Chemistry 9701/11 — May/June 2025

Background Concept

A balanced chemical equation must have equal numbers of atoms of each element on both sides. When an equation has several unknown coefficients, we write one atom-balance equation per element and solve them as a system of simultaneous equations.

This is a redox reaction: in $\text{ICl}_3$, iodine has oxidation state +3 (chlorine is more electronegative and takes -1 each). In the products, iodine appears as HI (oxidation state -1) and HIO$_3$ (oxidation state +5). Iodine is therefore simultaneously reduced (+3 $\rightarrow$ -1) and oxidised (+3 $\rightarrow$ +5) — a disproportionation. Chlorine stays at -1 throughout (in ICl$_3$, HCl and HClO$_3$), and oxygen is -2 in H$_2$O, HIO$_3$ and HClO$_3$.

Understanding the Question

We must find the smallest whole-number coefficients $v$, $x$ and $z$ that balance:

Note that the coefficient $z$ multiplies BOTH HIO$_3$ and HClO$_3$ — this is a fixed constraint of the equation as written. The options give candidate values for $v$, $x$ and $z$; we determine the correct set by balancing.

Approach

Write an atom-balance equation for each of the four elements (I, Cl, H, O). The oxygen balance immediately links $w$ and $z$ ($w = 6z$). The hydrogen balance then gives a relation between $x$, $y$ and $z$. The iodine and chlorine balances express $y$ and $x$ in terms of $v$ and $z$. Substituting into the hydrogen relation gives $v$ directly in terms of $z$, and the smallest integer solution follows.

Step-by-Step Reasoning

1. Iodine: left has $v$ atoms (one per ICl$_3$). Right: $y$ from HI + $z$ from HIO$_3$. So $v = y + z$.
2. Chlorine: left has $3v$. Right: $x$ from HCl + $z$ from HClO$_3$. So $3v = x + z$.
3. Hydrogen: left has $2w$. Right: $x$ (HCl) + $y$ (HI) + $z$ (HIO$_3$) + $z$ (HClO$_3$) = $x + y + 2z$. So $2w = x + y + 2z$.
4. Oxygen: left has $w$. Right: $3z$ (HIO$_3$) + $3z$ (HClO$_3$) = $6z$. So $w = 6z$.

From (3) and (4): $2w = 12z = x + y + 2z$, so $x + y = 10z$.

From (1): $y = v - z$. From (2): $x = 3v - z$.

Substitute: $(3v - z) + (v - z) = 10z$ $\rightarrow$ $4v - 2z = 10z$ $\rightarrow$ $4v = 12z$ $\rightarrow$ $v = 3z$.

The smallest whole-number solution is $z = 1$, giving $v = 3$. Then $x = 3v - z = 9 - 1 = 8$, and $y = v - z = 3 - 1 = 2$, $w = 6$.

Balanced equation: $3\text{ICl}_3 + 6\text{H}_2\text{O} \rightarrow 8\text{HCl} + 2\text{HI} + \text{HIO}_3 + \text{HClO}_3$.

Verification:

• I: $3 = 2 + 1$ ✓
• Cl: $9 = 8 + 1$ ✓
• H: $12 = 8 + 2 + 1 + 1$ ✓
• O: $6 = 3 + 3$ ✓

This matches option C ($v = 3$, $x = 8$, $z = 1$).

Distractors:

• A ($v=2$, $x=8$, $z=1$): $v = 2$ would require $z = 2/3$ from $v = 3z$, which is not a whole number.
• B ($v=2$, $x=5$, $z=1$): same problem — $v = 2$ is inconsistent with $v = 3z$.
• D ($v=3$, $x=5$, $z=1$): $v = 3$ is correct, but $x$ must be $3v - z = 8$, not 5.

Key Takeaways

• Balancing a multi-species equation is done systematically by writing one atom-balance equation per element and solving the resulting system.
• Recognise when a coefficient applies to more than one species ($z$ here multiplies both HIO$_3$ and HClO$_3$).
• In redox reactions, identify the element that is both oxidised and reduced (disproportionation) — here iodine in ICl$_3$.

Common Mistakes

• Forgetting that HClO$_3$ contains no iodine — miscounting iodine as $v = y + 2z$ instead of $v = y + z$.
• Forgetting that the coefficient $z$ multiplies both HIO$_3$ and HClO$_3$ in the hydrogen and oxygen balances.
• Arithmetic slips when solving the simultaneous equations.

Things to Be Careful About

• Count atoms element by element on both sides before writing each equation.
• The oxygen balance ($w = 6z$) is the key that unlocks the hydrogen relation.
• Always verify the final equation by re-counting every atom.
• Choose the smallest whole-number set of coefficients (here $z = 1$).

Background Concept

The rate of a reaction measures how quickly reactants are converted into products. For a reaction that produces a gas, the rate can be tracked by measuring the volume of gas collected over time. The average rate over a time interval is the change in the measured quantity divided by the length of the interval:

where $\Delta V$ is the change in gas volume and $\Delta t$ is the time interval. Because the metal is being consumed and the acid concentration falls as the reaction proceeds, the rate generally decreases with time — the volume-time graph becomes progressively less steep.

Understanding the Question

The table records the total (cumulative) volume of hydrogen produced at 30-second intervals over 3 minutes. The question defines P as the average rate during the first 30 seconds and Q as the average rate during the last 30 seconds, and asks for P − Q.

The word "total" is critical: each volume entry is the sum of all gas produced up to that time, so the volume produced within any interval is the difference between the two readings at its ends.

Approach

This is a data-processing question. For each interval, compute the average rate as (final volume − initial volume) ÷ 30 s, then subtract Q from P. No chemical reasoning beyond the definition of average rate is needed.

Step-by-Step Reasoning

1. First 30 s (0 to 30 s): The volume rises from 0 to $64 \text{ cm}^3$, so $\Delta V = 64 - 0 = 64 \text{ cm}^3$. Therefore $P = \frac{64}{30} = 2.13 \text{ cm}^3 \text{ s}^{-1}$.

