The number and arrangement of electrons in an atom decide how it behaves. The periodic table is organised so that those patterns repeat.
🔬 Inside the atom
An atom has a dense nucleus of protons (+1) and neutrons (0), surrounded by electrons (−1). The atomic number Z is the number of protons; the mass number A is protons + neutrons.
²³Na: Z = 11, A = 23 → neutrons = 23 − 11 = 12
Mg²⁺ (Z = 12): 12 protons, 12 − 2 = 10 electrons
Isotopes: same Z, different A
Cl-35 (75%) and Cl-37 (25%)
average mass = 0.75·35 + 0.25·37 = 35.5
Electrons fill shells in order. A shell n holds at most 2n² electrons (2, 8, 18, …), and the outer-shell valence electrons control bonding.
Ca (Z = 20): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² → 2 valence electrons
🖼️ See it
Atoms shrink across a period: more protons pull the same shell in tighter.
📈 Periodic trends
Moving across a period the nuclear charge grows while the shell stays the same, so electrons are pulled in more tightly. Moving down a group new shells are added, so the outer electrons are farther away and better shielded.
across a period → down a group ↓
atomic radius decreases increases
ionisation energy increases decreases
electronegativity increases decreases
Elements in the same group have the same number of valence electrons, which is why they behave alike.
🧪 Worked examples: configurations, isotopes and ionisation energy
Electron configuration. Fill the subshells in order of energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d. An s subshell holds 2 electrons, p holds 6, d holds 10.
S (Z = 16) 1s² 2s² 2p⁶ 3s² 3p⁴
Cl⁻ (Z = 17) 1s² 2s² 2p⁶ 3s² 3p⁶ (one extra electron in 3p)
Fe (Z = 26) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²
When a transition metal forms an ion the 4s electrons leave first:
Fe²⁺ … 3d⁶ Fe³⁺ … 3d⁵
Relative atomic mass from isotopes. Use a weighted average from a mass spectrum.
Successive ionisation energies. A big jump shows that the next electron comes from an inner shell, which tells you the group.
Ionisation energies (kJ/mol): 1st 578 2nd 1820 3rd 2750 4th 11 600
The big jump comes after the 3rd, so there are 3 outer electrons
→ group 13 (this is aluminium)
Two dips to remember. First ionisation energy falls from Mg to Al because the outer electron of Al is in a 3p subshell, which is higher in energy and slightly shielded by the 3s electrons. It falls from P to S because the S 3p electron is paired and repelled by its partner.
🎯 Exam focus: explaining trends
A full explanation of an ionisation energy trend needs three parts: nuclear charge, distance and shielding.
Down a group
• outer electron is in a higher shell, so farther from the nucleus
• more inner shells give more shielding
• these outweigh the larger nuclear charge → less energy needed
Across a period
• nuclear charge increases
• shielding stays about the same (same shell)
• outer electron is held more strongly → more energy needed
Common mistakes. Do not write “the electron is easier to remove because the atom is bigger” without mentioning shielding or distance from the nucleus. The mass number counts protons and neutrons, but isotopes differ only in neutrons. Cations are smaller than their atoms; anions are larger.
Definition to learn. The first ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.
✏️ Quick check · 1 of 11
Isotopes
Isotopes of an element have:
Same element means same Z.
✏️ Quick check · 2 of 11
Largest atom
Which atom has the largest atomic radius?
Think about period 3 from left to right.
✍️ Type your answer · 3 of 11
A sodium atom is ²³Na (atomic number 11).
How many neutrons does it have?
Enter an exact answer, without rounding.
neutrons = A − Z.
✍️ Type your answer · 4 of 11
A magnesium ion has the charge 2+ (Z = 12).
How many electrons does Mg²⁺ have?
Enter an exact answer, without rounding.
A positive charge means electrons were lost.
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Chlorine has two isotopes: Cl-35 (75%) and Cl-37 (25%).
What is its average atomic mass?
Enter an exact answer, without rounding.
Weighted average.
✍️ Type your answer · 6 of 11
Sulfur is in group 16.
How many valence electrons does it have?
Enter an exact answer, without rounding.
Units digit of the group number (for main groups).
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Use the 2n² rule.
What is the maximum number of electrons in shell n = 3?
Enter an exact answer, without rounding.
Max = 2n².
✍️ Type your answer · 8 of 11
Boron has two isotopes: ¹⁰B (20%) and ¹¹B (80%).
What is the relative atomic mass of boron?
Enter an exact answer, without rounding.
Weight each mass by its percentage.
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A chloride ion is written ³⁷Cl⁻ (atomic number 17).
How many neutrons does it contain?
Enter an exact answer, without rounding.
Mass number minus protons.
✏️ Quick check · 10 of 11
Mg to Al
Why is the first ionisation energy of aluminium lower than that of magnesium?
Compare the subshell of the outer electron.
✏️ Quick check · 11 of 11
Successive ionisation energies
The successive ionisation energies of an element show a very large jump between the 2nd and 3rd. In which group is the element?
Count the electrons before the jump.
📝 Lesson quiz · 5 questions
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Re-read the lesson, then give it another go.
Question 1 of 5
Moving down a group, ionisation energy generally:
Question 2 of 5
Noble gases are unreactive because they have:
Question 3 of 5
The maximum number of electrons in the 2p subshell is:
Question 4 of 5
The mass number of an atom equals:
Question 5 of 5
A positive ion (cation) forms when an atom:
Lesson 2 · Structure
🔗 Chemical Bonding & Molecular Shape
Atoms bond to reach stable electron arrangements. The type of bond and the shape of the molecule decide properties such as melting point, solubility and polarity.
