Chemistry 5070/12 — May/June 2025

Walkthrough

To determine the physical states of hexane and pentane at temperature X, we analyze the given properties using the kinetic particle theory:

1. Effect of pressure: Gases have large spaces between particles, making their volume highly dependent on pressure. In solids and liquids, particles are closely packed, so their volumes are almost completely unaffected by pressure changes between $1\text{ atm}$ and $2\text{ atm}$. This rules out the gas state for both substances, eliminating options B and D.
2. Hexane: Particles in a solid vibrate about fixed positions in a regular lattice. Since hexane molecules are in fixed positions, hexane must be a solid.
3. Pentane: Particles in a liquid are closely packed but are free to move and slide over one another. Since pentane molecules are free to move and cannot be a gas, pentane must be a liquid.

Matching hexane = solid and pentane = liquid gives option C.

Key Takeaways

• Solids: Particles are arranged in a fixed, regular pattern and can only vibrate about their fixed positions; they have a fixed volume and fixed shape.
• Liquids: Particles are close together in a random arrangement, free to move around and slide past each other; they have a fixed volume but take the shape of the container.
• Gases: Particles are far apart and move rapidly and randomly in all directions; they are easily compressed and fill any container.

Common Mistakes

• Confusing liquids with gases when interpreting "free to move": gas particles move freely and rapidly at high speeds with large gaps between them, whereas liquid particles are free to move past one another but remain closely packed.
• Forgetting that solids and liquids are both considered incompressible under moderate pressure changes.

Things to Be Careful About

• Check both criteria (compressibility and particle motion) for each substance to eliminate distractors systematically.

Walkthrough

When a gas is produced, its concentration is highest at the source of the reaction. Gas particles are in constant, random motion, colliding with other particles in the air. Over time, these random collisions cause the particles to spread out evenly across the entire room, moving from an area of higher concentration to an area of lower concentration. This process is called diffusion.

Let's evaluate the options:

• A condensation: The change of a substance from gas to liquid.
• B diffusion: The net movement of particles from a region of higher concentration to a region of lower concentration down a concentration gradient due to random thermal motion.
• C dissolving: A solute breaking down into individual particles in a liquid solvent to form a solution.
• D evaporation: The transition of liquid particles escaping into the gas phase from the surface of a liquid.

Therefore, option B is the correct choice.

Key Takeaways

• Diffusion describes how particles spread out naturally to fill the available space due to random movement.
• The rate of diffusion of a gas depends on its temperature (average kinetic energy) and its relative molecular mass ($M_r$).

Common Mistakes

• Confusing diffusion (particle movement/spreading) with changes of state such as evaporation or condensation.

Things to Be Careful About

• Ensure the definition used focuses on the random motion of particles leading to a net movement from high to low concentration.

Walkthrough

To determine the correct statement, let us evaluate each option based on core chemical definitions:

1. Option A: "All compounds are ionic."

   • A compound consists of two or more different elements chemically combined. These can be bonded ionically (between metals and non-metals) or covalently (between non-metal atoms). Therefore, this statement is false.

2. Option B: "All compounds conduct electricity when molten."

   • Only compounds containing mobile charged particles (such as ions in molten ionic compounds) conduct electricity. Covalent molecular substances (e.g. molten wax or liquid water) do not contain free ions or delocalised electrons to conduct electricity. Therefore, this statement is false.

3. Option C: "Each atom of an element contains the same number of protons."

   • An element is defined by its atomic (proton) number. While atoms of the same element can have different numbers of neutrons (isotopes), every single atom of that element must have the exact same number of protons. Therefore, this statement is correct.

4. Option D: "In a mixture of substances, the proportions of the substances are always the same."

   • A mixture contains two or more substances not chemically combined, and they can be present in any ratio or variable proportion (unlike compounds, which have fixed ratios by mass). Therefore, this statement is false.

Key Takeaways

• Proton number ($Z$): Defines the identity of an element. All atoms of the same element contain the identical number of protons.
• Compounds vs. Mixtures: Compounds have a fixed chemical composition and fixed ratios of atoms; mixtures have variable composition.
• Electrical conductivity: Requires mobile charged particles (delocalised electrons or free-moving ions).

Common Mistakes

• Confusing mass number with proton number: isotopes of the same element have different mass numbers (different numbers of neutrons), but always have the same proton number.
• Confusing the fixed composition of a compound with the variable composition of a mixture.

Things to Be Careful About

• Ensure you read broad generalisations like "All..." or "always..." carefully, as chemical counterexamples (like covalent bonding vs ionic bonding) readily disprove them.

Walkthrough

1. Identify the nature of the ion:
   Positive ions (cations) are formed when neutral atoms lose one or more valence electrons. Because the positive ion of $\text{Q}$ contains $10$ electrons, the neutral atom of $\text{Q}$ must have contained more than $10$ electrons (at least $11$ electrons).