2. Last 30 s (150 to 180 s): The volume rises from 161 to $167 \text{ cm}^3$, so $\Delta V = 167 - 161 = 6 \text{ cm}^3$. Therefore $Q = \frac{6}{30} = 0.20 \text{ cm}^3 \text{ s}^{-1}$.

3. Subtract: $P - Q = 2.13 - 0.20 = 1.93 \text{ cm}^3 \text{ s}^{-1}$, which matches option B.

Why the distractors are wrong:

• C (2.13): This is just P on its own — the candidate forgot to subtract Q.
• A (1.21): Close to P minus the overall average rate for the whole run ($167/180 \approx 0.93$), i.e. $2.13 - 0.93 \approx 1.20$. This mistakes the "last 30 s" rate for the whole-run average.
• D (3.43): Equals $(167 - 64)/30$ — taking the volume change from 30 s to 180 s but dividing by 30 s instead of 150 s. This is an interval/denominator mismatch.

Key Takeaways

• Average rate over an interval = change in measured quantity ÷ time interval.
• Cumulative data must be differenced to find the change within an interval.
• Rates fall as reactants are consumed — a shallower slope means a slower rate.
• Always identify the exact interval boundaries (here 150–180 s for the "last 30 seconds").

Common Mistakes

• Using the total volume at 180 s divided by 180 s as the "last" rate — this gives the overall average, not Q.
• Forgetting to subtract Q from P (choosing option C).
• Using the wrong interval for Q, e.g. $(167 - 64)/30 = 3.43$ (option D), which divides a 150-second volume change by 30 seconds.
• Mixing up units — the rates must be in $\text{cm}^3 \text{ s}^{-1}$, not $\text{cm}^3 \text{ min}^{-1}$.

Things to Be Careful About

• The table lists cumulative totals, not the volume produced in each 30 s window.
• The last 30 s runs from 150 s to 180 s, giving $\Delta V = 167 - 161 = 6 \text{ cm}^3$.
• Keep time in seconds throughout so the rates come out in $\text{cm}^3 \text{ s}^{-1}$.
• Round the final answer to two significant figures to match the options.

Background Concept

The rate of a chemical reaction in solution depends on the concentration of the reactants. A higher concentration means more frequent collisions between reactant particles, leading to a faster initial rate. The total volume of product formed in a reaction that goes to completion depends on the total number of moles of the limiting reactant, not its concentration. A catalyst speeds up the reaction by providing an alternative pathway with a lower activation energy, but it does not affect the total amount of product formed. Temperature increases the kinetic energy of particles, increasing both the frequency and the energy of collisions, thus increasing the rate of reaction.

Understanding the Question

The question provides a graph of the volume of oxygen gas produced over time during the decomposition of hydrogen peroxide. Curve X represents the original experiment using 100 cm³ of 1.0 mol dm⁻³ H₂O₂. Curve Y represents a modified experiment. We need to identify which modification produces curve Y. By observing the graph, curve Y has a lower initial slope (slower initial rate) and a higher final plateau (greater total volume of oxygen produced) compared to curve X.

Approach

We need to match the changes in the graph (lower initial rate, higher final volume) to the chemical effects of each proposed alteration.

1. Lower initial rate -> lower initial concentration of H₂O₂.
2. Higher final volume -> greater total moles of H₂O₂.
   Evaluate each option against these two criteria.

Step-by-Step Reasoning

• Curve analysis: The gradient of the volume-time graph at t=0 represents the initial rate of reaction. Curve Y is less steep initially than curve X, so the initial rate is lower. This happens if the initial concentration of the reactant is lower. The final horizontal asymptote represents the total volume of oxygen produced, which is proportional to the total moles of H₂O₂ decomposed. Curve Y levels off at a higher volume, so there must be more total moles of H₂O₂.
• Option A (adding more catalyst): A catalyst increases the rate of reaction (steeper initial gradient) but does not change the total moles of reactant, so the final volume remains the same. This would produce a curve steeper than X but with the same final level. Incorrect.
• Option B (adding some 0.1 mol dm⁻³ H₂O₂): Adding a solution of lower concentration (0.1 mol dm⁻³ < 1.0 mol dm⁻³) to the original solution increases the total volume, thereby decreasing the overall initial concentration of H₂O₂. This results in a slower initial rate (less steep gradient). However, it also adds extra moles of H₂O₂ to the mixture. Since there are more moles of reactant, more oxygen will be produced overall, leading to a higher final volume. This matches curve Y perfectly. Correct.
• Option C (adding water): Adding water dilutes the solution, lowering the concentration of H₂O₂ and thus decreasing the initial rate (less steep gradient). However, water does not add any H₂O₂, so the total moles of H₂O₂ remains unchanged. The final volume of oxygen would be the same as curve X. Incorrect.
• Option D (raising the temperature): Increasing the temperature increases the kinetic energy of the particles, leading to more frequent and more energetic collisions. This increases the rate of reaction (steeper initial gradient). It does not change the total moles of reactant, so the final volume remains the same. Incorrect.

Key Takeaways

• The initial gradient of a product vs. time graph indicates the initial rate, which is affected by concentration, temperature, and catalysts.
• The final volume (or mass) of product indicates the total amount (moles) of limiting reactant.
• Diluting a solution with a lower concentration reagent decreases the overall concentration (slowing the rate) but can increase the total moles if the reagent itself contains the reactant.

Common Mistakes

• Confusing concentration with total amount of substance: thinking that adding water or a dilute solution changes the total moles of reactant. Water adds volume but no moles of H₂O₂.
• Assuming a catalyst affects the final yield: catalysts only affect the rate, not the equilibrium position or total product formed from a given amount of reactant.