🧲 Types of bonding
The difference in electronegativity (ΔEN) between two atoms tells you what kind of bond to expect.
Ionic compounds conduct electricity only when molten or dissolved (the ions can move). Metals conduct in the solid state because of the delocalised electrons.
🖼️ See it
📐 VSEPR: predicting shape
Electron domains (bonds and lone pairs) around a central atom repel each other and spread out as far as possible.
Polarity: a molecule is polar only if it has polar bonds and an asymmetric shape. CO₂ has polar bonds but is non-polar because they cancel.
🧪 Worked examples: shapes, polarity and intermolecular forces
Predicting a shape. Count the bonding pairs and the lone pairs on the central atom, then use the domain table.
PCl₃ P has 5 outer electrons: 3 used in bonds, 1 lone pair
4 domains, 1 lone pair → trigonal pyramidal, about 107°
SO₂ S has 6 outer electrons: 2 double bonds and 1 lone pair
3 domains, 1 lone pair → bent, about 119°
SF₆ 6 bonding pairs, no lone pair → octahedral, 90°
Is the molecule polar? It needs polar bonds and an asymmetric shape, so the bond dipoles do not cancel.
CCl₄ polar bonds, tetrahedral and symmetric → non-polar
CHCl₃ polar bonds, not symmetric → polar
BF₃ polar bonds, trigonal planar and symmetric → non-polar
Forces between molecules. Melting and boiling points depend on the strength of the forces between molecules, not on the covalent bonds inside them.
strongest hydrogen bond H attached to N, O or F, plus a lone pair
permanent dipole–dipole polar molecules
weakest London (instantaneous dipole) forces all molecules, stronger
with more electrons
H₂O boils at 100 °C but H₂S at −60 °C, because water forms hydrogen bonds
🎯 Exam focus: structure and properties
giant ionic high m.p.; conducts only when molten or dissolved;
brittle (layers shift and like charges repel)
simple molecular low m.p.; no ions or free electrons so does not conduct;
weak forces between molecules
giant covalent very high m.p. (many strong covalent bonds)
diamond hard; no free electrons → insulator
graphite layers with weak forces; delocalised electrons
→ conducts and is soft
metallic lattice of cations in a sea of delocalised electrons;
conducts, malleable
Common mistakes. When water boils you break hydrogen bonds between molecules, not the covalent O–H bonds. Ionic solids do not conduct because their ions cannot move, not because they have no charge. A lone pair repels more strongly than a bonding pair, which is why the angle in NH₃ (107°) is less than in CH₄ (109.5°).
Explain questions. Always connect the property to the bonding: “high melting point because of strong electrostatic attraction between oppositely charged ions in the lattice, which needs a lot of energy to overcome”.
✏️ Quick check · 1 of 11
Ionic compound
Which substance is an ionic compound?
Metal + non-metal.
✏️ Quick check · 2 of 11
Shape of water
What is the shape of an H₂O molecule?
Count the lone pairs on oxygen.
✍️ Type your answer · 3 of 11
Count valence electrons: N has 5 and each H has 1.
How many valence electrons are in one NH₃ molecule?
Enter an exact answer, without rounding.
Add the valence electrons of each atom.
✍️ Type your answer · 4 of 11
Draw the Lewis structure of H₂O.
How many lone pairs are on the oxygen atom?
Enter an exact answer, without rounding.
6 − 2 bonding electrons = 4 electrons.
✍️ Type your answer · 5 of 11
Consider methane, CH₄.
How many electron domains surround the carbon atom?
Enter an exact answer, without rounding.
Count bonds and lone pairs on carbon.
✍️ Type your answer · 6 of 11
Consider carbon dioxide, CO₂.
What is the O=C=O bond angle, in degrees?
Enter an exact answer, without rounding.
°
Linear molecules have the largest possible angle.
✍️ Type your answer · 7 of 11
Consider N₂ with a triple bond.
How many electrons are shared in the N≡N bond?
Enter an exact answer, without rounding.
Each bond is a shared pair.
✍️ Type your answer · 8 of 11
Carbon dioxide has the structure O=C=O.
How many valence electrons are there in total in one CO₂ molecule? (C has 4 and each O has 6.)
Enter an exact answer, without rounding.
Add the outer electrons of all three atoms.
✍️ Type your answer · 9 of 11
A water molecule has a central oxygen atom with two O–H bonds.
How many lone pairs of electrons are on the oxygen atom?
Enter an exact answer, without rounding.
6 outer electrons, minus 2 shared.
✏️ Quick check · 10 of 11
Hydrogen bonding
Which compound can form hydrogen bonds between its molecules?
Look for N–H, O–H or F–H.
✏️ Quick check · 11 of 11
Graphite
Why does graphite conduct electricity?
Compare with diamond.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The larger the electronegativity difference between two atoms, the bond is more:
Question 2 of 5
Water has an unusually high boiling point for its size because of:
Question 3 of 5
Which statement about CO₂ is true?
Question 4 of 5
Metals conduct electricity because they have:
Question 5 of 5
Which intermolecular force is present in all molecules?
Lesson 3 · Quantities
🧪 The Mole & Stoichiometry
The mole lets chemists count particles by weighing. Balanced equations then tell you exactly how much of each substance reacts and forms.
⚖️ Moles and molar mass
One mole contains 6.02 × 10²³ particles (Avogadro's number). The molar mass M of a substance is the mass of one mole in g/mol, found by adding the atomic masses.
n = m / M
M(H₂O) = 2·1 + 16 = 18 g/mol
M(CO₂) = 12 + 2·16 = 44 g/mol
M(MgO) = 24 + 16 = 40 g/mol
36 g of H₂O → n = 36 / 18 = 2 mol
0.5 mol of CO₂ → m = 0.5 · 44 = 22 g
% Mg in MgO = 24 / 40 · 100 = 60%
🖼️ See it
Always go through moles: grams cannot be compared directly, but moles can be read from the balanced equation.
🧮 Using balanced equations
The coefficients in a balanced equation are mole ratios.
2 H₂ + O₂ → 2 H₂O
4 g of H₂: n = 4 / 2 = 2 mol
ratio 2 H₂ : 2 H₂O → 2 mol H₂O
m = 2 · 18 = 36 g
Limiting reactant: 3 mol H₂ with 1 mol O₂
O₂ needs 2 mol H₂ → H₂ is in excess, O₂ limits
product = 2 mol H₂O
% yield = actual / theoretical · 100
8 g obtained, 10 g possible → 80%
Always convert to moles first. Equations balance moles, not grams.
🧪 Worked examples: formulae, titrations and atom economy
Empirical formula from percentages. Divide each percentage by its atomic mass, then divide by the smallest answer.
A compound is 40.0% C, 6.7% H and 53.3% O
C: 40.0 ÷ 12 = 3.33 H: 6.7 ÷ 1 = 6.7 O: 53.3 ÷ 16 = 3.33
ratio ÷ 3.33 → C 1 : H 2 : O 1 → empirical formula CH₂O
M = 180, empirical mass = 12 + 2 + 16 = 30
180 ÷ 30 = 6 → molecular formula C₆H₁₂O₆
A titration calculation. Always follow the same four steps: moles of the known solution, ratio, moles of the unknown, concentration.
Atom economy. This shows how much of the reactant mass ends up in the product you want.
atom economy = M(desired product) ÷ sum of M(all products) × 100
CaCO₃ → CaO + CO₂
M(CaCO₃) = 40 + 12 + 48 = 100 M(CaO) = 40 + 16 = 56
atom economy = 56 ÷ 100 × 100 = 56%
Units. Convert cm³ to dm³ by dividing by 1000 before using n = c × V.
🎯 Exam focus: setting out calculations
Marks are given for method as well as the final answer. Write one line per step.
1 write the balanced equation
2 n = m ÷ M or n = c × V for the substance you know
3 use the mole ratio from the equation
4 convert back: m = n × M, c = n ÷ V, or V = n × 24 dm³ (gas at r.t.p.)
5 give the answer with the correct unit and significant figures
Common mistakes. Using grams in a ratio instead of moles. Forgetting that the coefficients refer to moles of each formula unit (2H₂O means 2 mol of water, so 4 mol of H atoms). Writing “moles of atoms” when the question asks about molecules. Giving percentage yield above 100%.
Significant figures. Give your answer to the same number of significant figures as the least precise data. Do not round in the middle of a calculation; round only at the end.
✏️ Quick check · 1 of 12
Coefficients
In a balanced chemical equation, the coefficients give the ratio of:
3 mol of H₂ is mixed with 1 mol of O₂ (2H₂ + O₂ → 2H₂O).
How many moles of water form at most?
Enter an exact answer, without rounding.
mol
Check which reactant runs out first.
✍️ Type your answer · 7 of 12
A reaction should give 10 g of product but only 8 g is obtained.
What is the percentage yield?
Enter an exact answer, without rounding.
%
actual ÷ theoretical.
✍️ Type your answer · 8 of 12
Use Mg = 24 and O = 16.
What is the percentage by mass of magnesium in MgO?
Enter an exact answer, without rounding.
%
part ÷ whole.
✍️ Type your answer · 9 of 12
A compound has the empirical formula CH₂O and a relative molecular mass of 180 (C 12, H 1, O 16).
How many carbon atoms are in one molecule?
Enter an exact answer, without rounding.
Find how many empirical units fit in 180.
✍️ Type your answer · 10 of 12
25.0 cm³ of 0.100 mol/dm³ NaOH is neutralised exactly by 20.0 cm³ of hydrochloric acid (NaOH + HCl → NaCl + H₂O).
What is the concentration of the acid, in mol/dm³?
Enter an exact answer, without rounding.
mol/dm³
The ratio is 1 : 1.
✍️ Type your answer · 11 of 12
Calcium carbonate decomposes: CaCO₃ → CaO + CO₂ (Ca 40, C 12, O 16).
What is the atom economy for making CaO, as a percentage?
Enter an exact answer, without rounding.
%
Mass of wanted product ÷ total mass of products.
✏️ Quick check · 12 of 12
Counting atoms
Which sample contains the greatest number of atoms?
Multiply by the number of atoms in each formula.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Avogadro's number is approximately:
Question 2 of 5
The molar mass of NaCl is (Na = 23, Cl = 35.5):
Question 3 of 5
Theoretical yield is:
Question 4 of 5
The reactant that is not completely used up is called the:
Question 5 of 5
2 mol of H₂O contains how many moles of hydrogen atoms?
Lesson 4 · Quantities
💨 Gases & Solutions
Gas behaviour follows simple laws, and the concentration of a solution tells you how much solute is dissolved in a given volume.
🎈 Gas laws
For a fixed amount of an ideal gas the combined law links pressure, volume and kelvin temperature.
P₁V₁ / T₁ = P₂V₂ / T₂
Avogadro: 1 mol of any gas at STP (0 °C, 1 atm) occupies 22.4 L
2 mol at STP → 2 · 22.4 = 44.8 L
Dalton: total pressure = sum of partial pressures
2 atm N₂ + 1 atm O₂ → 3 atm
1 atm, 2 L, 300 K → 2 atm, T = 600 K
V₂ = (1 · 2 · 600) / (300 · 2) = 2 L
Always use kelvin (T = °C + 273) in gas-law calculations.