2. Determine the number of protons:
   In any neutral atom, the number of protons equals the number of electrons. Therefore, the atomic number (proton number) of $\text{Q}$ must be at least $11$ ($Z \geq 11$). This immediately rules out C.

3. Calculate the mass number:

   $$\text{mass number } (A) = \text{number of protons} + \text{number of neutrons}$$

   Given that there are $14$ neutrons and at least $11$ protons:

   $$A \geq 11 + 14 = 25$$

   A value of $25$ or greater is strictly greater than $24$, which confirms statement D is correct (for instance, $\text{Q}$ could be magnesium-$26$, $^{26}_{12}\text{Mg}$, or aluminium-$27$, $^{27}_{13}\text{Al}$, etc.).

4. Evaluate the remaining statements:

   • A: With $11$ or more electrons, the electron configuration starts with 2,8,1... which occupies $3$ electron shells, not $2$.
   • B: Elements that readily lose electrons to form positive ions are metals, not non-metals.

Key Takeaways

• Positive ions are formed by the loss of electrons, so $\text{number of protons} > \text{number of electrons in a cation}$.
• Atomic number ($Z$) is the number of protons in the nucleus and defines the element.
• Mass number ($A$) is the sum of protons and neutrons in the nucleus ($A = Z + N$).

Common Mistakes

• Confusing the number of electrons in an ion with the atomic number of the element.
• Forgetting that positive ions lose electrons, incorrectly thinking a positive charge means gaining protons or electrons.

Things to Be Careful About

• Ensure you check whether the question refers to the atom or the ion at each stage of reasoning.

Walkthrough

To determine the number of electrons in any atom or ion:

1. Look up the atomic number (proton number, $Z$) of the element on the Periodic Table.
2. For a neutral atom, the number of electrons equals the number of protons.
3. For a positive ion (cation), subtract the magnitude of the positive charge from the proton number (electrons have been lost).
4. For a negative ion (anion), add the magnitude of the negative charge to the proton number (electrons have been gained).

Evaluating each option:

• $\text{Mg}^{2+}$: Proton number = $12$. Electrons = $12 - 2 = 10$.
• $\text{N}^{3-}$: Proton number = $7$. Electrons = $7 + 3 = 10$.
• $\text{Ne}$: Proton number = $10$. Electrons = $10$.
• $\text{S}^{2-}$: Proton number = $16$. Electrons = $16 + 2 = 18$.

Species A, B, and C are isoelectronic, each having $10$ electrons (a stable 2,8 noble gas configuration). Species D has $18$ electrons (a stable 2,8,8 configuration), which is the greatest.

Key Takeaways

• Cations have fewer electrons than protons.
• Anions have more electrons than protons.
• Species that share the same total number of electrons are described as isoelectronic.

Common Mistakes

• Inverting the sign when adjusting for charge (e.g., subtracting electrons for a negative charge or adding electrons for a positive charge).
• Using mass numbers instead of proton (atomic) numbers to determine electron counts.

Things to Be Careful About

• Ensure you locate the correct element on the Periodic Table and read the atomic number (smaller integer, number of protons) rather than the relative atomic mass.

Walkthrough

1. Recall the formula for relative atomic mass ($A_r$):
   The relative atomic mass of an element is calculated by finding the weighted average mass of naturally occurring isotopes taking into account their percentage abundances:

   $$A_r = \frac{\sum (\text{isotopic mass} \times \text{percentage abundance})}{100}$$

2. Substitute the given data:

   • Isotope 1: mass $= 150.92$, abundance $= 47.81\%$
   • Isotope 2: mass $= 152.92$, abundance $= 52.19\%$
   $$A_r = \frac{(150.92 \times 47.81) + (152.92 \times 52.19)}{100}$$

3. Perform the arithmetic:

   $$A_r = \frac{7215.4852 + 7980.8948}{100} = \frac{15196.38}{100} = 151.9638$$

   Rounding to 2 decimal places (matching the options) gives $151.96$, which corresponds to option B.

Key Takeaways

• Relative atomic mass ($A_r$) is a weighted mean of the masses of the isotopes based on their relative abundances.
• Because the heavier isotope ($152.92$) is slightly more abundant ($52.19\%$) than the lighter isotope ($47.81\%$), the relative atomic mass must lie slightly above the midpoint ($151.92$), ruling out values below $151.92$.

Common Mistakes

• Calculating the simple arithmetic average $\frac{150.92 + 152.92}{2} = 151.92$ instead of taking the weighted mean.
• Inverting the abundance percentages (multiplying $150.92$ by $52.19$ and $152.92$ by $47.81$).

Things to Be Careful About

• Ensure you divide by $100$ when using percentage abundances.
• Check that the total of the percentage abundances equals $100\%$ ($47.81 + 52.19 = 100.00\%$).

