Chemistry 5070/12 — October/November 2024

Walkthrough

According to the kinetic particle theory:

1. Gas particles (in this case, monoatomic noble gas atoms of helium, $\text{He}$, and argon, $\text{Ar}$) have negligible forces of attraction between them.
2. The particles possess high kinetic energy and move constantly, randomly, and rapidly in all directions.
3. Through the process of diffusion—the net movement of particles from a region of higher concentration to a region of lower concentration as a result of their random motion—the gases completely and uniformly mix with each other to form a homogeneous mixture.
4. Even though helium atoms have a smaller relative atomic mass ($A_r = 4$) and therefore diffuse faster than argon atoms ($A_r = 40$), the continuous collisions and random movement mean that both gases will spread out uniformly throughout the entire container rather than settling into separate layers.

Thus, option A is the correct statement.

Key Takeaways

• Gases expand to completely and uniformly fill any container they are placed in.
• Diffusion causes different gases to mix homogeneously regardless of differences in particle mass or density.
• The kinetic energy of gas particles at room temperature is far greater than the effect of gravity, preventing them from settling out.

Common Mistakes

• Confusing the rate of diffusion with the final equilibrium state: while lighter gas particles diffuse faster, both light and heavy gases ultimately mix evenly.
• Thinking that gases separate into layers like immiscible liquids of different densities.

Things to Be Careful About

• Relative atomic mass affects the rate (speed) of diffusion (lighter particles diffuse faster at the same temperature), but it does not prevent complete and even mixing once mixed.

Walkthrough

To determine the correct row, we evaluate the possible types of chemical classification (element vs. compound) for both structure types:

1. Substance X has a simple molecular structure:

   • A simple molecular substance consists of discrete molecules held together by covalent bonds, with weak intermolecular forces between the molecules.
   • Elements: Many non-metal elements exist as simple molecules, such as diatomic gases ($\text{H}_2$, $\text{O}_2$, $\text{N}_2$, $\text{Cl}_2$) and polyatomic molecules ($\text{P}_4$, $\text{S}_8$, buckminsterfullerene $\text{C}_{60}$).
   • Compounds: Many non-metal compounds exist as simple molecules (e.g. $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{CO}_2$, $\text{CH}_4$).
   • Hence, X can be either an element or a compound.

2. Substance Y has a giant covalent structure:

   • A giant covalent structure (macromolecule) consists of a three-dimensional lattice where all atoms are linked by strong covalent bonds.
   • Elements: Carbon exists in giant covalent allotropes including diamond and graphite.
   • Compounds: Silicon dioxide (silica, $\text{SiO}_2$) is a giant covalent compound composed of silicon and oxygen atoms.
   • Hence, Y can also be either an element or a compound.

Therefore, row D is the correct answer.

Key Takeaways

• Both simple molecular and giant covalent structures can exist as either pure elements (composed of only one type of atom) or chemical compounds (composed of two or more different elements chemically combined).
• Classic O Level examples to remember:
  • Simple molecular element: $\text{O}_2$, $\text{Cl}_2$
  • Simple molecular compound: $\text{H}_2\text{O}$, $\text{CO}_2$
  • Giant covalent element: diamond, graphite (carbon)
  • Giant covalent compound: silicon dioxide ($\text{SiO}_2$)

Common Mistakes

• Assuming giant covalent structures are only elements because diamond and graphite are the most frequently discussed examples, forgetting silicon dioxide (sand/quartz).
• Assuming simple molecular substances must be compounds because water and carbon dioxide are the most common examples, overlooking diatomic elemental gases like $\text{O}_2$ and $\text{N}_2$.

Things to Be Careful About

• Ensure you clearly distinguish between the structural type (simple molecular vs. giant covalent) and the chemical classification (element vs. compound).

Walkthrough

To find the correct chemical symbol (nuclide notation) $^{A}_{Z}\text{Z}$ for an atom, we need to determine its proton number ($Z$) and its mass number ($A$):

1. Determine the proton number ($Z$):
   An atom is electrically neutral, meaning the number of positive protons in the nucleus must equal the total number of negative electrons orbiting the nucleus.

   • Looking at the electron shells in the diagram: $2$ electrons in the first shell, $8$ in the second, and $3$ in the third shell.
   • Total number of electrons $= 2 + 8 + 3 = 13$.
   • Therefore, the number of protons $= 13$, which gives the bottom number (atomic number, $Z = 13$).

2. Determine the mass number ($A$):
   The mass number (nucleon number) is the total number of protons and neutrons in the nucleus.

   • The nucleus is labelled as containing $15\text{ neutrons}$.
   • $\text{Mass number } (A) = \text{protons} + \text{neutrons} = 13 + 15 = 28$.
   • This gives the top number ($A = 28$).

Combining these into the standard notation $^{A}_{Z}\text{Z}$ gives $^{28}_{13}\text{Z}$, which corresponds to option C.

Key Takeaways

• In a neutral atom: $\text{number of protons} = \text{number of electrons}$.
• Atomic number (proton number, $Z$) is the lower number in nuclide notation: $Z = \text{number of protons}$.
• Mass number (nucleon number, $A$) is the upper number in nuclide notation: $A = \text{number of protons} + \text{number of neutrons}$.

Common Mistakes

• Inverting the notation by putting the atomic number at the top and mass number at the bottom (as in option A).
• Forgetting to add the protons to the neutrons and writing the neutron number as the mass number (as in option D).
• Assuming the mass number is simply double the proton number ($26$) instead of using the provided neutron count (as in option B).

Things to Be Careful About

• Ensure you carefully count all the electrons across every shell without missing the unpaired electron on the outer shell.
• Remember that the mass number represents all nucleons (protons + neutrons), not just the neutrons.

Walkthrough

1. Identify the isotope masses and their relative abundances from the graph:

   • The peak at relative mass $20$ has a height of $1$.
   • The peak at relative mass $22$ has a height of $5$.

2. Apply the formula for relative atomic mass ($A_r$):
   Relative atomic mass is the weighted average mass of naturally occurring atoms of an element on a scale where an atom of $^{12}\text{C}$ has a mass of exactly $12$ units.

   $$A_r = \frac{\sum (\text{isotopic mass} \times \text{relative abundance})}{\text{total relative abundance}}$$

3. Substitute the values:

   $$A_r = \frac{(20 \times 1) + (22 \times 5)}{1 + 5} = \frac{130}{6} = 21.666... \approx 21.7$$

4. Check the result against intuition:
   Since the isotope of mass $22$ is much more abundant (5 out of 6 parts) than the isotope of mass $20$ (1 out of 6 parts), the average must lie much closer to $22$ than to $20$. A value of $21.7$ makes complete physical sense.

Key Takeaways

• The relative atomic mass ($A_r$) is a weighted mean that takes into account both the mass and the relative abundance of each isotope.
• When abundances are given as ratios or peak heights (not percentages), divide by the sum of the abundances rather than $100$.

Common Mistakes

• Dividing by $2$ instead of the total abundance ($6$): $\frac{20 + 22}{2} = 21.0$ (Option A), which incorrectly assumes equal abundance.
• Dividing by $100$ automatically without noticing that the abundances are given as simple ratios ($1$ and $5$), not percentages.
• Misreading the heights of the bars on the vertical axis.

Things to Be Careful About

• Always find the sum of the abundances first before carrying out the division.
• Round your final answer to the appropriate number of decimal places or significant figures as presented in the multiple-choice options (here, 1 decimal place).

Walkthrough

For a substance to conduct electricity, it must contain mobile charge carriers—either free-moving ions or delocalised electrons.

• In metals, conduction is caused by the movement of delocalised electrons through a regular lattice of fixed positive metal ions. Therefore, statement D is incorrect.
• In molten or aqueous ionic compounds, conduction is caused by the movement of mobile ions towards oppositely charged electrodes, not by electrons. Therefore, statement C is incorrect.
• In simple covalent compounds (such as glucose), there are no free ions or delocalised electrons; they consist of uncharged molecules and therefore cannot conduct electricity in the solid, liquid, or dissolved state. Therefore, statement A is incorrect.
• In aqueous solutions of acids (such as sulfuric acid, $\text{H}_2\text{SO}_4$), the acid dissociates in water into mobile hydrogen ions ($\text{H}^+$) and sulfate ions ($\text{SO}_4^{2-}$). Because these ions are free to move throughout the solution, dilute acids conduct electricity. Thus, statement B is correct.

Key Takeaways

• Electrical conductivity in metals and graphite is due to the movement of delocalised electrons.
• Electrical conductivity in molten ionic compounds and aqueous electrolytes (including acids and alkalis) is due to the movement of mobile ions.
• Solid ionic compounds and simple molecular covalent compounds do not conduct electricity because they have no free-moving charge carriers.

Common Mistakes

• Confusing the charge carriers: thinking that ionic compounds conduct via electrons, or that metals conduct by moving positive ions.
• Forgetting that acids ionise in aqueous solutions to produce mobile ions, making them good electrical conductors (electrolytes).

