Chemical Reactions

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1. Writing Chemical Equations

1.1 Word Equations

A word equation describes a chemical reaction using the names of the reactants and products.

Example: magnesium + hydrochloric acid $\to$ magnesium chloride + hydrogen

Word equations are useful as a first step but they lack precision. Different compounds can share the Same name in different contexts, and word equations convey no information about the stoichiometry of The reaction.

1.2 Symbol Equations

A symbol equation uses chemical formulae. It must be balanced — the same number of each type of Atom must appear on both sides (conservation of mass).

The conservation of mass is a consequence of the fact that atoms are neither created nor destroyed In a chemical reaction. The atoms are rearranged into different molecules. This is one of the Most fundamental principles in chemistry.

Worked Example. Balance the equation for the reaction between iron and oxygen to form iron(III) Oxide.

Unbalanced: $\mathrm{Fe + \mathrm{O_2 \to \mathrm{Fe_2\mathrm{O_3$

Count atoms: Fe: 1 on left, 2 on right. O: 2 on left, 3 on right.

To balance Fe: put 4 on the left. To balance O: put 3 O $_2$ on the left.

$4\mathrm{Fe + 3\mathrm{O_2 \to 2\mathrm{Fe_2\mathrm{O_3$

Check: Fe: 4 = 4, O: 6 = 6.

Worked Example. Balance the reaction of aluminium with hydrochloric acid.

Unbalanced: $\mathrm{Al + \mathrm{HCl \to \mathrm{AlCl_3 + \mathrm{H_2$

$2\mathrm{Al + 6\mathrm{HCl \to 2\mathrm{AlCl_3 + 3\mathrm{H_2$

Worked Example. Balance the combustion of ethane.

Unbalanced: $\mathrm{C_2\mathrm{H_6 + \mathrm{O_2 \to \mathrm{CO_2 + \mathrm{H_2\mathrm{O$

$2\mathrm{C_2\mathrm{H_6 + 7\mathrm{O_2 \to 4\mathrm{CO_2 + 6\mathrm{H_2\mathrm{O$

Worked Example. Balance the reaction of zinc with copper(II) sulfate.

$\mathrm{Zn + \mathrm{CuSO_4 \to \mathrm{ZnSO_4 + \mathrm{Cu$

This equation is already balanced. One atom of each element appears on both sides.

1.3 State Symbols

Symbol	State
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Aqueous (dissolved in water)

State symbols provide information about the physical state of each substance under the reaction Conditions. This is important because the state affects the energy change and the type of reaction That occurs.

Worked Example. Write a balanced symbol equation with state symbols for the reaction of calcium With hydrochloric acid.

$\mathrm{Ca(s) + 2\mathrm{HCl(aq) \to \mathrm{CaCl_2\mathrm{(aq) + \mathrm{H_2\mathrm{(g)$

Worked Example. Write the balanced equation for the thermal decomposition of copper(II) Carbonate.

$\mathrm{CuCO_3\mathrm{(s) \to \mathrm{CuO(s) + \mathrm{CO_2\mathrm{(g)$

Worked Example. Write the balanced equation for the reaction between sodium hydroxide and Sulfuric acid.

$2\mathrm{NaOH(aq) + \mathrm{H_2\mathrm{SO_4\mathrm{(aq) \to \mathrm{Na_2\mathrm{SO_4\mathrm{(aq) + 2\mathrm{H_2\mathrm{O(l)$

1.4 Ionic Equations

Ionic equations show only the species that actually participate in the reaction. Spectator ions (ions that do not change) are removed.

Worked Example. Write the ionic equation for the reaction of silver nitrate with sodium Chloride.

Full equation: $\mathrm{AgNO_3\mathrm{(aq) + \mathrm{NaCl(aq) \to \mathrm{AgCl(s) + \mathrm{NaNO_3\mathrm{(aq)$

Ionic: $\mathrm{Ag^+\mathrm{(aq) + \mathrm{NO_3^-\mathrm{(aq) + \mathrm{Na^+\mathrm{(aq) + \mathrm{Cl^-\mathrm{(aq) \to \mathrm{AgCl(s) + \mathrm{Na^+\mathrm{(aq) + \mathrm{NO_3^-\mathrm{(aq)$

Spectator ions: $\mathrm{Na^+$ and $\mathrm{NO_3^-$

Net ionic: $\mathrm{Ag^+\mathrm{(aq) + \mathrm{Cl^-\mathrm{(aq) \to \mathrm{AgCl(s)$

Worked Example. Write the ionic equation for the reaction of zinc with copper(II) sulfate.

