MYP Chemistry · Topic 6
Bonding
How atoms hold together decides how substances behave. Count the atoms, count the moles, and follow the energy and the collisions.
Topic: Bonding (structure and bonding, properties, chemical formulas, chemical reactions and the conservation of mass; balancing equations, the mole concept and chemical calculations; reaction kinetics [rates, and factors affecting rates/collision theory]; equilibria/reversible reactions; energy changes in reactions, endo- and exothermicity; combustion of fuels). This is the largest topic in the course. Part 1 explains the three types of bonding and the properties they produce; Part 2 turns formulas and equations into quantities with the mole; Part 3 explains how fast reactions go and why some are reversible; Part 4 follows the energy.
Prior knowledge: electron configuration and valency (Topic 4); metals and non-metals (Topic 1).
Learning objectives
Objectives and contextOverview
By the end of this chapter you should be able to:
- describe ionic, covalent and metallic bonding, draw Lewis (dot-and-cross) structures and explain the properties of each type of structure A
- deduce formulas of ionic compounds and balance equations, applying conservation of mass A
- calculate molar masses, amounts in moles and reacting quantities, in standard form and to a stated number of significant figures A C
- explain how concentration, temperature, surface area and catalysts affect rate using collision theory, and calculate rates A B C
- describe reversible reactions and dynamic equilibrium A
- classify reactions as exothermic or endothermic, interpret energy profiles and compare fuels using q = mcΔT A C D
Salt dissolves in water and the solution conducts electricity; sugar dissolves but the solution does not. Sand does not dissolve at all and melts only in a furnace. Copper bends without breaking and carries current. These are four white-to-brown solids that could look similar in a jar, yet they behave completely differently — because their particles are held together in different ways. This chapter starts from that idea and builds towards the quantitative chemistry of reactions: how much, how fast, how far and with how much energy.
02 / Ionic bonding
Ionic bondingStructure and bonding
When a metal reacts with a non-metal, electrons are transferred from the metal atoms to the non-metal atoms. The metal atoms lose their outer electrons to form positive ions; the non-metal atoms gain electrons to form negative ions. Both usually end up with a full outer shell, like a noble gas. The oppositely charged ions attract each other strongly in all directions: this electrostatic attraction is the ionic bond.
Magnesium (2,8,2) reacting with chlorine loses two electrons, one to each of two chlorine atoms, so magnesium chloride is MgCl2. In the solid, millions of ions are arranged in a regular three-dimensional giant ionic lattice, each ion surrounded by ions of the opposite charge.
Properties follow from the structure. Ionic compounds have high melting points, because a great deal of energy is needed to overcome the strong attractions between the many ions. They conduct electricity when molten or dissolved in water, because the ions are then free to move and carry charge, but not when solid, because the ions are fixed in the lattice. Many dissolve in water. They are brittle: a blow shifts the layers so that ions of the same charge line up and repel.
Two links are needed: sodium chloride is ionic, so dissolving it in water releases ions (Na+ and Cl−) → the ions are free to move and carry charge, so the solution conducts. Saying “salt water conducts” or “the salt has electrons” does not explain it.
03 / Covalent bonding
Covalent bondingStructure and bonding
Between non-metal atoms, electrons are shared rather than transferred. A covalent bond is a shared pair of electrons, held by the attraction of both nuclei to that pair. Each atom shares enough electrons to complete its outer shell. A double bond is two shared pairs (O=O, C=O); a triple bond is three (N≡N).
A Lewis structure (dot-and-cross diagram) shows the outer electrons of each atom; the electrons of one atom are drawn as dots and those of the other as crosses, so the shared pairs can be traced to their origin. Pairs of outer electrons not involved in bonding are lone pairs.
Covalent substances form two very different kinds of structure.
- Simple molecular substances (H2O, CO2, CH4, O2) consist of small molecules. The covalent bonds inside each molecule are strong, but the forces between molecules are weak, so little energy separates the molecules: melting and boiling points are low, and many are gases or liquids. They do not conduct electricity, because the molecules are neutral and there are no free charged particles.