Things to Be Careful About

• Always read both axes of a rate graph: the gradient gives the rate (affected by concentration, temperature, catalyst), while the final plateau gives the total yield (affected only by the total moles of limiting reactant).
• When adding a solution of a different concentration, remember that the new total volume dilutes the original solution, lowering its concentration even if more moles are added.

Background Concept

Successive ionisation energies are the energies required to remove each successive electron from an atom. When a large jump occurs between two successive ionisation energies, it signals that the next electron is being removed from a much lower, more stable shell — the noble-gas core. The number of electrons removed before the big jump equals the number of valence (outer-shell) electrons.

Understanding the Question

The question lists the first seven ionisation energies of an element between lithium and neon (a Period 2 element). Our task is to deduce the outer electronic configuration. The key clue is the dramatic jump between the 6th and 7th ionisation energies.

Approach

1. Scan the successive ionisation energies for the largest jump.
2. Count how many electrons were removed before that jump — this gives the number of valence electrons.
3. Match that count to the Period 2 configuration (2s/2p orbitals).

Step-by-Step Reasoning

1. The successive ionisation energies are 1310, 3390, 5320, 7450, 11,000, 13,300, 71,000 kJ mol⁻¹.
2. The jump from 13,300 to 71,000 is enormous (more than a fivefold increase).
3. This jump occurs after the 6th electron is removed, so the 7th electron must come from the stable 1s² core.
4. Therefore the element has 6 electrons in its outer shell.
5. A Period 2 element with 6 outer electrons has the configuration 2s² 2p⁴ — this is oxygen.
6. The correct option is C.

Key Takeaways

A large jump in successive ionisation energies marks the transition from the valence shell to a noble-gas core. Count the electrons removed before the jump to find the number of valence electrons.

Common Mistakes

• Miscounting which ionisation energy shows the jump (the jump is after the 6th, not the 7th).
• Confusing the number of valence electrons with the wrong group.

Things to Be Careful About

• The jump is between the 6th and 7th ionisation energies, meaning 6 electrons were removed before it.
• The element is between lithium and neon, so it is in Period 2 — configurations only involve 2s and 2p orbitals.
• Option D (2s² 2p⁶) would correspond to neon, which would show a jump after the 8th electron, not the 6th.

Background Concept

The standard enthalpy change of formation, $\Delta H_{\text{f}}^{\ominus}$, is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (298 K, 1 atm). The standard enthalpy change of combustion, $\Delta H_{\text{c}}^{\ominus}$, is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. A key skill in energetics is recognising when a given reaction is itself a formation reaction, so that its enthalpy change can be written directly in terms of $\Delta H_{\text{f}}^{\ominus}$ values without needing a Hess cycle.

Understanding the Question

This is a multiple-choice question asking which expression gives the standard enthalpy change of the reaction $2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{l})$. The four options are built from $\Delta H_{\text{f}}^{\ominus}(\text{H}_2\text{O})$ and $\Delta H_{\text{c}}^{\ominus}(\text{H}_2)$. The task is to decide which combination correctly represents the enthalpy change of this specific reaction.

Approach

Ask: what kind of reaction is this? Hydrogen gas and oxygen gas (both elements in their standard states) combine to form water. That is exactly the definition of a formation reaction for water. The only complication is that the equation forms two moles of water, not one. Since $\Delta H_{\text{f}}^{\ominus}$ is defined per mole of compound formed, the enthalpy change for forming two moles is simply twice the standard enthalpy of formation of water. No combustion enthalpies are involved because nothing is being burned — the reactants are already elements.

Step-by-Step Reasoning

1. Recognise that $\text{H}_2(\text{g})$ and $\text{O}_2(\text{g})$ are elements in their standard states. By definition, the standard enthalpy of formation of any element in its standard state is zero.
2. The product, $\text{H}_2\text{O}(\text{l})$, is a compound formed directly from its elements. Therefore the reaction is the formation reaction of water.
3. The definition of $\Delta H_{\text{f}}^{\ominus}$ refers to the formation of one mole of compound. The given equation produces $2\text{H}_2\text{O}$, i.e. two moles.
4. Therefore the enthalpy change of the reaction as written is $2 \times \Delta H_{\text{f}}^{\ominus}(\text{H}_2\text{O})$.
5. This matches option A.

Why the distractors are wrong:

• B uses combustion enthalpies, but combustion is the burning of a substance in oxygen. Here hydrogen is being reacted with oxygen to form water, not burned as a fuel in a combustion sense — and the expression shown does not correspond to any valid enthalpy relationship for this reaction.
• C gives only $\Delta H_{\text{c}}^{\ominus}(\text{H}_2)$, which is the enthalpy change for burning one mole of hydrogen to form one mole of water. This ignores the stoichiometric coefficient of 2 and, more fundamentally, uses the wrong type of enthalpy change.
• D incorrectly adds a combustion enthalpy term to the formation term. Once water has been formed, there is no further combustion step; adding $\Delta H_{\text{c}}^{\ominus}(\text{H}_2)$ double-counts or misassigns the energy change.

Key Takeaways

• Recognise a formation reaction on sight: a compound formed from its elements in their standard states.
• Always respect the stoichiometric coefficient — $\Delta H_{\text{f}}^{\ominus}$ is per mole, so multiply by the number of moles formed.
• Combustion enthalpies apply when a substance is burned in oxygen; they are not interchangeable with formation enthalpies.

Common Mistakes

• Forgetting the factor of 2 and choosing option C (which would be correct only for $\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}$).
• Confusing formation and combustion enthalpies, leading to options B or D.
• Thinking that $\Delta H_{\text{f}}^{\ominus}$ of an element must be included; it is zero by definition for elements in their standard states.

Things to Be Careful About

• The definition of $\Delta H_{\text{f}}^{\ominus}$ is strictly for one mole of compound — the coefficient in the balanced equation determines the multiplier.
• Note the physical state: water here is liquid, $\text{H}_2\text{O}(\text{l})$. The standard enthalpy of formation of liquid water differs slightly from that of water vapour, so the state symbol matters if values are quoted.
• In an MCQ, once you identify the reaction type and the correct multiplier, you can select the answer directly without computing any numbers.

Background Concept

An organic compound containing only carbon, hydrogen and oxygen burns in excess oxygen to give carbon dioxide and water. All of its carbon appears as $\text{CO}_2$ and all of its hydrogen appears as $\text{H}_2\text{O}$, so the masses of C and H in the original compound are known from the combustion data. If the compound itself contains oxygen, that oxygen mass is simply the remaining mass after subtracting the masses of C and H from the total sample mass.

The empirical formula is the smallest whole-number ratio of moles of atoms in the compound. To find it, convert each element's mass to moles and divide by the smallest number.

The other key clue is that carboxylic acids react with carbonates, releasing carbon dioxide:

The bubbles seen are $\text{CO}_2$. Aldehydes, ketones and alcohols do not normally react with sodium carbonate in this way.

Understanding the Question

The question gives the total mass of P, $2.20\text{ g}$, and tells you that this sample contains $1.20\text{ g}$ of carbon and $0.20\text{ g}$ of hydrogen. Because P burns to $\text{CO}_2$ and $\text{H}_2\text{O}$ only, it can contain C, H and possibly O. You must first calculate the mass of oxygen by difference.

It also tells you that P produces gas with sodium carbonate solution. This is a chemical test for a carboxylic acid: a compound with a $-\text{CO}_2\text{H}$ group. A candidate compound must therefore satisfy both the empirical-formula clue and the acid functional-group clue.

Approach

Work in three stages:

1. Find the mass of oxygen in 2.20 g of P by difference.
2. Convert the masses of C, H and O to moles, then reduce to the simplest whole-number ratio to get the empirical formula.
3. Use the sodium carbonate test to identify which candidate is a carboxylic acid, and check that its molecular formula is consistent with the empirical formula.

Step-by-Step Reasoning

Mass of oxygen:

Convert each mass to moles using the appropriate atomic masses:

Divide each mole value by the smallest, $0.050$:

So the empirical formula is $\text{C}_2\text{H}_4\text{O}$.

Now look at the options against two criteria:

• Option A, $\text{CH}_3\text{CHO}$, has molecular formula $\text{C}_2\text{H}_4\text{O}$. This does match the empirical formula, but ethanal is an aldehyde, not a carboxylic acid. It does not give bubbles with sodium carbonate.
• Option B, $\text{CH}_3\text{CO}_2\text{H}$, is ethanoic acid and does contain a carboxyl group, but its molecular formula is $\text{C}_2\text{H}_4\text{O}_2$, so its empirical formula is $\text{CH}_2\text{O}$, not $\text{C}_2\text{H}_4\text{O}$.
• Option C, $\text{CH}_3\text{COCH}_2\text{CH}_2\text{OH}$, has molecular formula $\text{C}_4\text{H}_8\text{O}_2$, so it has empirical formula $\text{C}_2\text{H}_4\text{O}$. But it contains a ketone carbonyl and an alcohol group, not a carboxylic acid group, so it does not react with sodium carbonate to give $\text{CO}_2$.
• Option D, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CO}_2\text{H}$ (butanoic acid), has molecular formula $\text{C}_4\text{H}_8\text{O}_2$. This is $2 \times \text{C}_2\text{H}_4\text{O}$, so its empirical formula is correct. It also contains a carboxylic acid group, so it reacts with sodium carbonate and releases carbon dioxide.

Therefore P is option D.

Key Takeaways

• Combustion data gives the masses of C and H; any O in the compound is found by difference.
• The empirical formula comes from the simplest mole ratio, not directly from the masses.
• The sodium carbonate test is specific for carboxylic acids: effervescence indicates $-\text{CO}_2\text{H}$.
• A correct answer must satisfy both quantitative composition and the chemical test.

Common Mistakes

• Forgetting to include oxygen. If you only use the C and H masses, you get a ratio that does not match any option and you miss the key step.
• Choosing A because its molecular formula is exactly C2H4O. The empirical formula is necessary but not sufficient; A is an aldehyde, not a carboxylic acid, so it fails the chemical test.
• Thinking any carbonyl compound reacts with carbonate. Aldehydes and ketones are not acidic enough to displace carbonic acid from sodium carbonate.
• Incorrectly matching option B. B is an acid, but its empirical formula is CH2O, not C2H4O.

Things to Be Careful About

• Use atomic masses C = 12, H = 1, O = 16.
• The phrase "burns in an excess of oxygen to give carbon dioxide and water only" tells you the compound contains C, H and possibly O, but no other elements such as N or S.
• Check the full molecular formula of each option, not just the displayed shorthand. For example, $\text{CH}_3\text{COCH}_2\text{CH}_2\text{OH}$ and $\text{CH}_3\text{CH}_2\text{CH}_2\text{CO}_2\text{H}$ both reduce to the same empirical formula, but only the one with the carboxyl group reacts with carbonate.
• State the final answer as the option letter; the working is what supports it.

Background Concept

Physical properties such as melting point, boiling point, electrical conductivity, and solubility can be used to deduce the type of structure and bonding in a substance.

• Ionic compounds: high melting/boiling points, poor conductivity as solids, good conductivity when molten or in aqueous solution, often soluble in water.
• Giant covalent substances: very high melting/boiling points, poor conductivity as solids and usually poor conductivity when molten, insoluble in water.
• Metals: high melting/boiling points, good conductivity as solids and when molten, insoluble in water.

Understanding the Question

The table gives four physical properties of substance W:

• m.p. = 2072 °C
• b.p. = 2980 °C
• solid conductivity = poor
• liquid conductivity = good
• conductivity in water = insoluble

The question asks which of the four substances matches all of these properties.

Approach

1. Use the high melting/boiling point to narrow the possibilities to giant structures.
2. Use electrical conductivity to distinguish between ionic, metallic, and covalent network structures.
3. Use solubility in water to choose between the remaining candidates.

Step-by-Step Reasoning

1. High m.p. and b.p.

   • Aluminium oxide, silicon dioxide, and iron all have very high melting/boiling points.
   • Sodium fluoride is ionic and also has a high melting point, but it is soluble in water.

2. Solid conductivity = poor

   • Iron is a metal, so it conducts electricity in the solid state. This eliminates iron.
   • Silicon dioxide is a giant covalent network; it does not conduct electricity as a solid or when molten. This would not match the “liquid conductivity = good” property.

3. Liquid conductivity = good

   • Aluminium oxide is ionic. In the solid state its ions are fixed in a lattice, so it does not conduct. When molten, the ions are free to move, so it conducts electricity.
   • Sodium fluoride is also ionic and would conduct when molten, but it is soluble in water.

4. Insoluble in water

   • Aluminium oxide is insoluble in water.
   • Sodium fluoride is soluble in water.

Therefore, substance W is aluminium oxide.

Key Takeaways

• Ionic compounds conduct electricity only when molten or in aqueous solution, not as solids.
• Giant covalent substances have very high melting points but generally do not conduct electricity even when molten.
• Solubility in water helps distinguish between ionic compounds that dissolve and those that do not.

Common Mistakes

• Choosing silicon dioxide because it has a high melting point, without checking that it would not conduct when molten.
• Choosing sodium fluoride because it is ionic, without checking that it is soluble in water.
• Forgetting that metals conduct electricity in the solid state.

Things to Be Careful About

• Read the conductivity columns carefully: solid, liquid, and in water are three separate tests.
• “Insoluble in water” is a key clue; many ionic compounds dissolve, but not all do.
• Aluminium oxide is an example of an ionic compound with a very high melting point and low solubility in water.

Background Concept

Sodium chloride is a giant ionic lattice (specifically, a rock salt structure). It is formed by the electrostatic attraction between positively charged sodium ions ($\text{Na}^+$) and negatively charged chloride ions ($\text{Cl}^-$). In such a lattice, ions are arranged in a regular, repeating 3D pattern where each ion is surrounded by ions of the opposite charge. This maximises attractive forces and minimises repulsive forces.

A key feature of ionic lattices is the relative size of the ions. The sodium ion ($\text{Na}^+$) has the electron configuration $1s^2 2s^2 2p^6$ (isoelectronic with neon, 2 electron shells). The chloride ion ($\text{Cl}^-$) has the electron configuration $1s^2 2s^2 2p^6 3s^2 3p^6$ (isoelectronic with argon, 3 electron shells). Therefore, the chloride ion is significantly larger than the sodium ion ($r(\text{Cl}^-) \approx 181 \text{ pm}$, $r(\text{Na}^+) \approx 102 \text{ pm}$).

Understanding the Question

The question asks to identify the correct 2D diagram representing the lattice structure of sodium chloride. We are given four diagrams (A, B, C, D) and a key defining the symbols:

• $\text{Cl}$ = chlorine atom
• $\text{Na}$ = sodium atom
• $-$ = chloride ion
• $+$ = sodium ion

We must evaluate each diagram based on: (1) whether it shows ions or atoms, (2) the arrangement (ordered vs. random), and (3) the relative sizes of the positive and negative species.

Approach

1. Check species: Sodium chloride is ionic, so the diagram must show ions ($+$ and $-$), not neutral atoms ($\text{Na}$ and $\text{Cl}$). This eliminates diagrams using atomic labels.
2. Check arrangement: Ionic lattices are highly ordered and regular. Random arrangements are incorrect.
3. Check relative sizes: The anion ($\text{Cl}^-$) must be larger than the cation ($\text{Na}^+$). Compare the sizes of the $+$ and $-$ circles in the remaining diagrams.

Step-by-Step Reasoning

• Diagram A: The key defines the labels as $\text{Na}$ (sodium atom) and $\text{Cl}$ (chlorine atom). Sodium chloride is an ionic compound consisting of $\text{Na}^+$ and $\text{Cl}^-$ ions, not neutral atoms. This diagram is incorrect.
• Diagram D: The ions are arranged randomly and irregularly. Ionic lattices have a regular, alternating arrangement of ions. This diagram likely represents a liquid or a gas, or simply a disordered mixture, not a crystal lattice. This is incorrect.
• Diagram C: This diagram shows an alternating arrangement of $+$ and $-$ ions, which is good. However, the positive ions ($+$, sodium) are drawn as large circles and the negative ions ($-$, chloride) as small circles. This implies the cation is larger than the anion. In reality, $\text{Cl}^-$ (3 shells) is much larger than $\text{Na}^+$ (2 shells). This diagram has the relative sizes wrong.
• Diagram B: This diagram shows an alternating arrangement of positive ($+$) and negative ($-$) ions, consistent with an ionic lattice. The key identifies $+$ as sodium ion and $-$ as chloride ion. The positive ions ($+$) are drawn as small circles and the negative ions ($-$) as large circles. This correctly represents the fact that the chloride ion ($\text{Cl}^-$) is larger than the sodium ion ($\text{Na}^+$). This is the correct representation.

Key Takeaways

• Ionic lattices are composed of ions, not neutral atoms. Always check the key and labels.
• Ionic lattices have a regular, alternating arrangement of cations and anions.
• For sodium chloride, the anion ($\text{Cl}^-$) is larger than the cation ($\text{Na}^+$) due to the extra electron shell.

Common Mistakes

• Choosing A: Students might see $\text{Na}$ and $\text{Cl}$ and think it's correct, forgetting that in the lattice they exist as ions ($\text{Na}^+$ and $\text{Cl}^-$). The key explicitly defines them as atoms here.
• Choosing C: Students might correctly identify the alternating pattern but fail to recall or apply the relative ionic radii ($\text{Cl}^- > \text{Na}^+$), leading them to pick the diagram with the wrong size ratio.
• Choosing D: Students might confuse the 2D lattice representation with a particle diagram of a liquid or gas where particles are close but not ordered.

Things to Be Careful About

• Read the key carefully: The key distinguishes between atoms ($\text{Na}$, $\text{Cl}$) and ions ($+$, $-$). Diagram A uses atomic labels, which is a common distractor.
• Relative sizes: Always verify the relative sizes of cations and anions. For Group 1 halides (like NaCl, KBr), the halide ion is always larger than the metal ion.
• Order vs. Random: Lattice structures are defined by their long-range order. Any diagram showing a random arrangement is incorrect for a solid lattice.

Background Concept

A weak acid such as $\text{HQ}$ only partially ionises in water:

At equilibrium the forward and reverse rates are equal. Adding a soluble salt that contains the conjugate base $\text{Q}^-$ introduces a large concentration of ions common to the equilibrium. This is the common ion effect: the added $\text{Q}^-$ disturbs the equilibrium, and by Le Chatelier's principle the position shifts to the side that reduces the disturbance — here to the left, consuming $\text{H}^+$ and $\text{Q}^-$ to form more $\text{HQ}$.

The pH is defined by

so a decrease in $[\text{H}^+]$ corresponds to an increase in pH.

Understanding the Question

The question gives a weak acid equilibrium and asks what happens to two linked quantities when the sodium salt of the acid, $\text{NaQ}$, is added to the solution. The key point is that $\text{NaQ}$ is a soluble ionic salt, so it dissociates completely into $\text{Na}^+$ and $\text{Q}^-$ ions. The $\text{Q}^-$ ion is the conjugate base of $\text{HQ}$ and is already present in the equilibrium mixture. Adding more of it is a classic common-ion situation. The command is essentially "predict and explain": decide the direction of the equilibrium shift, then translate the resulting $[\text{H}^+]$ change into a pH change.

Approach

1. Recognise that $\text{NaQ}$ supplies $\text{Q}^-$, a species on the right-hand side of the equilibrium.
2. Apply Le Chatelier's principle: increasing the concentration of a product shifts the equilibrium to the left.
3. Deduce that $[\text{H}^+]$ falls.
4. Use $\text{pH} = -\log_{10}[\text{H}^+]$ to determine that pH rises.

No calculation is needed; the question only tests qualitative prediction.

Step-by-Step Reasoning

• $\text{NaQ}$ is a soluble salt, so in aqueous solution it fully dissociates:

The $\text{Na}^+$ ion is a spectator and does not affect the acid equilibrium.

• The added $\text{Q}^-$ increases the concentration of a species on the right of the equilibrium:

By Le Chatelier's principle, the system responds by shifting in the direction that consumes some of the added $\text{Q}^-$ — that is, to the left.

• Shifting left converts $\text{H}^+$ and $\text{Q}^-$ into $\text{HQ}$, so $[\text{H}^+]$ decreases.

• Since $\text{pH} = -\log_{10}[\text{H}^+]$, a smaller $[\text{H}^+]$ gives a larger pH. Therefore pH increases.

This matches option B: $[\text{H}^+]$ decreases and pH increases.

Key Takeaways

• Adding a salt of the conjugate base of a weak acid suppresses the acid's ionisation: this is the common ion effect.
• Le Chatelier's principle applies to acid–base equilibria exactly as to other equilibria: adding a product shifts the equilibrium toward reactants.
• $[\text{H}^+]$ and pH move in opposite directions: lower $[\text{H}^+]$ means higher pH.
• A soluble salt is assumed to be fully dissociated, so only the relevant ion (here $\text{Q}^-$) matters.

Common Mistakes

• Choosing A: thinking pH decreases when $[\text{H}^+]$ decreases. This confuses the direction of pH with the direction of $[\text{H}^+]$.
• Thinking that adding $\text{NaQ}$ adds $\text{Na}^+$ and therefore changes pH directly. $\text{Na}^+$ is a spectator ion.
• Assuming the salt reacts with water to produce more $\text{H}^+$. Here the conjugate base $\text{Q}^-$ is the common ion and suppresses ionisation, not increases it.
• Forgetting that $\text{HQ}$ is weak, so the equilibrium is established and can be shifted; a strong acid would fully dissociate and the common ion effect would be irrelevant.

Things to Be Careful About

• pH is defined as $-\log_{10}[\text{H}^+]$, so a ten-fold decrease in $[\text{H}^+]$ raises pH by 1 unit.
• The equilibrium arrow is $\rightleftharpoons$, not $\rightarrow$; the system is dynamic and can shift.
• The salt $\text{NaQ}$ is stated to be soluble, so it dissociates completely; do not treat it as a weak electrolyte.
• In a buffer context, adding $\text{NaQ}$ to $\text{HQ}$ creates a buffer, but here the question only asks for the immediate direction of change, not the buffer capacity.

Background Concept

For a reversible reaction at equilibrium, the equilibrium constant $K_c$ depends only on temperature. Changing pressure or concentration does not change the value of $K_c$ (it only shifts the position of equilibrium).

For an endothermic reaction ($\Delta H > 0$), increasing the temperature favours the forward direction, so $K_c$ increases. For an exothermic reaction ($\Delta H < 0$), increasing the temperature favours the reverse direction, so $K_c$ decreases.

Understanding the Question

The reaction $\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$ is endothermic ($\Delta H = +58.0 \text{ kJ mol}^{-1}$). We are told that $K_c$ increases dramatically from $1.15 \times 10^{-1}$ to $1.70 \times 10^3 \text{ mol dm}^{-3}$. We need to identify which change in conditions caused this increase.

Approach

Recall that $K_c$ is only affected by temperature. Then determine whether an increase or decrease in temperature would increase $K_c$ for an endothermic reaction.

Step-by-Step Reasoning

1. Identify the reaction as endothermic: $\Delta H = +58.0 \text{ kJ mol}^{-1}$.
2. Note that $K_c$ has increased (from $10^{-1}$ to $10^3$).
3. For an endothermic reaction, increasing temperature shifts equilibrium to the right (towards products), increasing $K_c$.
4. Decreasing temperature would shift equilibrium to the left, decreasing $K_c$.
5. Pressure changes do not affect $K_c$ — they only shift the position of equilibrium to minimise the change, but the value of $K_c$ remains constant at a given temperature.
6. Therefore the correct change is an increase in temperature.

Key Takeaways

• $K_c$ is temperature-dependent only; pressure and concentration changes do not change its value.
• For an endothermic reaction, $K_c$ increases with increasing temperature.
• For an exothermic reaction, $K_c$ decreases with increasing temperature.

Common Mistakes

• Confusing the effect of pressure: pressure shifts the position of equilibrium but does not change $K_c$.
• Mixing up endothermic and exothermic: remember "endothermic absorbs heat, so adding heat (increasing temperature) favours the forward reaction."

Things to Be Careful About

• Always check the sign of $\Delta H$ before deciding how temperature affects $K_c$.
• The units of $K_c$ (mol dm$^{-3}$) confirm the stoichiometry ($\Delta n = 1$), but this is not needed to answer the question.
• In a closed container, changing pressure (e.g. by changing volume) does not change $K_c$ at constant temperature.

Background Concept

The enthalpy of neutralisation is the heat evolved when one mole of water is formed from the reaction between an acid and a base. For a strong acid and a strong base it is typically about -57 kJ mol⁻¹. In this experiment the heat released warms the whole aqueous mixture, so we use calorimetry: q = VcΔT, where c is given per unit volume.

Understanding the Question

50 cm³ of 1.0 mol dm⁻³ H₂SO₄ and 100 cm³ of 1.0 mol dm⁻³ NaOH are mixed. Both start at 20 °C and the highest temperature reached is 29 °C. The task is to find the enthalpy of neutralisation from this temperature rise. The key is to calculate the heat released and then divide by the number of moles of water actually formed.

Approach

1. Find the total volume of the mixture and the temperature rise.
2. Calculate the heat released using q = VcΔT.
3. Calculate the moles of H₂SO₄ and NaOH.
4. Use the balanced neutralisation equation to find the moles of water formed.
5. Divide the heat by the moles of water and give a negative sign because the reaction is exothermic.

Step-by-Step Reasoning

• Total volume = 50 + 100 = 150 cm³.
• Temperature rise = 29 - 20 = 9 K.
• Heat released = 150 × 4.2 × 9 = 5670 J = 5.67 kJ.
• n(H₂SO₄) = 0.050 × 1.0 = 0.050 mol.
• n(NaOH) = 0.100 × 1.0 = 0.100 mol.
• Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
• 0.050 mol H₂SO₄ requires 0.100 mol NaOH, so the acid and base are exactly in the stoichiometric ratio and 0.100 mol water is formed.
• ΔH = -5.67 kJ / 0.100 mol = -56.7 kJ mol⁻¹.

Key Takeaways

• Enthalpy of neutralisation is expressed per mole of water formed, not per mole of acid or base.
• Sulfuric acid is diprotic, so one mole of H₂SO₄ reacts with two moles of NaOH and produces two moles of water.
• When the heat capacity is given per cm³, use volume in q = VcΔT.

Common Mistakes

• Using only the volume of one solution instead of the total volume.
• Dividing by moles of H₂SO₄ instead of moles of water formed.
• Forgetting that the reaction is exothermic and omitting the negative sign.
• Using ΔT = 29 K instead of 29 - 20 = 9 K.

Things to Be Careful About

• The temperature rise is 9 K, not 29 K.
• The unit J cm⁻³ K⁻¹ means q = VcΔT, not mcΔT with mass.
• Check the stoichiometry: H₂SO₄ provides two H⁺ ions, so it needs two moles of NaOH.
• The final answer must be negative because heat is released.

Background Concept

This question tests core stoichiometry: converting between mass and moles using molar mass, applying a stoichiometric ratio from a balanced equation, and expressing moles as a concentration in mol dm⁻³. The key chemical insight is that the decomposition of the dithionate ion produces sulfate ion, which immediately precipitates with barium ion as insoluble barium sulfate. Barium sulfate is famously insoluble — it is the basis of the gravimetric sulfate analysis and the classic sulfate test — while barium dithionate is, as stated, soluble.

Understanding the Question

We start with 3.513 g of the hydrated salt BaS₂O₆·2H₂O dissolved in 100 cm³ of acidified solution. The dithionate ion slowly decomposes:

$\text{S}_2\text{O}_6^{2-}(\text{aq}) \rightarrow \text{SO}_2(\text{g}) + \text{SO}_4^{2-}(\text{aq})$

The SO₄²⁻ produced reacts with Ba²⁺ to form a white precipitate of BaSO₄. At time x, 0.661 g of precipitate has formed. We need the concentration of undecomposed BaS₂O₆ remaining at that moment. The command is a calculation — "What is the concentration" — so full numerical working is required.

The traps are: (1) the initial mass refers to the hydrated salt, so the water of crystallisation must be included in the molar mass; (2) the precipitate is a different compound (BaSO₄), not BaS₂O₆; (3) the volume must be in dm³ for a concentration in mol dm⁻³.

Approach

1. Find the molar mass of the hydrated salt and convert the initial mass to initial moles.
2. Recognise the precipitate is BaSO₄, find its molar mass, and convert the precipitate mass to moles.
3. Use the 1:1 stoichiometry of the decomposition to find moles of dithionate decomposed.
4. Subtract to find moles remaining.
5. Divide by the volume in dm³ to get concentration.

Step-by-Step Reasoning

Step 1 — Initial moles.
M_r(BaS₂O₆·2H₂O) = 137.3 + 2(32.1) + 6(16.0) + 2(18.0) = 333.5 g mol⁻¹.

$n = \frac{3.513}{333.5} = 0.01053\ \text{mol}$

Step 2 — The precipitate.
The decomposition produces SO₄²⁻, which reacts with Ba²⁺:

$\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})$

M_r(BaSO₄) = 137.3 + 32.1 + 4(16.0) = 233.4 g mol⁻¹.

$n = \frac{0.661}{233.4} = 0.00283\ \text{mol}$

Step 3 — Stoichiometry.
Each mole of S₂O₆²⁻ that decomposes produces one mole of SO₄²⁻, which precipitates as one mole of BaSO₄. So moles of dithionate decomposed = moles of BaSO₄ = 0.00283 mol.

Step 4 — Remaining moles.

$n = 0.01053 - 0.00283 = 0.00770\ \text{mol}$

Step 5 — Concentration.
Volume = 100 cm³ = 0.100 dm³.

$[\text{BaS}_2\text{O}_6] = \frac{0.00770}{0.100} = 0.0770\ \text{mol dm}^{-3}$

This matches option C.

Why the distractors are wrong:

• A (0.0077): This is the moles remaining, not divided by the volume — forgetting the 0.100 dm³ factor.
• B (0.0090): Likely from using an incorrect molar mass (e.g. omitting the water of crystallisation) or an arithmetic slip in the subtraction.
• D (0.090): A factor-of-ten error on C, or using the wrong initial moles.

Key Takeaways

• Always include water of crystallisation in the molar mass of a hydrated salt.
• Recognise that a precipitate in such problems is often the insoluble salt of the cation with the product anion (here BaSO₄).
• The 1:1 stoichiometry of the decomposition links moles of reactant decomposed to moles of product formed.
• Concentration requires dividing by volume in dm³ (100 cm³ = 0.100 dm³).

Common Mistakes

• Forgetting the 2H₂O in the molar mass of the hydrated salt, giving a wrong initial mole count.
• Not recognising that the precipitate is BaSO₄ and trying to relate the precipitate mass directly to BaS₂O₆.
• Forgetting to convert 100 cm³ to 0.100 dm³ before computing concentration.
• Confusing moles remaining with concentration — option A is exactly the moles without dividing by volume.

Things to Be Careful About

• Molar masses: use accurate values (Ba 137.3, S 32.1, O 16.0, H 1.0).
• The volume is 100 cm³ = 0.100 dm³.
• The precipitate mass is for BaSO₄, not for BaS₂O₆.
• The decomposition is slow — the question is about the state at a particular time, not completion.
• State symbols: SO₂ is a gas and escapes, so it does not affect the precipitate mass.

Background Concept

The ideal gas equation, $PV = nRT$, relates the pressure ($P$), volume ($V$), amount of substance in moles ($n$), and absolute temperature ($T$) of an ideal gas. The gas constant $R$ has a value of $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$, which requires pressure to be in pascals (Pa), volume in cubic metres (m$^3$), and temperature in kelvin (K). When gases from multiple containers are mixed, the total amount of gas is the sum of the moles in each container, and the total volume available is the sum of the individual container volumes (assuming the connecting tube volume is negligible).

Understanding the Question

We are given two containers, each with a volume of $1.00 \text{ m}^3$, containing methane ($\text{CH}_4$).

• Container 1: $16 \text{ g}$ methane at $100 \text{ kPa}$.
• Container 2: $96 \text{ g}$ methane at $200 \text{ kPa}$.
  The tap connecting them is opened, allowing the gas to occupy the combined volume. The system is then heated to $800 \text{ K}$ and reaches a new constant pressure. We need to calculate this final pressure.

Approach

1. Determine the total number of moles of methane in the system by converting the given masses to moles using the molar mass of methane.
2. Determine the total volume available to the gas after the tap is opened.
3. Use the ideal gas equation ($P = nRT/V$) with the total moles, total volume, and final temperature to find the final pressure.

Step-by-Step Reasoning

Step 1: Calculate total moles of methane.
The molar mass ($M_r$) of methane ($\text{CH}_4$) is $12 + 4(1.0) = 16 \text{ g mol}^{-1}$.
Moles in container 1: $n_1 = 16 \text{ g} / 16 \text{ g mol}^{-1} = 1.0 \text{ mol}$.
Moles in container 2: $n_2 = 96 \text{ g} / 16 \text{ g mol}^{-1} = 6.0 \text{ mol}$.
Total moles: $n_{\text{total}} = 1.0 + 6.0 = 7.0 \text{ mol}$.

Step 2: Determine total volume.
When the tap is opened, the gas expands to fill both containers.
$V_{\text{total}} = 1.00 \text{ m}^3 + 1.00 \text{ m}^3 = 2.00 \text{ m}^3$.

Step 3: Calculate final pressure.
Given: $n = 7.0 \text{ mol}$, $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$, $T = 800 \text{ K}$, $V = 2.00 \text{ m}^3$.

Convert pascals to kilopascals ($1 \text{ kPa} = 1000 \text{ Pa}$):

This matches option A.

Key Takeaways

• When mixing gases in connected containers, add the moles and add the volumes.
• Always check units in the ideal gas equation: use Pa, m$^3$, K, and mol when $R = 8.314$.
• Converting between Pa and kPa is a common step that requires attention to decimal places.

Common Mistakes

• Forgetting to add volumes: Using $V = 1.00 \text{ m}^3$ instead of $2.00 \text{ m}^3$ would give a pressure of $46.6 \text{ kPa}$ (Option B), which is incorrect because the gas expands into both containers.
• Using wrong units for R: If $R = 8.314$ is used with volume in dm$^3$ or pressure in kPa directly without conversion, the result will be wrong.
• Ignoring the mass of one container: Calculating moles using only one of the masses (e.g., just 16 g or just 96 g) leads to incorrect total moles and thus wrong pressure.