Full: $\mathrm{Zn(s) + \mathrm{CuSO_4\mathrm{(aq) \to \mathrm{ZnSO_4\mathrm{(aq) + \mathrm{Cu(s)$

Ionic: $\mathrm{Zn(s) + \mathrm{Cu^{2+}\mathrm{(aq) + \mathrm{SO_4^{2-}\mathrm{(aq) \to \mathrm{Zn^{2+}\mathrm{(aq) + \mathrm{SO_4^{2-}\mathrm{(aq) + \mathrm{Cu(s)$

Net ionic: $\mathrm{Zn(s) + \mathrm{Cu^{2+}\mathrm{(aq) \to \mathrm{Zn^{2+}\mathrm{(aq) + \mathrm{Cu(s)$

Spectator ion: SO $_4^{2-}$ .

1.5 Half Equations for Redox Reactions

Redox reactions can be split into two half equations: one for oxidation and one for reduction.

Example: The reaction between magnesium and copper(II) sulfate.

Oxidation: $\mathrm{Mg \to \mathrm{Mg^{2+} + 2e^-$ (magnesium loses electrons)

Reduction: $\mathrm{Cu^{2+} + 2e^- \to \mathrm{Cu$ (copper gains electrons)

Worked Example. Write half equations for the reaction of chlorine with potassium bromide.

Oxidation: $2\mathrm{Br^- \to \mathrm{Br_2 + 2e^-$ (bromide ions are oxidised to bromine)

Reduction: $\mathrm{Cl_2 + 2e^- \to 2\mathrm{Cl^-$ (chlorine is reduced to chloride ions)

1.6 Derivation: Conservation of Mass from Atomic Theory

Since atoms are neither created nor destroyed in a chemical reaction (Law of Conservation of Mass, Lavoisier), the total mass of the reactants equals the total mass of the products. This is why Equations must be balanced: the number of atoms of each element must be the same on both sides.

In a closed system, this is always true. In an open system, apparent mass changes can occur (e.g. Gas escaping), but the atoms are still conserved — they have left the system.

2. Types of Chemical Reaction

2.1 Exothermic and Endothermic Reactions

Exothermic reactions release energy to the surroundings (temperature increases). Examples: Combustion, neutralisation, respiration, oxidation.

Endothermic reactions absorb energy from the surroundings (temperature decreases). Examples: Thermal decomposition, photosynthesis.

Activation energy: The minimum energy that particles must have for a reaction to occur. All Reactions require activation energy, even exothermic ones. The activation energy is the energy Barrier that must be surmounted for the reaction to proceed.

Why do exothermic reactions need activation energy?

An exothermic reaction releases energy overall, but bonds must still be broken before new bonds can Form. Breaking bonds always requires an energy input. The activation energy is the energy needed to Break the initial bonds. Once the reaction has started, the energy released from forming new bonds More than compensates for the initial input.

Consider a ball rolling down a hill. The ball releases energy (potential energy converts to kinetic Energy), but it first needs to be pushed over a small bump at the top. The bump is the activation Energy.

2.2 Energy Level Diagrams

Exothermic:

$\mathrm{Reactants \xrightarrow{\mathrm{activation energy} \mathrm{Products + \mathrm{energy$

Products are at a lower energy level than reactants. The difference in energy levels is the energy Released.

Endothermic:

$\mathrm{Reactants + \mathrm{energy \xrightarrow{\mathrm{activation energy} \mathrm{Products$

Products are at a higher energy level than reactants. The difference in energy levels is the energy Absorbed.

2.3 Bond Energies

The bond energy is the energy required to break one mole of a particular bond.

During a reaction:

Energy is absorbed to break bonds (endothermic)
Energy is released when new bonds form (exothermic)

$\Delta H = \mathrm{energy absorbed (breaking) - \mathrm{energy released (forming)$

If $\Delta H$ is negative, the reaction is exothermic. If $\Delta H$ is positive, the reaction is Endothermic.

This is a consequence of energy conservation. If more energy is released by forming new bonds than Is absorbed by breaking old bonds, the excess is released to the surroundings (exothermic).

Worked Example. Calculate the enthalpy change for the reaction: $\mathrm{H_2 + \mathrm{Cl_2 \to 2\mathrm{HCl$

Given: H-H bond energy = 436 kJ/mol, Cl-Cl bond energy = 242 kJ/mol, H-Cl bond energy = 431 kJ/mol.

$\mathrm{Energy absorbed = 436 + 242 = 678 \mathrm{ kJ$ $\mathrm{Energy released = 2 \times 431 = 862 \mathrm{ kJ$ $\Delta H = 678 - 862 = -184 \mathrm{ kJ/mol$

The reaction is exothermic ( $\Delta H \lt 0$ ).

