Chemistry 9701/12 — February/March 2025

Background Concept

The Boltzmann distribution shows the distribution of kinetic energies among particles in a gas at a given temperature. The y-axis represents the number of particles, and the x-axis represents their kinetic energy. The curve starts at the origin, rises to a peak (the most probable energy), and then tails off towards higher energies, never quite reaching the x-axis. The area under the curve represents the total number of particles, which is constant if the amount of gas doesn't change. The activation energy $E_{\text{a}}$ is a fixed energy threshold on the x-axis; particles with energy $\ge E_{\text{a}}$ can react.

Understanding the Question

We are given a Boltzmann distribution at an initial temperature, with a point $\times$ marking the number of particles that have exactly the activation energy $E_{\text{a}}$ (or more precisely, the height of the curve at $E_{\text{a}}$). The temperature is then decreased. We need to find which point (A, B, C, or D) represents the new number of particles at the same activation energy $E_{\text{a}}$ on the new, lower-temperature curve.

Approach

1. Recall how a Boltzmann distribution curve changes when temperature decreases: the curve shifts to the left, the peak lowers, and the tail at high energies becomes shorter.
2. Recognize that the activation energy $E_{\text{a}}$ is an intrinsic property of the reaction and does not change with temperature; it remains at the same x-coordinate.
3. Determine the new y-value at $x = E_{\text{a}}$ on the new curve. Since the curve has shifted left and down, the number of particles at $E_{\text{a}}$ must be lower than before.
4. Identify the point that lies at $x = E_{\text{a}}$ with a lower y-value.

Step-by-Step Reasoning

• Effect of decreasing temperature on the curve: When temperature decreases, the average kinetic energy of the particles decreases. The curve's peak moves to a lower energy (left), the peak height decreases (because the area under the curve, representing total particles, is constant and the curve is broader), and the right-hand tail drops significantly.
• Position of $E_{\text{a}}$: The activation energy $E_{\text{a}}$ is marked on the x-axis. It is a fixed energy value for the reaction and does not change when the temperature changes. Therefore, we must look at the new curve at the exact same x-coordinate ($E_{\text{a}}$).
• Number of particles at $E_{\text{a}}$: At a lower temperature, fewer particles have enough energy to overcome the activation barrier. The number of particles with energy $\ge E_{\text{a}}$ (the area under the curve to the right of $E_{\text{a}}$) is smaller. Consequently, the height of the curve at $E_{\text{a}}$ (the number of particles with that energy) is lower.
• Identifying the point: Point $\times$ is on the original curve at $E_{\text{a}}$. Point D is at the same x-coordinate ($E_{\text{a}}$) but with a lower y-value. This matches the expected change: at the same activation energy, there are fewer particles at the lower temperature. Points A, B, and C are at different x-coordinates or higher y-values, which do not fit the new distribution at $E_{\text{a}}$.

Key Takeaways

• The activation energy $E_{\text{a}}$ is independent of temperature; it stays at the same position on the energy axis.
• Decreasing temperature lowers the number of particles with energy $\ge E_{\text{a}}$, shifting the curve left and down.
• When comparing points on a Boltzmann distribution at a fixed energy, a lower temperature means a lower number of particles (lower y-value at the same x-coordinate).

Common Mistakes

• Confusing $E_{\text{a}}$ with temperature: Thinking that $E_{\text{a}}$ changes or shifts with temperature. $E_{\text{a}}$ is a fixed energy threshold.
• Misreading the axes: Looking for a point on the new curve that represents a shift in energy rather than a change in the number of particles at a fixed energy.
• Choosing a point with a higher y-value: Thinking that decreasing temperature increases the number of particles at high energies (which is the opposite of what happens).

Things to Be Careful About

• The question asks for the number of particles whose energy is the same as the activation energy. This means we look along the vertical line $x = E_{\text{a}}$.
• Ensure you understand that the y-axis is "number of particles" (or probability density), and at a lower temperature, the curve is lower at high energies.
• Point D is directly below $\times$, indicating the same energy but fewer particles, which is exactly what happens when temperature decreases.

Background Concept

This question is a limiting-reagent stoichiometry calculation. The balanced chemical equation tells you the mole ratio in which reactants combine. For a solution, the amount of solute is found from $n = cV$, where $V$ must be in $\text{dm}^3$. The reactant that provides fewer moles than the equation requires is the limiting reagent; the other reactant is in excess. The mass of excess reactant left over is the initial mass minus the mass consumed in reacting with the limiting reagent.

Malachite is a basic carbonate with formula $\text{Cu(OH)}_2\cdot\text{CuCO}_3$. It contains two basic units: one $\text{Cu(OH)}_2$ and one $\text{CuCO}_3$. Each unit neutralises two moles of $\text{HNO}_3$, so one mole of malachite requires four moles of nitric acid:

Understanding the Question

You are given $12.0\ \text{g}$ of malachite and $30.0\ \text{cm}^3$ of $1.50\ \text{mol dm}^{-3}$ nitric acid. The question asks for the mass of malachite left unreacted when the reaction is complete. This is not asking for the mass consumed; it asks for the excess remaining after the acid has been used up. The command word 'which' implies a calculation, and the options include both the consumed mass and the remaining mass, so you must be careful to subtract.

Approach

1. Convert the volume of acid from $\text{cm}^3$ to $\text{dm}^3$.
2. Calculate $n(\text{HNO}_3)$ using $n = cV$.
3. Use the 1 : 4 stoichiometric ratio to find the moles of malachite that react.
4. Convert those moles to mass using $M_{\text{r}} = 221.0$.
5. Subtract the consumed mass from the initial $12.0\ \text{g}$.

Step-by-Step Reasoning

First, the volume of nitric acid:

So:

From the balanced equation, $1$ mol of malachite reacts with $4$ mol of $\text{HNO}_3$. Therefore the moles of malachite consumed are:

Convert this to mass:

The initial mass was $12.0\ \text{g}$, so the mass left unreacted is:

This corresponds to option D. As a check, the initial moles of malachite are $12.0/221.0 = 0.0543$ mol, which is much larger than $0.01125$ mol, confirming that malachite is in excess and nitric acid is the limiting reagent.

Option B, $2.49\ \text{g}$, is the mass of malachite consumed, not the mass remaining. Option A is close to a value obtained if the wrong stoichiometric ratio is used; option C is also a distractor based on incorrect ratios or subtraction.