Walkthrough

Magnesium is a Group II metal and oxygen is a Group VI non-metal. When they react:

• A magnesium atom transfers its two outer-shell electrons to an oxygen atom.
• This forms a magnesium ion, $\text{Mg}^{2+}$ (a cation, because it has a positive charge), and an oxide ion, $\text{O}^{2-}$ (an anion, because it has a negative charge).
• The oppositely charged ions attract each other strongly via electrostatic forces to form a giant ionic lattice.

Properties of ionic compounds tested here:

1. Melting/Boiling Points: They have high melting and boiling points due to strong electrostatic forces of attraction between oppositely charged ions in the lattice.
2. Electrical Conductivity:
   • In the solid state (e.g. at r.t.p.), the ions are locked in fixed positions within the lattice and can only vibrate; with no mobile ions or delocalised electrons, solid ionic compounds are electrical insulators.
   • In the molten (liquid) or aqueous state, the lattice structure is destroyed and the ions become free to move, enabling the substance to conduct electricity.

Evaluating the options:

• A: False because $\text{MgO}$ cannot conduct electricity at room temperature and pressure (solid state).
• B: False because magnesium forms cations and oxygen forms anions.
• C: True because it is an ionic compound and conducts electricity when molten due to mobile $\text{Mg}^{2+}$ and $\text{O}^{2-}$ ions.
• D: False because shared electron pairs describe covalent bonding, not ionic bonding.

Key Takeaways

• Metal + non-metal typically forms ionic bonding via electron transfer.
• Cations are positive ions (metals lose electrons), and anions are negative ions (non-metals gain electrons).
• Ionic compounds conduct electricity only when molten or in aqueous solution because the ions are free to move.

Common Mistakes

• Confusing the terms cation (positive ion) and anion (negative ion).
• Thinking ionic solids conduct electricity due to electrons — electrical conductivity in ionic substances is always due to mobile ions, not electrons.
• Confusing ionic bonding (electrostatic attraction between oppositely charged ions) with covalent bonding (shared pairs of electrons).

Things to Be Careful About

• Always verify the state of the substance when asked about conductivity: solids have fixed ions (no conductivity), while liquids/melts and aqueous solutions have mobile ions (good conductivity).

Walkthrough

Silicon(IV) oxide ($ext{SiO}_2$), commonly known as silica or quartz, is a giant covalent substance (macromolecule). In this structure:

• Each silicon atom is covalently bonded to four oxygen atoms in a tetrahedral arrangement.
• Each oxygen atom is bonded to two silicon atoms.

Evaluating each option:

• A is incorrect: All valence electrons are fixed in localized single covalent bonds; there are no delocalised electrons or mobile ions to carry an electric current.
• B is correct: It exists as a continuous 3D network of covalent bonds, which defines a giant covalent structure.
• C is incorrect: An alloy is a mixture of a metal with other elements; $ext{SiO}_2$ is a chemical compound of non-metals.
• D is incorrect: The bonding between silicon and oxygen is covalent (shared pairs of electrons), not ionic.

Key Takeaways

• Silicon(IV) oxide ($ext{SiO}_2$), diamond, and graphite are the standard giant covalent substances studied in Cambridge O Level Chemistry.
• Giant covalent substances have very high melting and boiling points due to the extensive network of strong covalent bonds.

Common Mistakes

• Confusing silicon(IV) oxide with simple covalent molecules like carbon dioxide ($ext{CO}_2$), which are gases at room temperature with weak intermolecular forces.
• Assuming that compounds containing polyatomic non-metal combinations might be ionic.

Things to Be Careful About

• Ensure you recall the coordination numbers in silica: each $ext{Si}$ is bonded to 4 $ext{O}$ atoms, and each $ext{O}$ is bonded to 2 $ext{Si}$ atoms, leading to the empirical formula $ext{SiO}_2$.

Walkthrough

To find the compound with the highest relative formula mass ($M_r$):

1. Write the correct chemical formula for each ionic compound by balancing the charges of the constituent ions:

   • Calcium carbonate: $\text{Ca}^{2+}$ and $\text{CO}_3^{2-}$ give $\text{CaCO}_3$.
   • Lithium sulfate: $\text{Li}^+$ and $\text{SO}_4^{2-}$ give $\text{Li}_2\text{SO}_4$.
   • Magnesium chloride: $\text{Mg}^{2+}$ and $\text{Cl}^-$ give $\text{MgCl}_2$.
   • Potassium nitrate: $\text{K}^+$ and $\text{NO}_3^-$ give $\text{KNO}_3$.