🖼️ See it
Doubling the volume halves the pressure, so pV stays at 60 kPa·dm³ for this sample of gas.
🧂 Concentration and dilution
Molarity (mol/L) = moles of solute ÷ litres of solution.
M = n / V
0.5 mol in 250 mL (0.25 L): M = 0.5 / 0.25 = 2 M
Dilution: the moles of solute do not change
M₁V₁ = M₂V₂
50 mL of 2 M diluted to 200 mL:
M₂ = 2 · 50 / 200 = 0.5 M
Mass for 1 L of 0.5 M NaCl:
m = 0.5 · 58.5 = 29.25 g
🧪 Worked examples: mixtures, concentration and non-ideal gases
Mole fraction and partial pressure. The partial pressure of a gas is its mole fraction × the total pressure.
A mixture has 2 mol N₂ and 3 mol O₂ at a total pressure of 5 atm
mole fraction of N₂ = 2 ÷ 5 = 0.4
partial pressure of N₂ = 0.4 × 5 = 2 atm
partial pressure of O₂ = 0.6 × 5 = 3 atm (they add up to 5 atm)
The ideal gas equation. pV = nRT with p in Pa, V in m³, T in K and R = 8.31 J mol⁻¹ K⁻¹.
How many moles of gas are in 2.0 × 10⁻³ m³ (2.0 dm³) at 1.0 × 10⁵ Pa and 300 K?
n = pV ÷ RT = (1.0 × 10⁵ × 2.0 × 10⁻³) ÷ (8.31 × 300)
= 200 ÷ 2493 = 0.080 mol
From mass to concentration. Convert the mass to moles, then divide by the volume in dm³.
5.85 g of NaCl is dissolved to make 500 cm³ of solution (M = 58.5)
n = 5.85 ÷ 58.5 = 0.100 mol
V = 500 ÷ 1000 = 0.500 dm³
c = 0.100 ÷ 0.500 = 0.200 mol/dm³
Real gases. Real gases behave most like an ideal gas at low pressure and high temperature. At high pressure the particles are close, so their own volume matters and the attractions between them are not negligible.
🎯 Exam focus: assumptions and practical skills
Assumptions of an ideal gas
• particles have negligible volume compared with the container
• no forces of attraction between particles
• collisions are perfectly elastic
• random motion; kinetic energy ∝ temperature in kelvin
Making a standard solution. Weigh the solid by difference, dissolve it in a beaker with some distilled water, transfer it (with rinsings) to a volumetric flask, fill to the mark with the bottom of the meniscus on the line, then invert to mix.
Common mistakes. Using °C instead of K in a gas-law calculation. Using the volume of the solvent instead of the volume of the solution. Forgetting to convert cm³ to dm³. Using 22.4 dm³ at room temperature: that is the value at s.t.p.; at r.t.p. the molar volume is 24 dm³.
Dilution. The number of moles does not change when you add water, so c₁V₁ = c₂V₂. Use any volume unit, as long as it is the same on both sides.
✏️ Quick check · 1 of 12
Dalton's law
The total pressure of a mixture of gases equals:
Dalton's law of partial pressures.
✏️ Quick check · 2 of 12
Diluting
When water is added to a solution, the amount (moles) of solute:
You added water, not solute.
✍️ Type your answer · 3 of 12
Use 22.4 L/mol at STP.
What volume do 2 mol of gas occupy at STP, in L?
Enter an exact answer, without rounding.
L
V = n × 22.4.
✍️ Type your answer · 4 of 12
A gas mixture contains N₂ at 2 atm and O₂ at 1 atm.
What is the total pressure, in atm?
Enter an exact answer, without rounding.
atm
Add the partial pressures.
✍️ Type your answer · 5 of 12
A gas is at 1 atm, 2 L and 300 K, then changed to 2 atm and 600 K.
What is the new volume, in L?
Enter an exact answer, without rounding.
L
Use P₁V₁/T₁ = P₂V₂/T₂.
✍️ Type your answer · 6 of 12
0.5 mol of solute is dissolved to make 250 mL of solution.
What is the molarity, in mol/L?
Enter an exact answer, without rounding.
M
Convert mL to L first.
✍️ Type your answer · 7 of 12
50 mL of 2 M solution is diluted to 200 mL.
What is the new concentration, in M?
Enter an exact answer, without rounding.
M
M₁V₁ = M₂V₂.
✍️ Type your answer · 8 of 12
Use M(NaCl) = 58.5 g/mol.
What mass of NaCl is needed for 1 L of 0.5 M solution, in g?
Enter an exact answer, without rounding.
g
Find the moles first.
✍️ Type your answer · 9 of 12
A gas mixture contains 2 mol of N₂ and 3 mol of O₂ at a total pressure of 5 atm.
What is the partial pressure of N₂, in atm?
Enter an exact answer, without rounding.
atm
Mole fraction × total pressure.
✍️ Type your answer · 10 of 12
5.85 g of sodium chloride (M = 58.5 g/mol) is dissolved in water to make 500 cm³ of solution.
What is the concentration, in mol/dm³?
Enter an exact answer, without rounding.
mol/dm³
Convert to moles, and cm³ to dm³.
✍️ Type your answer · 11 of 12
A gas occupies 4.0 dm³ at 1 atm. The temperature stays constant and the pressure is increased to 2 atm.
What is the new volume, in dm³?
Enter an exact answer, without rounding.
dm³
At constant temperature, pV is constant.
✏️ Quick check · 12 of 12
Real gases
Under which conditions does a real gas behave most like an ideal gas?