Key Takeaways

• Always write and balance the full equation before using mole ratios.
• Use $n = cV$ with volume in $\text{dm}^3$.
• Identify the limiting reagent; the other reactant is in excess.
• For 'left unreacted', subtract the consumed mass from the initial mass.

Common Mistakes

• Using $1 : 2$ instead of $1 : 4$ because malachite contains two copper-containing units; each unit needs two $\text{HNO}_3$.
• Forgetting to convert $30.0\ \text{cm}^3$ to $0.0300\ \text{dm}^3$; using $30.0$ directly gives a wrong number of moles.
• Selecting $2.49\ \text{g}$ (option B), which is the mass consumed, not the mass remaining.
• Omitting the subtraction step and giving the consumed mass as the final answer.

Things to Be Careful About

• The balanced equation must be correct; the formula of malachite contains both hydroxide and carbonate, so the acid requirement is $4$ mol per formula unit.
• Keep units consistent: $\text{cm}^3 \rightarrow \text{dm}^3$ by dividing by $1000$.
• Use the given $M_{\text{r}} = 221.0$ and give the final answer to three significant figures, consistent with the data.
• In an MCQ, check whether the option matches 'mass consumed' or 'mass remaining'.

Background Concept

When electrons are removed one by one from an atom, the successive ionisation energies increase steadily at first, then jump sharply. The steady increase arises because each electron is removed from the same outer shell — as the positive charge on the ion grows, the remaining electrons are held more tightly. The sharp jump occurs when the next electron must come from a completely new, inner shell, where the electrons are much closer to the nucleus and experience far less shielding.

For a Group 17 element (halogen), the electron configuration is $n\text{s}^2 n\text{p}^5$ — 7 electrons in the outermost shell. Removing all 7 of these gives IE₁ to IE₇, which increase steadily. The 8th ionisation energy removes an electron from the inner shell, so there is a dramatic jump between IE₇ and IE₈. For a noble gas (Group 18), the configuration is $n\text{s}^2 n\text{p}^6$ — 8 outer electrons — so the jump comes between IE₈ and IE₉.

Understanding the Question

We are given the 5th to 9th ionisation energies of two elements X and Y that belong to the same group. The task is to identify them from four options pairing halogens (fluorine/chlorine) and noble gases (neon/argon). The key skill is reading the data: the position of the large jump reveals how many electrons occupy the outer shell, and the relative magnitudes of the values reveal which element is higher in the group.

Approach

1. Look at each element's data and locate the largest jump between successive ionisation energies.
2. The jump position tells you the number of outer-shell electrons, which identifies the group.
3. Compare the relative magnitudes of the ionisation energies: the element higher in the group (smaller atom) has higher ionisation energies.
4. Match to the options.

Step-by-Step Reasoning

Element X: 5th = 11 020, 6th = 15 160, 7th = 17 870, 8th = 92 040, 9th = 106 437 kJ mol⁻¹. The jump from 17 870 to 92 040 (between the 7th and 8th) is enormous. This means the 8th electron is removed from an inner shell, so X has 7 outer electrons → Group 17 (a halogen).

Element Y: 5th = 6 540, 6th = 9 360, 7th = 11 020, 8th = 33 360, 9th = 38 600 kJ mol⁻¹. The jump from 11 020 to 33 360 (between the 7th and 8th) is also the largest. So Y also has 7 outer electrons → also a halogen.

Both are halogens. Among the options, only B and C pair halogens. X has higher ionisation energies than Y at every stage (e.g. 5th IE: 11 020 vs 6 540 kJ mol⁻¹). A smaller atom holds its electrons more tightly, so the element higher in the group has higher ionisation energies. Fluorine sits above chlorine in Group 17, so X = fluorine and Y = chlorine. Answer C.

Why the noble gas options fail: neon and argon have 8 outer electrons ($n\text{s}^2 n\text{p}^6$), so their big jump would occur between the 8th and 9th ionisation energies, not between the 7th and 8th. Options A and D are therefore wrong.

Why B is wrong: it reverses the relative magnitudes — chlorine, being lower in the group and larger, would have lower ionisation energies than fluorine, so chlorine cannot be X.

Key Takeaways

• A sharp jump in successive ionisation energies signals the start of a new electron shell.
• The position of the jump reveals the number of outer-shell electrons and hence the group of the element.
• Within a group, ionisation energies decrease down the group as atomic radius increases.
• Successive ionisation energy data is a powerful tool for identifying elements and their positions in the Periodic Table.

Common Mistakes

• Assuming the jump is between the 8th and 9th ionisation energies for all elements — this is only true for noble gases, which have 8 outer electrons.
• Confusing the direction of the trend: higher ionisation energies mean a smaller atom higher in the group, not lower.
• Miscounting outer electrons: for a halogen, 7 electrons ($n\text{s}^2 n\text{p}^5$) are removed before the jump.

Things to Be Careful About

• Read the table carefully: the 5th to 9th ionisation energies are given, so the 7th → 8th jump is directly visible in the data.
• Note the magnitude of the jump: for fluorine it is very large (17 870 → 92 040 kJ mol⁻¹) because the inner shell is the 1s shell, extremely close to the nucleus.
• Units are kJ mol⁻¹ — not needed for the deduction, but keep them consistent if quoting values.

Background Concept

The nucleon number (mass number) is the total number of protons and neutrons in the nucleus. The charge on an ion tells you the difference between the number of protons and electrons: a $2-$ ion has two more electrons than protons.

Understanding the Question

We are told an ion has a charge of $2-$, contains 10 electrons and 14 neutrons, and we must find its nucleon number.

Approach

1. Work out the number of protons from the ion charge and the electron count.
2. Add protons and neutrons to get the nucleon number.

Step-by-Step Reasoning

• A $2-$ ion has two more electrons than protons.
• Therefore protons = $10 - 2 = 8$.
• Nucleon number = protons + neutrons = $8 + 14 = 22$.

Key Takeaways

The charge on an ion always tells you how many electrons have been gained (negative) or lost (positive) relative to the neutral atom. The nucleon number is always protons + neutrons, never electrons.

Common Mistakes

• Forgetting to correct the electron count for the charge, e.g. taking protons = 10 instead of 8.
• Adding electrons instead of protons when calculating the nucleon number.