- Giant covalent substances (diamond, graphite, silicon dioxide in sand and glass) are huge networks in which every atom is covalently bonded to its neighbours. Melting means breaking many strong covalent bonds, so melting points are very high; they are hard and insoluble in water. Silicon dioxide, SiO2, is covalently bonded — electrons are shared between silicon and oxygen atoms — and its giant structure will not dissolve.
Carbon and silicon are both in group 14, so each forms four covalent bonds, and their hydrides have the same type of formula: CH4 (methane) and SiH4 (silane). But silane contains no carbon, so it is not an alkane. Likewise CO2 is simple molecular while SiO2 is giant covalent — similar formulas, completely different structures. Always reason from the bonds: a covalent bond forms when electrons are shared, and here there are four in each molecule.
04 / Metallic bonding
Metallic bonding and propertiesProperties
In a metal the atoms give up their outer electrons into a shared “sea” of delocalised electrons that move freely through a lattice of positive metal ions. The metallic bond is the attraction between the positive ions and the delocalised electrons. The delocalised electrons carry charge and energy, so metals conduct electricity and heat. The layers of ions can slide over each other without breaking the bonding, because the electron sea moves with them — so metals are malleable and ductile (Topic 5 explains why alloys are harder).
| Structure | Particles and bonding | Melting point | Conducts electricity? | Examples |
|---|---|---|---|---|
| Giant ionic | Positive and negative ions; strong electrostatic attraction | High | Only when molten or dissolved | NaCl, MgO, CaCO3 |
| Simple molecular | Small molecules; strong covalent bonds inside, weak forces between molecules | Low | No | H2O, CO2, CH4, I2 |
| Giant covalent | Network of atoms, all joined by covalent bonds | Very high | No (graphite is an exception) | Diamond, SiO2 |
| Metallic | Positive ions in a sea of delocalised electrons | Usually high | Yes, solid and liquid | Cu, Fe, Al |
Identify the structure from data
Each row gives measured properties. Decide the type of structure; the feedback explains which property settles it.
05 / Formulas
Chemical formulasChemical formulas
A formula shows the elements in a compound and the ratio of their atoms. For an ionic compound the ratio is fixed by the charges: the compound as a whole is neutral, so the total positive charge must equal the total negative charge.
| Positive ions | Negative ions |
|---|---|
| Na+, K+, H+, NH4+ (ammonium) | Cl−, Br−, I−, F−, OH− (hydroxide), NO3− (nitrate) |
| Mg2+, Ca2+, Sr2+, Ba2+, Cu2+, Fe2+, Zn2+ | O2−, S2−, SO42− (sulfate), CO32− (carbonate) |
| Al3+, Fe3+ | N3−, PO43− (phosphate) |
Find: the formula of magnesium phosphate.
Reasoning: Mg2+ and PO43−. The lowest common multiple of 2 and 3 is 6: three Mg2+ give +6, two PO43− give −6.
Answer: Mg3(PO4)2. The brackets are essential — they show that the subscript 2 multiplies the whole phosphate ion. Without them, “Mg3PO42” would be meaningless.
When a question gives a word equation, the salt formed can be determined from the ions it contains: calcium carbonate reacting with sulfuric acid gives Ca2+ and SO42−, so the salt is CaSO4. A question that asks you to determine the formula of the salt wants that formula identified, not only a full equation. Subscripts must be written as subscripts.
Ionic formula builder
Choose a positive ion and a negative ion. The model finds the smallest whole-number ratio that makes the total charge zero, adds brackets around a polyatomic ion when there is more than one, and calculates the molar mass.
06 / Equations
Balanced equationsFormulas and equations
A chemical equation is a record of what happens to atoms. In a reaction the atoms are rearranged — bonds break and new bonds form — but no atom is created or destroyed. The equation must therefore contain the same number of each type of atom on both sides. This is the particle-level meaning of the conservation of mass: if every atom is still there after the reaction, the total mass cannot change.
Balancing is done by changing the numbers in front of formulas (the coefficients), never the subscripts inside them. Changing a subscript changes the substance: H2O is water, H2O2 is hydrogen peroxide.
