiliser industry. Ammonium sulphate, \ce{(NH4)2SO4}, is made by neutralising ammonia with sulphuric acid. It supplies both nitrogen and sulphur to crops, and sulphur helps plants make amino acids and vitamins.
\ce{2NH3(aq) + H2SO4(aq) -> (NH4)2SO4(aq)}
Detergent, paint, dye and fibre industries. Sulphuric acid is the starting chemical for making detergents, paints, pigments, dyes, synthetic fibres and plastics.
Food industry. Sulphur dioxide and sulphites such as \ce{Na2SO3} are food preservatives. They are reducing agents, so they stop the food from being oxidised.

Sulphuric acid is also used in car batteries, in tanning leather and in pickling of metals, and sulphur itself is used for vulcanising rubber and in gunpowder.

2025 Annual I, Set D, MCQ 111 mark

Which compound of sulphur is used as fertilizer?

A\ce{(NH4)2SO4}Answer
B\ce{CuSO4}
C\ce{Na2SO4}
D\ce{BaSO4}

Why

Ammonium sulphate, \ce{(NH4)2SO4}, is used as a fertiliser. It is made by neutralising ammonia with sulphuric acid, \ce{2NH3 + H2SO4 -> (NH4)2SO4}, and it supplies both nitrogen and sulphur to the plants.

Copper sulphate (option B), sodium sulphate (option C) and barium sulphate (option D) contain no nitrogen and are not used as the fertiliser compound of sulphur in the book.

2025 Annual I, Set D, Q.2 (viii) OR3 marks

Write down any three harmful effects of acid rain.

Solution

Acid rain is formed when the oxides of sulphur and nitrogen dissolve in rain water to give sulphuric acid and nitric acid. Its harmful effects (1 mark each):

Damage to buildings, statues and metals. The acids attack marble and limestone (\ce{CaCO3}) and corrode iron and steel structures.
\ce{CaCO3 + H2SO4 -> CaSO4 + H2O + CO2}
Damage to water life. Acid rain lowers the pH of lakes and rivers, so fish and other water life die.
Damage to soil, plants and forests. The acids wash out nutrients from the soil and harm the leaves and roots of trees and crops, so growth is reduced.

Nitrogen oxides, the cause of the nitric acid, are also lung irritants.

Exam focus

What is acid rain? Describe its formation with equations, and give three harmful effects. (short)
2025 Set B Q.2 (vi) OR (definition and formation) and Set D Q.2 (viii) OR (harmful effects); not asked in 2026.
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Uses of sulphur and its compounds (fertiliser, food preservatives, detergents, paints, rubber). (short)
2025 Set B Q.2 (x) (three industries) and Set D MCQ 11 (ammonium sulphate as fertiliser).
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Photochemical smog and the formation of PAN. (short)
2025 Set B MCQ 9 (PAN forms from oxides of nitrogen and unburned hydrocarbons); the three stages of the mechanism are in the section.
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Concentrated sulphuric acid as an oxidising agent, for example with halides and copper. (mcq)
2025 Set B MCQ 10 (NaI reduces \ce{H2SO4} to \ce{H2S}).
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Contact process: stages, catalyst, temperature, pressure and why \ce{SO3} is absorbed in sulphuric acid. (long)
Book exercise MCQs 1 and 7 and long question 5; not in the 2025 or 2026 paper.
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Catalytic converter and the control of exhaust gases from a petrol engine. (long)
Book exercise MCQ 5 and long question 3; not in the 2025 or 2026 paper.
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Haber process and the unreactivity of nitrogen. (long)
Book short question 2 and long question 4; not in the 2025 or 2026 paper.
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Nitrogen cycle: fixation, nitrification and denitrification, with the bacteria and equations. (mcq)
Book exercise MCQs 2, 3, 6 and 8; not in the 2025 or 2026 paper.
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Sulphur as an oxidising agent and as a reducing agent. (short)
Book short question 6; not in the 2025 or 2026 paper.
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