Things to Be Careful About

• A $2-$ ion has MORE electrons than protons (10 vs 8), so the neutral atom has 8 electrons.
• Neutrons are unaffected by ion formation; only electrons are gained or lost.

Background Concept

VSEPR (Valence Shell Electron Pair Repulsion) theory predicts molecular geometry based on the number of electron domains (bonding pairs and lone pairs) around a central atom. Electron domains repel each other and arrange themselves to minimise repulsion.

• 2 electron domains → linear → 180°
• 3 electron domains → trigonal planar → 120°
• 4 electron domains → tetrahedral → 109.5°

When lone pairs are present, they occupy more space than bonding pairs (lone pair–lone pair repulsion > lone pair–bonding pair repulsion > bonding pair–bonding pair repulsion), compressing the bond angles slightly below the ideal value. For example, in NH$_3$ (3 bonding pairs + 1 lone pair), the H–N–H bond angle is approximately 107°, less than the ideal tetrahedral angle of 109.5°.

Hybridisation is related to the number of electron domains:

• 4 domains → sp$^3$ hybridisation → tetrahedral geometry
• 3 domains → sp$^2$ hybridisation → trigonal planar geometry
• 2 domains → sp hybridisation → linear geometry

In VSEPR, a double or triple bond counts as a single electron domain (one region of electron density) because the electron pairs in multiple bonds are confined to the same region between the two atoms.

Understanding the Question

The question asks to order three labelled bond angles in histamine from smallest to largest:

• Angle $x$: the H–N–H angle at the terminal amine nitrogen (–NH$_2$) in the side chain.
• Angle $y$: the H–C–H angle at a methylene carbon (–CH$_2$–) in the side chain.
• Angle $z$: the C–C=C angle at a ring carbon in the imidazole ring that is part of a C=C double bond.

The command word is implicit in the multiple-choice format: identify the correct ordering. The student must determine the electron domain geometry at each labelled atom and deduce the approximate bond angle.

Approach

For each labelled angle, identify the central atom, count its electron domains (bonding pairs + lone pairs, treating multiple bonds as one domain), determine the electron arrangement, and assign the approximate bond angle. Then order the three angles.

Step-by-Step Reasoning

Angle $x$ (H–N–H at the –NH$_2$ nitrogen):

The nitrogen atom in the terminal amine group is bonded to two hydrogen atoms and one carbon atom, and carries one lone pair. This gives 3 bonding pairs + 1 lone pair = 4 electron domains. The electron arrangement is tetrahedral (sp$^3$ hybridised). The ideal tetrahedral angle is 109.5°, but the lone pair repels the bonding pairs more strongly, compressing the H–N–H angle to approximately 107° (similar to ammonia, NH$_3$).

Angle $y$ (H–C–H at the –CH$_2$– carbon):

This carbon atom is bonded to four atoms: two hydrogens, one carbon (towards the –NH$_2$), and one carbon (towards the ring). There are 4 bonding pairs and 0 lone pairs = 4 electron domains. The arrangement is tetrahedral (sp$^3$ hybridised). With no lone pairs to compress the angle, the bond angle is approximately 109.5°.

Angle $z$ (C–C=C at the ring carbon):

This carbon atom is part of the imidazole ring and is involved in a C=C double bond. It is bonded to: one carbon in the side chain (single bond), one carbon in the ring (double bond, counts as 1 domain), and one carbon in the ring (single bond). There are 3 bonding regions and 0 lone pairs = 3 electron domains. The arrangement is trigonal planar (sp$^2$ hybridised). The bond angle is approximately 120°.

Ordering:

$x \approx 107° < y \approx 109.5° < z \approx 120°$

This matches row A: $x$, $y$, $z$ from smallest to largest.

Key Takeaways

• Lone pairs compress bond angles below the ideal tetrahedral value (107° vs 109.5° for NH$_3$ vs CH$_4$).
• sp$^3$ atoms with 4 bonding pairs have bond angles of ~109.5°; sp$^3$ atoms with 1 lone pair have angles of ~107°.
• sp$^2$ atoms with 3 bonding regions have bond angles of ~120°.
• In VSEPR, multiple bonds count as a single electron domain.

Common Mistakes

• Counting the electrons in a double bond as two separate domains instead of one region of electron density. This would incorrectly give the ring carbon 4 domains and predict a tetrahedral angle, making $z$ appear smaller than it is.
• Forgetting the lone pair on the amine nitrogen, which would predict a 109.5° angle for $x$ and eliminate the distinction between $x$ and $y$.
• Confusing the nitrogen in the ring (which has a lone pair in an sp$^2$ orbital and is part of the aromatic system) with the terminal amine nitrogen. The angle $x$ is at the terminal –NH$_2$, not at a ring nitrogen.
• Assuming all tetrahedral angles are exactly 109.5° and missing the lone-pair compression effect.

Things to Be Careful About

• Always count electron domains (regions of electron density), not total electrons or total bonds. A double bond is one domain.
• Distinguish between electron-domain geometry and molecular geometry. The nitrogen in –NH$_2$ has a tetrahedral electron arrangement but a trigonal pyramidal molecular shape; the bond angle is still determined by the tetrahedral arrangement compressed by the lone pair.
• State symbols and exact angle values are not required for this question; the approximate values (107°, 109.5°, 120°) and the correct ordering are what matter.
• The ring nitrogen bearing the H (–NH– in the imidazole ring) is not one of the labelled angles, so its geometry does not need to be considered here. (That nitrogen is sp$^2$ hybridised with its lone pair in an sp$^2$ orbital in the plane of the ring, contributing to aromaticity.)

Background Concept

Esterification is a classic reversible reaction: a carboxylic acid reacts with an alcohol to form an ester and water. Because the reaction is reversible, it reaches a dynamic equilibrium in which the forward and reverse reactions proceed at equal rates. The equilibrium constant, $K_{\text{c}}$, measures the position of this equilibrium — the ratio of product concentrations to reactant concentrations at equilibrium.

For the general reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$, the equilibrium constant is:

$K_{\text{c}} = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$

For esterification, all stoichiometric coefficients are 1, so:

$K_{\text{c}} = \frac{[\text{Ester}][\text{Water}]}{[\text{Acid}][\text{Alcohol}]}$

A crucial simplification: when all species are in the same container (same volume) and the total number of moles is the same on both sides of the equation (2 products vs 2 reactants here), the volume cancels out of the $K_{\text{c}}$ expression. This means moles can be used directly in place of concentrations.