Unbalanced: C3H8 + O2 → CO2 + H2O
Step 1 — carbon: 3 C on the left, so 3CO2.
Step 2 — hydrogen: 8 H on the left, so 4H2O.
Step 3 — oxygen last (it appears in two products): 3 × 2 + 4 × 1 = 10 O atoms on the right, so 5O2.
Result: C3H8 + 5O2 → 3CO2 + 4H2O. Check: C 3 = 3, H 8 = 8, O 10 = 10.
Balance the element that appears in only one substance on each side first; leave elements that appear in several places (usually oxygen in combustion) until last. A balanced equation can also carry state symbols: (s) solid, (l) liquid, (g) gas and (aq) aqueous — dissolved in water. State symbols describe the substances at the temperature of the reaction; at 25 °C water is (l), not (aq), because water cannot be dissolved in itself.
Examination feedback reports that some students could not balance an equation because they could not turn a name into a formula — for example nitrogen and oxygen as the diatomic molecules N2 and O2. Write every formula correctly first; balancing a wrong formula can never give the right equation. In on-screen tasks, select a number for every box: a box left as “?” scores nothing.
Equation balancer
Choose a reaction and set the coefficients. The model counts every atom on each side and shows which elements are still unbalanced.
07 / The mole
The mole conceptMole concept
Atoms are far too small to count one by one, so chemists count them in batches. One mole is the amount of substance containing 6.02 × 1023 particles (the Avogadro constant). The number is chosen so that one mole of an element has a mass in grams equal to its relative atomic mass: 12 g of carbon, 24 g of magnesium, 56 g of iron.
Relative formula mass (Mr): the sum of the relative atomic masses of all the atoms in the formula. It has no unit.
Molar mass (M): the mass of one mole of a substance, numerically equal to Mr, with the unit g mol−1.
Amount of substance (n): measured in moles, mol.
n = m ÷ M amount (mol) = mass (g) ÷ molar mass (g mol−1)
The equation links the macroscopic quantity you can measure on a balance (mass) to the number of particles you cannot see (moles). Rearranged, m = n × M gives the mass of a known amount, and M = m ÷ n identifies a substance from a measured mass and amount. The mass must be in grams: a mass given in kilograms is multiplied by 1000 first.
Find: the molar mass of calcium carbonate, CaCO3.
Relationship: add the relative atomic masses of every atom.
Calculation: 40 + 12 + (3 × 16) = 100.
Answer: 100 g mol−1. The unit is part of the answer — examination feedback reports that most students calculated molar masses correctly but lost the mark by leaving out the unit.
Given: 12 kg of lithium (Ar = 6.9). Find: the amount in moles.
Conversion: 12 kg = 12 000 g.
Substitution: n = 12 000 ÷ 6.9 = 1739.1… mol.
Answer: 1739.1 mol (1.7 × 103 mol).
Reasonableness: about 7 g per mole and twelve thousand grams — well over a thousand moles is sensible. Forgetting to convert gives 1.74 mol, an answer 1000 times too small.
Using a balanced equation
The coefficients in a balanced equation give the ratio in which particles, and therefore moles, react. They do not give the ratio of masses. A calculation from one substance to another therefore always passes through moles:
mass of A → moles of A → (mole ratio from the equation) → moles of B → mass of B
Equation: 4Fe + 3O2 → 2Fe2O3. Find: the amount of oxygen that reacts with 0.600 g of iron.
Step 1: n(Fe) = 0.600 ÷ 56 = 0.010 71 mol.
Step 2 — ratio: 3 mol O2 react with every 4 mol Fe, so n(O2) = 0.010 71 × 3 ÷ 4 = 0.008 04 mol.
Answer: 8.04 × 10−3 mol of O2 (3 s.f.). To find its mass, multiply by 32 g mol−1: 0.257 g.
Very large or very small answers are written in standard form: 0.00313 mol is 3.13 × 10−3 mol. Keep full calculator values through the working and round only the final answer to the number of significant figures asked for.