Worked Example. Calculate the enthalpy change for the reaction: $\mathrm{N_2 + 3\mathrm{H_2 \to 2\mathrm{NH_3$

Given: N $\equiv$ N = 945 kJ/mol, H-H = 436 kJ/mol, N-H = 391 kJ/mol.

$\mathrm{Energy absorbed = 945 + 3(436) = 945 + 1308 = 2253 \mathrm{ kJ$ $\mathrm{Energy released = 6(391) = 2346 \mathrm{ kJ$ $\Delta H = 2253 - 2346 = -93 \mathrm{ kJ/mol$

Worked Example. Calculate the enthalpy change for the combustion of methane: $\mathrm{CH_4 + 2\mathrm{O_2 \to \mathrm{CO_2 + 2\mathrm{H_2\mathrm{O$

Given: C-H = 413 kJ/mol, O=O = 495 kJ/mol, C=O = 805 kJ/mol, O-H = 464 kJ/mol.

$\mathrm{Energy absorbed = 4(413) + 2(495) = 1652 + 990 = 2642 \mathrm{ kJ$ $\mathrm{Energy released = 2(805) + 4(464) = 1610 + 1856 = 3466 \mathrm{ kJ$ $\Delta H = 2642 - 3466 = -824 \mathrm{ kJ/mol$

2.4 Required Practical: Investigating Temperature Changes

Measure the temperature change when a known mass of solid is added to a known volume of solution.

The energy change can be calculated using:

$q = mc\Delta T$

Where $m$ is the mass of the solution (g), $c$ is the specific heat capacity of water ( $4.18 \mathrm{ J/g^{\circ}\mathrm{C$ ), and $\Delta T$ is the temperature change.

Worked Example. When 2.0 g of calcium chloride is dissolved in 50 g of water, the temperature Rises by $8.5^{\circ}\mathrm{C$ . Calculate the energy change.

$q = 50 \times 4.18 \times 8.5 = 1776.5 \mathrm{ J = 1.78 \mathrm{ kJ$

Since the temperature increased, the reaction is exothermic, so $\Delta H = -1.78 \mathrm{ kJ$ .

3. Rates of Reaction

3.1 Collision Theory

For a reaction to occur, particles must:

Collide with each other
Collide with sufficient energy (at least the activation energy)
Collide with the correct orientation

The rate of reaction is the speed at which reactants are converted into products. It can be Measured by monitoring the change in concentration of a reactant or product over time.

Collision theory provides a microscopic explanation for the macroscopic factors that affect rate. Any change that increases the frequency of successful collisions (those with sufficient energy and Correct orientation) will increase the rate.

3.2 Factors Affecting Rate of Reaction

Factor	Effect on Rate	Reason
Increasing concentration	Increases	More particles per unit volume, more frequent collisions
Increasing temperature	Increases	Particles have more kinetic energy; more collisions exceed activation energy
Increasing surface area	Increases	More particles exposed for collision
Adding a catalyst	Increases	Provides an alternative pathway with lower activation energy
Increasing pressure (gases)	Increases	More particles per unit volume

Temperature has a particularly strong effect because it changes the distribution of molecular Energies. A small increase in temperature shifts a much larger fraction of molecules above the Activation energy threshold. The rate approximately doubles for every $10^{\circ}\mathrm{C$ increase In temperature.

Surface area matters for solid reactants. A lump of calcium carbonate has a small surface area, But the same mass as a powder has a much larger surface area. Since reactions occur at the surface, The powder reacts much faster.

3.3 Catalysts

A catalyst is a substance that increases the rate of a reaction without being used up. It works By providing an alternative reaction pathway with a lower activation energy.

Examples:

Iron in the Haber process
Platinum in catalytic converters
Enzymes (biological catalysts)

Advantages of catalysts:

Lower temperature and pressure can be used (saves energy and money)
Do not get used up (only small amounts needed)
Reduce waste

A catalyst does not change the products of a reaction, the position of equilibrium, or the enthalpy Change. It only provides a lower-energy pathway from reactants to products.

3.4 Required Practical: Investigating Rate

Method 1: Measuring gas volume

React marble chips (CaCO $_3$ ) with hydrochloric acid
Collect the CO $_2$ gas in a gas syringe
Measure the volume at regular time intervals
Plot a graph of volume against time

Method 2: Measuring mass loss

React marble chips with hydrochloric acid on a balance
Record the mass at regular time intervals
The mass decreases as CO $_2$ escapes

Method 3: Measuring the time for a precipitate

React sodium thiosulfate with hydrochloric acid
Time how long it takes for a cross drawn under the flask to be obscured by the sulfur precipitate

3.5 Interpreting Rate Graphs

A graph of amount of product against time:

Starts steep (fast initial rate)
Curves and levels off (rate decreases as reactants are used up)
Reaches a maximum when one reactant is fully consumed

The gradient at any point gives the rate at that time. The steepest gradient is at the start (initial rate).

Comparing rate graphs: When comparing two experiments on the same graph, the curve with the Steeper initial gradient corresponds to the faster reaction. Both curves should level off at the Same maximum if the same total amount of reactant is used (assuming complete reaction).

3.6 Higher Tier: Calculating Rate from a Graph

The rate at a particular time is equal to the gradient of the tangent to the curve at that point.

$\mathrm{Rate = \frac{\Delta y}{\Delta x}$

Where $\Delta y$ is the change in the quantity measured (volume, mass, concentration) and $\Delta x$ Is the change in time.

Worked Example. In an experiment, 20 cm $^3$ of gas is collected in the first 30 seconds. Calculate the mean rate of reaction.

$\mathrm{Rate = \frac{20}{30} = 0.67 \mathrm{ cm^3/\mathrm{s$

4. Reversible Reactions and Equilibrium

4.1 Reversible Reactions

A reversible reaction is one that can proceed in both directions. It is written with a Reversible arrow ( $\rightleftharpoons$ ).

Example: Ammonium chloride decomposes on heating and reforms on cooling.

$\mathrm{NH_4\mathrm{Cl(s) \rightleftharpoons \mathrm{NH_3\mathrm{(g) + \mathrm{HCl(g)$

4.2 Dynamic Equilibrium

When a reversible reaction takes place in a closed system, the forward and reverse reactions Both occur. Eventually, the rates of the forward and reverse reactions become equal, and the Concentrations of reactants and products remain constant. This state is called dynamic Equilibrium.

The word “dynamic” is important: at equilibrium, both reactions are still occurring, but at the same Rate. The concentrations do not change because the forward and reverse rates are balanced. If the System is not closed (e.g., a gas escapes), equilibrium cannot be established.

Conditions for equilibrium:

The system must be closed (no substances can enter or leave)
The forward and reverse rates must be equal
The concentrations remain constant (but not necessarily equal)

4.3 Le Chatelier’s Principle

Le Chatelier’s principle: If a system at equilibrium is subjected to a change, the equilibrium Shifts in the direction that counteracts the change.

Change	Effect on equilibrium
Increase concentration of a reactant	Equilibrium shifts to the right (to use up the added reactant)
Increase temperature	Shifts in the endothermic direction
Increase pressure	Shifts to the side with fewer gas molecules
Add a catalyst	No effect on position; reaches equilibrium faster

Key points:

Only temperature changes affect the value of the equilibrium constant
Pressure changes affect the position but not the value of the equilibrium constant (for a given temperature)
Concentration changes affect the position but not the value of the equilibrium constant

4.4 The Haber Process

$\mathrm{N_2\mathrm{(g) + 3\mathrm{H_2\mathrm{(g) \rightleftharpoons 2\mathrm{NH_3\mathrm{(g) \quad \Delta H = -92 \mathrm{ kJ/mol$

Conditions used:

Temperature: 450 $^{\circ}$ C (compromise: lower temperature favours product, but reaction would be too slow)
Pressure: 200 atm (high pressure favours product, 4 moles $\to$ 2 moles; but high pressure is expensive and dangerous)
Catalyst: iron (speeds up the rate)

The conditions chosen represent an economic compromise. The theoretically optimal conditions (low Temperature, very high pressure) are not practical because the reaction would be too slow or the Equipment too expensive. The chosen conditions give a reasonable yield at an acceptable rate and Cost.

Worked Example. Explain the effect of increasing temperature on the Haber process equilibrium.

The forward reaction is exothermic ( $\Delta H = -92$ kJ/mol). Increasing temperature favours the Endothermic (reverse) reaction, so the equilibrium shifts to the left, producing more nitrogen and Hydrogen and less ammonia. The yield of ammonia decreases, but the rate of reaction increases.

4.5 The Contact Process

$2\mathrm{SO_2\mathrm{(g) + \mathrm{O_2\mathrm{(g) \rightleftharpoons 2\mathrm{SO_3\mathrm{(g) \quad \Delta H = -197 \mathrm{ kJ/mol$

Conditions: 450 $^{\circ}$ C, 1-2 atm, vanadium(V) oxide catalyst.