2. Sum the relative atomic masses ($A_r$) of all the atoms in each formula unit using the given values:

   • $\text{CaCO}_3$: $40 + 12 + (3 \times 16) = 100$
   • $\text{Li}_2\text{SO}_4$: $(2 \times 7) + 32 + (4 \times 16) = 14 + 32 + 64 = 110$
   • $\text{MgCl}_2$: $24 + (2 \times 35.5) = 95$
   • $\text{KNO}_3$: $39 + 14 + (3 \times 16) = 101$

3. Compare the values: $110 > 101 > 100 > 95$, which corresponds to lithium sulfate (B).

Key Takeaways

• Determining formula mass requires first finding the correct chemical formula from the names of ionic compounds.
• Ionic compounds must have an overall neutral charge, so the charges on positive and negative ions must balance.

Common Mistakes

• Forgetting the stoichiometry in compound formulae, such as writing $\text{LiSO}_4$ instead of $\text{Li}_2\text{SO}_4$ or $\text{MgCl}$ instead of $\text{MgCl}_2$.
• Multiplication errors when multiplying the number of oxygen atoms (e.g., $4 \times 16 = 64$).

Things to Be Careful About

• Ensure you check the formula valencies: lithium has a $+1$ charge and sulfate has a $-2$ charge, requiring two lithium ions per sulfate ion.

Walkthrough

1. Write and balance the combustion equation: $$\text{C}_2\text{H}_4(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$$
2. Use the mole-to-volume relationship: By Avogadro's law, molar ratios of gases correspond directly to volume ratios at r.t.p.
3. Determine the volume of oxygen reacted and unreacted:
   • $100\text{ cm}^3$ of $\text{C}_2\text{H}_4$ requires $100 \times 3 = 300\text{ cm}^3$ of $\text{O}_2$.
   • Since $400\text{ cm}^3$ was supplied, oxygen is in excess by $400 - 300 = 100\text{ cm}^3$.
4. Determine the gaseous products formed:
   • Complete combustion of $100\text{ cm}^3$ of $\text{C}_2\text{H}_4$ produces $100 \times 2 = 200\text{ cm}^3$ of $\text{CO}_2$.
   • Water is in the liquid state at room temperature and pressure (r.t.p.), so its volume does not count toward the gas volume.
5. Sum the remaining gas volumes:
   • Final gas volume = unreacted $\text{O}_2$ ($100\text{ cm}^3$) + produced $\text{CO}_2$ ($200\text{ cm}^3$) = $300\text{ cm}^3$.

Hence, the correct option is C.

Key Takeaways

• For reactions involving gases at constant temperature and pressure, volume ratios are identical to the stoichiometric mole ratios in the balanced equation.
• Always identify whether a reactant is in excess and include the unreacted volume in the final total.
• Water ($\text{H}_2\text{O}$) formed during combustion is a liquid at r.t.p., so it does not contribute to the final gas volume.

Common Mistakes

• Forgetting the unreacted excess oxygen: Only counting the $200\text{ cm}^3$ of $\text{CO}_2$ gives option A ($200\text{ cm}^3$).
• Treating water as a gas at r.t.p.: Adding $200\text{ cm}^3$ for steam would give $500\text{ cm}^3$ (option D).

Things to Be Careful About

• Ensure the hydrocarbon formula is correct: ethene is $\text{C}_2\text{H}_4$, not ethane ($\text{C}_2\text{H}_6$).
• Check the state symbols at the specified conditions (r.t.p. means $20^\circ\text{C}$ to $25^\circ\text{C}$ and $1\text{ atm}$, where water is liquid).

Walkthrough

To find the percentage composition by mass of an element in a compound:

1. Determine the relative formula mass ($M_r$) of the hydrated salt:

   • $\text{Na}_2\text{CO}_3$: $(2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106$
   • $10\text{H}_2\text{O}$: $10 \times [ (2 \times 1) + 16 ] = 10 \times 18 = 180$
   • Total $M_r = 106 + 180 = 286$

2. Determine the total mass contribution of oxygen:

   • There are $3$ oxygen atoms in the carbonate group and $10$ oxygen atoms in the ten water molecules of crystallisation, giving a total of $13$ oxygen atoms.
   • Total relative mass of oxygen $= 13 \times 16 = 208$.

3. Calculate the percentage:

   • $\frac{208}{286} \times 100\% = 72.727\% \approx 72.7\%$

Hence, option D is the correct answer.

Key Takeaways

• For hydrated compounds, remember to count atoms from both the anhydrous salt and the waters of crystallisation.
• The multiplier in front of $\text{H}_2\text{O}$ applies to both hydrogen and oxygen inside the water molecule.

Common Mistakes

• Only counting the oxygen in the carbonate ion ($3 \times 16 = 48$), which leads to $\frac{48}{286} \times 100\% = 16.8\%$ (Option A).
• Forgetting the water of crystallisation when calculating the $M_r$ and getting $\frac{48}{106} \times 100\% = 45.3\%$, or including only the oxygen in water ($10 \times 16 = 160$), which gives $\frac{160}{286} \times 100\% = 55.9\%$ (Option C).

Things to Be Careful About

• Ensure you multiply the oxygen in water by $10$ and add the $3$ oxygen atoms from the carbonate anion.

Walkthrough

To determine the charge carriers at each numbered position, we must consider the type of conductor present:

1. External Circuit (Wires): Metallic wires conduct electricity through the movement of free/delocalised electrons. Arrow 1 and Arrow 2 are both in the external wiring, so the charge carriers for both are electrons. Electrons travel from the negative terminal of the power supply to the cathode (Arrow 2), and from the anode back to the positive terminal of the power supply (Arrow 1).
2. Electrolyte (Molten $\text{PbBr}_2$): In a molten ionic compound, charge is carried by mobile ions rather than electrons. The left electrode is connected to the $+$ terminal of the power supply, making it the positive anode. Oppositely charged anions are attracted to it. In molten lead(II) bromide, the anions are bromide ions ($\text{Br}^-$). Arrow 3 points towards the positive electrode, so it represents the migration of bromide ions.
3. Electrode: Inert electrodes used in electrolysis are typically made of graphite or platinum. These solid conductors conduct electricity through the movement of electrons (Arrow 4) as bromide ions give up electrons at the anode surface:

These released electrons travel up through the anode and around the external circuit.

Combining these points gives the sequence: electrons, electrons, bromide ions, electrons, which matches option C.

Key Takeaways

• In metallic conductors and electrodes (like wires and graphite rods), electric current is the flow of delocalised electrons.
• In electrolytes (molten ionic compounds or aqueous solutions), electric current is the movement of mobile ions (cations towards the cathode, anions towards the anode).
• Electrons never flow through the electrolyte itself.

Common Mistakes

• Confusing the direction of ion flow: thinking lead ions ($\text{Pb}^{2+}$) move to the anode (left) instead of the cathode (right).
• Assuming electrons travel through the electrolyte solution (as suggested in option D).
• Believing ions travel into the external wire (as suggested in option A).

Things to Be Careful About

• Always check the $+/-$ signs on the power supply to correctly identify the anode ($+$) and cathode ($-$).
• Anions (negative ions) move to the Anode (positive electrode).
• Cations (positive ions) move to the Cathode (negative electrode).

Walkthrough

In the electrolysis of aqueous copper(II) sulfate ($\text{CuSO}_4(\text{aq})$):

1. Cathode (negative electrode): Both $\text{Cu}^{2+}$ and $\text{H}^+$ cations migrate to the cathode. In the reactivity series, copper is less reactive than hydrogen, so $\text{Cu}^{2+}$ ions are preferentially discharged by gaining electrons (reduction) to form solid copper metal: $$\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})$$
2. Anode (positive electrode): When inert electrodes (such as carbon/graphite or platinum) are used, hydroxide ions ($\text{OH}^-$) from water are discharged to form oxygen gas ($\text{O}_2$). However, this question specifies copper electrodes (active/reactive electrodes). The copper anode itself oxidises and dissolves into the electrolyte as aqueous copper(II) ions: $$\text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^-$$

Thus, the product formed at the anode is $\text{Cu}^{2+}(\text{aq})$ and at the cathode is copper metal ($\text{Cu}(\text{s})$).

Key Takeaways

• Inert vs. Active Electrodes: Inert electrodes (graphite, platinum) do not take part in the reaction. Active electrodes (such as copper) dissolve at the anode.
• Purification/Electroplating Principle: Using copper electrodes in $\text{CuSO}_4(\text{aq})$ transfers copper from the anode to the cathode, keeping the concentration of copper(II) ions in the solution constant.

Common Mistakes

• Confusing this with the electrolysis of $\text{CuSO}_4(\text{aq})$ using carbon/graphite electrodes, which produces oxygen gas ($\text{O}_2$) at the anode (Option C).
• Assuming hydrogen gas is evolved at the cathode; $\text{Cu}^{2+}$ is discharged in preference to $\text{H}^+$ because copper is lower in the reactivity series.

Things to Be Careful About

• Always check whether the electrodes are described as inert (carbon/graphite, platinum) or reactive/active (copper, silver).

Walkthrough

A hydrogen-oxygen fuel cell generates electricity by combining hydrogen and oxygen chemically through electrochemical reactions at the electrodes, rather than by burning the fuel.

1. Overall reaction:

Because water is the only product formed, statement A is completely correct.

1. Evaluating the distractors:

• B: Fuel cells do not involve direct combustion/burning; they operate via controlled oxidation and reduction half-reactions at electrodes.
• C: Fuel cells produce electricity (output) from chemical energy; they require a continuous supply of fuel and oxidant, not a continuous input of electrical energy (which is true for electrolysis, not cells).
• D: Fuel cells convert chemical energy directly to electrical energy without thermal energy losses associated with heat engines, making them more efficient than petrol engines.

Key Takeaways

• In a hydrogen-oxygen fuel cell, hydrogen is the fuel, oxygen is the oxidant, and water ($\text{H}_2\text{O}$) is the only chemical product.
• Fuel cells are non-polluting at the point of use and have higher energy conversion efficiencies compared to conventional combustion engines.

Common Mistakes

• Confusing electrochemical oxidation in a fuel cell with simple combustion (burning in a flame).
• Confusing fuel cells (which produce electricity) with electrolysis cells (which consume electricity).

Things to Be Careful About

• Ensure you distinguish between the continuous input required by fuel cells (hydrogen and oxygen fuels) versus the continuous output produced (electrical current and water).

Walkthrough

1. Analyse the reaction pathway diagram:

   • The horizontal level for the reactants is lower than the horizontal level for the products.
   • This means overall energy is absorbed from the surroundings into the chemical system, so the reaction is endothermic ($\Delta H$ is positive).

2. Evaluate each option to find the endothermic change:

   • Bond breaking is endothermic (requires energy intake).
   • Bond making is exothermic (releases energy).
   • In A ($2\text{H(g)} + \text{O(g)} \rightarrow \text{H}_2\text{O(g)}$), two $\text{O}-\text{H}$ covalent bonds are formed from separate atoms. Since only bond formation occurs, it is an exothermic process.
   • In B ($\text{H}_2\text{(g)} \rightarrow 2\text{H(g)}$), the covalent bond in an $\text{H}_2$ molecule is broken to produce isolated hydrogen atoms. Bond breaking absorbs energy, so this is an endothermic process.
   • In C ($\text{H}_2\text{O(g)} \rightarrow \text{H}_2\text{O(l)}$), water vapour condenses into liquid water, forming intermolecular forces, which is an exothermic physical change.
   • In D ($2\text{O(g)} \rightarrow \text{O}_2\text{(g)}$), a double covalent bond $\text{O}=\text{O}$ is formed from separate oxygen atoms, which is an exothermic process.

Thus, only reaction B corresponds to the diagram.

Key Takeaways

• In an endothermic profile, the energy of the products is higher than that of the reactants ($\Delta H > 0$).
• In an exothermic profile, the energy of the products is lower than that of the reactants ($\Delta H < 0$).
• "MEXO BENDO": Making bonds is EXOthermic; Breaking bonds is ENDOthermic.

Common Mistakes

• Confusing bond breaking and bond making (thinking that forming a bond requires energy rather than releasing it).
• Misreading the diagram as exothermic by focusing only on the downward slope from the top of the activation energy hump to the products level, rather than comparing the initial reactant level to the final product level.

Things to Be Careful About

• Always compare the baseline of the reactants on the left directly with the baseline of the products on the right to determine the overall enthalpy change ($\Delta H$). The peak between them represents the activation energy ($E_a$), not the net energy change.

Walkthrough

1. A physical change is a change in which no new chemical substances are produced. Changes of state (such as melting, freezing, boiling, condensing, evaporating) and separation methods (such as filtration, distillation, chromatography) are physical processes because the chemical identity of the particles remains unchanged.
2. A chemical change (or chemical reaction) involves the breaking and making of chemical bonds to form one or more new substances with different chemical and physical properties.
3. Evaluating the given choices:
   • Option A (distillation of aqueous ethanol) separates ethanol from water using differences in their boiling points. Both substances retain their original chemical formulas.
   • Option B (evaporation of water from aqueous sodium chloride) removes water vapor physically, leaving solid $\text{NaCl}$ crystals behind. No chemical bonds within the compounds are permanently altered into new chemical species.
   • Option C (melting of wax) only alters the arrangement and kinetic energy of the hydrocarbon molecules from a solid lattice to a liquid state; no new substance is formed.
   • Option D (rusting of iron) is a redox reaction in which metallic iron ($\text{Fe}$) reacts with oxygen ($\text{O}_2$) and moisture ($\text{H}_2\text{O}$) from the air to produce hydrated iron(III) oxide, $\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}$. A completely new substance is formed, making it a chemical change.

Therefore, option D is correct.

Key Takeaways

• In a physical change, no new substance is made and the process is usually easily reversible.
• In a chemical change, one or more new substances are formed, typically accompanied by detectable energy changes (temperature change, light), colour changes, or gas evolution.
• Rusting is a chemical oxidation process that forms hydrated iron(III) oxide.

Common Mistakes

• Confusing the melting of wax (physical change) with the burning/combustion of a wax candle (chemical change).
• Believing that dissolving or evaporation is a chemical reaction because the appearance of the solution changes.

Things to Be Careful About

• Be clear on the distinction between phase changes (melting, freezing, boiling) and chemical reactions (rusting, combustion, neutralisation, thermal decomposition).

Walkthrough

According to collision theory, for a chemical reaction to occur, reactant particles must collide with energy greater than or equal to the activation energy ($E_a$).

1. Effect of Concentration:
   Increasing the concentration of solutions (such as aqueous sodium thiosulfate and hydrochloric acid) increases the number of particles per unit volume. As the particles are crowded closer together, they collide more frequently, increasing the rate of successful collisions per unit time (collision frequency).

2. Effect of Pressure:
   Increasing the pressure on a gaseous mixture (such as nitrogen and hydrogen) decreases the volume available, thereby increasing the number of gas particles per unit volume. This also leads to an increased frequency of collisions between reacting molecules.

3. Activation Energy:
   Activation energy is an intrinsic property of the reaction pathway. Neither changing concentration nor changing pressure alters the activation energy; only the addition of a catalyst provides an alternative pathway with a lower activation energy.

Hence, row A is the correct option.

Key Takeaways

• Concentration and Pressure: Both increase the number of particles per unit volume, leading to a higher frequency of collisions between reacting particles.
• Catalysts: The only factor that changes (lowers) the activation energy of a reaction.

Common Mistakes

• Confusing factors that affect collision frequency (concentration, pressure, surface area) with factors that affect activation energy (catalysts) or the fraction of particles with $E \geq E_a$ (temperature).

Things to Be Careful About

• Ensure you describe collision changes as an increase in the frequency or rate of collisions (collisions per second/unit time), rather than just the total number of collisions.

Walkthrough

To choose the correct graph, two independent features must be determined:

1. The final volume of gas (the plateau level):

   • $\text{Moles of Mg} = \frac{24.0\text{ g}}{24\text{ g / mol}} = 1.0\text{ mol}$.
   • $\text{Moles of }\text{H}_2\text{SO}_4\text{ in exp 1} = \frac{100}{1000} \times 1.0 = 0.10\text{ mol}$.
   • $\text{Moles of }\text{H}_2\text{SO}_4\text{ in exp 2} = \frac{50}{1000} \times 2.0 = 0.10\text{ mol}$.

   Since $1\text{ mol}$ of $\text{Mg}$ reacts with $1\text{ mol}$ of $\text{H}_2\text{SO}_4$, the magnesium is in large excess in both experiments. The acid is the limiting reactant. Because both experiments supply the same number of moles of acid ($0.10\text{ mol}$), each reaction produces the same total volume of hydrogen gas ($0.10\text{ mol} \times 24\text{ dm}^3\text{ / mol} = 2.4\text{ dm}^3$). Therefore, both curves must level off at the identical height (eliminating C and D).

2. The initial rate of reaction (the initial steepness / gradient):

   • Experiment 2 uses $2.0\text{ mol / dm}^3$ acid, whereas experiment 1 uses $1.0\text{ mol / dm}^3$ acid.
   • A higher concentration contains more acid particles per unit volume, leading to a greater collision frequency between acid particles and the magnesium surface.
   • Hence, experiment 2 has a faster initial rate, corresponding to a steeper initial gradient and reaching the plateau in a shorter time (eliminating B).

This matches Graph A.

Key Takeaways

• The initial gradient of a volume-against-time graph represents the initial rate of reaction, which increases with higher reactant concentration, smaller particle size (surface area), higher temperature, or the addition of a catalyst.
• The final plateau (horizontal line) represents the total amount of product formed and is determined strictly by the moles of the limiting reactant, not by the rate of reaction.

Common Mistakes

• Mistaking a smaller total volume of solution ($50\text{ cm}^3$ vs $100\text{ cm}^3$) for less product formed, without calculating the moles ($n = c \times V$).
• Confusing the solid and dashed lines in the key and selecting B instead of A.

Things to Be Careful About

• Always convert volumes in $\text{cm}^3$ to $\text{dm}^3$ by dividing by $1000$ when using the formula $n = c \times V$.
• Check for excess reactants by comparing the mole ratio of the balanced chemical equation.

Walkthrough

In Cambridge O Level Chemistry, candidates are expected to recall the essential conditions and catalysts for key industrial processes:

1. The Contact Process produces sulfuric acid ($\text{H}_2\text{SO}_4$). In stage 2, sulfur dioxide is converted to sulfur trioxide:

This reaction requires:

• A catalyst of vanadium(V) oxide, $\text{V}_2\text{O}_5$
• A temperature of around $450\text{ }^\circ\text{C}$
• A pressure of $1\text{ to }2\text{ atm}$

1. The Haber Process produces ammonia ($\text{NH}_3$) from nitrogen and hydrogen:

This reaction requires:

• An iron catalyst (finely divided)
• A temperature of around $450\text{ }^\circ\text{C}$
• A pressure of around $200\text{ atm}$

Matching these two catalysts gives row C.