Feedback across recent sessions identifies three recurring losses: masses given in kilograms not converted to grams (answers out by a factor of 1000); calculations that stop after finding the moles of one substance instead of using the equation's ratio to reach the substance asked about; and difficulty converting numbers to standard form.
Mole calculator
Type a formula and a mass. The model works out the molar mass from relative atomic masses, then the amount in moles and the number of particles, showing each step. Relative atomic masses are rounded values used in these notes.
08 / Kinetics
Rates of reactionReaction kinetics
The rate of a reaction is how quickly reactants are used up or products are formed. It is found from a quantity that can be measured as the reaction proceeds, divided by time:
rate = change in quantity ÷ time taken e.g. cm3 s−1, g min−1
Rate is calculated, not measured directly. In an investigation the dependent variable is therefore the quantity you actually measure — volume of gas collected, loss of mass, time for a colour to change or a glow to stop — and the rate is worked out from it afterwards. Examination feedback repeatedly reports students naming “rate of reaction” as the dependent variable; it was not accepted.
Given: 44 cm3 of oxygen collected in 30 s. Relationship: rate = volume ÷ time.
Calculation: 44 ÷ 30 = 1.47 cm3 s−1.
Unit: volume per unit time — cm3 s−1 (equivalently 88 cm3 min−1). A rate without a unit is incomplete.
Collision theory
Particles can only react when they collide. Not every collision leads to a reaction: the particles must collide with enough energy to break the existing bonds — at least the activation energy, Ea — and in a suitable orientation. A collision that meets these conditions is a successful collision. The rate depends on how many successful collisions happen per second, which gives four factors to control:
| Change | Effect on particles | Result |
|---|---|---|
| Increase concentration (or gas pressure) | More particles in the same volume | More frequent collisions → faster rate |
| Increase surface area of a solid (smaller pieces, powder) | More particles exposed at the surface | More frequent collisions → faster rate |
| Increase temperature | Particles have more kinetic energy and move faster | More frequent collisions and a larger fraction with energy ≥ Ea → faster rate |
| Add a catalyst | Provides an alternative pathway with a lower activation energy | A larger fraction of collisions are successful → faster rate |
Temperature has the largest effect because it acts in two ways at once. The more important of the two is energy: at a higher temperature a much larger proportion of collisions carry at least the activation energy.
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. Because it is not consumed, a small amount can be used again and again. Transition metals and their compounds are frequently catalysts (Topic 1).
Students generally knew that catalysts speed reactions up and lower the activation energy, but feedback notes two parts that were often missing: that the catalyst is not used up, and that the lower energy comes from an alternative pathway. A complete answer carries all three ideas.
A catalyst, a higher temperature or a finer powder makes the same amount of product appear sooner. None of them changes how much product is made from a fixed amount of reactant; that is set by the limiting reactant. On a graph of gas volume against time the faster reaction has the steeper initial gradient, but both curves level off at the same final volume.
Collision model
Particles move with speeds drawn from the chosen temperature. A collision counts as successful only when the collision energy is at least the activation energy. Change the temperature, the number of particles and the catalyst, and watch the count of successful collisions per second. The model is qualitative — it shows the trends, not real rates.
Planning a rate investigation
Rate investigations appear often in criteria B and C. A good design changes one factor (the independent variable), measures one quantity with a named instrument (the dependent variable: gas syringe or inverted measuring cylinder for volume, balance for mass, stopwatch for time), and controls the others — volume and concentration of acid, mass and particle size of the solid, temperature (water bath), and the same catalyst mass. The hypothesis should give the scientific reason using collision theory: “If the temperature increases, then the glow stick will stop glowing sooner, because particles have more kinetic energy, so more collisions per second have at least the activation energy and the reaction is faster.”
09 / Equilibria
Reversible reactions and equilibriumEquilibria
Some reactions can go in both directions: the products can react to re-form the reactants. These are reversible reactions, shown with the double arrow ⇌. Heating blue hydrated copper(II) sulfate drives off water and leaves white anhydrous copper(II) sulfate; adding water turns it blue again. Cobalt(II) chloride paper works the same way, turning from blue to pink when water is added — which is why it is used as a test for water.