The moderate temperature is again a compromise between equilibrium yield and rate. The pressure is Only slightly above atmospheric because the equilibrium already favours the product side (3 moles $\to$ 2 moles of gas), so very high pressure is not economically justified.

4.6 Equilibrium Summary Table

Process	Equation	Favourable Conditions
Haber process	$\mathrm{N_2 + 3\mathrm{H_2 \rightleftharpoons 2\mathrm{NH_3$	High pressure, moderate temp, iron catalyst
Contact process	$2\mathrm{SO_2 + \mathrm{O_2 \rightleftharpoons 2\mathrm{SO_3$	Moderate pressure, moderate temp, V $_2$ O $_5$ catalyst

5. Acids, Bases, and Salts

5.1 The pH Scale

The pH scale measures how acidic or alkaline a solution is:

pH	Description
0—6	Acidic (lower = more acidic)
7	Neutral
8—14	Alkaline (higher = more alkaline)

Acids produce H $^+$ ions in solution.

Bases neutralise acids. A base that dissolves in water is called an alkali (produces OH $^-$ ions).

The pH scale is logarithmic: each unit decrease in pH represents a tenfold increase in hydrogen ion Concentration. A solution of pH 3 has ten times the $[\mathrm{H^+]$ of a solution of pH 4.

5.2 Strong and Weak Acids

Strong acids fully dissociate in water:

$\mathrm{HCl \to \mathrm{H^+ + \mathrm{Cl^-$

Weak acids partially dissociate in water:

$\mathrm{CH_3\mathrm{COOH \rightleftharpoons \mathrm{H^+ + \mathrm{CH_3\mathrm{COO^-$

A weak acid has a higher pH than a strong acid of the same concentration. This is because the weak Acid produces fewer H $^+$ ions per unit volume.

“Strong” and “weak” refer to the degree of dissociation, not the concentration. A dilute solution of A strong acid can have a higher pH than a concentrated solution of a weak acid.

5.3 Reactions of Acids

Reaction	General Equation	Example
Acid + metal $\to$ salt + hydrogen	acid + metal $\to$ salt + H $_2$	2HCl + Zn $\to$ ZnCl $_2$ + H $_2$
Acid + base $\to$ salt + water	acid + base $\to$ salt + H $_2$ O	HCl + NaOH $\to$ NaCl + H $_2$ O
Acid + carbonate $\to$ salt + water + CO $_2$	acid + carbonate $\to$ salt + H $_2$ O + CO $_2$	2HCl + CaCO $_3$ $\to$ CaCl $_2$ + H $_2$ O + CO $_2$

5.4 Making Salts

Soluble salts: Made by reacting an acid with a base (or metal, or carbonate), then evaporating The water to crystallise the salt.

Required practical:

Add the acid to the base slowly (with a Bunsen burner) until the base is in excess
Filter to remove the excess base
Evaporate the filtrate to obtain the salt crystals

Insoluble salts: Made by precipitation — mixing two solutions that contain the ions of the Desired salt.

5.5 Titrations

A titration is used to find the exact volume of acid needed to neutralise a known volume of Alkali.

Method:

Use a pipette to measure a known volume of alkali into a conical flask
Add a few drops of indicator (e.g. Phenolphthalein or universal indicator)
Fill a burette with acid of known concentration
Slowly add the acid from the burette, swirling the flask, until the indicator changes colour
Record the volume of acid used (the titre)
Repeat to obtain concordant results (within 0.10 cm $^3$ )

Worked Example. 25.0 cm $^3$ of 0.100 mol/dm $^3$ NaOH is neutralised by HCl. The average titre is 22.5 cm $^3$ . Calculate the concentration of the HCl.

$n(\mathrm{NaOH) = 0.100 \times 0.0250 = 0.00250 \mathrm{ mol$

Mole ratio 1:1, so $n(\mathrm{HCl) = 0.00250 \mathrm{ mol$ .

$c(\mathrm{HCl) = \frac{0.00250}{0.0225} = 0.111 \mathrm{ mol/dm^3$

5.6 Tests for Gases

Gas	Test	Result
Hydrogen	Lighted splint	Squeaky pop
Oxygen	Glowing splint	Relights
Carbon dioxide	Limewater	Turns cloudy/milky
Chlorine	Damp litmus paper	Bleaches the litmus (turns white)
Ammonia	Damp red litmus paper	Turns blue

6. Electrolysis

6.1 Principles

Electrolysis is the decomposition of an ionic compound using electricity. It requires a liquid (molten or aqueous) to allow ions to move and carry charge.

Components:

Anode: Positive electrode (attracts anions)
Cathode: Negative electrode (attracts cations)
Electrolyte: The ionic compound being decomposed

6.2 Electrolysis of Molten Ionic Compounds

When a molten ionic compound is electrolysed, the metal is produced at the cathode and the non-metal Is produced at the anode.

Example: Electrolysis of molten lead(II) bromide.

Cathode (negative): $\mathrm{Pb^{2+} + 2e^- \to \mathrm{Pb$ (lead metal)

Anode (positive): $2\mathrm{Br^- \to \mathrm{Br_2 + 2e^-$ (bromine gas)

6.3 Electrolysis of Aqueous Solutions

When an aqueous solution is electrolysed, water can also be electrolysed, producing hydrogen and/or Oxygen.

At the cathode (negative): Metal ions or hydrogen are discharged (reduced).

At the anode (positive): Non-metal ions (except in aqueous solution where OH $^-$ may be Discharged) are discharged (oxidised).

Rules for discharge at electrodes:

Cathode: If the metal is more reactive than hydrogen, hydrogen is produced. If less reactive, the Metal is produced.

Anode (aqueous): Halides produce the halogen. Otherwise, oxygen is produced from hydroxide ions.

6.4 Worked Examples

Worked Example. Predict the products of the electrolysis of copper(II) sulfate solution using Inert electrodes.

Cathode: Copper is less reactive than hydrogen, so copper is produced:

$\mathrm{Cu^{2+} + 2e^- \to \mathrm{Cu$

Anode: Sulfate is not a halide, so oxygen is produced:

$4\mathrm{OH^- \to 2\mathrm{H_2\mathrm{O + \mathrm{O_2 + 4e^-$

Worked Example. Predict the products of the electrolysis of aqueous sodium chloride using inert Electrodes.

Cathode: Sodium is more reactive than hydrogen, so hydrogen is produced:

$2\mathrm{H^+ + 2e^- \to \mathrm{H_2$

Anode: Chloride is a halide, so chlorine is produced:

$2\mathrm{Cl^- \to \mathrm{Cl_2 + 2e^-$

Worked Example. Predict the products of the electrolysis of aqueous potassium iodide.

Cathode: Potassium is more reactive than hydrogen, so hydrogen is produced from water:

$2\mathrm{H_2\mathrm{O + 2e^- \to \mathrm{H_2 + 2\mathrm{OH^-$

Anode: Iodide is a halide, so iodine is produced:

$2\mathrm{I^- \to \mathrm{I_2 + 2e^-$

6.5 Electrolysis in Industry

Extraction of aluminium: Aluminium oxide (bauxite) is dissolved in molten cryolite and Electrolysed. Aluminium is produced at the cathode and oxygen at the anode. The carbon anodes react With the oxygen and must be replaced periodically.

Cathode: $\mathrm{Al^{3+} + 3e^- \to \mathrm{Al$

Anode: $2\mathrm{O^{2-} \to \mathrm{O_2 + 4e^-$

Electroplating: The cathode is the object to be plated. The anode is made of the plating metal. The electrolyte contains ions of the plating metal. For example, to copper-plate a key, the key is The cathode, a copper bar is the anode, and the electrolyte is copper(II) sulfate solution.

7.1 Classification Table

Type	Description	Example
Combustion	Burning in oxygen	$\mathrm{CH_4 + 2\mathrm{O_2 \to \mathrm{CO_2 + 2\mathrm{H_2\mathrm{O$
Thermal decomposition	Breaking down with heat	$\mathrm{CaCO_3 \to \mathrm{CaO + \mathrm{CO_2$
Neutralisation	Acid + base $\to$ salt + water	$\mathrm{HCl + \mathrm{NaOH \to \mathrm{NaCl + \mathrm{H_2\mathrm{O$
Displacement	More reactive element displaces a less reactive one	$\mathrm{Zn + \mathrm{CuSO_4 \to \mathrm{ZnSO_4 + \mathrm{Cu$
Oxidation/reduction	Transfer of electrons	$\mathrm{Fe^{2+} \to \mathrm{Fe^{3+} + e^-$

7.2 Oxidation States

Oxidation state rules:

Elements in their standard state: 0
Monatomic ions: equal to their charge
Oxygen: $-2$ (except in peroxides: $-1$ )
Hydrogen: $+1$ (except in metal hydrides: $-1$ )
The sum of oxidation states in a neutral compound is 0

Worked Example. Determine the oxidation state of manganese in KMnO $_4$ .

K = $+1$ O = $-2$ (four oxygens: $4 \times -2 = -8$ ).

$+1 + \mathrm{Mn + (-8) = 0$ So Mn = $+7$ .