Key Takeaways

• Contact Process catalyst: Vanadium(V) oxide ($\text{V}_2\text{O}_5$).
• Haber Process catalyst: Iron ($\text{Fe}$).
• Hydrogenation of alkenes catalyst: Nickel ($\text{Ni}$) — often included as a distractor in these questions.

Common Mistakes

• Confusing the iron catalyst of the Haber process with nickel (nickel is used in the hydrogenation of vegetable oils/alkenes to make margarine/alkanes).
• Swapping the catalysts between the two processes.

Things to Be Careful About

• Ensure you check both columns carefully before selecting your answer to avoid swapping row pairings.

Walkthrough

To determine the correct colours of the indicators:

1. Read the pH from the titration curve:
   Locate $40\text{ cm}^3$ on the horizontal axis ("volume of strong alkali / $\text{cm}^3$"). Follow the line up to the curve and read across to the vertical axis. At $40\text{ cm}^3$, the $\text{pH}$ is well into the alkaline plateau region, approximately $\text{pH} = 12.4$.

2. Determine the colour of methyl orange:

   • Methyl orange is an acid-base indicator that changes colour over the $\text{pH}$ range of roughly $3.1$ to $4.4$.
   • In acidic solution below $\text{pH } 3.1$, it is red.
   • In neutral and alkaline solutions (above $\text{pH } 4.4$), it is yellow.
   • At $\text{pH } 12.4$, methyl orange is yellow.

3. Determine the colour of thymolphthalein:

   • Thymolphthalein changes colour in the alkaline region (roughly $\text{pH } 9.3$ to $10.5$).
   • In acidic and neutral solutions (below $\text{pH } 9.3$), it is colourless.
   • In strongly alkaline solutions (above $\text{pH } 10.5$), it is blue.
   • At $\text{pH } 12.4$, thymolphthalein is blue.

Matching these two observations gives yellow for methyl orange and blue for thymolphthalein, corresponding to option D.

Key Takeaways

• For Cambridge O Level (5070), candidates should know the colours of standard indicators in both acidic and alkaline conditions:
  • Litmus: red in acid, blue in alkali
  • Methyl orange: red in acid, yellow in alkali
  • Thymolphthalein: colourless in acid, blue in alkali
  • Phenolphthalein: colourless in acid, pink in alkali
• When alkali is added beyond the equivalence point (here around $25\text{ cm}^3$), the mixture becomes strongly alkaline with high $\text{pH}$.

Common Mistakes

• Confusing the colours of methyl orange: candidates often mistakenly think methyl orange turns orange or red in alkali.
• Confusing thymolphthalein with phenolphthalein: phenolphthalein turns pink in alkali, whereas thymolphthalein turns blue.
• Looking at the start of the titration curve (pH 2.5–3) rather than at the specified volume of $40\text{ cm}^3$.

Things to Be Careful About

• Ensure you check the exact volume of titrant stated in the question stem ($40\text{ cm}^3$) and not the equivalence point ($25\text{ cm}^3$) or initial point ($0\text{ cm}^3$).

Walkthrough

An acid is defined as a proton ($\text{H}^+$) donor that dissociates in water to produce hydrogen ions.

1. Weak acids:

   • Methanoic acid ($\text{HCOOH}$) is a weak organic acid (a carboxylic acid).
   • In aqueous solution, weak acids ionise only partially. Most of the molecules remain undissociated, and an equilibrium is established between the un-ionised molecules and the ions.
   • Because this process is reversible, it is represented with a reversible reaction arrow ($\rightleftharpoons$): $$\text{HCOOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{HCOO}^-\text{(aq)}$$

2. Strong acids:

   • Nitric acid ($\text{HNO}_3$) is a strong mineral acid.
   • Strong acids ionise completely in aqueous solution. Essentially no undissociated $\text{HNO}_3$ molecules remain in solution.
   • Because the ionisation goes to completion, it is represented with a single forward arrow ($\rightarrow$): $$\text{HNO}_3\text{(aq)} \rightarrow \text{H}^+\text{(aq)} + \text{NO}_3^-\text{(aq)}$$

Comparing with the options in the table:

• Option A incorrectly uses a single arrow for methanoic acid.
• Option B incorrectly uses a single arrow for methanoic acid and a reversible arrow for nitric acid.
• Option C incorrectly uses a reversible arrow for nitric acid.
• Option D correctly uses $\rightleftharpoons$ for methanoic acid and $\rightarrow$ for nitric acid.

Key Takeaways

• Strong acid: An acid that fully/completely dissociates (ionises) in aqueous solution (represented by $\rightarrow$). Common examples include $\text{HCl}$, $\text{HNO}_3$, and $\text{H}_2\text{SO}_4$.
• Weak acid: An acid that only partially dissociates (ionises) in aqueous solution (represented by $\rightleftharpoons$). Common examples include carboxylic acids (e.g., ethanoic acid, methanoic acid) and carbonic acid ($\text{H}_2\text{CO}_3$).