Understanding the Question

The question gives the initial amounts of acid (0.40 mol) and alcohol (0.30 mol), and states that at equilibrium, 0.20 mol of ester has formed. We are asked to find the value of $K_{\text{c}}$ under these conditions.

The key insight is that the amount of ester formed tells us exactly how much of each reactant has been consumed and how much water has been produced, because the stoichiometry is 1:1:1:1. The temperature is stated (25°C) because $K_{\text{c}}$ is temperature-dependent, but since all data are at the same temperature, we simply use the given equilibrium amounts.

Approach

Use an ICE (Initial-Change-Equilibrium) table:

1. Write the balanced equation.
2. Record the initial amounts of all four species.
3. Use the given equilibrium amount of ester (0.20 mol) to determine the change row: ester +0.20, water +0.20, acid −0.20, alcohol −0.20.
4. Compute the equilibrium amounts by adding the initial and change rows.
5. Substitute the equilibrium amounts into the $K_{\text{c}}$ expression and evaluate.

Step-by-Step Reasoning

Step 1: Write the balanced equation.

$\text{Acid} + \text{Alcohol} \rightleftharpoons \text{Ester} + \text{Water}$

Step 2: Set up the ICE table.

• Initial: acid 0.40 mol, alcohol 0.30 mol, ester 0 mol, water 0 mol.
• Change: since 0.20 mol of ester forms, the change in ester is +0.20 mol. By the 1:1 stoichiometry, water increases by +0.20 mol, and both acid and alcohol decrease by −0.20 mol each.
• Equilibrium:
  • Acid: 0.40 − 0.20 = 0.20 mol
  • Alcohol: 0.30 − 0.20 = 0.10 mol
  • Ester: 0 + 0.20 = 0.20 mol
  • Water: 0 + 0.20 = 0.20 mol

Step 3: Write the $K_{\text{c}}$ expression.

$K_{\text{c}} = \frac{[\text{Ester}][\text{Water}]}{[\text{Acid}][\text{Alcohol}]}$

Step 4: Substitute and evaluate.

$K_{\text{c}} = \frac{0.20 \times 0.20}{0.20 \times 0.10} = \frac{0.04}{0.02} = 2.0$

The correct option is C.

Why the distractors are wrong:

• A (0.33): Could arise from using the initial amount of alcohol (0.30) instead of the equilibrium amount, e.g. $(0.20 \times 0.20)/(0.20 \times 0.30) = 0.33$, or from a ratio like $0.10/0.30$.
• B (0.50): Could come from swapping the equilibrium amounts of acid and alcohol, e.g. $(0.20 \times 0.10)/(0.20 \times 0.20) = 0.50$, or from $0.10/0.20$.
• D (10): Could come from an arithmetic slip such as evaluating $(0.20 \times 0.20)/(0.20 \times 0.10)$ incorrectly, or from $0.20/0.02 = 10$.

Key Takeaways

• The ICE table is the standard tool for equilibrium calculations: it organises initial amounts, the changes dictated by stoichiometry, and the equilibrium amounts.
• The equilibrium amount of any one species (here, ester = 0.20 mol) fixes the changes for all species through the stoichiometric coefficients.
• When the total moles of species are equal on both sides of the equation, the container volume cancels, and moles can be substituted directly into $K_{\text{c}}$.
• Always check that the $K_{\text{c}}$ expression has products over reactants, each raised to the power of its stoichiometric coefficient.

Common Mistakes

• Using initial amounts instead of equilibrium amounts in the $K_{\text{c}}$ expression — the most frequent error. Substituting 0.40 and 0.30 would give a wrong value.
• Forgetting that both reactants decrease: the alcohol drops from 0.30 to 0.10 mol, not by 0.20 from 0.40.
• Inverting the $K_{\text{c}}$ expression — putting reactants over products.
• Treating the reaction as going to completion, ignoring reversibility and the equilibrium amounts.
• Confusing the limiting reagent: the alcohol (0.30 mol) is limiting here since only 0.20 mol reacts, but this is not needed for the calculation.

Things to Be Careful About

• The volume of the container is not given, but it cancels because all four species share the same volume and the mole count is equal on both sides. So using moles directly is valid.
• All stoichiometric coefficients are 1, so no powers appear in the $K_{\text{c}}$ expression.
• $K_{\text{c}}$ is dimensionless here because the number of moles of products equals the number of moles of reactants (2 on each side).
• The temperature (25°C) is stated because $K_{\text{c}}$ is temperature-dependent; since all data are at the same temperature, we simply use the given equilibrium amounts.

Background Concept

The structure and bonding of a substance determine its observable properties:

• Electrical conductivity depends on whether mobile charged particles exist. In a giant metallic structure, delocalised electrons are free to move, so metals conduct in both the solid and molten states. In a giant ionic lattice, the ions are fixed in position in the solid, so an ionic compound does not conduct when solid but does conduct when molten or dissolved in water. In a simple molecular substance, the molecules are electrically neutral and there are no free electrons or ions, so it never conducts.
• Melting point reflects the strength of the forces holding the particles together. Giant metallic and giant ionic structures have strong bonds (metallic bonding or electrostatic attraction) and generally high melting points; simple molecular substances are held together only by weak intermolecular forces (van der Waals, etc.), so they melt at low temperatures.
• Reaction with water gives a chemical clue to the element type: metals react with water to form metal hydroxides, which are alkaline; many covalent non-metal compounds (oxides, chlorides) hydrolyse to give acidic solutions.

Understanding the Question

The table gives three pieces of evidence for two unknown substances, P and Q: electrical conductivity in different states, the effect of adding them to water, and their melting points. The question asks which row (A–D) correctly identifies the structure and bonding of P and Q. The command is essentially "deduce" — use each property as a clue and combine them to reach a confident identification.

Approach

Start with the most discriminating property: electrical conductivity in the solid state. Only a metal conducts when solid, so this single observation identifies P. Then use the reaction with water and the melting point to confirm. For Q, the fact that it never conducts rules out metallic and ionic structures, leaving simple molecular; the low melting point and acidic reaction with water confirm it.

Step-by-Step Reasoning

Substance P

1. P conducts electricity when solid and when molten. Of the four structure types, only a giant metallic structure conducts in the solid state, because metals have a sea of delocalised electrons free to move. A giant ionic solid does not conduct when solid (ions are trapped in the lattice), so P cannot be ionic.
2. P reacts vigorously with water to produce an alkaline solution. Metals react with water to form metal hydroxides, e.g. $\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \tfrac{1}{2}\text{H}_2$, and NaOH(aq) is alkaline. This confirms P is a metal.
3. The melting point of 454 K is consistent with a metal (for comparison, sodium melts at 371 K, magnesium at 923 K). So P is giant metallic.

Substance Q

1. Q does not conduct in any state. This rules out giant metallic (conducts when solid and molten) and giant ionic (conducts when molten or aqueous). The only remaining option is simple molecular: the molecules are neutral and carry no mobile charge carriers.
2. Q reacts vigorously with water to produce an acidic solution. This is typical of covalent non-metal compounds, e.g. $\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}$, or $\text{SO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_3$. The acidic product confirms a non-metal covalent substance.
3. The low melting point of 317 K is exactly what is expected for a simple molecular substance, where only weak intermolecular forces hold the molecules together. So Q is simple molecular.

Therefore P = giant metallic and Q = simple molecular, which is option A.

Why the distractors fail:

• B (P simple molecular, Q giant metallic): wrong because P conducts when solid — simple molecular substances never conduct, and a giant metallic substance would conduct (so Q could not be metallic).
• C (P giant ionic, Q simple molecular): wrong for P — giant ionic solids do not conduct electricity when solid.
• D (P giant metallic, Q giant ionic): wrong for Q — a giant ionic substance conducts when molten and has a high melting point, not 317 K.

Key Takeaways

• Conductivity in the solid state is the key discriminator: only metals conduct when solid; ionic compounds conduct only when molten or aqueous; simple molecular substances never conduct.
• Melting point and reaction with water are supporting evidence that confirm the structure type.
• A metal + water gives an alkaline solution; a covalent non-metal compound + water gives an acidic solution.

Common Mistakes

• Confusing the conduction behaviour of metals and ionic compounds: a giant ionic solid does not conduct when solid because its ions are fixed in the lattice, whereas a metal conducts in both solid and molten states.
• Assuming ionic compounds have low melting points — they have high melting points due to strong electrostatic forces between ions, so a low melting point (317 K) points to simple molecular.
• Forgetting that "does not conduct in any state" eliminates both metallic and ionic structures immediately.

Things to Be Careful About

• Read the conductivity data carefully: "good when solid and when molten" is the signature of a metal; "does not conduct in any state" is the signature of a simple molecular substance.
• The melting points are given in kelvin; 454 K and 317 K are both relatively low, but the key discriminator here is conductivity, not melting point alone.
• Match the deduced structures to the exact wording of the options — "giant metallic" and "simple molecular" are the precise terms the mark scheme expects.

Background Concept

This question combines two ideas: the ideal gas equation and stoichiometry.

The ideal gas equation is

where $p$ is pressure, $V$ is volume, $n$ is the amount of gas in moles, $R$ is the gas constant ($8.314\text{ J K}^{-1}\text{ mol}^{-1}$), and $T$ is temperature in kelvin. It lets us find the number of moles of a gas from its measured volume, pressure and temperature.

Stoichiometry then connects that amount of gas to the amount of acid that reacted. The balanced equation for zinc reacting with hydrochloric acid is

so 1 mol of hydrogen gas is produced from 2 mol of HCl. Because the zinc is in excess, all the hydrochloric acid is used up, so the moles of HCl can be found directly from the moles of hydrogen produced.

Understanding the Question

The question gives a volume of hydrochloric acid, $x\text{ cm}^3$, at a concentration of $2.00\text{ mol dm}^{-3}$. An excess of zinc is added, so the acid is the limiting reagent. The hydrogen gas produced is dried and collected, and its volume is measured as $1.534\text{ dm}^3$ at $101\,000\text{ Pa}$ and $293\text{ K}$. We are asked to find $x$, the volume of acid used.

The key steps are:

1. Use $pV = nRT$ to find the moles of hydrogen gas.
2. Use the balanced equation to find the moles of HCl that reacted.
3. Use $n = cV$ to find the volume of acid solution.

Approach

Start with the gas measurement. The pressure is in Pa, so the volume must be in m$^3$ when using $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$:

Then substitute into $pV = nRT$ and solve for $n$.

Next, look at the stoichiometry. The equation shows that 2 mol of HCl produce 1 mol of H$_2$, so:

Finally, since concentration is moles per unit volume,

and convert the volume from dm$^3$ to cm$^3$ by multiplying by 1000.

Step-by-Step Reasoning

1. Convert the gas volume to m$^3$

   Since $1\text{ dm}^3 = 10^{-3}\text{ m}^3$,

   $$V = 1.534\text{ dm}^3 = 1.534 \times 10^{-3}\text{ m}^3$$

2. Find moles of hydrogen using $pV = nRT$

   $$n(\text{H}_2) = \frac{pV}{RT}$$ $$n(\text{H}_2) = \frac{101\,000 \times 1.534 \times 10^{-3}}{8.314 \times 293}$$ $$n(\text{H}_2) = \frac{154.934}{2436.0} = 0.0636\text{ mol}$$

   Alternatively, using $p = 101\text{ kPa}$ and $V = 1.534\text{ dm}^3$ with the same $R$ value also works because $1\text{ kPa dm}^3 = 1\text{ J}$.

3. Use the stoichiometric ratio

   The balanced equation is:

   $$\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$$

   Therefore,

   $$n(\text{HCl}) = 2 \times n(\text{H}_2) = 2 \times 0.0636 = 0.1272\text{ mol}$$

4. Find the volume of acid solution

   The concentration is $2.00\text{ mol dm}^{-3}$, so:

   $$V(\text{HCl}) = \frac{n}{c} = \frac{0.1272}{2.00} = 0.0636\text{ dm}^3$$

   Convert to cm$^3$:

   $$0.0636\text{ dm}^3 = 63.6\text{ cm}^3$$

   This matches option C.

Key Takeaways

• The ideal gas equation is used to convert a measured gas volume into moles when pressure and temperature are known.
• Always check the stoichiometric ratio in the balanced equation before relating amounts of different substances.
• When pressure is in Pa, volume must be in m$^3$ with $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$; alternatively, kPa with dm$^3$ gives the same result.
• Concentration, moles and volume are linked by $n = cV$.

Common Mistakes

• Forgetting the 2:1 ratio: Some candidates use $n(\text{HCl}) = n(\text{H}_2)$ instead of $n(\text{HCl}) = 2n(\text{H}_2)$. This gives $31.8\text{ cm}^3$, option A.
• Using the wrong units in the gas equation: If $p$ is in Pa but $V$ is left in dm$^3$, the calculation gives a wrong value. Convert volume to m$^3$ first, or use kPa with dm$^3$.
• Confusing dm$^3$ and cm$^3$: The final volume is $0.0636\text{ dm}^3$, which is $63.6\text{ cm}^3$, not $0.0636\text{ cm}^3$.
• Ignoring the limiting reagent idea: The question says zinc is in excess, so all the acid reacts. This is why the moles of HCl can be found from the moles of H$_2$.

Things to Be Careful About

• Use the correct gas constant units consistently.
• Include the factor of 2 from the balanced equation.
• Convert the final volume to the unit asked for in the question, cm$^3$.
• The gas is described as ideal, so $pV = nRT$ applies exactly; no correction for real gas behaviour is needed.

Background Concept

The average rate of a reaction is the change in the amount of a reactant or product divided by the time taken. For this question the rate is defined using the reactant, $\text{H}_2\text{O}_2$, so the numerator must be the moles of hydrogen peroxide that have decomposed. The only measured quantity is a gas volume, so the calculation begins by converting volume to moles. At room temperature and pressure, the molar gas volume is about $24\text{ dm}^3\text{ mol}^{-1}$, which is $24000\text{ cm}^3\text{ mol}^{-1}$. The balanced equation then links the moles of oxygen produced to the moles of hydrogen peroxide consumed.

Understanding the Question

The equation shows $2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})$. A total of $180\text{ cm}^3$ of oxygen is collected in 90 seconds, and we are asked for the average rate over that whole time in $\text{mol min}^{-1}$. The word 'average' means we use the total change in moles divided by the total time, not a gradient at a particular instant. The unit tells us the time must be in minutes.

Approach

1. Convert the volume of oxygen to moles using the molar gas volume.
2. Use the 2:1 stoichiometric ratio from the equation to find the moles of $\text{H}_2\text{O}_2$ that decomposed.
3. Convert 90 seconds to 1.5 minutes.
4. Divide the moles of $\text{H}_2\text{O}_2$ by the time in minutes.

Step-by-Step Reasoning

First, convert $180\text{ cm}^3$ to moles of oxygen. Since $1\text{ dm}^3 = 1000\text{ cm}^3$, $180\text{ cm}^3 = 0.180\text{ dm}^3$. Using the room-condition molar volume:

Equivalently, using $24000\text{ cm}^3\text{ mol}^{-1}$ gives the same result.

Next, apply the stoichiometry. The equation shows that 2 mol of $\text{H}_2\text{O}_2$ produce 1 mol of $\text{O}_2$. Therefore the moles of $\text{H}_2\text{O}_2$ decomposed are twice the moles of oxygen produced:

Now convert the time to minutes: $90\text{ s} = 1.5\text{ min}$.

Finally, substitute into the rate equation:

This matches option D. Option C, $0.0050$, is what you would get if you used the moles of oxygen directly instead of converting to moles of hydrogen peroxide. Options A and B are the corresponding values expressed per second ($8.33 \times 10^{-5}$ and $1.67 \times 10^{-4}$), so they arise from not converting seconds to minutes.

Key Takeaways

• Always identify which species the rate is defined on; here it is the reactant, $\text{H}_2\text{O}_2$.
• Convert gas volumes to moles with the correct molar volume: $24\text{ dm}^3\text{ mol}^{-1}$ at room conditions.
• Use the stoichiometric ratio from the balanced equation before calculating rate.
• Check the unit requested; here $\text{mol min}^{-1}$ means time must be in minutes.

Common Mistakes

• Using the moles of oxygen as if they were the moles of hydrogen peroxide. The 2:1 ratio means the moles of $\text{H}_2\text{O}_2$ are twice the moles of $\text{O}_2$.
• Using 90 seconds directly in the denominator, giving a rate in $\text{mol s}^{-1}$ rather than $\text{mol min}^{-1}$.
• Using $22.4\text{ dm}^3$ (the molar volume at s.t.p.) instead of $24\text{ dm}^3$ for room conditions.
• Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ or to use $24000\text{ cm}^3$.

Things to Be Careful About

• The molar gas volume depends on conditions: $24\text{ dm}^3\text{ mol}^{-1}$ at room temperature and pressure, not $22.4\text{ dm}^3\text{ mol}^{-1}$ which is at s.t.p.
• The rate is defined as change in moles of $\text{H}_2\text{O}_2$ per time, so the stoichiometric factor of 2 is essential.
• Always include the unit $\text{mol min}^{-1}$ in the final answer.
• If you use the volume in $\text{cm}^3$, use $24000\text{ cm}^3\text{ mol}^{-1}$; if you convert to $\text{dm}^3$ first, use $24\text{ dm}^3\text{ mol}^{-1}$.

Background Concept

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions (concentration, pressure, or temperature), the position of equilibrium shifts to counteract that change and restore equilibrium.

Three changes matter here:

• Pressure: only affects equilibria involving gases. Increasing pressure shifts the equilibrium towards the side with fewer moles of gas molecules.
• Temperature: depends on the enthalpy change. For an exothermic forward reaction (negative $\Delta H$), increasing temperature shifts the equilibrium to the left (favouring the endothermic reverse reaction); decreasing temperature shifts it to the right.
• Catalyst: lowers the activation energy for both the forward and reverse reactions equally, so it speeds up the attainment of equilibrium but does not change the position of equilibrium or the equilibrium yield.

Understanding the Question

The reaction is:

We are asked which of four changes increases the equilibrium yield of methanol. The key is to apply Le Chatelier's principle to each option in turn, paying attention to the gas mole counts and the sign of $\Delta H$.

Approach

For each option, identify the disturbance and predict the direction of the equilibrium shift:

• A (increasing pressure): count the total moles of gas on each side.
• B (adding excess steam): recognise that steam, $\text{H}_2\text{O}$, is a product.
• C (adding a catalyst): recall that a catalyst does not affect the position of equilibrium.
• D (increasing temperature): use the sign of $\Delta H$ to decide which direction is favoured.

Step-by-Step Reasoning

Option A — increasing the pressure

Count gas moles:

• Left (reactants): 1 mol $\text{CO}_2$ + 3 mol $\text{H}_2$ = 4 mol gas.
• Right (products): 1 mol $\text{CH}_3\text{OH}$ + 1 mol $\text{H}_2\text{O}$ = 2 mol gas.

Increasing the pressure shifts the equilibrium towards the side with fewer gas molecules, i.e. to the right. This increases the yield of methanol. A is correct.

Option B — adding an excess of steam

Steam is $\text{H}_2\text{O}$, a product. Adding a product increases its concentration, so the equilibrium shifts to the left to reduce it, decreasing the yield of methanol. B is incorrect.

Option C — adding a catalyst

A catalyst provides an alternative pathway with lower activation energy for both the forward and reverse reactions equally. It speeds up the rate at which equilibrium is reached but does not change the position of equilibrium or the equilibrium yield. C is incorrect.

Option D — increasing the temperature

The forward reaction is exothermic ($\Delta H = -49\text{ kJ mol}^{-1}$). Increasing the temperature favours the endothermic direction, which here is the reverse reaction. This decreases the yield of methanol. D is incorrect.

Therefore, the only change that increases the equilibrium yield of methanol is increasing the pressure.

Key Takeaways

• For gaseous equilibria, increasing pressure shifts equilibrium towards the side with fewer moles of gas.
• Adding a product shifts equilibrium to the left; adding a reactant shifts it to the right.
• A catalyst never changes the position of equilibrium or the equilibrium yield — it only affects how quickly equilibrium is reached.
• For an exothermic forward reaction, increasing temperature shifts equilibrium to the left; for an endothermic forward reaction, it shifts to the right.

Common Mistakes

• Thinking a catalyst increases yield: a catalyst increases the rate of both forward and reverse reactions equally, so the equilibrium position is unchanged.
• Confusing steam with a reactant: $\text{H}_2\text{O}$ is a product here, so adding it shifts equilibrium to the left, not the right.
• Assuming higher temperature always increases yield: higher temperature speeds up the reaction but shifts the equilibrium against an exothermic forward reaction.
• Forgetting to count all gas moles: both $\text{CH}_3\text{OH}$ and $\text{H}_2\text{O}$ are gaseous, so the product side has 2 mol of gas, not 1.

Things to Be Careful About

• Always check the physical states: the pressure argument only applies to gases, and here all four species are (g).
• Note that $\Delta H$ is negative, meaning the forward reaction is exothermic — this determines the temperature effect.
• When counting gas moles, include every gaseous species on each side, even products such as $\text{H}_2\text{O}$.
• A catalyst affects rate, not yield — this is a common trap in equilibrium questions.

Background Concept

Oxidation number is a bookkeeping number assigned to an atom in a compound to track electron transfer in redox reactions. For a neutral compound, the sum of all oxidation numbers is zero. Common rules: Group 1 metals such as K are always $+1$ in compounds, and oxygen is usually $-2$.

A disproportionation reaction is a redox reaction in which the same element in one oxidation state is simultaneously oxidised to a higher oxidation state and reduced to a lower oxidation state. The reactant therefore acts as both the oxidising agent and the reducing agent.

Understanding the Question

This multiple-choice question gives the reaction

and asks two things: whether this is a disproportionation reaction, and what the oxidation number of chlorine is in $\text{KClO}_4$. The options combine these two judgements, so both must be determined correctly to select the right row.

Approach

First, assign oxidation numbers to chlorine in the reactant and in both products using the standard rules. Then compare the oxidation numbers of chlorine: if chlorine appears in both a higher and a lower oxidation state in the products than in the reactant, the reaction is a disproportionation. Finally, read the correct combination from the table.

Step-by-Step Reasoning

1. In $\text{KClO}_3$, let the oxidation number of Cl be $x$.
   $+1 + x + 3(-2) = 0$
   $x = +5$
2. In $\text{KCl}$, let the oxidation number of Cl be $y$.
   $+1 + y = 0$
   $y = -1$
3. In $\text{KClO}_4$, let the oxidation number of Cl be $z$.
   $+1 + z + 4(-2) = 0$
   $z = +7$
4. Chlorine changes from $+5$ in $\text{KClO}_3$ to both $-1$ in $\text{KCl}$ and $+7$ in $\text{KClO}_4$. It is therefore both reduced and oxidised in the same reaction.
5. This is exactly the definition of disproportionation: one element in a single reactant forms a product with a higher oxidation state and another product with a lower oxidation state.
6. The oxidation number of chlorine in $\text{KClO}_4$ is $+7$, not $+4$.

The correct row is therefore B: yes, $+7$.

The other options are wrong because:

• A gives the correct disproportionation answer but the wrong oxidation number, $+4$.
• C gives the wrong answer for both.
• D gives the correct oxidation number but incorrectly says the reaction is not a disproportionation.

Key Takeaways

• Oxidation numbers are assigned using fixed rules for K and O, then solving for the unknown element.
• A disproportionation reaction involves the same element being both oxidised and reduced.
• In this reaction, chlorine in $\text{KClO}_3$ is the only source of chlorine, so it must be both the oxidising and reducing agent.

Common Mistakes

• Calculating the oxidation number of Cl in $\text{KClO}_4$ as $+4$ instead of $+7$. This usually comes from forgetting to multiply the oxidation number of oxygen by four atoms.
• Thinking that a reaction is not redox because no obvious single product is formed. Disproportionation is still a redox reaction.
• Confusing disproportionation with a simple decomposition. Here decomposition occurs, but the key feature is the simultaneous oxidation and reduction of chlorine.

Things to Be Careful About

• Always include the sign when writing oxidation numbers: $+7$, not 7.
• Remember that oxygen is usually $-2$; this is essential for the calculation.
• Check the electron bookkeeping: one Cl goes from $+5$ to $-1$ (gain of 6 electrons), while three Cl atoms go from $+5$ to $+7$ (loss of 2 electrons each, total loss of 6 electrons). This confirms the redox change is balanced.

Background Concept

When an acid and a base react in aqueous solution, the neutralisation reaction releases heat. In a simple calorimetry experiment, the heat released warms the solution, and we measure the temperature rise. The heat absorbed by the solution is given by

where $m$ is the mass of solution being warmed (in g), $c$ is the specific heat capacity of the solution (taken as $4.18\text{ J g}^{-1}\text{ K}^{-1}$, the value for water), and $\Delta T$ is the temperature rise (in K or $^\circ\text{C}$, which are numerically equal for a temperature difference).

The heat released by the reaction is equal in magnitude to the heat gained by the solution. To express this as an enthalpy change per mole of a chosen reactant, we divide the heat by the number of moles of that reactant that actually reacted. Because the reaction is exothermic, the sign of $\Delta H$ is negative.

Understanding the Question

We mix $25.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ NaOH with $25.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ HCl and observe a temperature rise of $2.50^\circ\text{C}$. We are asked for the enthalpy change of the neutralisation per mole of NaOH.

Key given data:

• Volume of NaOH solution $= 25.0\text{ cm}^3 = 0.0250\text{ dm}^3$
• Concentration of NaOH $= 0.100\text{ mol dm}^{-3}$
• Temperature rise $\Delta T = 2.50^\circ\text{C}$
• Total volume of solution $= 50.0\text{ cm}^3$ (both solutions combined)

We must convert the measured heat to a per-mole quantity. The units in the options (kJ mol⁻¹ vs J mol⁻¹) are a deliberate trap, so we must keep track of them carefully.

Approach

1. Find the total mass of solution warmed. Since the solutions are aqueous and dilute, assume density $1\text{ g cm}^{-3}$, so $50.0\text{ cm}^3$ has mass $50.0\text{ g}$.
2. Calculate the heat absorbed by the solution using $q = mc\Delta T$.
3. Calculate the moles of NaOH present using $n = cV$.
4. Divide the heat by the moles and apply a negative sign (exothermic) to obtain $\Delta H$ in J mol⁻¹, then convert to kJ mol⁻¹.

The acid and base are present in equimolar amounts (both $0.0025$ mol), so NaOH is not in excess or limiting — all of it reacts, and dividing by the NaOH moles is valid.

Step-by-Step Reasoning

Step 1 — Mass of solution.
Both solutions are added together, so the total volume is

Assuming the density of the dilute solution is $1\text{ g cm}^{-3}$,

Step 2 — Heat released.
Using $q = mc\Delta T$ with $c = 4.18\text{ J g}^{-1}\text{ K}^{-1}$:

This is the heat gained by the solution, equal to the heat released by the reaction.

Step 3 — Moles of NaOH.

Step 4 — Enthalpy change per mole.
The reaction is exothermic, so:

Converting to kJ:

This matches option A.

Why the other options are wrong:

• B ($-104.5\text{ kJ mol}^{-1}$) — this is exactly half of the correct value. It arises from dividing by $0.005$ mol (using the total moles of both acid and base) instead of the moles of NaOH alone, or from using only one solution's volume for the mass. The enthalpy change is asked per mole of NaOH, so we divide by $0.0025$ mol.
• C ($-209\text{ J mol}^{-1}$) — the number is right but the unit is wrong: the calculation gives J mol⁻¹ before conversion to kJ. $-209\text{ J mol}^{-1}$ would be $209\text{ J}$ per mole, which is far too small.
• D ($-522.5\text{ J mol}^{-1}$) — this is simply the raw heat value $q = 522.5\text{ J}$ with a negative sign, with no division by the number of moles. It is the total heat released by the whole sample, not the enthalpy change per mole.

Key Takeaways

• In calorimetry, the heat absorbed by the solution is $q = mc\Delta T$, using the total mass of all solutions mixed.
• The enthalpy change per mole is $\Delta H = -q/n$, where $n$ is the moles of the substance specified in the question.
• Always convert J to kJ (divide by 1000) and check the units against the answer options.
• The negative sign is essential: an exothermic reaction (temperature rise) has a negative $\Delta H$.

Common Mistakes

• Forgetting to use the total volume — using only $25.0\text{ cm}^3$ as the mass gives $q = 261.25\text{ J}$ and a wrong answer. The heat warms the whole $50.0\text{ cm}^3$ of mixed solution.
• Dividing by the wrong number of moles — dividing by $0.005$ mol (total acid + base) instead of $0.0025$ mol of NaOH gives option B, $-104.5\text{ kJ mol}^{-1}$.
• Unit confusion — reporting $209\text{ J mol}^{-1}$ instead of $209\text{ kJ mol}^{-1}$ (option C). Always convert J to kJ when the options are in kJ.
• Omitting the negative sign — the temperature rises, so the reaction is exothermic and $\Delta H$ must be negative.
• Using the raw $q$ value as the answer — option D is just $q$ with a sign; the question asks for a per-mole value, so division by moles is mandatory.

Things to Be Careful About

• The specific heat capacity is that of the solution, taken as $4.18\text{ J g}^{-1}\text{ K}^{-1}$ (same as water) — this value is usually given in the question or assumed.
• A temperature difference in $^\circ\text{C}$ is numerically identical to one in K, so $\Delta T = 2.50$ is used directly.
• Ensure concentration is in $\text{mol dm}^{-3}$ and volume in $\text{dm}^3$ when calculating moles: $25.0\text{ cm}^3 = 0.0250\text{ dm}^3$.
• State the sign of $\Delta H$ explicitly; the mark scheme for such calculations requires the negative sign and the correct unit.

Background Concept

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. Enthalpy is a state function, so the overall enthalpy change of a reaction depends only on the initial and final states, not on the route taken. This is Hess's law.

For a reaction in which reactants and products can each be burned to the same combustion products, the combustion enthalpies can be arranged in a cycle. The enthalpy change of the target reaction is obtained by adding the combustion enthalpies of the reactants and subtracting the combustion enthalpy of the product:

Understanding the Question

The reaction is $\text{CO}(\text{g}) + \text{CH}_3\text{OH}(\text{l}) \rightarrow \text{CH}_3\text{CO}_2\text{H}(\text{l})$. The table gives $\Delta H_{\text{c}}^{\ominus}$ for each of the three species. The question asks for $\Delta H_{\text{r}}^{\ominus}$ of this reaction. The options include the simple sum of all three values and both signs of the correct value, so the main challenge is to set up the Hess cycle correctly and keep the sign right.

Approach

Write the combustion equations for the two reactants and for the product, then combine them so that the combustion products cancel. Equivalently, apply the formula above. Since the product ethanoic acid is on the right-hand side of the target equation, its combustion reaction must be reversed in the cycle, which changes the sign of its enthalpy change.

Step-by-Step Reasoning

1. Add the combustion enthalpies of the reactants: