Chemistry 5070/21 — October/November 2025

Walkthrough

Methanol is the alcohol with one carbon atom, represented by the formula $\text{CH}_3\text{OH}$. Looking at the provided options, compound A is $\text{CH}_3\text{OH}$.

Key Takeaways

• Alcohols contain the hydroxyl functional group ($-\text{OH}$).
• The prefix 'meth-' indicates a continuous chain of 1 carbon atom.

Common Mistakes

• Confusing methanol ($ext{CH}_3 ext{OH}$, compound A) with ethanol ($ext{CH}_3 ext{CH}_2 ext{OH}$, compound B).

Things to Be Careful About

• Ensure you match the single-carbon alcohol correctly to the formula.

Walkthrough

The fermentation of aqueous glucose using yeast produces ethanol ($ext{CH}_3 ext{CH}_2 ext{OH}$) and carbon dioxide:

Compound B is ethanol ($ext{CH}_3\text{—}\text{CH}_2\text{OH}$).

Key Takeaways

• Ethanol is produced industrially by either the fermentation of glucose or the catalytic hydration of ethene.

Common Mistakes

• Selecting methanol (A) instead of ethanol (B).

Things to Be Careful About

• Fermentation specifically yields ethanol, not other alcohols.

Walkthrough

Carboxylic acids are characterised by the carboxyl functional group ($-\text{COOH}$). Among the given compounds, compound F ($ext{CH}_3\text{—}\text{CH}_2\text{—}\text{CH}_2\text{—}\text{CH}_2\text{—}\text{COOH}$, pentanoic acid) contains the $-\text{COOH}$ group.

Key Takeaways

• Carboxylic acids have the general formula $\text{C}_n\text{H}_{2n+1}\text{COOH}$ and contain the $-\text{COOH}$ functional group.

Common Mistakes

• Confusing the ester group ($-\text{COO}-$) in E or G with the carboxylic acid group ($-\text{COOH}$) in F.

Things to Be Careful About

• Check the ending of the structural formula carefully: $-\text{COOH}$ vs. $-\text{COOCH}_3$.

Walkthrough

A hydrocarbon is a compound that contains hydrogen and carbon only. A saturated hydrocarbon contains only single carbon-carbon bonds (no $\text{C}=\text{C}$ double bonds).

• Compounds A, B, E, F, and G contain oxygen, so they are not hydrocarbons.
• Compound D ($ext{CH}_3\text{—}\text{CH}=\text{CH}_2$) is an alkene containing a $\text{C}=\text{C}$ double bond, making it unsaturated.
• Compound C (cyclopropane, a 3-membered ring of $\text{CH}_2$ groups) contains only carbon and hydrogen, and all carbon-carbon bonds are single bonds, so it is a saturated hydrocarbon.

Key Takeaways

• Saturated: contains only single $\text{C}-\text{C}$ bonds.
• Hydrocarbon: contains carbon and hydrogen only.

Common Mistakes

• Selecting D by confusing saturated with unsaturated.
• Forgetting that cycloalkanes are saturated hydrocarbons even though they have the general formula $\text{C}_n\text{H}_{2n}$.

Things to Be Careful About

• Ensure the chosen compound contains NO oxygen atoms.

Walkthrough

When methanol ($ext{CH}_3 ext{OH}$) reacts with propanoic acid ($ext{CH}_3 ext{CH}_2 ext{COOH}$) in an esterification reaction, the ester formed is methyl propanoate:

Compound E is $\text{CH}_3\text{—}\text{CH}_2\text{—}\text{COOCH}_3$, which is methyl propanoate.

Key Takeaways

• Esterification: $\text{carboxylic acid} + \text{alcohol} \rightleftharpoons \text{ester} + \text{water}$.
• The acid provides the alkanoate part ($ext{CH}_3 ext{CH}_2\text{COO}-$) and the alcohol provides the alkyl group ($-\text{CH}_3$).

Common Mistakes

• Choosing G ($ext{CH}_3\text{—}\text{CH}=\text{CH—}\text{COOCH}_3$), which contains a $\text{C}=\text{C}$ double bond and is formed from an unsaturated acid (but-2-enoic acid), not propanoic acid.

Things to Be Careful About

• Count the number of carbon atoms in the acid chain (propanoic acid has 3 carbons: $\text{CH}_3\text{CH}_2\text{CO}-$).

Walkthrough

The diagram plots energy on the y-axis against the progress of the reaction. Arrow A points from the reactant energy level up to the peak of the curve. This peak represents the transition state, and the energy difference between the reactants and the peak is the activation energy ($E_a$). Arrow B points from the reactant energy level up to the product energy level. This difference represents the overall enthalpy change ($\Delta H$) of the reaction.

Key Takeaways

On a reaction pathway diagram, the height from reactants to the peak is the activation energy, and the height from reactants to products is the enthalpy change.

Common Mistakes

Confusing activation energy with enthalpy change, or writing $\Delta H$ for arrow A. Activation energy is always measured from the reactants to the peak.

Things to Be Careful About

Both 'activation energy' and '$E_a$' are accepted for A. Both 'enthalpy change' and '$\Delta H$' are accepted for B. Ensure you match the label to the correct arrow.

Walkthrough

An endothermic reaction is one where the products have more energy than the reactants, meaning energy has been absorbed from the surroundings. On the diagram, the horizontal line for the products ($\text{Na}_2\text{CO}_3(\text{s}) + \text{H}_2\text{O}(\text{g}) + \text{CO}_2(\text{g})$) is positioned higher on the energy axis than the horizontal line for the reactants ($2\text{NaHCO}_3(\text{s})$).

Key Takeaways

If the product line is above the reactant line on an energy profile, the reaction is endothermic. If it is below, the reaction is exothermic.

Common Mistakes

Saying 'energy is absorbed' without referencing the diagram. The question specifically asks to 'use the reaction pathway diagram', so you must describe the relative positions of the energy levels.

Things to Be Careful About

Keep the explanation concise. One clear statement comparing the product and reactant energy levels is all that is required for the mark.

Walkthrough

In the liquid state:

1. Arrangement: Particles (or molecules in molecular substances such as bromine, $\text{Br}_2$) are close together with most touching, but they are arranged randomly without a regular, fixed repeating pattern.
2. Motion: Particles have sufficient kinetic energy to overcome fixed positions, so they slide, flow, or move past each other continuously.

Giving both the random/touching arrangement and the sliding/flowing motion secures the 2 marks.

Key Takeaways

• In a liquid, particles are closely packed and touching (unlike a gas) but irregularly/randomly arranged (unlike a regular lattice in a solid).
• Liquid particles are free to slide and move over each other, allowing liquids to flow and take the shape of their container.

Common Mistakes

• Describing particles in a liquid as "far apart" (which is only true for gases).
• Describing movement merely as "vibrating about fixed positions" (which describes solids).

Things to Be Careful About

• Ensure you describe both arrangement (random / touching) and motion (sliding / moving over each other) clearly.

Walkthrough

Bromine exists as simple covalent molecules ($\text{Br}_2$). When liquid bromine boils, the strong covalent bonds within the bromine molecules are not broken. Instead, it is only the weak intermolecular forces (forces between separate $\text{Br}_2$ molecules) that are overcome. Because these intermolecular attractions are weak, very little heat/thermal energy is required to separate the molecules, resulting in a low boiling point.

Key Takeaways

• Simple molecular substances have low melting and boiling points because the intermolecular forces between molecules are weak.
• Covalent bonds within the molecules are strong and remain intact during boiling.

Common Mistakes

• Stating that "covalent bonds are weak and break easily" — covalent bonds are very strong; boiling does not break them.
• Omitting the key term "intermolecular".

Things to Be Careful About

• Always specify "weak forces between molecules" or "weak intermolecular forces" rather than simply "weak bonds".

Walkthrough

Potassium fluoride ($\text{KF}$) is an ionic compound composed of a giant ionic lattice of alternating $\text{K}^+$ and $\text{F}^-$ ions held together by strong electrostatic attractions.

Characteristic physical properties of ionic compounds include:

1. High melting and boiling points (due to strong electrostatic attractions between oppositely charged ions throughout the giant lattice).
2. Electrical conductivity when molten or aqueous (ions become free to move and carry charge; does not conduct when solid because ions are in fixed positions).
3. Solubility in water (most simple ionic compounds dissolve in water).

Giving any two standard properties scores full marks.

Key Takeaways

• Giant ionic structures exhibit high melting/boiling points, conduct electricity only when liquid/molten or dissolved in water, and are usually solid at room temperature.

Common Mistakes

• Writing "conducts electricity" without specifying when molten or in aqueous solution (ionic solids do not conduct electricity).

Things to Be Careful About

• Always qualify electrical conductivity for ionic compounds: specify "molten" or "in aqueous solution".

Walkthrough

Potassium is a Group I metal with 1 valence electron. To achieve a stable electron configuration, a potassium atom loses 1 electron to form a potassium cation with a $1+$ charge:

Key Takeaways

• Metals lose valence electrons during oxidation to form positive cations.
• The lost electron(s) are written on the product side (right-hand side) of the half-equation.

Common Mistakes

• Writing $\text{K} - \text{e}^- \rightarrow \text{K}^+$ (Cambridge conventions require electrons to be added on the appropriate side using $+$, not subtracted).
• Incorrect charge on the potassium ion (e.g., $\text{K}^{2+}$).

Things to Be Careful About

• Ensure the number of atoms and total charges balance on both sides (LHS: charge 0; RHS: $+1 + (-1) = 0$).

Walkthrough

Fluorine gas exists as diatomic molecules, $\text{F}_2$. Each fluorine atom needs to gain 1 electron to complete its outer shell and form a fluoride ion ($\text{F}^-$). Therefore, one $\text{F}_2$ molecule gains 2 electrons to form 2 fluoride ions:

Key Takeaways

• Non-metals gain electrons (reduction) to form negative anions.
• Halogens exist as diatomic molecules ($\text{X}_2$), so 2 electrons are needed per halogen molecule to yield 2 halide ions ($2\text{X}^-$).

Common Mistakes

• Writing $\text{F} + \text{e}^- \rightarrow \text{F}^-$ when the question specifically asked for the formation from fluorine molecules ($\text{F}_2$).
• Writing $\text{F}_2^-$ instead of $2\text{F}^-$.

Things to Be Careful About

• Check charge and atom balance: LHS charge is $0 + 2(-1) = -2$; RHS charge is $2(-1) = -2$.

Walkthrough

Chlorine is higher in Group VII than iodine, meaning chlorine is more reactive than iodine. A more reactive halogen displaces a less reactive halide from its aqueous solution.

The word equation is:

The chemical equation is:

Therefore, the two products formed are potassium chloride and iodine.

Key Takeaways

• Halogen reactivity decreases down Group VII.
• A more reactive halogen displaces a less reactive halide ion from solution, liberating the less reactive halogen element and forming a new halide salt.

Common Mistakes

• Writing "potassium chlorate" or confusing the names "iodine" and "iodide".

Things to Be Careful About

• The question asks for the names of the products, not their formulae.

Walkthrough

Aqueous potassium iodide ($\text{KI}$) is a colourless solution. When chlorine gas is bubbled into it, chlorine oxidises iodide ions ($\text{I}^-$) to form aqueous iodine ($\text{I}_2$). Aqueous iodine in the presence of iodide forms a brown/red-brown (or yellow/orange-brown) solution.

Therefore, the visible observation is that the colourless solution turns brown.

Key Takeaways

• Aqueous iodide ions ($\text{I}^-$) are colourless.
• Aqueous iodine ($\text{I}_2$) appears yellow, orange, or brown depending on its concentration.

Common Mistakes

• Stating that a "purple gas" is seen (iodine vapour is purple, but in aqueous solution at room temperature, it dissolves to form a brown solution).
• Giving a conclusion rather than an observation (e.g., writing "iodine is formed" instead of describing what is seen).

Things to Be Careful About

• An observation requires what you actually see (a colour change), not the chemical name of the product.

Walkthrough

Metals consist of a giant lattice of positive ions surrounded by a 'sea' of delocalised valence electrons. When a potential difference is applied, these delocalised electrons are free to move throughout the metallic lattice, conducting electric charge.

Key Takeaways

• Electrical conductivity in metals is due to mobile (delocalised) electrons.
• In ionic compounds when molten or aqueous, conductivity is due to mobile ions, NOT electrons.

Common Mistakes

• Saying "ions can move" (ions are fixed in the lattice; only electrons move in solid metals).
• Simply saying "it contains electrons" without mentioning that they are delocalised or free to move.

Things to Be Careful About

• Always mention that the electrons can move / are mobile.

Walkthrough

In pure metals, atoms/ions are arranged in regular layers. When a tensile force is applied (pulling into a wire, i.e., ductility), these layers of positive ions can slide past one another without disrupting the non-directional metallic bonding holding the structure together.

Key Takeaways

• Ductility (drawn into wires) and malleability (hammered into sheets) occur because layers of positive metal ions can slide over one another.
• The metallic bond is non-directional, so it remains intact as layers slide.

Common Mistakes

• Referring to "layers of molecules" or "layers of atoms sliding" (the mark scheme specifically expects "layers of ions" or "layers of metal atoms/ions").
• Forgetting to mention the sliding of layers.

Things to Be Careful About

• Clearly state that it is the layers of ions that slide.

Walkthrough

1. Identify the Full Equation: Nickel reacts with hydrochloric acid:
   $\text{Ni}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{NiCl}_2(\text{aq}) + \text{H}_2(\text{g})$
2. Separate Aqueous Species into Ions:
   $\text{Ni}(\text{s}) + 2\text{H}^+(\text{aq}) + 2\text{Cl}^-(\text{aq}) \rightarrow \text{Ni}^{2+}(\text{aq}) + 2\text{Cl}^-(\text{aq}) + \text{H}_2(\text{g})$
3. Cancel Spectator Ions: Chloride ions ($\text{Cl}^-$) are spectators:
   $\text{Ni}(\text{s}) + 2\text{H}^+(\text{aq}) \rightarrow \text{Ni}^{2+}(\text{aq}) + \text{H}_2(\text{g})$
4. Check Charges and State Symbols: Charge on LHS is $+2$, charge on RHS is $+2$. States are $(\text{s})$, $(\text{aq})$, $(\text{aq})$, and $(\text{g})$.

Key Takeaways

• Full acid + metal reactions give a salt + hydrogen.
• The ionic equation for a metal reacting with dilute acid involves the oxidation of the metal to metal ions and reduction of $\text{H}^+$ ions to $\text{H}_2$ gas.

Common Mistakes

• Writing a full molecular equation containing $\text{Cl}^-$ ions instead of the net ionic equation.
• Omitting state symbols or writing incorrect ones (e.g. $\text{Ni}^{2+}(\text{s})$).
• Forgetting to balance the hydrogen ions ($2\text{H}^+$).

Things to Be Careful About

• Ensure the charge balance is satisfied in addition to atom balance.

Walkthrough

In part (d)(i), we learned that nickel reacts with dilute acid to produce hydrogen gas, which shows that nickel is above hydrogen in the reactivity series. Copper is below hydrogen in the reactivity series and does not react with dilute acids. Therefore, nickel is more reactive than copper and will displace $\text{Cu}^{2+}$ ions from aqueous solution:
$\text{Ni}(\text{s}) + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Ni}^{2+}(\text{aq}) + \text{Cu}(\text{s})$

Key Takeaways

• A more reactive metal displaces a less reactive metal from a solution of its ions/salt.
• Since nickel reacts with dilute acid (releasing $\text{H}_2$), it is more reactive than hydrogen and copper.

Common Mistakes

• Stating that copper is more reactive than nickel.
• Predicting 'no' based on nickel reacting 'slowly' with acid without comparing it to copper.

Things to Be Careful About

• Both prediction ('yes') and explanation ('nickel is more reactive than copper') are required to secure the mark.

Walkthrough

First, calculate the relative molecular mass ($M_r$) of ethyne ($\text{C}_2\text{H}_2$). Carbon has $A_r = 12$ and hydrogen has $A_r = 1$, so $M_r = (2 \times 12) + (2 \times 1) = 26$. Next, use the formula $n = m / M_r$ to find the number of moles: $0.130 / 26 = 0.005$ mol. Finally, multiply the number of moles by Avogadro's constant ($6.02 \times 10^{23}$) to find the number of molecules: $0.005 \times 6.02 \times 10^{23} = 3.01 \times 10^{21}$.

Key Takeaways

• $M_r$ is the sum of the relative atomic masses of all atoms in a molecule.
• The mole connects mass to number of particles via Avogadro's constant.

Common Mistakes

• Using the wrong $M_r$ (e.g., forgetting to multiply by 2 for carbon or hydrogen).
• Forgetting to multiply moles by Avogadro's constant.
• Incorrect handling of scientific notation.

Things to Be Careful About

• Ensure the final answer is in the correct form (scientific notation with appropriate significant figures).

Walkthrough

Ethyne has the formula $\text{C}_2\text{H}_2$, meaning each molecule contains 2 carbon atoms and 2 hydrogen atoms, for a total of 4 atoms per molecule. Multiply the number of molecules calculated in part (i) by 4 to get the total number of atoms: $4 \times 3.01 \times 10^{21} = 1.204 \times 10^{22}$.

Key Takeaways

• Total atoms = (number of molecules) × (atoms per molecule).

Common Mistakes

• Forgetting to count all atoms in the molecule (e.g., only counting carbon atoms).
• Carrying forward an error from part (i) without correction.

Things to Be Careful About

• Use the unrounded value from part (i) if possible, though $3.01 \times 10^{21}$ is exact here.

Walkthrough

Cracking is the thermal decomposition process used in the petrochemical industry to break down large, less useful long-chain hydrocarbon molecules (such as $\text{C}_{12}\text{H}_{26}$) into smaller, more useful short-chain alkanes and alkenes like ethene ($\text{C}_2\text{H}_4$).

Key Takeaways

• Cracking breaks covalent bonds in larger hydrocarbon molecules to produce smaller molecules, including alkenes containing $\text{C}=\text{C}$ double bonds.

Common Mistakes

• Confusing cracking with 'fractional distillation' (which separates hydrocarbons based on boiling points without breaking chemical bonds).

Things to Be Careful About

• Ensure the term 'cracking' (or 'catalytic / thermal cracking') is spelled correctly.

Walkthrough

Catalytic cracking requires strong heating (high temperatures, typically $450^\circ\text{C}$ to $700^\circ\text{C}$) to provide the energy needed to break strong $\text{C}-\text{C}$ covalent bonds, along with a solid catalyst (such as $\text{Al}_2\text{O}_3$, $\text{SiO}_2$, or zeolite) to speed up the process and allow it to occur at lower operating temperatures.

Key Takeaways

• Industrial catalytic cracking conditions: high temperature / heat and a catalyst (or steam/thermal cracking at very high temperature).

Common Mistakes

• Writing 'high pressure' — cracking does not require high pressure (it is typically carried out at atmospheric or slight positive pressure).

Things to Be Careful About

• Make sure to give both distinct conditions required for the two marks: temperature (heat) and catalyst.

Walkthrough

When temperature is increased:

1. The gas particles gain kinetic energy and move faster.
2. A much higher fraction of colliding particles possess energy equal to or exceeding the activation energy ($E_a$).
3. Consequently, the frequency of successful (effective) collisions increases, causing the rate of the forward reaction to increase.

Key Takeaways

• Higher temperature increases particle speed / kinetic energy.
• The crucial factor for the rate increase is that more collisions possess energy $\geq E_a$ (effective collisions).

Common Mistakes

• Stating only that 'collisions increase' without mentioning kinetic energy or that collisions are 'successful' / have energy $\geq E_a$.
• Confusing the effect on reaction rate with the shift in equilibrium position.

Things to Be Careful About

• Ensure both the effect (rate increases) and the collision theory reasoning are clearly stated.

Walkthrough

1. Look at the equation:

1. Count the moles of gas on each side:

• Reactants: $1 + 1 = 2\text{ moles of gas}$
• Products: $1\text{ mole of gas}$

1. According to Le Chatelier's principle, increasing the pressure favours the side with fewer moles of gas in order to reduce the pressure. Therefore, the equilibrium shifts to the right (favouring the formation of ethanol).

Key Takeaways

• Increasing pressure shifts equilibrium towards the side with fewer moles of gas (smaller gas volume).

Common Mistakes

• Forgetting that $\text{H}_2\text{O}(\text{g})$ is a gas in this industrial hydration reaction and not counting it as a gaseous mole.

Things to Be Careful About

• State both the direction of shift ('shifts to the right' or 'forward direction') and the comparison of moles/volumes of gas.

Walkthrough

During the addition polymerisation of ethene, the double bond ($\text{C}=\text{C}$) in each monomer breaks open to form single bonds ($\text{C}-\text{C}$) linking the units into a long carbon chain.

1. One repeat unit consists of $-\text{CH}_2-\text{CH}_2-$.
2. Two repeat units require a continuous chain of 4 carbon atoms joined by single bonds ($-\text{C}-\text{C}-\text{C}-\text{C}-$).
3. Each carbon atom is bonded to 2 hydrogen atoms.
4. Continuation bonds (extension single bonds) must extend outwards through each end of the chain to show that the polymer continues.

Key Takeaways

• Addition polymerisation converts $\text{C}=\text{C}$ double bonds into single bonds in the polymer backbone.
• Two repeat units of ethene contain 4 carbon atoms in the main chain.

Common Mistakes

• Leaving double bonds between the carbons in the polymer.
• Omitting continuation/extension bonds at the ends of the chain.
• Drawing brackets with $n$ instead of showing the required two repeat units explicitly.

Things to Be Careful About

• Ensure every carbon atom forms exactly 4 single bonds.

Walkthrough

Poly(ethene) is a non-biodegradable polymer because its strong $\text{C}-\text{C}$ and $\text{C}-\text{H}$ single bonds make it chemically unreactive and resistant to breakdown by microorganisms.
This leads to major environmental challenges:

1. Landfill issues: Takes up large amounts of landfill space without decomposing.
2. Marine pollution: Accumulates in oceans and waterways, choking or entangling marine organisms.
3. Incineration issues: Burning plastics can produce toxic fumes (e.g. carbon monoxide, acidic gases, particulates).

Key Takeaways

• Plastics are non-biodegradable, leading to persistent environmental accumulation in landfills and oceans, and produce toxic emissions upon incineration.

Common Mistakes

• Vague answers like 'causes pollution' without specifying how (e.g. landfills, oceans, toxic gases).
• Stating that plastics deplete the ozone layer.

Things to Be Careful About

• Give two clearly distinct environmental issues as prompted by 1 and 2.

Walkthrough

An addition reaction is defined chemically as a reaction in which two (or more) molecules combine across a double (or multiple) bond to form a single product with no other by-products.

Key Takeaways

• In addition reactions: $\text{Reactant A} + \text{Reactant B} \rightarrow \text{Single Product}$.

Common Mistakes

• Confusing addition with substitution (where an atom is replaced and two products are formed) or condensation (where a small molecule like water is eliminated).

Things to Be Careful About

• State clearly that only one product is formed.

Walkthrough

When ethene reacts with hydrogen ($\text{H}_2$) in the presence of a nickel catalyst (hydrogenation), the $\text{C}=\text{C}$ double bond is converted to a $\text{C}-\text{C}$ single bond, and one hydrogen atom adds to each carbon atom.
The product is ethane, $\text{C}_2\text{H}_6$.
A displayed formula must show all atoms and all covalent bonds explicitly:

• Two carbon atoms linked by a single line ($\text{C}-\text{C}$).
• Each carbon atom bonded to 3 hydrogen atoms via single lines.

Key Takeaways

• Hydrogenation of an alkene produces the corresponding alkane.
• A displayed formula shows all atoms and all bonds.

Common Mistakes

• Writing a structural formula (e.g. $\text{CH}_3\text{CH}_3$) instead of a displayed formula with explicit bond lines.
• Leaving a double bond between the carbons.

Things to Be Careful About

• Ensure every $\text{C}-\text{H}$ and $\text{C}-\text{C}$ bond is drawn as a line.