Things to Be Careful About

• Ensure you clearly distinguish between ions (which conduct in electrolytes/liquids) and delocalised electrons (which conduct in metallic and graphite structures).

Walkthrough

To find the correct substance, evaluate each option against the two conditions given:

1. Conducts electricity by the movement of electrons through a lattice of atoms:

   • In aqueous sodium chloride ($\text{NaCl}(\text{aq})$), the charge carriers are mobile aqueous ions ($\text{Na}^+$ and $\text{Cl}^-$), not electrons.
   • In solid sodium chloride ($\text{NaCl}(\text{s})$), the ions are locked in fixed positions in the lattice, so it cannot conduct electricity.
   • Both gold and graphite conduct electricity using delocalised electrons moving through a lattice.

2. Is not malleable:

   • Gold is a typical metal consisting of layers of cations that can slide over each other without breaking the metallic bond, making it highly malleable.
   • Graphite has a giant covalent structure consisting of layers of carbon atoms held together by weak intermolecular forces. Applying a force causes the layers to flake off or break rather than deform plastically; hence, graphite is non-malleable (it is brittle/soft).

Thus, graphite fits both criteria.

Key Takeaways

• Electrical conduction can occur via delocalised electrons (in metals and graphite) or via mobile ions (in molten or aqueous ionic compounds).
• Graphite has carbon atoms each bonded to three other carbon atoms in hexagonal layers, leaving one delocalised electron per carbon atom free to move and carry charge throughout the layers.
• Metals are malleable because regular layers of ions can slide over each other without disrupting the metallic bond, whereas giant covalent structures like graphite are not malleable.

Common Mistakes

• Confusing electrical conduction by ions with conduction by electrons (e.g., selecting aqueous sodium chloride).
• Overlooking the condition "not malleable" and incorrectly choosing gold.
• Forgetting that solid ionic compounds do not conduct electricity because their ions are fixed in place.

Things to Be Careful About

• Ensure you check both requirements stated in the stem: mechanical property (not malleable) and electrical conduction mechanism (movement of electrons through a lattice of atoms).

Walkthrough

1. Identify the chemical formula of ethene:
   • Ethene belongs to the homologous series of alkenes with the general formula $\text{C}_n\text{H}_{2n}$.
   • With $n = 2$, ethene has the molecular formula $\text{C}_2\text{H}_4$.
2. Determine how relative molecular mass ($M_r$) is calculated:
   • The $M_r$ of a covalent molecule is calculated by adding together the relative atomic masses ($A_r$) of all the individual atoms in its molecular formula.
   • In $\text{C}_2\text{H}_4$, there are $2$ carbon atoms and $4$ hydrogen atoms.
   • Therefore, $M_r = 2 \times A_r(\text{C}) + 4 \times A_r(\text{H})$.
3. Match with the correct option:
   • Option C accurately describes this calculation.

Key Takeaways

• Molecular formula of ethene: $\text{C}_2\text{H}_4$.
• Relative molecular mass ($M_r$): The sum of the relative atomic masses ($A_r$) of all the atoms shown in the molecular formula.

Common Mistakes

• Confusing ethene ($\text{C}_2\text{H}_4$) with ethane ($\text{C}_2\text{H}_6$), which leads to choosing option D.
• Confusing atomic number (proton number) with relative atomic mass ($A_r$).

Things to Be Careful About

• Always double-check prefixes: "eth-" means 2 carbons, and the "-ene" suffix indicates an alkene ($\text{C}_n\text{H}_{2n}$), not an alkane ($\text{C}_n\text{H}_{2n+2}$).

Walkthrough

To find the relative molecular mass ($M_r$) of a compound, sum the relative atomic masses ($A_r$) of all the atoms present in its chemical formula:

1. Identify the constituent elements and the number of atoms of each from the formula $\text{N}_2\text{O}$:

   • $2$ nitrogen ($\text{N}$) atoms
   • $1$ oxygen ($\text{O}$) atom

2. Look up the relative atomic masses ($A_r$) on the Periodic Table:

   • $A_r(\text{N}) = 14$
   • $A_r(\text{O}) = 16$

3. Multiply each $A_r$ by the number of atoms and sum the values:

Therefore, option C is correct.

Key Takeaways

• The relative molecular mass ($M_r$) is the sum of the relative atomic masses of all atoms shown in the formula.
• Subscripts denote the quantity of each specific atom (e.g. the subscript $2$ in $\text{N}_2\text{O}$ applies only to nitrogen).

Common Mistakes

• Confusing $\text{N}_2\text{O}$ (dinitrogen monoxide) with $\text{NO}_2$ (nitrogen dioxide, $M_r = 46$) or $\text{NO}$ (nitrogen monoxide, $M_r = 30$).
• Forgetting to multiply the $A_r$ of nitrogen by $2$.

Things to Be Careful About

• Always double-check which atom the subscript applies to before calculating.

Walkthrough

To find which sample contains the greatest mass of oxygen, we need to calculate the quantity of oxygen atoms present in each option:

1. Identify the number of oxygen atoms per formula unit:

   • In $\text{Al}(\text{NO}_3)_3$, the bracket subscript multiplies everything inside the bracket: there are $3 \times 3 = 9$ oxygen atoms per formula unit.
   • In $\text{K}_2\text{SO}_4$, there are $4$ oxygen atoms per formula unit.
   • In $\text{NaNO}_3$, there are $3$ oxygen atoms per formula unit.
   • In $\text{MgCO}_3$, there are $3$ oxygen atoms per formula unit.

2. Multiply by the number of moles of compound:

   • Option A: $0.2\text{ mol} \times 9 = 1.8\text{ mol}$ of $\text{O}$
   • Option B: $0.3\text{ mol} \times 4 = 1.2\text{ mol}$ of $\text{O}$
   • Option C: $0.4\text{ mol} \times 3 = 1.2\text{ mol}$ of $\text{O}$
   • Option D: $0.5\text{ mol} \times 3 = 1.5\text{ mol}$ of $\text{O}$

3. Relate moles to mass:
   Since $\text{mass} = \text{moles} \times A_r$, and the relative atomic mass of oxygen ($A_r = 16$) is constant for all options, the sample with the largest number of moles of oxygen atoms automatically has the greatest mass of oxygen.

Option A gives $1.8\text{ mol}$ of $\text{O}$ ($28.8\text{ g}$), which is greater than option D ($24.0\text{ g}$), option B ($19.2\text{ g}$), and option C ($19.2\text{ g}$).

Key Takeaways

• When brackets are present in a chemical formula, multiply the subscript inside the bracket by the subscript outside the bracket to get the total number of atoms of that element.
• Comparing masses of the same element across different compounds only requires comparing the total moles of that element.

Common Mistakes

• Forgetting to multiply subscripts inside brackets by the outside subscript (e.g., counting only $3$ or $6$ oxygen atoms in $\text{Al}(\text{NO}_3)_3$ instead of $3 \times 3 = 9$).
• Selecting the option with the largest number of moles of compound ($0.5\text{ mol}$ in D) without accounting for the number of oxygen atoms in the formula unit.

Things to Be Careful About

• Ensure you calculate the moles of oxygen atoms ($ext{O}$) rather than assuming oxygen gas ($ext{O}_2$), though relative comparisons yield the same result either way.

Walkthrough

To find the empirical formula of a compound from percentage composition by mass:

1. Determine the percentage of all elements:
   The question states that compound Z contains only carbon, hydrogen, and oxygen. The sum of all mass percentages must be $100\%$.

   $$\%\text{ O} = 100 - 48.65 - 8.11 = 43.24\%$$

2. Convert percentages to moles:
   Assuming a $100\text{ g}$ sample of the compound, the mass of each element in grams is equal to its percentage. Divide each mass by the respective relative atomic mass ($A_r$):

   • $\text{Moles of C} = \frac{48.65}{12} \approx 4.054\text{ mol}$
   • $\text{Moles of H} = \frac{8.11}{1} = 8.110\text{ mol}$
   • $\text{Moles of O} = \frac{43.24}{16} \approx 2.703\text{ mol}$

3. Find the simplest molar ratio:
   Divide each mole value by the smallest number of moles calculated, which is $2.703$:

   • $\text{C} : \frac{4.054}{2.703} = 1.50$
   • $\text{H} : \frac{8.110}{2.703} = 3.00$
   • $\text{O} : \frac{2.703}{2.703} = 1.00$

4. Convert non-integers to integers:
   Since the ratio contains a half-integer ($1.50$), multiply all numbers in the ratio by $2$:

   • $\text{C} = 1.50 \times 2 = 3$
   • $\text{H} = 3.00 \times 2 = 6$
   • $\text{O} = 1.00 \times 2 = 2$

This gives an empirical formula of $\text{C}_3\text{H}_6\text{O}_2$, which corresponds to option B.

Key Takeaways

• The empirical formula is the simplest whole-number ratio of the atoms of each element present in a compound.
• When a percentage is missing for a compound of known elemental composition, subtract the given percentages from $100\%$.
• If dividing by the smallest mole value yields a decimal ending close to $.5$, multiply all ratios by $2$ (or by $3$ if ending in $.33$ or $.67$) to reach the lowest whole-number ratio.

Common Mistakes

• Forgetting to calculate the percentage of oxygen by subtracting the given percentages from $100\%$.
• Rounding $1.5$ directly to $1$ or $2$ instead of multiplying all ratios by $2$ to maintain the true proportions.
• Dividing by the relative molecular mass of diatomic gases (e.g., using $32$ for $\text{O}_2$ or $2$ for $\text{H}_2$) rather than atomic masses ($A_r$).

Things to Be Careful About

• Use relative atomic masses directly from the Periodic Table ($A_r(\text{C}) = 12$, $A_r(\text{H}) = 1$, $A_r(\text{O}) = 16$).
• Keep intermediate calculations to at least three significant figures to prevent rounding errors that could distort the molar ratio.

Walkthrough

To find which fertiliser contains the highest percentage of nitrogen by mass, calculate the percentage of nitrogen in each compound:

1. In $\text{NH}_4\text{NO}_3$ (ammonium nitrate), there are 2 nitrogen atoms:
   $\frac{2 \times 14}{80} \times 100\% = 35.0\%$
2. In $(\text{NH}_4)_3\text{PO}_4$ (ammonium phosphate), there are 3 nitrogen atoms:
   $\frac{3 \times 14}{149} \times 100\% = 28.2\%$
3. In $(\text{NH}_4)_2\text{SO}_4$ (ammonium sulfate), there are 2 nitrogen atoms:
   $\frac{2 \times 14}{132} \times 100\% = 21.2\%$
4. In $(\text{NH}_2)_2\text{CO}$ (urea), there are 2 nitrogen atoms:
   $\frac{2 \times 14}{60} \times 100\% = 46.7\%$

Urea (option D) gives the highest percentage of nitrogen by mass ($46.7\%$).

Key Takeaways

• Percentage by mass depends on the ratio of the total atomic mass of the target element to the formula mass ($M_r$) of the whole compound.
• When comparing fractions with identical numerators (such as $\frac{28}{60}$, $\frac{28}{80}$, and $\frac{28}{132}$), the fraction with the smallest denominator gives the largest value.

Common Mistakes

• Counting only 1 nitrogen atom instead of 2 in $\text{NH}_4\text{NO}_3$ or $(\text{NH}_2)_2\text{CO}$.
• Assuming the compound with the most nitrogen atoms per formula unit (e.g., $(\text{NH}_4)_3\text{PO}_4$ with 3 nitrogens) must have the highest percentage, without taking its larger formula mass into account.

Things to Be Careful About

• Ensure all subscripts and bracket multipliers are correctly accounted for when counting the number of nitrogen atoms in the formula unit.

Walkthrough

During electrolysis:

1. The power supply sets up the polarity of the electrodes: the long line of the battery symbol represents the positive terminal (connected to the anode) and the short line represents the negative terminal (connected to the cathode).
2. Current flows through the electrolyte via the migration of mobile ions:
   • Positive ions (cations) migrate towards the negatively charged cathode.
   • Negative ions (anions) migrate towards the positively charged anode.
3. This directly matches statement C.

Let's verify why the other options are incorrect:

• A is incorrect: Electrons leave the anode (where oxidation produces electrons) and travel through the external circuit wires to the cathode (where reduction consumes electrons).
• B is incorrect: Hydrogen ions ($ext{H}^+$) are positively charged, so they are attracted to the cathode, where they gain electrons according to $2 ext{H}^+ + 2 ext{e}^- ightarrow ext{H}_2$.
• D is incorrect: Hydroxide ions ($ext{OH}^-$) are negatively charged anions and therefore migrate towards the positive anode.

Key Takeaways

• Remember the mnemonic: PANIC (Positive Anode, Negative Is Cathode) for electrolytic cells.
• Cations $ightarrow$ Cathode; Anions $ightarrow$ Anode.
• Charge is carried by electrons in the external metallic circuit and by mobile ions in the electrolyte solution.

Common Mistakes

• Confusing the direction of electron flow with the movement of ions.
• Mixing up the polarity of the cathode and anode, or which ions migrate to which electrode.

Things to Be Careful About

• Ensure you distinguish between an electrolytic cell (driven by an external power supply) and a simple electrochemical cell (which produces voltage). In an electrolytic cell, the anode is positive and the cathode is negative.

Walkthrough

In electrolysis, the anode is the positive electrode where oxidation (the loss of electrons) takes place.

When inert electrodes (such as carbon/graphite or platinum) are used, anions in the aqueous solution (such as $\text{OH}^-$ or halide ions) are discharged. However, when an active metal electrode such as copper is used as the anode, the metal atoms of the anode itself lose electrons more readily than any anions in the solution.

Therefore, copper atoms at the anode lose two electrons each to form aqueous copper(II) ions:

Evaluating the given options:

• A: $\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-$ is the correct oxidation reaction occurring at the copper anode.
• B: $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$ represents reduction (gain of electrons), which occurs at the cathode.
• C: $\text{Ag} \rightarrow \text{Ag}^+ + \text{e}^-$ would represent the oxidation of silver, but the electrodes are specified as pure copper.
• D: $\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}$ is a reduction half-equation occurring at the cathode.

Thus, A is the correct answer.

Key Takeaways

• Anode vs. Cathode: Oxidation occurs at the anode (positive electrode in electrolysis); reduction occurs at the cathode (negative electrode).
• Active vs. Inert Electrodes: If an active metal (like copper) is used as the anode, the electrode dissolves ($ext{M} \rightarrow \text{M}^{n+} + n\text{e}^-$) rather than water or anions discharging.

Common Mistakes

• Confusing oxidation with reduction and picking B.
• Forgetting that the electrode material dictates the anode reaction when using non-inert electrodes.
• Selecting C due to the presence of silver nitrate in the electrolyte, forgetting that the anode itself is pure copper.

Things to Be Careful About

• Ensure you check the material of the electrodes: "pure copper electrodes" indicates that the anode participates chemically in the reaction.

Walkthrough

A hydrogen-oxygen fuel cell generates electricity through the reaction between hydrogen and oxygen:

Let us analyse the options to identify the disadvantage of using a fuel cell:

• Option A: While gasoline is a non-renewable fossil fuel, this is an issue related to internal combustion engines, not a drawback of using a hydrogen fuel cell.
• Option B: Hydrogen must be stored as a compressed gas or liquid at very high pressures. In a car accident, a ruptured fuel tank could release hydrogen, which ignites and explodes very easily in air. This is a well-known safety disadvantage.
• Option C: The only product of the overall cell reaction is water, which is non-polluting and completely non-toxic.
• Option D: Obtaining oxygen directly from the surrounding air is an advantage as it requires no separate oxygen storage on board the vehicle.

Therefore, B is the correct answer.

Key Takeaways

• The main advantages of hydrogen-oxygen fuel cells are that they produce no greenhouse gases or toxic pollutants (only water is formed) and they are highly efficient.
• The major disadvantages include the difficulty and safety risks of storing high-pressure flammable hydrogen, the lack of refuelling infrastructure, and the fact that hydrogen is often manufactured from fossil fuels.

Common Mistakes

• Confusing the combustion/reaction products of hydrogen with fossil fuel emissions (hydrogen produces only water, not carbon monoxide or oxides of nitrogen).
• Selecting statements that describe disadvantages of conventional petrol vehicles rather than fuel cells.

Things to Be Careful About

• Ensure you distinguish clearly between advantages (e.g. non-toxic product $\text{H}_2\text{O}$, abundance of $\text{O}_2$ in air) and disadvantages (flammability/storage hazards of $\text{H}_2$).

Walkthrough

A reaction that gives out thermal energy is exothermic. The word 'exothermic' means heat is released to the surroundings.

In any chemical reaction, bonds in the reactants must be broken, and this requires energy. New bonds are then formed in the products, and this releases energy.

• If the energy released when bonds form is greater than the energy needed to break bonds, the overall reaction releases energy: exothermic.
• If the energy needed to break bonds is greater than the energy released when bonds form, the overall reaction absorbs energy: endothermic.

Here, thermal energy is given out, so the reaction is exothermic and the correct explanation is that less energy is required to break the bonds than is released when the bonds are formed. That is row D.

Key Takeaways

• Exothermic reactions release thermal energy; endothermic reactions absorb thermal energy.
• Bond breaking always requires energy.
• Bond forming always releases energy.
• The difference between these two amounts decides whether a reaction is exothermic or endothermic.

Common Mistakes

• Choosing A or B by thinking that 'energy is given out' means endothermic. Endothermic reactions take in energy.
• Choosing C because it says exothermic but then giving the endothermic explanation. If more energy is needed to break bonds than is released on forming bonds, the reaction is endothermic, not exothermic.
• Thinking that bond breaking releases energy. Bond breaking requires energy; bond forming releases energy.

Things to Be Careful About

• Read the phrase 'thermal energy is given out' as a clear sign of an exothermic reaction.
• Compare the two energy terms in the correct order: energy required to break bonds versus energy released when bonds are formed.
• The mark scheme accepts only the exact idea that less energy is required to break bonds than is released when bonds are formed for an exothermic reaction.

Walkthrough

A physical change is a change in which no new chemical substances are formed. The chemical identity and composition of the particles remain unchanged, but their arrangement, energy, movement, or appearance may change.

Let us evaluate each option:

• Option A: Physical changes are typically easily reversible (e.g., heating ice melts it into water, and cooling the water freezes it back into ice). Chemical changes, on the other hand, are often difficult to reverse.
• Option B: In a physical change, physical properties such as appearance, state of matter (solid, liquid, gas), or shape can certainly change. For example, crushing a crystal, dissolving salt in water, or boiling water alters the appearance.
• Option C: The formation of one or more new chemical substances is the defining feature of a chemical change, not a physical change.
• Option D: Energy changes accompany physical changes. For instance, melting and boiling are endothermic (absorb heat energy), while freezing and condensing are exothermic (release heat energy).

Therefore, option B is the correct statement.

Key Takeaways

• In a physical change, no new chemical substances are produced, and the change is usually easy to reverse.
• In a chemical change, new substances with different chemical properties are formed, bonds are broken and made, and the change is generally difficult to reverse.
• Physical changes involve energy changes (e.g., latent heat during changes of state).

Common Mistakes

• Confusing the characteristics of physical and chemical changes (thinking new substances are formed in physical changes).
• Believing that no energy is transferred during physical changes; state changes always absorb or release thermal energy.

Things to Be Careful About

• Do not confuse changes of appearance (physical) with evidence of a chemical reaction (such as unexpected colour change due to a new compound forming, effervescence/gas production, or precipitation).

Walkthrough

1. Finding the ratio of $x$ to $y$:
   Under the same conditions of temperature and pressure, the volume of a gas is directly proportional to the number of moles (Avogadro's law / molar gas volume concept). Looking at the vertical axis of either graph:

   • Volume of gas $\text{P}$ used up $= 4\text{ grid squares}$.
   • Volume of gas $\text{Q}$ formed $= 6\text{ grid squares}$.

   The ratio of moles reacting is therefore:

   $$x : y = 4 : 6 = 2 : 3$$

   Thus, $x = 2$ and $y = 3$.

2. Comparing temperatures $T_1$ and $T_2$:

   • The graph for $T_1$ shows a steeper initial slope and reaches the plateau (reaction completion) earlier than the graph for $T_2$.
   • A steeper curve indicates a higher rate of reaction.
   • Increasing temperature increases the kinetic energy of the particles, leading to more frequent collisions and a greater proportion of particles having energy greater than or equal to the activation energy, thus increasing the rate of reaction.
   • Since the rate is faster at $T_1$, $T_1$ must be higher than $T_2$.

Matching these deductions ($x = 2$, $y = 3$, $T_1 > T_2$) gives option A.

Key Takeaways

• For reactions involving gases at identical conditions, the ratio of reacting and produced volumes directly reflects the stoichiometric coefficients in the balanced chemical equation.
• On a volume-time or mass-time graph, a steeper initial gradient and earlier plateau indicate a faster rate of reaction.
• Higher temperature increases the rate of reaction.

Common Mistakes

• Inverting the stoichiometric ratio (thinking $x : y = 3 : 2$ because $\text{Q}$ ends up higher than $\text{P}$ started).
• Confusing the effect of temperature: assuming a longer time to complete means a higher temperature.

Things to Be Careful About

• Ensure you count grid units accurately from $(0,0)$. The volume of $\text{P}$ starts at $4$ units and drops to $0$, while $\text{Q}$ starts at $0$ and rises to $6$.

Walkthrough

In the given reversible reaction:

• The forward reaction has a negative enthalpy change ($\Delta H < 0$), meaning it is exothermic (releases heat).
• The backward or reverse reaction is endothermic (absorbs heat).

When the temperature of an equilibrium mixture is increased, the system responds by shifting in the direction that takes in heat (endothermic) to counteract the increase in temperature. This shifts the position of equilibrium to the left.

• Because equilibrium shifts to the left, more $\text{NH}_3$ decomposes into $\text{N}_2$ and $\text{H}_2$, so the amount of product decreases. This makes option B correct and option A incorrect.
• Once the new equilibrium is established, the rate of the forward reaction equals the rate of the reverse reaction by definition of dynamic equilibrium. Therefore, options C and D are incorrect.

Key Takeaways

• For an exothermic forward reaction, increasing temperature favours the reverse reaction (shifts equilibrium to the left), reducing the yield of products.
• At dynamic equilibrium, the rates of the forward and reverse reactions are always equal, even though the amounts of reactants and products may be different.

Common Mistakes

• Confusing the temporary state during the shift with the newly established equilibrium (e.g., thinking one rate stays permanently higher than the other at equilibrium).
• Forgetting that $\Delta H$ refers specifically to the forward reaction as written.

Things to Be Careful About

• Ensure you identify the sign of $\Delta H$: a negative value indicates an exothermic reaction, while a positive value indicates an endothermic reaction.

Walkthrough

The Contact process is used industrially to make sulfuric acid ($\text{H}_2\text{SO}_4$). It involves three main stages:

1. Burning sulfur in air to produce sulfur dioxide: $$\text{S}(\text{s}) + \text{O}_2(\text{g}) \rightarrow \text{SO}_2(\text{g})$$
2. The oxidation of sulfur dioxide to sulfur trioxide (reaction R shown in the diagram): $$2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})$$ The essential conditions for this equilibrium reaction are:
   • Catalyst: Vanadium(V) oxide, $\text{V}_2\text{O}_5$
   • Temperature: $450\,^\circ\text{C}$
   • Pressure: $1-2\text{ atm}$ (just above atmospheric pressure)
3. Converting sulfur trioxide to sulfuric acid by dissolving it in concentrated $\text{H}_2\text{SO}_4$ to form oleum ($\text{H}_2\text{S}_2\text{O}_7$), followed by dilution with water.

Matching the diagram, the two gases entering reactor R are oxygen ($\text{O}_2$) and gas Z, which must be sulfur dioxide ($\text{SO}_2$). Thus, option A is the correct statement.

Key Takeaways

• Be familiar with the key stage of the Contact process: $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$.
• Remember the standard industrial conditions for the Contact process ($450\,^\circ\text{C}$, $1-2\text{ atm}$, $\text{V}_2\text{O}_5$ catalyst).

Common Mistakes

• Confusing the conditions of the Contact process with those of the Haber process (iron catalyst, $200\text{ atm}$, $450\,^\circ\text{C}$).
• Misreading $45\,^\circ\text{C}$ as $450\,^\circ\text{C}$.

Things to Be Careful About

• High pressure is not required in the Contact process because the equilibrium yield of $\text{SO}_3$ at $1-2\text{ atm}$ and $450\,^\circ\text{C}$ is already very high (around $98\%$), making expensive high-pressure equipment unnecessary.

Walkthrough

To solve this question, we apply the definitions of oxidation, reduction, and oxidising/reducing agents using electron transfer (OIL RIG: Oxidation Is Loss, Reduction Is Gain):

1. Determine what happens to iron:
   Iron changes from $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$.
   $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$
   Each iron(II) ion loses one electron to become an iron(III) ion. Loss of electrons is oxidation, so iron is oxidised.

2. Determine the element that is reduced:
   The question states that the only elements either oxidised or reduced are sulfur and iron. Since iron is oxidised, sulfur must be the element that is reduced.

3. Determine the behaviour of $\text{S}_2\text{O}_8^{2-}$:

   • An oxidising agent accepts electrons from another substance, causing that substance to be oxidised while the agent itself is reduced.
   • Since $\text{S}_2\text{O}_8^{2-}$ causes $\text{Fe}^{2+}$ to be oxidised to $\text{Fe}^{3+}$, and is itself reduced, $\text{S}_2\text{O}_8^{2-}$ acts as an oxidising agent.

Combining these two conclusions corresponds to row C.

Key Takeaways

• Oxidation is the loss of electrons (or an increase in oxidation state).
• Reduction is the gain of electrons (or a decrease in oxidation state).
• An oxidising agent oxidises another species and undergoes reduction in the process.
• A reducing agent reduces another species and undergoes oxidation in the process.

Common Mistakes

• Confusing the element being reduced with the role of the substance: students often forget that a substance being reduced is the oxidising agent, not the reducing agent.
• Misidentifying the change in charge: thinking going from $+2$ to $+3$ is a gain of electrons instead of a loss of negative charge.