Think of widely spaced, fast particles.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Standard temperature and pressure (STP) is:
Question 2 of 5
In a solution, the substance that is dissolved is called the:
Question 3 of 5
If a gas's volume is halved at constant temperature, its pressure:
Question 4 of 5
Which is an assumption of the ideal gas model?
Question 5 of 5
How many moles of NaOH are in 2 L of a 0.25 M solution?
Lesson 5 · Energy
🔥 Thermochemistry
Chemical reactions absorb or release energy. Enthalpy change ΔH measures that heat at constant pressure.
🌡️ Exothermic, endothermic and calorimetry
An exothermic reaction releases heat to the surroundings (ΔH < 0, temperature rises). An endothermic reaction absorbs heat (ΔH > 0, temperature falls).
q = m c ΔT c(water) = 4.2 J/(g·K)
100 g of water warmed by 20 K:
q = 100 · 4.2 · 20 = 8400 J
q = 4200 J into 100 g of water:
ΔT = 4200 / (100 · 4.2) = 10 K
CH₄ + 2O₂ → CO₂ + 2H₂O ΔH = −890 kJ/mol
16 g CH₄ = 1 mol → 890 kJ released
🖼️ See it
Products sit lower than reactants, so energy is released to the surroundings.
🧩 Hess's law and bond energies
Hess's law: the total enthalpy change is the same whichever route is taken, so you can add the ΔH values of steps.
Why the real value differs. A calorimeter loses heat to the surroundings, so the measured temperature rise is too small and the measured |ΔH| is too low. Bond energies are averages, so a bond-energy ΔH is only approximate.
🎯 Exam focus: definitions and energy profiles
standard enthalpy change of formation, ΔHf°
the enthalpy change when 1 mol of a compound is formed from its
elements in their standard states, at 298 K and 100 kPa
standard enthalpy change of combustion, ΔHc°
the enthalpy change when 1 mol of a substance is completely
burned in oxygen under standard conditions
On an energy profile label the reactants, the products, the activation energy Eₐ (reactants up to the peak), and ΔH (reactants to products). For an exothermic reaction the products are lower than the reactants.
Common mistakes. Forgetting the sign: exothermic ΔH is negative. Dividing by the mass of the solution instead of the moles that reacted. Using the mass of the solid instead of the mass of the solution in q = mcΔT. Writing ΔHf for an element other than zero.
Hess cycle tip. Draw the cycle with the reactants and products at the top and the elements (or combustion products) at the bottom, and follow the arrows with the sign of each step.
✏️ Quick check · 1 of 11
Sign of ΔH
For an exothermic reaction, ΔH is:
Heat is released.
✏️ Quick check · 2 of 11
Hess's law
Hess's law states that the enthalpy change of a reaction:
Only start and end states matter.
✍️ Type your answer · 3 of 11
100 g of water is heated by 20 K (c = 4.2 J/(g·K)).
How much heat is absorbed, in J?
Enter an exact answer, without rounding.
J
q = mcΔT.
✍️ Type your answer · 4 of 11
4200 J of heat is absorbed by 100 g of water (c = 4.2 J/(g·K)).
How much energy is released, in kJ, for the equation as written?
Enter an exact answer, without rounding.
kJ
Formed minus broken.
✏️ Quick check · 11 of 11
Standard states
What is the standard enthalpy change of formation of O₂(g)?
It is formed from itself.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
A reaction vessel gets colder during a reaction. The reaction is:
Question 2 of 5
Breaking chemical bonds:
Question 3 of 5
The standard enthalpy of formation of an element in its standard state is:
Question 4 of 5
Specific heat capacity is the energy needed to raise the temperature of:
Question 5 of 5
In an energy profile for an exothermic reaction, the products are:
Lesson 6 · Kinetics
⏱️ Reaction Rates
Some reactions are over in an instant, others take years. Collision theory explains what controls the speed and how to change it.
💥 Collision theory
For particles to react they must collide with enough energy (at least the activation energy Ea) and with the right orientation.
rate = change in concentration / time
[product] rises from 0.20 M to 0.60 M in 20 s
rate = 0.40 / 20 = 0.02 M/s
Ways to speed up a reaction
higher concentration or pressure → more collisions per second
higher temperature → more particles exceed Eₐ
larger surface area → more exposed particles
catalyst → alternative path with lower Eₐ
🖼️ See it
The lower peak means more particles have enough energy to react, so the rate rises. ΔH is unchanged.
📉 Rate laws and half-life
A rate law links the rate to concentrations: rate = k[A]ⁿ, where the order n must be found by experiment.
rate = k[A]²
double [A] → rate × 2² = 4
Rule of thumb: rate roughly doubles for every +10 K
300 K → 330 K: three doublings, 2³ = 8 times faster
First-order half-life: constant
half-life 20 s
time to fall to 1/8 = 3 half-lives = 60 s
80 g after 3 half-lives: 80 → 40 → 20 → 10 g
A catalyst speeds up a reaction but is not used up and does not change the equilibrium position or ΔH.
🧪 Worked examples: initial rates, order and half-life
Rate from a graph. The rate at any moment is the gradient of the concentration–time (or volume–time) curve. For a straight line, divide the change by the time.
gas collected: 0 → 40 cm³ in the first 20 s
initial rate = 40 ÷ 20 = 2.0 cm³/s
For a curve draw a tangent and find its gradient:
rate = Δy ÷ Δx
Finding the order from experiments. Change one concentration at a time and see how the rate responds.
