Chemistry 5070/11 — October/November 2025

Walkthrough

1. Effect of pressure on volume: For a fixed mass of gas at constant temperature, pressure and volume are inversely related. Reducing the external pressure allows the gas particles to push outward until the internal and external pressures balance, causing the gas to expand. Therefore, the volume of the air increases.
2. Effect on particle separation: The total number of particles (molecules of nitrogen, oxygen, etc.) remains constant. When these particles spread out to occupy a larger total volume, the mean distance separating individual particles must increase.

Combining both conclusions identifies row D as the correct answer.

Key Takeaways

• For gases at constant temperature, decreasing pressure leads to an increase in volume (gas expansion).
• In an expanded gas volume, particles are spread further apart, increasing the average inter-particle distance.

Common Mistakes

• Confusing direct and inverse relationships: thinking a decrease in pressure causes a decrease in volume.
• Believing that changing pressure alters the size of individual particles rather than the empty space (distance) between them.

Things to Be Careful About

• Check whether the question states pressure is increased or reduced, and whether temperature is kept constant.

Walkthrough

Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration down a concentration gradient, as a result of their random thermal motion.

Two factors determine the speed at which a gas diffuses:

1. Mass of the particles ($M_r$ or $A_r$): Lighter particles have higher average velocities than heavier particles at the same temperature. Comparing the two gases:

   • For carbon dioxide ($ext{CO}_2$): $M_r = 12 + 2(16) = 44$
   • For neon ($ext{Ne}$): $A_r = 20$
     Because neon has a significantly lower relative atomic mass than carbon dioxide's relative molecular mass, neon particles diffuse faster.

2. Temperature: As temperature increases, gas particles gain kinetic energy, causing them to move faster. Therefore, diffusion is faster at $25\,^\circ\text{C}$ than at $0\,^\circ\text{C}$.

To find the substance that diffuses most quickly, we choose the lighter gas at the higher temperature: neon at $25\,^\circ\text{C}$ (Option D).

Key Takeaways

• Rate of diffusion $\propto \frac{1}{\sqrt{M_r}}$: Lower relative molecular/atomic mass results in a faster rate of diffusion.
• Rate of diffusion $\propto \text{Temperature}$: Higher temperature provides greater kinetic energy, resulting in faster particle movement and higher rate of diffusion.

Common Mistakes

• Forgetting that noble gases like neon exist as monatomic gases ($ext{Ne}$, $A_r = 20$) rather than diatomic molecules.
• Assuming higher density or heavier molecules have more momentum and therefore diffuse faster (a common physics misconception; in reality, smaller/lighter particles move faster at the same kinetic energy).

Things to Be Careful About

• Check both variables (mass and temperature) systematically before selecting the option.
• Ensure you calculate $M_r$ accurately using values from the Periodic Table ($C = 12$, $O = 16$, $ext{Ne} = 20$).

Walkthrough

To find the correct row, we evaluate both statements in turn:

1. Statement 1: Compound vs Mixture

   • A compound is a single substance formed from two or more elements chemically combined in fixed proportions.
   • A mixture contains two or more different substances (elements or compounds) that are not chemically bonded together.
   • Because mineral X consists of calcium carbonate ($\text{CaCO}_3$) containing grains of silicon(IV) oxide ($\text{SiO}_2$), it is composed of two separate chemical compounds. This makes mineral X a mixture.
   • This eliminates options A and B.

2. Statement 2: Number of different elements

   • Break down each chemical formula into its constituent elements:
     • $\text{CaCO}_3$ contains calcium ($\text{Ca}$), carbon ($\text{C}$), and oxygen ($\text{O}$).
     • $\text{SiO}_2$ contains silicon ($\text{Si}$) and oxygen ($\text{O}$).
   • Count the total number of distinct elements present: $\text{Ca}$, $\text{C}$, $\text{Si}$, and $\text{O}$. Since oxygen is present in both compounds, it is only counted once.
   • Thus, there are exactly $4$ different elements present in X.
   • This eliminates option D.

Therefore, row C is correct.

Key Takeaways

• A mixture consists of two or more substances that can be present in varying proportions and are not chemically bonded together.
• When counting the number of different elements in a mixture or complex substance, identify each unique chemical symbol and avoid double-counting elements shared between compounds.