CuSO4·5H2O(s) ⇌ CuSO4(s) + 5H2O(l)
In a closed system — where nothing can enter or leave — a reversible reaction reaches dynamic equilibrium. At the start the forward reaction is fast because the reactant concentrations are high. As products build up, the reverse reaction speeds up. Eventually the two rates become equal. From then on the concentrations of reactants and products stay constant, even though both reactions are still happening.
The state of a reversible reaction in a closed system in which the forward and reverse reactions occur at the same rate, so the concentrations of reactants and products remain constant. “Dynamic” means the reactions have not stopped; “equilibrium” means the overall amounts do not change.
An important example is the behaviour of a weak acid in water. Methanoic (formic) acid only partly dissociates: most molecules stay intact, and an equilibrium is set up between the molecules and the ions they form.
HCOOH(aq) + H2O(l) ⇌ H3O+(aq) + HCOO−(aq) or HCOOH(aq) ⇌ H+(aq) + HCOO−(aq)
Writing this equation well means four separate things: the reactants, the products, the state symbols and the equilibrium arrow. Each carries marks. A strong acid such as hydrochloric acid dissociates completely and is written with a single arrow.
If the conditions of a system at equilibrium change — a concentration, the pressure of a gas, the temperature — the position of equilibrium shifts to oppose the change. Removing a product, for example, makes the forward reaction run faster until equilibrium is restored, so more product forms. This idea explains why industry removes products as they form, and why a reversible reaction in an open container (where a gas can escape) can go to completion.
10 / Energetics
Energy changes and fuelsEnergy changes
Breaking a chemical bond requires energy; forming a bond releases energy. Every reaction involves both, and the overall energy change depends on which is larger.
- In an exothermic reaction more energy is released by forming bonds than is needed to break bonds. Energy is transferred to the surroundings, so the temperature of the surroundings rises. Combustion, neutralisation and the reactions in fireworks and hand-warmers are exothermic. ΔH is negative.
- In an endothermic reaction more energy is needed to break bonds than is released by forming them. Energy is taken from the surroundings, so the temperature falls. Thermal decomposition, photosynthesis and dissolving ammonium nitrate (in cold packs) are endothermic. ΔH is positive.
“Exothermic” alone is a classification. A justified answer says what happens to the energy: heat (energy) is released to the surroundings, so the reaction is exothermic. When a question gives a temperature change, use it as evidence: a rise in the temperature of the water means energy was released.
Measuring the energy released by a fuel
The energy released when a fuel burns can be estimated by using it to heat a known mass of water and measuring the temperature rise:
q = m c ΔT
Here q is the energy transferred (J), m is the mass of water heated (g), c is the specific heat capacity of water (4.18 J g−1 °C−1: the energy needed to raise 1 g by 1 °C) and ΔT is the temperature change (°C). The method assumes all the energy from the fuel goes into the water. In practice heat is lost to the air and the container and combustion may be incomplete, so the measured value is lower than the true value.
Given: 1.00 g of ethanol releases 30.0 kJ; 400 g of water is heated from 10.0 °C to 98.0 °C. Find: the mass of ethanol needed.
Step 1: ΔT = 98.0 − 10.0 = 88.0 °C.
Step 2: q = 400 × 4.18 × 88.0 = 147 136 J = 147 kJ. (With c = 4.19 J g−1 °C−1 this is 147.5 kJ; use the value the question gives.)
Step 3: mass = 147 ÷ 30.0 = 4.90 g of ethanol (3 s.f.).
Check: about 5 g of fuel to bring 400 g of cold water near boiling is reasonable.
Comparing fuels
Fuels are compared fairly by the energy released per gram (important when fuel must be carried) or per mole. Butane releases 2877 kJ mol−1; with a molar mass of 58 g mol−1 that is 2877 ÷ 58 = 49.6 kJ g−1, compared with 30.0 kJ g−1 for ethanol, so butane releases more energy per gram. A full evaluation also weighs the products: burning fossil fuels releases carbon dioxide (a greenhouse gas) and, with too little oxygen, carbon monoxide and soot, while hydrogen burns to give only water (Topic 3).
Fuel comparison
Enter the mass of fuel burned, the mass of water heated and the temperature rise. The model applies q = mcΔT with c = 4.18 J g−1 °C−1 and reports the energy released per gram, assuming no heat loss.
11 / Practice
Examination practicePast examination tasks
The tasks below come from past on-screen examinations and are grouped by the criterion they assess. Criterion A items test bonding, formulas, moles, rates, equilibria and energy directly; criteria B and C use rate investigations; criterion D applies mole and rate calculations to real contexts. Attempt each one before opening the marking guidance.
A Criterion A · Knowing and understanding
State the type of bond that would form in a reaction between magnesium and chlorine.
Marking guidance
- ionic bond Accept electrovalent
Scientists studying space wonder whether life would be possible on planets other than the Earth. Most believe that, if molecular oxygen is present on a planet, then life is possible on this planet. But molecular oxygen itself is not a sign of life on a planet, because it might have formed from water. Samples collected from the Moon contain several isotopes of oxygen.
Calculate the relative molecular mass of carbon dioxide containing one atom of oxygen-16 and one atom of oxygen-18.
Marking guidance
- 46 Ignore units if present
Fireworks contain different salts which emit a characteristic colour at very high temperatures. The colours are shown in the table.
| Salt | Colour |
|---|---|
| copper chloride | blue-green |
| calcium chloride | orange-red |
| iron chloride | red-brown |
| strontium chloride | bright red |
| barium chloride | green |
State if the reactions happening in fireworks are endothermic or exothermic. Justify your answer.
Marking guidance
- energy/heat is released or release (much) more energy/heat than they absorb WTTE
- (so) the reaction is exothermic Only award second mark if first is awarded
Glow sticks produce light in a chemical reaction. They can be used by the military, by divers or just for fun. Glow sticks work by combining two chemicals to produce light in a chemical reaction known as chemiluminescence. One of the chemicals in the glow stick is a catalyst. The catalyst is separated from the other reactants inside a sealed tube. Keeping the catalyst separate prevents the reaction starting until light is needed. Once the tube containing the catalyst is broken, the reaction will produce light.
Outline the function of a catalyst.
Marking guidance
- a chemical which speeds up the rate of a reaction or lowers the activation energy WTTE
- without itself being used up WTTE
In 1671 John Ray, an English naturalist, distilled a liquid collected from ants. He produced a strong-smelling acidic liquid that he named formic acid from the Latin word formica meaning ant. Formic acid is the simplest carboxylic acid.
Formic acid partially dissociates in water to form an equilibrium. Write down a balanced equation for this equilibrium including state symbols.
Marking guidance
- HCOOH (aq) + H2O(l) ⇌ H3O+(aq) + HCOO-(aq) Allow HCOOH(aq) ⇌ H+(aq) + HCOO-(aq)
- reactant(s) correct Accept incorrect order for example,
- products correct H2CO2, CHOOH
- states all correct Accept l for HCOOH
- correct use of equilibrium arrow Allow any double-headed arrow
Portable heaters need fuel that is both lightweight and that burns without producing a lot of smoke. Butane and ethanol are both commonly used as fuels in portable heaters.
The energy released on combustion can be calculated using the equation:
q = mc∆T
where q = energy released, m is the mass of water heated and ∆T is the temperature change.
If the combustion of 1.00 g of ethanol releases 30.0 kJ of energy, calculate the mass of ethanol that is required to heat 400 g of water at 10.0℃ to 98.0℃. The specific heat capacity of water is 4.19 Jg−1℃−1. Give your answer to three significant figures.
Marking guidance
- attempt to substitute into equation
- (energy released =) 147.488 kJ or 147 488 J Seen or implied
- (mass of ethanol = 147.488/30.0 = ) 4.91626… Accept incorrect use of kJ for 3rd marking point
- answer given correctly to 3 sig figs 4.92 (g) Unit required
- unit of g/gram Award unit mark separately
The molar mass of butane is 58 g mol−1 and the combustion of one mole of butane releases 2877 kJ. Use this information and the information from part (c) to identify whether ethanol or butane is the best choice of fuel. Use calculations to justify your answer.
Marking guidance
- so 1 g releases 49.6 kJ or ethanol releases 1380 kJmol-1 and butane release 2877kJmol-1 ECF from 1c
- butane is a better fuel for portable heaters because it releases more energy per gram or butane is a better fuel as it releases more energy per mole WTTE
B Criterion B · Inquiring and designing
Identify the variables for this investigation.
Independent variable: ____________
Dependent variable: ____________
Control variable one: Control variable two:
Marking guidance
- temperature
- time taken for the glow stick to stop glowing WTTE
- Any two reasonable control variables [max 2] Do not accept amount of water
- type or colour of glow stick
- volume of water
- time to equilibrate
Formulate a suitable hypothesis for the investigation.
If: ____________
Then: ____________
Because: ____________
Marking guidance
- if the temperature increases
- then the length of time the glow stick will glow will decrease
- because the rate of the reaction increases
H2O2 decomposes faster when a catalyst is added. A catalyst is a substance which increases the rate of reaction, without being used up itself. A student has a catalyst in two states: a solid and a solution. Formulate a hypothesis to test which state will give the greatest increase in rate of reaction. You should use collision theory in your answer.
If: ____________
Then: ____________
Because: ____________
Marking guidance
- If the catalyst is a solution ORA for solid catalyst
- Then the rate of decomposition will be higher Do not award the first marking point without correct link to the second marking point
- Third marking point from the list [max 1]
- because the number of collisions is higher
- greater chance of collision
- higher frequency of collision
- particles can move more freely
C Criterion C · Processing and evaluating
The student formulated the following hypothesis:
“The best catalyst will produce the flame that burns the longest because more oxygen is released.”
State and justify whether this hypothesis is valid.
Marking guidance
- First explanation linked to rate of production, for example [max 1]
- as the rate of decomposition would be faster for the best catalyst WTTE
- (so) the burn time would be shorter for the best catalyst
- a longer burn would come from a smaller rate of reaction
- Second explanation linked to oxygen, for example [max 1]
- total volume of oxygen would the same for all catalysts
- steady burning does not necessarily mean it is the fastest rate of oxygen
- production
- a good catalyst might produce oxygen too quickly for it to be burnt
- Final mark: (so) the hypothesis is invalid Do not award final mark unless at least one correct explanation is given
A second student has suggested an alternative method to determine which catalyst would decompose H2O2 the fastest, collecting the oxygen produced. The volume of gas collected after 30 seconds for three trials is shown in the table.
| Volume from trial one / cm3 | Volume from trial two / cm3 | Volume from trial three / cm3 |
|---|---|---|
| 43 | 44 | 45 |
Calculate the average volume of oxygen collected for this experiment and determine the rate of oxygen production.
Marking guidance
- Average volume = 44 (cm3)
- Value of rate: 44/30=1.47 (accept 1.5) ECF from first marking point
- Unit of rate: cm3s−1 Accept cm3/s or ml/s, 88.2 cm3min−1
D Criterion D · Reflecting on the impacts of science
A glow stick contains 6.58 × 105 molecules of phenyl oxalate. The time for the reaction to stop at 20 °C is 260 minutes. Calculate the rate at which the phenyl oxalate molecules are used up at this temperature. You should include appropriate units in your answer.
Marking guidance
- rate = molecules ÷ time Seen or implied
- = 6.58 × 105 ÷ 260
- = 2530(.0769…)
- molecules min−1 or 42.2 molecules s−1 Award the unit mark separately
Stained glass windows are one example of artwork using glass.
A 35.0 g sample of commercial glass contains 26.3 g of SiO2 and 3.4 g of CaO; the remainder is Na2O. Calculate the number of moles of sodium oxide in the sample. Give your answer to two significant figures.
Marking guidance
- Mass of Na2O=5.3 (g)
- Molar mass of Na2O = 62
- (Moles of Na2O in sample =) 0.0854838… ECF if molar mass is incorrect
- 0.085
The original Slinky was made from 24.4 m of high carbon steel wire. The mass of a Slinky is 0.405 kg. Calculate the number of moles of iron needed to make a Slinky assuming that all the wire is made of iron. Give your answer to 2 significant figures.