Experiment [A] mol/dm³ initial rate (mol dm⁻³ s⁻¹)
1 0.10 0.0020
2 0.20 0.0080
[A] doubled and the rate × 4 → second order in A: rate = k[A]²
k = 0.0020 ÷ 0.10² = 0.20 mol⁻¹ dm³ s⁻¹
Half-life. For a first-order reaction the half-life is constant, so the fraction left depends only on the number of half-lives.
t½ = 10 min. After 30 min = 3 half-lives
fraction left = (½)³ = 1/8 = 12.5%
start 80 g → 40 → 20 → 10 g
Why a catalyst helps. A catalyst gives the reaction an alternative route with a lower activation energy. More particles then have at least Eₐ at the same temperature, but the distribution of energies is unchanged.
🎯 Exam focus: Maxwell–Boltzmann curves
The curve shows the number of particles with each energy. It starts at the origin, is skewed to the right, and never touches the axis again. The area under the curve is the total number of particles.
raising the temperature
• peak moves right and gets lower (area stays the same)
• many more particles have energy ≥ Eₐ
adding a catalyst
• curve unchanged
• Eₐ marked further left, so more particles to the right of it
Common mistakes. Saying “particles collide more often” as the only reason for a faster reaction at a higher temperature: the main reason is that a much greater fraction of collisions has energy ≥ Eₐ. Writing that a catalyst “gives the particles more energy”. Calculating the order from the balanced equation: it must come from experimental data.
Sketch tip. On a concentration–time graph of a reactant the curve falls steeply and flattens as it is used up. The gradient at t = 0 is the initial rate.
✏️ Quick check · 1 of 11
Catalyst
A catalyst increases the reaction rate by:
Think about Eₐ.
✏️ Quick check · 2 of 11
Temperature
Raising the temperature increases the rate mainly because:
Particle energy, not Eₐ, changes.
✍️ Type your answer · 3 of 11
A product's concentration rises from 0.20 M to 0.60 M in 20 s.
What is the average rate, in M/s?
Enter an exact answer, without rounding.
M/s
Change in concentration ÷ time.
✍️ Type your answer · 4 of 11
For rate = k[A]², the concentration of A is doubled.
By what factor does the rate increase?
Enter an exact answer, without rounding.
Square the factor.
✍️ Type your answer · 5 of 11
A first-order reaction has a half-life of 20 s.
How long does it take for the amount to fall to 1/8, in s?
Enter an exact answer, without rounding.
s
Count half-lives: 1/2, 1/4, 1/8.
✍️ Type your answer · 6 of 11
A rule of thumb says the rate doubles for every 10 K rise.
By what factor does the rate increase from 300 K to 330 K?
Enter an exact answer, without rounding.
How many 10 K steps are there?
✍️ Type your answer · 7 of 11
A first-order reaction starts with 80 g of reactant.
How many grams remain after 3 half-lives?
Enter an exact answer, without rounding.
g
Halve it 3 times.
✍️ Type your answer · 8 of 11
In an experiment 40 cm³ of gas is collected in the first 20 s, at a steady rate.
What is the initial rate, in cm³/s?
Enter an exact answer, without rounding.
cm³/s
Volume divided by time.
✍️ Type your answer · 9 of 11
When [A] is doubled from 0.10 to 0.20 mol/dm³ the initial rate increases from 0.0020 to 0.0080 mol dm⁻³ s⁻¹.
What is the order of the reaction with respect to A?
Enter an exact answer, without rounding.
Rate factor = concentration factor raised to the order.
✍️ Type your answer · 10 of 11
A first-order reaction has a half-life of 10 minutes.
What percentage of the reactant is left after 30 minutes?
Enter an exact answer, without rounding.
%
How many half-lives fit into 30 min?
✏️ Quick check · 11 of 11
Catalyst and energies
How does a catalyst increase the rate of a reaction at constant temperature?
The temperature has not changed.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Activation energy is:
Question 2 of 5
Crushing a solid into powder speeds up a reaction because it:
Question 3 of 5
The order of a reaction with respect to a reactant is found:
Question 4 of 5
For a zero-order reaction, the rate:
Question 5 of 5
Which of these does a catalyst change?
Lesson 7 · Equilibrium
⚖️ Chemical Equilibrium
Many reactions do not go to completion: they reach a balance where the forward and reverse reactions continue at equal rates.
🔄 Dynamic equilibrium and K
At equilibrium the rates of the forward and reverse reactions are equal, so concentrations stay constant. Nothing has stopped: it is dynamic.
K ≫ 1 means products are favoured; K ≪ 1 means reactants are favoured. K changes only with temperature. Pure solids and liquids are left out of the expression.
🖼️ See it
At equilibrium the concentrations stop changing because forward and reverse rates are equal.
🎛️ Le Chatelier's principle
If a system at equilibrium is disturbed, it shifts to oppose the change.
Add reactant / remove product → shifts right (more product)
Increase pressure → shifts to the side with fewer gas moles
Increase temperature → shifts in the endothermic direction
Add a catalyst → no shift (equilibrium reached faster)
N₂ + 3H₂ ⇌ 2NH₃: 4 mol gas on the left, 2 on the right
higher pressure shifts right
ICE table: A ⇌ B, K_c = 1, start [A] = 1.0 M, [B] = 0
change: −x, +x
K_c = x / (1 − x) = 1 → x = 0.5
at equilibrium [A] = [B] = 0.5 M
🧪 Worked examples: ICE tables and K for different reactions
ICE method: write the Initial amounts, the Change, and the Equilibrium amounts.
A + B ⇌ C start: [A] = [B] = 1.0 mol/dm³, [C] = 0
at equilibrium [C] = 0.5, so A and B each fell by 0.5
[A] = [B] = 0.5, [C] = 0.5
K_c = [C] ÷ ([A][B]) = 0.5 ÷ (0.5 × 0.5) = 2 mol⁻¹ dm³
Temperature and K. K is changed only by temperature. For an exothermic forward reaction a higher temperature shifts the position to the left and K decreases. For an endothermic forward reaction, K increases.
🎯 Exam focus: industrial compromise
The Haber process N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ/mol) shows the compromise between yield and rate.
pressure high pressure favours ammonia (4 mol gas → 2 mol gas)
but costs more and needs stronger equipment: about 200 atm
temperature low temperature favours ammonia (exothermic)
but the rate is too slow: about 450 °C
catalyst iron: speeds up both directions equally,
so it gives the same yield, sooner
recycling unreacted N₂ and H₂ are returned to the reactor
Common mistakes. Saying the reaction “stops” at equilibrium: both reactions continue at equal rates. Saying a catalyst changes the position of equilibrium or K. Including solids or pure liquids in the K_c expression. Forgetting the units of K_c, which depend on the powers in the expression.
Le Chatelier answers. Always give the direction of shift and the reason: “increasing the pressure shifts the position to the right because there are fewer moles of gas on the right-hand side”.
✏️ Quick check · 1 of 11
Catalyst
Adding a catalyst to a reaction at equilibrium will:
It does not favour either direction.
✏️ Quick check · 2 of 11
Le Chatelier
Adding more reactant to a system at equilibrium causes the equilibrium to:
Oppose the change.
✍️ Type your answer · 3 of 11
For A + B ⇌ C: [A] = 0.5 M, [B] = 0.2 M and [C] = 0.4 M.
What is K_c?
Enter an exact answer, without rounding.
Products over reactants.
✍️ Type your answer · 4 of 11
For 2NO₂ ⇌ N₂O₄: [NO₂] = 0.2 M and [N₂O₄] = 0.08 M.
What is K_c?
Enter an exact answer, without rounding.
Square the concentration of NO₂.
✍️ Type your answer · 5 of 11
For A + B ⇌ C with K_c = 4, [A] = [B] = 0.5 M at equilibrium.
What is [C], in M?
Enter an exact answer, without rounding.
M
Rearrange the K_c expression.
✍️ Type your answer · 6 of 11
For N₂ + 3H₂ ⇌ 2NH₃.
How many moles of gas are on the product side per equation?
Enter an exact answer, without rounding.
Read the coefficient.
✍️ Type your answer · 7 of 11
A ⇌ B with K_c = 1. Start with [A] = 1.0 M and [B] = 0.
What is [B] at equilibrium, in M?
Enter an exact answer, without rounding.
M
Use an ICE table.
✍️ Type your answer · 8 of 11
For A + B ⇌ C the equilibrium concentrations are [A] = 0.5, [B] = 0.5 and [C] = 0.5 mol/dm³.
What is the value of K_c?
Enter an exact answer, without rounding.
Products over reactants.
✍️ Type your answer · 9 of 11
For N₂O₄ ⇌ 2NO₂ the equilibrium concentrations are [NO₂] = 0.4 mol/dm³ and [N₂O₄] = 0.2 mol/dm³.
What is the value of K_c?
Enter an exact answer, without rounding.
Remember to square [NO₂].
✍️ Type your answer · 10 of 11
A ⇌ B has K_c = 3. The starting concentration of A is 1.0 mol/dm³ and there is no B at the start.
What is [B] at equilibrium, in mol/dm³?
Enter an exact answer, without rounding.
mol/dm³
Let [B] = x and [A] = 1 − x.
✏️ Quick check · 11 of 11
Temperature and K
For an exothermic forward reaction, what happens to K_c when the temperature is raised?
The equilibrium moves in the endothermic direction.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
A very large K_c means the equilibrium mixture contains mostly:
Question 2 of 5
K_c for a reaction changes when:
Question 3 of 5
If the reaction quotient Q is greater than K, the reaction proceeds:
Question 4 of 5
At equilibrium:
Question 5 of 5
Pure solids are left out of the K_c expression because:
Lesson 8 · Redox
🔋 Redox Reactions
Redox reactions transfer electrons. They power batteries, rust iron and release energy in your cells.
🔢 Oxidation numbers
OIL RIG:Oxidation Is Loss of electrons, Reduction Is Gain. Oxidation numbers help you track the electrons.
Rules: elements = 0; simple ion = its charge; O is usually −2; H is usually +1
the numbers in a compound add up to its charge
KMnO₄: +1 + Mn + 4(−2) = 0 → Mn = +7
H₂SO₄: 2(+1) + S + 4(−2) = 0 → S = +6
Fe₂O₃: 2Fe + 3(−2) = 0 → Fe = +3
K₂Cr₂O₇: 2(+1) + 2Cr + 7(−2) = 0 → Cr = +6
H₂O₂ (peroxide): O = −1
🖼️ See it
⚡ Half-reactions and agents
Split a redox reaction into two half-reactions, one for oxidation and one for reduction. The electrons lost must equal the electrons gained.
Zn + Cu²⁺ → Zn²⁺ + Cu
oxidation: Zn → Zn²⁺ + 2e⁻ (Zn is the reducing agent)
reduction: Cu²⁺ + 2e⁻ → Cu (Cu²⁺ is the oxidising agent)
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Mn goes from +7 to +2: 5 electrons gained
The reducing agent is the thing that gets oxidised (it gives away electrons); the oxidising agent is the thing that gets reduced.
🧪 Worked examples: half-equations, titration and cell potential
Combining half-equations. Balance atoms, then charge with electrons, then multiply so the electrons cancel.
Standard electrode potentials. The more positive E° belongs to the species that is more easily reduced.
Zn²⁺ + 2e⁻ ⇌ Zn E° = −0.76 V
Cu²⁺ + 2e⁻ ⇌ Cu E° = +0.34 V
E°cell = E°(positive electrode) − E°(negative electrode)
= +0.34 − (−0.76) = +1.10 V
E°cell > 0 → the reaction Zn + Cu²⁺ → Zn²⁺ + Cu is feasible
Disproportionation. A single element is oxidised and reduced at the same time, for example Cl₂ + 2NaOH → NaCl + NaClO + H₂O. Chlorine goes from 0 to −1 and from 0 to +1.
🎯 Exam focus: oxidation numbers and agents
Use oxidation numbers to decide whether a reaction is redox: look for any element whose number changes.
rules uncombined element 0; simple ion = its charge;
O is −2 (−1 in peroxides); H is +1 (−1 in metal hydrides);
the numbers add up to the charge on the species
Cr₂O₇²⁻ 2Cr + 7(−2) = −2 → 2Cr = +12 → Cr = +6
SO₄²⁻ S + 4(−2) = −2 → S = +6
Common mistakes. Mixing up “oxidising agent” and “oxidised”: the oxidising agent is the species that gets reduced. Forgetting to write the roman numeral in names: iron(II), manganate(VII). Mixing up the two styles: oxidation numbers are written with the sign first (+6, −2) and ion charges with the sign last (2−, 3+). Reversing the E°cell subtraction.
Remember. A positive E°cell means the reaction is feasible under standard conditions, but feasible does not mean fast: a high activation energy can still stop it happening.
✏️ Quick check · 1 of 12
OIL RIG
In a redox reaction, reduction is:
The second word of the mnemonic.
✏️ Quick check · 2 of 12
Reducing agent
A reducing agent is a substance that:
It donates electrons.
✍️ Type your answer · 3 of 12
Consider KMnO₄.
What is the oxidation number of Mn?
Enter an exact answer, without rounding.
K is +1 and O is −2.
✍️ Type your answer · 4 of 12
Consider H₂SO₄.
What is the oxidation number of S?
Enter an exact answer, without rounding.
The total must be 0.
✍️ Type your answer · 5 of 12
Consider Fe₂O₃.
What is the oxidation number of Fe?
Enter an exact answer, without rounding.
Two iron atoms share +6.
✍️ Type your answer · 6 of 12
Consider K₂Cr₂O₇.
What is the oxidation number of Cr?
Enter an exact answer, without rounding.
Two chromium atoms together must be +12.
✍️ Type your answer · 7 of 12
Zn → Zn²⁺ + ?e⁻
How many electrons are lost per Zn atom?
Enter an exact answer, without rounding.
Match the charge.
✍️ Type your answer · 8 of 12
Consider hydrogen peroxide, H₂O₂.
What is the oxidation number of oxygen?
Enter an exact answer, without rounding.
It is a peroxide, not an ordinary oxide.
✍️ Type your answer · 9 of 12
25.0 cm³ of an Fe²⁺ solution needs 20.0 cm³ of 0.0200 mol/dm³ KMnO₄ for complete oxidation. The ratio is MnO₄⁻ : Fe²⁺ = 1 : 5.
What is the concentration of the Fe²⁺ solution, in mol/dm³?
Enter an exact answer, without rounding.
mol/dm³
Use the 1 : 5 ratio between the moles.
✍️ Type your answer · 10 of 12
E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V.
What is the standard cell potential E°cell of the zinc–copper cell, in volts?
Enter an exact answer, without rounding.
V
More positive minus more negative.
✍️ Type your answer · 11 of 12
The dichromate ion is Cr₂O₇²⁻ and oxygen is −2.
What is the oxidation number of chromium?
Enter an exact answer, without rounding.
The oxidation numbers add up to the charge on the ion.
✏️ Quick check · 12 of 12
Disproportionation
Which statement describes disproportionation?
One substance, two changes.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The oxidation number of an uncombined element (such as O₂ or Fe) is:
Question 2 of 5
In 2Mg + O₂ → 2MgO, magnesium is:
Question 3 of 5
The oxidation number of oxygen in most compounds is:
Question 4 of 5
In a redox reaction, the number of electrons lost:
Question 5 of 5
Rusting of iron is an example of:
Final Challenge
🏆 16-Question Challenge Quiz
Questions from every topic in this course. Pick an answer for each one, then hit Check My Answers.
0 / 16
💪 Keep practicing — go back through the lessons and retake the quiz!
Q1 · Atomic Structure & Periodic Trends
Which element has the highest electronegativity?
Q2 · Atomic Structure & Periodic Trends
A positive ion (cation) forms when an atom:
Q3 · Chemical Bonding & Molecular Shape
The shape of a methane (CH₄) molecule is:
Q4 · Chemical Bonding & Molecular Shape
Which intermolecular force is present in all molecules?
Q5 · The Mole & Stoichiometry
How many moles are in 88 g of CO₂ (M = 44 g/mol)?
Q6 · The Mole & Stoichiometry
2 mol of H₂O contains how many moles of hydrogen atoms?
Q7 · Gases & Solutions
1 mol of an ideal gas at STP occupies:
Q8 · Gases & Solutions
How many moles of NaOH are in 2 L of a 0.25 M solution?
Q9 · Thermochemistry
In an exothermic reaction the products have:
Q10 · Thermochemistry
In an energy profile for an exothermic reaction, the products are:
Q11 · Reaction Rates
A catalyst makes a reaction faster by:
Q12 · Reaction Rates
Which of these does a catalyst change?
Q13 · Chemical Equilibrium
For N₂ + 3H₂ ⇌ 2NH₃, increasing the pressure shifts the equilibrium:
Q14 · Chemical Equilibrium
Pure solids are left out of the K_c expression because: