Chemistry 9701/11 — May/June 2024

Background Concept

When a solution is diluted by adding water, the amount of solute in moles does not change. Only the total volume changes. Therefore, the number of moles of HCl in the original sample must equal the number of moles of HCl in the final 200 cm³ solution.

The key relationship is:
n = c × V
where n is the amount in moles, c is the concentration in mol dm⁻³, and V is the volume in dm³.

Understanding the Question

The question gives the final concentration and final volume of an HCl solution made by adding water to sample X. We need to find which sample contains exactly the right number of moles of HCl so that, after dilution to 200 cm³, the concentration becomes 0.100 mol dm⁻³.

This is a dilution problem: the final moles of HCl must come entirely from sample X.

Approach

1. Convert the final volume from cm³ to dm³.
2. Calculate the moles of HCl required in the final solution using n = c × V.
3. For each option, convert its volume to dm³ and calculate the moles of HCl it contains.
4. Select the option whose moles match the required moles.

Step-by-Step Reasoning

Final volume = 200 cm³ = 0.200 dm³.

Required moles of HCl:
n = 0.100 × 0.200 = 0.0200 mol

Now check each option:

Option A: 10 cm³ = 0.010 dm³
n = 1.00 × 0.010 = 0.0100 mol
This is too little.

Option B: 30 cm³ = 0.030 dm³
n = 0.90 × 0.030 = 0.0270 mol
This is too much.

Option C: 50 cm³ = 0.050 dm³
n = 0.40 × 0.050 = 0.0200 mol
This matches exactly.

Option D: 100 cm³ = 0.100 dm³
n = 0.30 × 0.100 = 0.0300 mol
This is too much.

Only option C gives the correct amount of HCl.

Key Takeaways

• Dilution does not change the number of moles of solute.
• Always use volume in dm³ when applying n = c × V.
• The final volume is the total volume after water is added, not the volume of water added.

Common Mistakes

• Forgetting to convert cm³ to dm³ before calculating moles.
• Comparing concentrations instead of moles. Since the final volume is the same for all options, comparing moles is the correct method.
• Thinking that a larger volume always means more moles; concentration also matters.

Things to Be Careful About

• The phrase "made up to 200 cm³" means the final total volume is 200 cm³.
• Use the same units throughout the calculation.
• Check every option carefully, even if one seems obvious at first.

Background Concept

This question tests the stoichiometric relationship between gas volumes in a chemical reaction. Avogadro's law states that equal volumes of gases, measured at the same temperature and pressure, contain the same number of molecules. Consequently, the volumes of gases reacting and produced are in the same ratio as the coefficients in the balanced equation. This allows volume ratios to be read directly from a balanced equation, provided all volumes are measured under identical conditions (as stated in the question).

The second key idea is that aqueous potassium hydroxide, KOH(aq), is a strong alkali that absorbs acidic gases, specifically carbon dioxide, $\text{CO}_2$:

$$2\text{KOH}(\text{aq}) + \text{CO}_2(\text{g}) \rightarrow \text{K}_2\text{CO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l})$$

Oxygen and other neutral gases are not absorbed by KOH.

Understanding the Question

The question gives a mixture of $10\text{ cm}^3$ of methane ($\text{CH}_4$) and $10\text{ cm}^3$ of ethane ($\text{C}_2\text{H}_6$), sparked with an excess of oxygen. After the reaction, the mixture is cooled (so water vapour condenses to liquid) and the remaining gas is passed through aqueous KOH. We are asked which volume of gas is absorbed — i.e., the volume of $\text{CO}_2$ produced by complete combustion of both hydrocarbons.

The command word is implicit: "Which volume..." — a calculation. All gas volumes are measured at the same temperature and pressure, so Avogadro's law applies directly.

Approach

1. Write the balanced equation for the complete combustion of methane.
2. Write the balanced equation for the complete combustion of ethane.
3. Use the coefficients to convert each hydrocarbon volume into the volume of $\text{CO}_2$ produced.
4. Add the two $\text{CO}_2$ volumes.
5. Recognise that KOH absorbs $\text{CO}_2$ (and not the excess $\text{O}_2$), so the volume absorbed equals the total $\text{CO}_2$ volume.

Step-by-Step Reasoning

Step 1 — Combustion of methane.
Complete combustion of an alkane produces $\text{CO}_2$ and $\text{H}_2\text{O}$:

$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$

The coefficients show 1 volume of $\text{CH}_4$ gives 1 volume of $\text{CO}_2$. So $10\text{ cm}^3$ of methane gives $10\text{ cm}^3$ of $\text{CO}_2$.

Step 2 — Combustion of ethane.

$$2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}$$

Here 2 volumes of $\text{C}_2\text{H}_6$ give 4 volumes of $\text{CO}_2$, i.e. a 1:2 ratio. So $10\text{ cm}^3$ of ethane gives $20\text{ cm}^3$ of $\text{CO}_2$.

Step 3 — Total $\text{CO}_2$.

$$10\text{ cm}^3 + 20\text{ cm}^3 = 30\text{ cm}^3$$

Step 4 — What KOH absorbs.
After cooling, water condenses to liquid and is no longer a gas. The residual gas is a mixture of $\text{CO}_2$ and excess $\text{O}_2$. Aqueous KOH absorbs $\text{CO}_2$ but not $\text{O}_2$. Therefore the volume absorbed is $30\text{ cm}^3$, which is option C.

Why the distractors are wrong:

• A ($15\text{ cm}^3$) — would arise from incorrectly assuming a 1:1 ratio for ethane too, or averaging the two volumes.
• B ($20\text{ cm}^3$) — would arise from ignoring the methane contribution entirely, or from a wrong ethane:CO₂ ratio.
• D ($40\text{ cm}^3$) — would arise from incorrectly doubling the methane contribution (e.g. assuming 2 moles of $\text{CO}_2$ per mole of $\text{CH}_4$), or from adding the hydrocarbon volumes directly.

Key Takeaways

• Gas volume ratios in a reaction equal the mole ratios from the balanced equation (Avogadro's law), provided temperature and pressure are constant.
• Complete combustion of alkanes always produces $\text{CO}_2$ and $\text{H}_2\text{O}$; the $\text{CO}_2$ volume per volume of alkane equals the number of carbon atoms in the alkane (1 for methane, 2 for ethane).
• Aqueous KOH is a standard reagent for absorbing $\text{CO}_2$ from a gas mixture; it does not absorb $\text{O}_2$ or other neutral gases.
• "After cooling" signals that water vapour has condensed and should not be counted as gas.

Common Mistakes

• Using the wrong stoichiometric ratio for ethane. Ethane combustion has a 1:2 ratio of $\text{C}_2\text{H}_6$ to $\text{CO}_2$, not 1:1. Forgetting this gives $10\text{ cm}^3$ instead of $20\text{ cm}^3$.
• Counting water vapour as gas. The question says "after cooling", so water is liquid and not part of the gas volume. Including $\text{H}_2\text{O}$ as gas would give a wrong total.
• Assuming KOH absorbs oxygen. KOH is an alkali and absorbs the acidic gas $\text{CO}_2$ only; excess $\text{O}_2$ passes through unabsorbed.
• Adding hydrocarbon volumes directly. $10 + 10 = 20\text{ cm}^3$ is not the $\text{CO}_2$ volume; the stoichiometry of each combustion must be applied.

Things to Be Careful About

• Ensure both combustion equations are correctly balanced before reading off volume ratios — especially ethane, where the coefficient of $\text{CO}_2$ is not equal to the coefficient of the alkane.
• Note that "excess oxygen" guarantees complete combustion of both alkanes; no limiting-reagent calculation is needed.
• State symbols matter conceptually: water is (l) after cooling, so it is excluded from gas volumes.
• The same temperature and pressure condition is essential for using volume ratios directly — the question explicitly provides this.
• In the final answer, include the unit ($\text{cm}^3$) when stating the volume.

Background Concept

Ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms or ions. Successive ionisation energies always increase because the remaining electrons are held more tightly by the same nuclear charge. A very large jump occurs when the next electron to be removed comes from a complete inner shell rather than the valence shell. For a main-group element, the number of electrons removed before this jump equals the number of valence electrons, so it identifies the group.

First ionisation energy generally decreases down a group because atomic radius increases and inner-shell electrons shield the nuclear charge, so the outermost electron is less strongly attracted. This means the element with the largest first ionisation energy in a group is usually at the top of the group, and the second largest is the next element down.

Understanding the Question

The question gives two clues for each element: the position of a large ionisation-energy jump and the rank of its first ionisation energy within its group. We must use these clues to identify X and Y, then choose the compound that contains both elements.

Approach

1. Use the large jump to count valence electrons and deduce the group.
2. Use the first-ionisation-energy rank to identify the exact element in that group.
3. Check the options for a formula containing both identified elements.

Step-by-Step Reasoning

For X, the large increase between the 5th and 6th ionisation energies means that after five electrons have been removed, the next electron must come from a lower inner shell. X therefore has five valence electrons and is in Group 15. In Group 15, the order of first ionisation energies is N > P > As > Sb > Bi, so nitrogen has the largest first ionisation energy and phosphorus has the second largest. Thus X is phosphorus, P.

For Y, the large increase between the 6th and 7th ionisation energies means that six electrons have been removed from the valence shell before the next electron comes from an inner shell. Y therefore has six valence electrons and is in Group 16. In Group 16, the order is O > S > Se > Te > Po, so oxygen has the largest first ionisation energy. Thus Y is oxygen, O.

Now check the options:

• A, $\text{NO}_2$, contains nitrogen and oxygen. Nitrogen is in Group 15, but it has the largest first ionisation energy in Group 15, not the second largest, so X would not be nitrogen.
• B, $\text{PCl}_5$, contains phosphorus and chlorine. Chlorine is in Group 17, not Group 16, so Y would not be chlorine.
• C, $\text{P}_4\text{O}_{10}$, contains phosphorus and oxygen. This matches X = P and Y = O.
• D, $\text{SF}_6$, contains sulfur and fluorine. Sulfur is in Group 16, but it does not have the largest first ionisation energy in Group 16; oxygen does. Fluorine is in Group 17, not Group 16.

Therefore the correct compound is $\text{P}_4\text{O}_{10}$, option C.

Key Takeaways

A large jump in successive ionisation energies reveals how many valence electrons an atom has and therefore its group. Combining this with the trend in first ionisation energy down a group identifies the element. Finally, the identified elements must be matched to a formula.

Common Mistakes

• Counting the jump incorrectly: a large jump between the 5th and 6th ionisation energies means five valence electrons, not six.
• Confusing "second largest" with "second smallest" or "second from the bottom" of the group.
• Choosing sulfur in $\text{SF}_6$ because sulfur has six valence electrons, while forgetting that oxygen, not sulfur, has the largest first ionisation energy in Group 16.
• Choosing $\text{NO}_2$ because both nitrogen and oxygen are top-of-group elements, while ignoring that nitrogen is the largest, not the second largest, in Group 15.

Things to Be Careful About

• Ionisation energy trends are based on gaseous atoms; the general decrease down a group is due to increasing atomic radius and shielding.
• The group number for a main-group element equals the number of valence electrons.
• Check that the chosen formula contains both identified elements; $\text{P}_4\text{O}_{10}$ is the molecular formula of phosphorus(V) oxide, with empirical formula $\text{P}_2\text{O}_5$.
• No state symbols are needed in this multiple-choice question.

Background Concept

Nuclide notation is written as $^{A}_{Z}\text{X}$, where:

• $A$ is the mass number (total number of protons + neutrons),
• $Z$ is the proton number or atomic number (number of protons).

The number of neutrons is therefore $A - Z$.

In a neutral atom, the number of electrons equals the number of protons, $Z$. When an atom forms an ion, the number of protons and neutrons does not change; only the number of electrons changes. A negative ion has gained electrons, so it has more electrons than protons. A positive ion has lost electrons, so it has fewer electrons than protons.

Iodine is in Group 17 and typically forms a 1- ion, $\text{I}^-$, by gaining one electron.

Understanding the Question

This question gives the nuclide $^{131}_{53}\text{I}$ and asks which statement about a negative ion of this isotope is correct. The four options differ only in the numbers of neutrons and electrons stated. To answer, we need to:

• read the mass number and proton number from the notation,
• calculate the number of neutrons,
• adjust the electron count for the ion's negative charge.

Approach

Start by extracting the two key numbers from the nuclide notation:

• $A = 131$,
• $Z = 53$.

Then calculate neutrons using $N = A - Z$. Finally, find the electron count for a negative ion: start from the neutral atom's electron count and add one electron for each unit of negative charge.

Step-by-Step Reasoning

1. Identify the mass number and proton number.
   For $^{131}_{53}\text{I}$:

   • mass number $A = 131$,
   • proton number $Z = 53$.

2. Calculate the number of neutrons.
   $N = A - Z = 131 - 53 = 78$
   So the nucleus contains 78 neutrons.

3. Determine the number of electrons in the neutral atom.
   A neutral iodine atom has 53 electrons, equal to its proton number.