Worked Example. Determine the oxidation state of chromium in Cr $_2$ O $_7^{2-}$ .

O = $-2$ (seven oxygens: $7 \times -2 = -14$ ).

$2\mathrm{Cr + (-14) = -2$ So $2\mathrm{Cr = +12$ Giving Cr = $+6$ .

Common Pitfalls

Not balancing equations correctly. Always count atoms of each element on both sides. Check that the total charge is balanced in ionic equations.
Confusing exothermic and endothermic. Exothermic = releases heat ( $\Delta H \lt 0$ ); endothermic = absorbs heat ( $\Delta H \gt 0$ ).
Forgetting state symbols in equations. These are often required in exams and provide useful information.
Misapplying Le Chatelier’s principle. The equilibrium shifts to COUNTERACT the change, not to reinforce it.
Confusing a catalyst with a reactant. A catalyst speeds up the reaction but is not consumed. It does not change the yield.
Writing the wrong products in electrolysis. Remember the rules for discharge at each electrode. In aqueous solution, hydrogen may be produced at the cathode if the metal is reactive, and oxygen may be produced at the anode if the anion is not a halide.
Forgetting that the pH scale is logarithmic. A change of 1 pH unit means a tenfold change in $[\mathrm{H^+]$ .
Using the wrong reagent to identify a gas. Know each test precisely: limewater for CO $_2$ glowing splint for O $_2$ Lighted splint for H $_2$ .
Confusing the oxidation state of oxygen in peroxides. In H $_2$ O $_2$ Oxygen has oxidation state $-1$ Not $-2$ .
Not including the correct charges in half equations. Always balance both atoms and charge.

Practice Questions

Balance the equation: $\mathrm{Al + \mathrm{HCl \to \mathrm{AlCl_3 + \mathrm{H_2$
Using the bond energies H-H = 436 kJ/mol, O=O = 498 kJ/mol, and O-H = 464 kJ/mol, calculate the enthalpy change for $2\mathrm{H_2 + \mathrm{O_2 \to 2\mathrm{H_2\mathrm{O$ .
Explain how increasing the temperature increases the rate of a reaction, with reference to collision theory.
State Le Chatelier’s principle and use it to predict the effect of increasing pressure on the equilibrium: $\mathrm{N_2\mathrm{O_4\mathrm{(g) \rightleftharpoons 2\mathrm{NO_2\mathrm{(g)$ .
Write the ionic equation for the reaction between zinc and copper(II) sulfate solution.
25.0 cm $^3$ of sodium hydroxide solution (concentration 0.100 mol/dm $^3$ ) is neutralised by hydrochloric acid. Calculate the volume of 0.200 mol/dm $^3$ hydrochloric acid needed.
Describe the products formed at each electrode during the electrolysis of molten lead(II) bromide.
Explain why a catalyst increases the rate of reaction but does not change the position of equilibrium.
Describe how you would prepare a pure, dry sample of copper(II) sulfate crystals.
Explain the difference between a strong acid and a weak acid of the same concentration.
A student investigates the rate of reaction between marble chips and hydrochloric acid at two different temperatures. Sketch the two graphs on the same axes and explain the differences.
Explain why the Haber process uses a temperature of 450 $^{\circ}$ C rather than room temperature or 100 $^{\circ}$ C.
Write the balanced equation for the reaction between sulfuric acid and sodium hydroxide, including state symbols.
Describe a test to identify (a) hydrogen gas and (b) carbon dioxide gas.
Explain, in terms of electrode reactions, why the mass of the anode decreases during the electroplating of a metal object.
Balance the equation: $\mathrm{C_4\mathrm{H_{10} + \mathrm{O_2 \to \mathrm{CO_2 + \mathrm{H_2\mathrm{O$
Predict the products of electrolysis of aqueous sodium sulfate using inert electrodes.
Calculate the pH of a solution made by diluting 10 cm $^3$ of 0.1 mol/dm $^3$ HCl to 100 cm $^3$ .
Write half equations for the reaction between magnesium and silver nitrate.
Explain, using Le Chatelier’s principle, why increasing the pressure increases the yield of ammonia in the Haber process.

Practice Problems

Question 1: Balancing equations

Balance the following chemical equations: (a) $\mathrm{Fe + \mathrm{O_2 \to \mathrm{Fe_2\mathrm{O_3$ (b) $\mathrm{Al + \mathrm{HCl \to \mathrm{AlCl_3 + \mathrm{H_2$ (c) $\mathrm{C_3\mathrm{H_8 + \mathrm{O_2 \to \mathrm{CO_2 + \mathrm{H_2\mathrm{O$

Answer

(a) $4\mathrm{Fe + 3\mathrm{O_2 \to 2\mathrm{Fe_2\mathrm{O_3$ (b) $2\mathrm{Al + 6\mathrm{HCl \to 2\mathrm{AlCl_3 + 3\mathrm{H_2$ (c) $\mathrm{C_3\mathrm{H_8 + 5\mathrm{O_2 \to 3\mathrm{CO_2 + 4\mathrm{H_2\mathrm{O$

Question 2: Types of chemical reactions

Classify each of the following reactions as combination, decomposition, displacement, or combustion: (a) $2\mathrm{Mg + \mathrm{O_2 \to 2\mathrm{MgO$ (b) $2\mathrm{H_2\mathrm{O_2 \to 2\mathrm{H_2\mathrm{O + \mathrm{O_2$ (c) $\mathrm{Zn + \mathrm{CuSO_4 \to \mathrm{ZnSO_4 + \mathrm{Cu$ (d) $\mathrm{CH_4 + 2\mathrm{O_2 \to \mathrm{CO_2 + 2\mathrm{H_2\mathrm{O$

Answer

(a) Combination (two elements combine to form one compound). (b) Decomposition (one compound breaks down into simpler substances). (c) Displacement (zinc displaces copper from copper sulfate). (d) Combustion (hydrocarbon burns in oxygen to produce carbon dioxide and water).

Question 3: Reactivity series predictions

A student places pieces of zinc, copper, and magnesium into separate test tubes containing iron(II) sulfate solution. Predict what will happen in each test tube and explain your reasoning.

Answer

Using the reactivity series: potassium, sodium, calcium, magnesium, aluminium, (carbon), zinc, iron, (hydrogen), copper.

Zinc: zinc is above iron in the reactivity series, so it will displace iron: $\mathrm{Zn + \mathrm{FeSO_4 \to \mathrm{ZnSO_4 + \mathrm{Fe$ . The blue solution fades as iron is displaced.

Copper: copper is below iron, so no reaction occurs. Copper cannot displace iron.

Magnesium: magnesium is above iron, so it displaces iron: $\mathrm{Mg + \mathrm{FeSO_4 \to \mathrm{MgSO_4 + \mathrm{Fe$ . This reaction is more vigorous than zinc’s because magnesium is more reactive.

Question 4: Exothermic and endothermic reactions

Explain the difference between exothermic and endothermic reactions. Give one example of each and describe how the temperature changes.

Answer

Exothermic: energy is released to the surroundings. Temperature increases. Example: combustion of methane — the products ( $\mathrm{CO_2$ and $\mathrm{H_2\mathrm{O$ ) have less chemical energy than the reactants.

Endothermic: energy is absorbed from the surroundings. Temperature decreases. Example: thermal decomposition of calcium carbonate — energy is needed to break the bonds in $\mathrm{CaCO_3$ .

Question 5: Le Chatelier's principle

The Haber process: $\mathrm{N_2(g) + 3\mathrm{H_2(g) \rightleftharpoons 2\mathrm{NH_3(g)$ , $\Delta H = -92 \mathrm{ kJ/mol$ . Explain the effect on the yield of ammonia of (a) increasing pressure, (b) increasing temperature, and (c) removing ammonia as it is formed.

Answer

(a) Increasing pressure: equilibrium shifts to the side with fewer gas molecules (4 mol reactants $\to$ 2 mol products). Yield of ammonia increases.

(b) Increasing temperature: the forward reaction is exothermic, so increasing temperature shifts equilibrium in the endothermic (backward) direction. Yield of ammonia decreases, but the rate of reaction is faster.

(c) Removing ammonia: reducing the concentration of a product shifts equilibrium to the right (forward direction) to replace the removed ammonia. Yield increases.

Worked Examples

Example 1: $K_c$ calculation

For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ , at equilibrium the concentrations are $[\text{N}_2] = 0.50\,\text{mol\,dm}^{-3}$ , $[\text{H}_2] = 1.50\,\text{mol\,dm}^{-3}$ , $[\text{NH}_3] = 0.25\,\text{mol\,dm}^{-3}$ .

Solution:

$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.25)^2}{(0.50)(1.50)^3} = \frac{0.0625}{1.6875} \approx 0.0370\,\text{mol}^{-2}\,\text{dm}^6$

Summary

This topic covers the key concepts of Chemical Reactions for GCSE Chemistry. Focus on understanding the fundamental principles, practising with exam-style questions, and applying your knowledge to unfamiliar contexts.

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Chemical Reactions