Common Mistakes

• Counting the total number of element symbols across both formulae ($3 + 2 = 5$) instead of counting unique elements, forgetting that oxygen ($\text{O}$) is present in both $\text{CaCO}_3$ and $\text{SiO}_2$.
• Confusing the terms mixture and compound, mistaking the mineral as a whole for a single chemical compound.

Things to Be Careful About

• Element symbols consist of either a single capital letter (e.g., $\text{C}$, $\text{O}$) or a capital letter followed by a lowercase letter (e.g., $\text{Ca}$, $\text{Si}$). Ensure that two-letter symbols like $\text{Ca}$ are recognised as one element (calcium) rather than two separate elements.

Walkthrough

To find the correct row, we evaluate both columns separately:

1. Total number of electrons in the molecule:

   • A neutral atom contains a number of electrons equal to its proton number (atomic number).
   • From the Periodic Table:
     • Nitrogen ($\text{N}$) has atomic number 7 $\rightarrow 7\text{ electrons}$
     • Hydrogen ($\text{H}$) has atomic number 1 $\rightarrow 1\text{ electron}$ each ($3 \times 1 = 3\text{ electrons}$)
     • Oxygen ($\text{O}$) has atomic number 8 $\rightarrow 8\text{ electrons}$
   • Total electrons in $\text{NH}_2\text{OH} = 7 + 3 + 8 = 18\text{ electrons}$.
   • This eliminates options A and B (which only counted valence electrons: $5 + 3 + 6 = 14$).

2. Number of electrons in the bonds of the molecule:

   • Each single covalent bond is formed by a shared pair of electrons (2 electrons).
   • Hydroxylamine contains:
     • $2 \times \text{N}-\text{H}$ bonds
     • $1 \times \text{N}-\text{O}$ bond
     • $1 \times \text{O}-\text{H}$ bond (the bond between $\text{O}$ and $\text{H}$ in the hydroxyl group is often written condensed as $-\text{OH}$, but is still a single covalent bond)
   • Total single bonds $= 2 + 1 + 1 = 4\text{ bonds}$.
   • Number of bonding electrons $= 4 \times 2 = 8\text{ electrons}$.

Matching both results (18 total electrons, 8 bonding electrons) leads directly to option C.

Key Takeaways

• The total number of electrons in a neutral molecule is the sum of the atomic numbers (proton numbers) of all constituent atoms, not just the outer-shell (valence) electrons.
• A single covalent bond consists of 2 electrons (one shared pair).
• Groups written in condensed form such as $-\text{OH}$ contain an internal covalent bond (the $\text{O}-\text{H}$ single bond) that must be included when counting bonds.

Common Mistakes

• Confusing the total number of electrons with the total number of valence (outer-shell) electrons ($5 + 3 + 6 = 14$), which leads incorrectly to option A or B.
• Overlooking the covalent bond between oxygen and hydrogen in the $-\text{OH}$ group, counting only 3 visible bond lines in the diagram and concluding there are $3 \times 2 = 6$ bonding electrons, leading to option D.

Things to Be Careful About

• Ensure all bonds are counted when functional groups are written as $-\text{OH}$, $-\text{NH}_2$, or $-\text{CH}_3$.
• Double-check atomic numbers from the Periodic Table rather than using relative atomic masses ($A_r$).

Walkthrough

When a metal reacts with a non-metal to form an ionic compound, the metal atom transfers electrons to the non-metal atom:

1. Calcium ($ext{Ca}$) has an atomic number of $20$ and an electronic configuration of $2,8,8,2$. It loses its $2$ outer-shell electrons to form a $ext{Ca}^{2+}$ ion with a stable noble gas configuration ($2,8,8$).
2. Hydrogen ($ext{H}$) has $1$ electron in its first electron shell. To complete this shell (which has a maximum capacity of $2$ electrons, like helium), each hydrogen atom accepts $1$ electron from calcium.
3. The resulting ion is the hydride ion, $ext{H}^-$, which possesses $2$ electrons in its outer (and only) shell.

Therefore, each hydride ion in $ext{CaH}_2$ has $2$ electrons in its outer shell, corresponding to option C.

Key Takeaways

• When hydrogen acts as a non-metal in metal hydrides, it gains an electron to form the hydride ion, $ext{H}^-$.
• The first electron shell holds a maximum of $2$ electrons (a complete duplet, matching the electron configuration of helium).