Marking guidance
- kg converted to g
- n = m/ram or n = 405/56 Seen or implied. Accept correct answers using 98.15% high carbon steel
- 7.23 (moles) Award 3 marks if only this answer is seen
- 7.2 (moles) Correct answer expressed to 2 sig figs
Quick check: bonding, moles and rates
Original practice questions for retrieval — not past examination items.
12 / Examination feedback
Examiner's overall observationEvidence from examination feedback
Examiner's overall observation
Balancing equations, identifying exothermic reactions, recognising Lewis structures, calculating molar masses and explaining how a catalyst works were areas in which students were generally well prepared. Almost all students identified the bonding in carbon dioxide as covalent. The weaknesses were consistent across sessions. In covalent bonding, students recognised Lewis diagrams but were less secure about what a covalent bond is — electrons shared between atoms — and in one task assumed that methane and silane were both alkanes rather than counting four covalent bonds in each. Mole calculations were the most common source of lost marks: masses given in kilograms were not converted to grams, so answers were out by a factor of 1000; many students found the moles of one substance and then could not use the balanced equation to reach the moles and mass of another; and conversion to standard form was weak. Molar masses were frequently given without a unit. When determining the formula of a salt, several students wrote the full equation without identifying the formula asked for, and some did not write subscripts correctly. Some students could not translate the name of a substance into its formula, which prevented them from balancing the equation. On catalysts, the ideas that a catalyst is not used up and that it works by an alternative pathway were often missing, and in investigations students named a calculated quantity such as rate as the dependent variable instead of a measured one.
13 / Summary
Summary and knowledge organiserRevision
Essential knowledge
- Ionic bonding: electrons transferred from metal to non-metal; oppositely charged ions held by strong electrostatic attraction in a lattice. Covalent bonding: pairs of electrons shared between non-metal atoms. Metallic bonding: positive ions in a sea of delocalised electrons.
- Ionic formulas balance the charges; brackets go round a polyatomic ion used more than once.
- Balance equations with coefficients, never by changing subscripts; atoms are conserved, so mass is conserved.
- n = m ÷ M with m in grams; the equation's coefficients give the mole ratio, not the mass ratio.
- Rate increases with concentration, surface area, temperature and a catalyst — explained by the frequency and energy of collisions.
- A reversible reaction in a closed system reaches dynamic equilibrium: equal forward and reverse rates, constant concentrations.
- Exothermic: energy released, ΔH negative, surroundings warm up. Endothermic: energy absorbed, ΔH positive, surroundings cool. q = mcΔT.
Definitions
- Mole — 6.02 × 1023 particles
- Molar mass — mass of 1 mol, g mol−1
- Activation energy — minimum collision energy for reaction
- Catalyst — speeds reaction, not used up, alternative pathway
- Dynamic equilibrium — equal rates, constant concentrations
Equations and units
- n = m ÷ M (mol = g ÷ g mol−1)
- rate = change ÷ time (cm3 s−1)
- q = mcΔT (J; c = 4.18 J g−1 °C−1)
- energy per gram = energy per mole ÷ M
Must-remember distinctions
- Coefficient ≠ subscript
- Mole ratio ≠ mass ratio
- Rate (calculated) ≠ dependent variable (measured)
- Rate ≠ extent
- Exothermic (classification) vs justified answer (energy released)
Common errors
- kg not converted to g
- Stopping after the first mole step
- Molar mass with no unit
- Full equation instead of the salt's formula
- “Catalyst lowers energy” with no pathway or not-used-up
Examination checklist
- Write formulas before balancing
- Show n = m ÷ M with units at every step
- Round only the final answer; use standard form
- Explain rate changes with collisions and energy
- State all four parts of an equilibrium equation
Other chapters: Criteria A–D · 1 · Periodic table · 2 · IUPAC naming · 3 · Atmosphere · 4 · Matter · 5 · Pure and impure · 6 · Bonding · 7 · Types of reaction
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