Common Mistakes

• Confusing the hydride ion ($ext{H}^-$, with $2$ electrons) with the hydrogen ion/proton ($ext{H}^+$, with $0$ electrons).
• Thinking the outer shell always requires an octet ($8$ electrons), forgetting that hydrogen's outer shell is the first shell, which is full with $2$ electrons.

Things to Be Careful About

• Pay attention to the prefix/suffix: "hydride" refers to the negative anion $ext{H}^-$, whereas "hydrogen ion" commonly refers to $ext{H}^+$.

Walkthrough

Diamond is an allotrope of carbon with a giant covalent macromolecular structure:

• Each carbon atom forms four strong covalent bonds with four other carbon atoms in a rigid, tetrahedral arrangement.
• Because breaking this rigid network requires a vast amount of energy, diamond is extremely hard and has a very high melting and boiling point.
• All valence electrons are held in single covalent bonds, meaning there are no free or delocalised electrons to carry an electrical current (unlike graphite, which has delocalised electrons and acts as a good conductor).
• The lubricant property also belongs to graphite, where weak intermolecular forces between layers allow the layers to slide over one another easily.

Therefore, the only correct property listed for diamond is that it is extremely hard (D).

Key Takeaways

• Diamond: Giant covalent structure; 4 covalent bonds per carbon; tetrahedral lattice; extremely hard; high melting point; electrical insulator.
• Graphite: Giant covalent structure; 3 covalent bonds per carbon (forming hexagonal layers) with 1 delocalised electron per carbon; conducts electricity; soft and slippery (used as a lubricant).

Common Mistakes

• Confusing the properties of diamond with those of graphite (such as electrical conductivity and use as a lubricant).
• Thinking non-metals universally have low melting points; giant covalent structures (diamond, graphite, silicon dioxide) have very high melting points.

Things to Be Careful About

• Ensure you attribute electrical conductivity and softness/lubrication specifically to graphite due to its layered structure and delocalised electrons, whereas diamond's 4-bond tetrahedral arrangement makes it hard and an electrical insulator.

Walkthrough

Metallic bonding is found in pure metals like copper ($\text{Cu}$) and in alloys. In a metallic lattice, metal atoms lose their outer shell electrons to form positively charged ions (cations). These delocalised outer-shell electrons are free to move throughout the entire structure, forming a mobile 'sea' of electrons. The metallic bond itself is defined as the strong electrostatic attraction between the positively charged metal ions and the negatively charged delocalised electrons.

• Statement A is wrong because the species in the lattice are positive ions, having already released their outer valence electrons into the delocalised sea, not neutral atoms.
• Statements B and D describe covalent bonding (the attraction of nuclei to a shared pair of electrons), which occurs between non-metal atoms, not in metals.
• Statement C accurately describes metallic bonding.

Key Takeaways

• Metallic bonding involves a regular lattice of positive metal ions embedded in a 'sea' of delocalised electrons.
• Covalent bonding involves a shared pair of electrons between non-metal atoms.
• Ionic bonding involves electrostatic attraction between oppositely charged ions.

Common Mistakes

• Confusing metal atoms with metal ions; in a metal crystal lattice, valence electrons are delocalised, leaving positive metal cations.
• Mixing up the terminology of covalent bonding ('shared pair of electrons') with metallic bonding ('delocalised electrons').

Things to Be Careful About

• Pay close attention to keywords in definition questions: distinguish clearly between atoms, ions, and molecules.

Walkthrough

1. Determine the ionic formula:

   • Aluminium forms $\text{Al}^{3+}$ ions.
   • Sulfate is a polyatomic ion with the formula $\text{SO}_4^{2-}$.
   • In any stable ionic compound, the sum of all positive and negative charges must equal zero.
   • The lowest common multiple of $3$ and $2$ is $6$. Thus, we need two $\text{Al}^{3+}$ ions ($2 \times +3 = +6$) and three $\text{SO}_4^{2-}$ ions ($3 \times -2 = -6$).
   • Written showing the ions, this is $\text{Al}^{3+}_2(\text{SO}_4^{2-})_3$.

2. Determine the empirical formula:

   • The empirical formula is the simplest whole-number ratio of the elements in the compound.
   • Expanding the formula unit $\text{Al}_2(\text{SO}_4)_3$ gives $2\text{ Al}$, $3\text{ S}$, and $3 \times 4 = 12\text{ O}$.
   • The simplest integer ratio of $\text{Al} : \text{S} : \text{O}$ is $2 : 3 : 12$, making the empirical formula $\text{Al}_2\text{S}_3\text{O}_{12}$.

3. This correctly corresponds to option C.

Key Takeaways

• For an ionic compound to be electrically neutral, the total positive charge must equal the total negative charge.
• The empirical formula gives the simplest whole-number ratio of all atoms of each element in the substance.
• When expanding polyatomic groups with subscripts (like $(\text{SO}_4)_3$), multiply the subscript inside the brackets by the subscript outside the brackets ($4 \times 3 = 12$ oxygen atoms).

Common Mistakes

• Forgetting to multiply the number of oxygen atoms by the outer bracket subscript (e.g., mistakenly thinking there are only $4$ or $8$ oxygen atoms).
• Swapping the charges of the cation and anion (e.g., thinking $\text{Al}$ is $+2$ and $\text{SO}_4$ is $-3$ as in distractor D).
• Omitting the need to balance charges (e.g., $1:1$ ratio in distractor A).

Things to Be Careful About

• Ensure subscripts outside the brackets apply to every atom inside the bracket: in $(\text{SO}_4)_3$, there are $3$ sulfur atoms and $12$ oxygen atoms.

Walkthrough

Let us evaluate each option systematically:

• Option A: Concentration is defined as the amount of solute per unit volume of solution. Therefore, its standard units are $\text{mol / dm}^3$ or $\text{g / dm}^3$. The given unit $\text{dm}^3 / \text{mol}$ is inverted (it represents molar volume).

• Option B: This statement gives the definition of the molecular formula, not the empirical formula. The empirical formula is defined as the simplest whole-number ratio of the elements in a compound.

• Option C: The molecular formula is an integer multiple ($n = 1, 2, 3, \dots$) of the empirical formula. When $n = 1$ (such as in $\text{H}_2\text{O}$, $\text{NH}_3$, and $\text{CH}_4$), the molecular formula is identical to the empirical formula and contains the exact same number of atoms, so it does not always contain more atoms.

• Option D: This correctly states the standard IUPAC and Cambridge O Level definition of relative atomic mass ($A_r$), which is the ratio of the average mass of one atom of the element to $\frac{1}{12}$ of the mass of one carbon-12 atom.

Hence, statement D is the correct choice.

Key Takeaways

• Relative Atomic Mass ($A_r$): The average mass of naturally occurring atoms of an element on a scale where the carbon-12 atom has a mass of exactly 12 units.
• Empirical Formula: The simplest whole-number ratio of atoms of each element in a compound.
• Molecular Formula: The actual number of atoms of each element present in one molecule of a compound.
• Concentration: The amount of substance in moles (or mass in grams) dissolved in $1\text{ dm}^3$ of solution ($ext{mol / dm}^3$ or $\text{g / dm}^3$).

Common Mistakes

• Confusing the definitions of empirical formula and molecular formula.
• Assuming the molecular formula is always larger than the empirical formula, forgetting cases where the molecular formula cannot be simplified further (i.e. $n = 1$).
• Forgetting that the carbon-12 standard is based on $\frac{1}{12}$ the mass of a single $^{12}\text{C}$ atom.

Things to Be Careful About

• Watch out for absolute words like always in multiple choice options, as counter-examples often disprove them.
• Keep units clear: volume is in $\text{dm}^3$, moles are in $\text{mol}$, so concentration is $\text{mol / dm}^3$.

Walkthrough

To find the total number of atoms in a given mass of a compound, follow these steps:

1. Find the molar mass ($M_r$) of water ($\text{H}_2\text{O}$):

   • Relative atomic mass of $\text{H} = 1.0$
   • Relative atomic mass of $\text{O} = 16.0$
   • $M_r = (2 \times 1.0) + 16.0 = 18.0$

2. Find the number of moles of water molecules:

   • $\text{Moles} = \frac{\text{mass}}{M_r} = \frac{27.0\text{ g}}{18.0\text{ g / mol}} = 1.50\text{ mol}$

3. Count the atoms per formula unit:

   • Each water molecule contains two hydrogen atoms and one oxygen atom, which is a total of $3$ atoms per molecule.
   • Therefore, the number of moles of atoms is $1.50\text{ mol} \times 3 = 4.50\text{ mol}$.

4. Calculate the total number of individual atoms:

   • Using the Avogadro constant ($L \approx 6.02 \times 10^{23}\text{ mol}^{-1}$):
   • $\text{Number of atoms} = 4.50 \times 6.02 \times 10^{23} = 2.709 \times 10^{24}$.

Hence, option C is correct.

Key Takeaways

• Always distinguish between molecules and atoms when dealing with mole calculations for covalent compounds.
• To find the number of particles, multiply the number of moles by the Avogadro constant ($6.02 \times 10^{23}$). If finding total atoms, multiply further by the number of atoms per molecule.

Common Mistakes

• Option A ($9.030 \times 10^{23}$): Calculates only the number of $\text{H}_2\text{O}$ molecules ($1.50 \times 6.02 \times 10^{23}$) without multiplying by $3$ atoms per molecule.
• Option B ($1.806 \times 10^{24}$): Accounts for only the $2$ hydrogen atoms ($1.50 \times 2 \times 6.02 \times 10^{23}$) and forgets the oxygen atom.
• Option D ($3.612 \times 10^{24}$): Multiplies incorrectly by $4$ instead of $3$.

Things to Be Careful About

• Ensure you read whether the question asks for the number of molecules or the number of atoms.
• Be precise with scientific notation powers of $10$.

Walkthrough

1. Identify what causes the mass loss: In the thermal decomposition of sodium hydrogencarbonate, the products are solid $\text{Na}_2\text{CO}_3$, gaseous $\text{CO}_2$, and gaseous $\text{H}_2\text{O}$. Since sodium chloride does not react or decompose, the loss in mass ($1.5\text{ g}$) corresponds entirely to the total mass of the escaped gases, $\text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{g})$.

2. Determine the mass relationship from stoichiometry:

   • $2\text{ mol}$ of $\text{NaHCO}_3$ has a mass of $2 \times 84 = 168\text{ g}$.
   • Decomposing $2\text{ mol}$ of $\text{NaHCO}_3$ yields $1\text{ mol}$ of $\text{CO}_2$ ($44\text{ g}$) and $1\text{ mol}$ of $\text{H}_2\text{O}$ ($18\text{ g}$), giving a total mass loss of $44 + 18 = 62\text{ g}$.

3. Calculate the mass of $\text{NaHCO}_3$ in the mixture:

1. Calculate percentage by mass:

This matches option C.

Key Takeaways

• When a reaction mixture loses mass upon heating in an open container, the loss in mass is equal to the mass of all gaseous products formed.
• You must account for all gases released (here both $\text{CO}_2$ and $\text{H}_2\text{O}$), not just one of them.
• Percentage by mass in a mixture is calculated as $\frac{\text{mass of component}}{\text{total initial mass}} \times 100\%$.

Common Mistakes

• Forgetting that water is produced as steam (gas) and assuming only $\text{CO}_2$ escapes, which leads to using $44\text{ g}$ instead of $62\text{ g}$ and finding an incorrect mass.
• Forgetting the stoichiometric coefficient of $2$ in front of $\text{NaHCO}_3$, using $84\text{ g}$ instead of $168\text{ g}$ (which yields $34\%$, corresponding to distractor A).

Things to Be Careful About

• Check state symbols carefully: $\text{H}_2\text{O}(\text{g})$ means it is in the gas phase and escapes the container along with $\text{CO}_2(\text{g})$.
• Keep unrounded intermediate numbers in your calculator before converting to the final percentage to avoid rounding discrepancies.

Walkthrough

To solve this question, we must apply two fundamental concepts of electroplating:

1. Role of the electrodes in electroplating:
   • Electroplating involves depositing a layer of metal onto an object.
   • The metal to be deposited exists in the electrolyte as positive cations (here, $\text{Ag}^+$).
   • Positive ions (cations) are attracted to the negative electrode, which is the cathode.
   • At the cathode, reduction takes place as silver ions gain electrons to form solid silver atoms:

• Therefore, the object to be plated (the statue) must be connected to the negative terminal as the cathode.

1. Choice of a suitable electrolyte:
   • The electrolyte must be an aqueous solution containing the ions of the plating metal ($\text{Ag}^+$).
   • From the solubility rules, all common nitrates are soluble in water, making $\text{AgNO}_3(\text{aq})$ an ideal aqueous electrolyte.
   • Silver chloride ($\text{AgCl}$) is an insoluble precipitate, so it cannot provide a suitable aqueous electrolyte solution.

Matching these two requirements gives row B.

Key Takeaways

• In electroplating:
  • The object to be plated is made the cathode (negative electrode).
  • The plating metal is made the anode (positive electrode) so it can dissolve and replenish the solution.
  • The electrolyte must be a soluble salt containing ions of the plating metal.
• All nitrates are soluble in water; halides of silver ($\text{AgCl}$, $\text{AgBr}$, $\text{AgI}$) are insoluble.

Common Mistakes

• Confusing the cathode and anode: candidates often mistakenly think the object is the anode because they confuse which electrode attracts positive ions.
• Overlooking the solubility of silver salts: selecting $\text{AgCl(aq)}$ without realising that $\text{AgCl}$ is insoluble in water.

Things to Be Careful About

• Always check both columns in electroplating tables: the electrode assignment (cathode vs anode) and the chemical validity of the electrolyte (soluble vs insoluble compound).

Walkthrough

A fuel cell is an electrochemical device that converts chemical energy from a fuel directly into electrical energy through redox reactions.

In a hydrogen–oxygen fuel cell:

• Hydrogen gas ($\text{H}_2$) is supplied at the anode, where it is oxidised.
• Oxygen gas ($\text{O}_2$) is supplied at the cathode, where it is reduced.
• The electrons flow through an external circuit, generating an electric current (electrical energy).
• The hydrogen and oxygen combine to produce pure water ($\text{H}_2\text{O}$) as the only chemical product:

Because its primary function is the continuous generation of electricity and its sole reaction product is water, row D is the correct answer.

Key Takeaways

• A fuel cell's main purpose is to produce electrical energy from chemical energy.
• The only chemical product formed in a hydrogen–oxygen fuel cell is water ($\text{H}_2\text{O}$), making it non-polluting at the point of use.
• Unlike rechargeable batteries, a fuel cell does not store electrical energy; it produces electricity continuously as long as fuel and oxygen are supplied.

Common Mistakes

• Confusing reactants with products: confusing hydrogen (the fuel reactant) with the product (water).
• Confusing fuel cells with waste management or recycling systems.

Things to Be Careful About

• Ensure you distinguish between simple chemical cells, rechargeable secondary cells, and fuel cells: fuel cells require a continuous external supply of reactants.

Walkthrough

To calculate the overall enthalpy change ($\Delta H$) of a reaction from bond energies, follow these three steps:

1. Calculate the energy required to break all the bonds in the reactants:

   • Bond breaking is an endothermic process (takes in energy, $+ve$).
   • Looking at hydrazine ($\text{N}_2\text{H}_4$), there are $4 \times \text{N–H}$ single bonds and $1 \times \text{N–N}$ single bond: $$4 \times 390\text{ kJ / mol} + 160\text{ kJ / mol} = 1720\text{ kJ / mol}$$
   • Oxygen gas ($\text{O}_2$) contains one double bond ($\text{O=O}$): $$1 \times 496\text{ kJ / mol} = 496\text{ kJ / mol}$$
   • Total energy absorbed $= 1720 + 496 = +2216\text{ kJ / mol}$.

2. Calculate the energy released when all new bonds form in the products:

   • Bond making is an exothermic process (releases energy, $-ve$).
   • Two water molecules ($2\text{H}_2\text{O}$) contain a total of $4 \times \text{O–H}$ single bonds: $$4 \times 460\text{ kJ / mol} = 1840\text{ kJ / mol}$$
   • A nitrogen molecule ($\text{N}_2$) contains one triple bond ($\text{N}\equiv\text{N}$): $$1 \times 944\text{ kJ / mol} = 944\text{ kJ / mol}$$
   • Total energy released $= 1840 + 944 = 2784\text{ kJ / mol}$.

3. Calculate $\Delta H$:

   $$\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds formed})$$ $$\Delta H = 2216 - 2784 = -568\text{ kJ / mol}$$

Thus, option B is correct.

Key Takeaways

• Bond breaking is endothermic (requires energy input); bond making is exothermic (releases energy).
• $\Delta H = \text{Energy absorbed to break bonds} - \text{Energy released to form bonds}$.
• Be careful to identify the correct bond order: $\text{O}_2$ has a double bond ($\text{O=O}$), and $\text{N}_2$ has a triple bond ($\text{N}\equiv\text{N}$).

Common Mistakes

• Selecting the wrong bond from the table (e.g., using $\text{O–O}$ instead of $\text{O=O}$ for $\text{O}_2$, or $\text{N–N}$ / $\text{N=N}$ instead of $\text{N}\equiv\text{N}$ for $\text{N}_2$).
• Forgetting that $2\text{H}_2\text{O}$ contains $4 \times \text{O–H}$ bonds in total, not 2.
• Inverting the formula to calculate $\text{bonds formed} - \text{bonds broken}$, which gives $+568\text{ kJ / mol}$ (wrong sign for an exothermic reaction).

Things to Be Careful About

• Always count all individual bonds using the displayed formula and stoichiometric balancing coefficients.
• Check the sign: since more energy is released in bond formation ($2784\text{ kJ / mol}$) than is required for bond breaking ($2216\text{ kJ / mol}$), the overall reaction must be exothermic (negative $\Delta H$).

Walkthrough

When calcium carbonate reacts with hydrochloric acid, carbon dioxide gas is produced and escapes through the cotton wool plug:

Because gas escapes the flask, the reading on the mass balance decreases over time.

Two key features of the graph allow us to assign each curve:

1. The final horizontal level (extent of reaction):

   • In Experiment 1 and Experiment 2, acid is in excess, so all $2\text{ g}$ of $\text{CaCO}_3$ reacts. They produce the same total mass of $\text{CO}_2$, so both curves must finish at the exact same lowest mass balance reading. These are curves Y and Z.
   • In Experiment 3, acid is limiting (less acid is used), so less $\text{CO}_2$ escapes. The final mass of the flask remains higher. Thus, Experiment 3 corresponds to curve X.

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

   • Experiment 1 uses powder, giving a greater surface area of contact between solid particles and acid particles. This increases collision frequency, leading to a faster rate (steeper curve, leveling off sooner). Thus, Experiment 1 corresponds to curve Z.
   • Experiment 2 uses lumps, which have a smaller surface area, resulting in a lower collision frequency and a slower rate (less steep curve). Thus, Experiment 2 corresponds to curve Y.

Matching with the given options confirms that C is the correct statement.

Key Takeaways

• Surface area: Smaller particle size (powder) increases total surface area exposed, increasing the collision frequency and the rate of reaction (steeper initial gradient).
• Limiting reactant: The amount of the limiting reactant determines the total amount of product formed and hence the final plateau of the curve.
• On a mass loss curve, a faster reaction is represented by a steeper downward slope, and a smaller total mass loss is represented by a curve leveling off higher up on the y-axis.

Common Mistakes

• Confusing a mass loss graph with a gas volume collection graph: in mass loss, curves slope downwards, and a faster reaction drops more steeply towards the bottom.
• Overlooking the limiting reactant in Experiment 3 and assuming it would reach the same mass level as the other two experiments.

Things to Be Careful About

• Ensure you identify which substance is in excess and which is limiting in each experiment to predict the final vertical height of the horizontal plateau.

Walkthrough

To determine the correct statement when the temperature is increased, we consider the two separate effects of temperature in chemical reactions:

1. Effect on Reaction Rates (Collision Theory):

   • When temperature increases, particles gain kinetic energy and move faster.
   • A greater proportion of colliding particles have energy equal to or greater than the activation energy ($E_a$).
   • Consequently, the frequency of effective collisions increases, which increases the rate of both the forward and the backward reactions.
   • The decomposition of ammonia is the reverse reaction ($\text{2NH}_3 \rightarrow \text{N}_2 + 3\text{H}_2$), so its rate increases. This makes D correct, and rules out C.

2. Effect on Equilibrium (Le Chatelier's Principle):

   • The forward reaction is exothermic (releases heat).
   • According to Le Chatelier's principle, increasing the temperature causes the equilibrium to shift in the direction that absorbs heat (the endothermic direction, which is to the left).
   • This decreases the yield of ammonia (less $\text{NH}_3$ is formed), ruling out B.

3. Effect on Activation Energy:

   • The activation energy ($E_a$) is an inherent characteristic of the reaction pathway and does not change with temperature. Only the addition of a catalyst changes activation energy. This rules out A.

Key Takeaways

• Temperature always increases reaction rates: Both the forward and reverse reaction rates increase as temperature rises because collisions occur more frequently and with greater energy.
• Equilibrium position vs. Rate: An increase in temperature shifts the equilibrium in the endothermic direction, but the rates of both forward and reverse reactions increase simultaneously (though the endothermic rate increases more until a new equilibrium is reached).

Common Mistakes

• Confusing the shift in equilibrium position with the change in reaction rate (e.g., thinking that because the forward reaction is not favoured at high temperatures, its rate must decrease).
• Thinking that increasing temperature changes the activation energy.

Things to Be Careful About

• Make sure to distinguish between "amount/yield of ammonia" (an equilibrium concept) and "rate of reaction" (a kinetic concept).

Walkthrough

An acid is defined by its ability to dissociate and produce hydrogen ions ($\text{H}^+$) in aqueous solution.

• Strong acids (such as $\text{HCl}$, $\text{HNO}_3$, and $\text{H}_2\text{SO}_4$) dissociate completely into ions when dissolved in water:
  $\text{HCl}(\text{aq}) \rightarrow \text{H}^+(\text{aq}) + \text{Cl}^-(\text{aq})$
• Weak acids (such as carboxylic acids like $\text{CH}_3\text{COOH}$) only partially dissociate in water, establishing a dynamic equilibrium where most molecules remain intact:
  $\text{CH}_3\text{COOH}(\text{aq}) \rightleftharpoons \text{CH}_3\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq})$

Evaluating the options:

• Option A is incorrect because neutralisation is a complete reaction; adding excess base drives the equilibrium to the right until all the acid has reacted.
• Option B correctly describes the definition of acid strength: ethanoic acid is weaker because a smaller proportion of its molecules ionise in solution.
• Option C confuses the pH scale (lower pH means more acidic / higher $[ ext{H}^+]$) and ignores that pH depends on concentration as well as strength.
• Option D confuses basicity (the number of replaceable hydrogen ions per molecule, i.e., diprotic) with acid strength (completeness of dissociation).

Key Takeaways

• Acid strength is a measure of the extent to which an acid dissociates/ionises in aqueous solution.
• Acid concentration refers to the number of moles of acid per unit volume ($\text{mol / dm}^3$).
• Basicity (or proticity) refers to the number of ionisable $\text{H}^+$ ions per acid molecule (e.g., $\text{H}_2\text{SO}_4$ is dibasic, $\text{HCl}$ is monobasic).

Common Mistakes

• Confusing strength (extent of ionisation) with concentration (amount of solute dissolved in water).
• Confusing dibasic/diprotic nature (number of $\text{H}^+$ per molecule) with being a strong acid.
• Forgetting that on the pH scale, lower numbers indicate higher acidity.

Things to Be Careful About

• Even though weak acids are partially dissociated in pure water, adding an alkali shifts the dissociation equilibrium until the weak acid reacts completely.

Walkthrough

Oxides are classified based on their acid-base behaviour:

1. Basic oxides: Oxides of metals (e.g. $\text{MgO}$, $\text{Na}_2\text{O}$) that react with acids to form a salt and water. They do not react with alkalis.
2. Acidic oxides: Oxides of non-metals (e.g. $\text{P}_4\text{O}_{10}$, $\text{SO}_2$, $\text{CO}_2$) that react with bases/alkalis to form a salt and water. They do not react with acids.
3. Amphoteric oxides: Oxides of specific metals (specifically aluminium oxide, $\text{Al}_2\text{O}_3$, and zinc oxide, $\text{ZnO}$) that exhibit both basic and acidic properties. They react with acids to form a salt and water, and also react with alkalis to form complex salts (aluminates or zincates) and water.
4. Neutral oxides: Oxides of non-metals (e.g. $\text{CO}$, $\text{NO}$) that do not react with either acids or bases.

The question gives three clues:

• Reacts with acid: indicates basic or amphoteric character.
• Insoluble in water.
• Reacts with alkali ($\text{NaOH}$): indicates acidic or amphoteric character.

Since it reacts with both an acid and an alkali, it must be amphoteric, which uniquely matches aluminium oxide (A).

Key Takeaways

• Amphoteric oxides (notably $\text{Al}_2\text{O}_3$ and $\text{ZnO}$ in O Level Chemistry) react with both acids and alkalis.
• Most metal oxides are basic, whereas most non-metal oxides are acidic.

Common Mistakes

• Confusing magnesium oxide with aluminium oxide: magnesium oxide is purely basic and will not react with sodium hydroxide.
• Thinking that non-metal oxides react with acids.

Things to Be Careful About

• Ensure you check both reaction clues (reaction with acid and reaction with base) together to identify amphoteric behavior rather than basic or acidic behavior alone.