4. Adjust for the negative ion.
   A negative ion of iodine is $\text{I}^-$, meaning it has gained one electron.
   $\text{electrons in } \text{I}^- = 53 + 1 = 54$

5. Match with the options.
   The correct combination is 78 neutrons and 54 electrons, which is option D.

   • Option A: 53 neutrons and 52 electrons — wrong neutrons and wrong electrons.
   • Option B: 53 neutrons and 54 electrons — correct electrons but wrong neutrons.
   • Option C: 78 neutrons and 52 electrons — correct neutrons but wrong electrons.
   • Option D: 78 neutrons and 54 electrons — correct.

Key Takeaways

• In nuclide notation $^{A}_{Z}\text{X}$, neutrons = $A - Z$.
• A neutral atom has the same number of electrons as protons.
• Ion formation only changes the electron count: a negative ion has gained electrons, a positive ion has lost electrons.
• The number of protons and neutrons is unchanged when an ion forms.

Common Mistakes

• Confusing the mass number with the number of neutrons. The mass number includes protons and neutrons.
• Forgetting to adjust the electron count for the ion's charge. A negative ion has more electrons than a neutral atom, not fewer.
• Thinking that ion formation changes the number of protons or neutrons. It does not.
• Misreading the notation and using 53 as the neutron number because it appears as the subscript.

Things to Be Careful About

• Read the superscript and subscript in nuclide notation carefully: superscript = mass number, subscript = proton number.
• A negative charge means electrons are gained; a positive charge means electrons are lost.
• The number of neutrons is unaffected by ionisation.
• No state symbols or charges on the nucleus are needed here; the question is purely about counting sub-atomic particles.

Background Concept

Aluminium is in Group 13 and has three valence electrons. In $\text{AlCl}_3$, each Al atom forms three covalent bonds using all three of its electrons, leaving only six electrons in its valence shell — an incomplete octet. Such a species is called electron-deficient.

To achieve a full octet, two $\text{AlCl}_3$ molecules combine into the dimer $\text{Al}_2\text{Cl}_6$. In this dimer, two chlorine atoms act as bridges. Each bridging chlorine already has its own lone pairs; it donates one lone pair to one of the aluminium atoms, forming a coordinate (dative covalent) bond. The other bond from that chlorine to the other aluminium is a normal covalent bond.

Understanding the Question

The question asks which types of bonding are present in the dimer $\text{Al}_2\text{Cl}_6$. You must identify both the ordinary covalent bonds (terminal Al–Cl bonds and one of each bridging Al–Cl bond) and the coordinate bonds (the donated lone pairs from bridging chlorines).

Approach

1. Recognise that $\text{AlCl}_3$ is electron-deficient.
2. Recall that dimerisation is the way aluminium trichloride satisfies the octet.
3. Visualise the structure: two Al atoms joined by two bridging Cl atoms, with each bridging Cl forming one normal covalent bond and one coordinate bond.
4. Check that no ionic bonding is present — all bonds are shared-electron bonds.

Step-by-Step Reasoning

1. Write the formula of the monomer: $\text{AlCl}_3$.
2. Count valence electrons: Al has 3, each Cl has 7, so each $\text{AlCl}_3$ has $3 + 3 \times 7 = 24$ valence electrons.
3. In $\text{Al}_2\text{Cl}_6$, the two Al atoms share a total of 48 valence electrons. The terminal Cl atoms form ordinary covalent bonds; the bridging Cl atoms each form one ordinary covalent bond to one Al and one coordinate bond to the other Al.
4. Since every bond is a shared-electron pair (either contributed by both atoms or donated by one atom), the bonding is covalent plus coordinate — not ionic.

Key Takeaways

• $\text{AlCl}_3$ is a covalent molecular compound, not ionic, despite being a metal–non-metal combination.
• Electron-deficient species form coordinate bonds to complete their octet.
• The dimer $\text{Al}_2\text{Cl}_6$ is the classic example of this behaviour.

Common Mistakes

• Assuming AlCl3 is ionic because it contains a metal and a non-metal.
• Forgetting that the bridging chlorine atoms donate a lone pair, so the bond is coordinate, not just covalent.
• Confusing the bridging bonds with ionic interactions.

Things to Be Careful About

• The terminal Al–Cl bonds are ordinary covalent bonds.
• Only the bridging Al–Cl bonds involve coordinate character.
• The dimer formula is $\text{Al}_2\text{Cl}_6$, not $\text{Al}_2\text{Cl}_3$.

Background Concept

Valence Shell Electron Pair Repulsion (VSEPR) theory states that electron domains (bonding pairs, lone pairs, or multiple bonds treated as a single domain) around a central atom will arrange themselves as far apart as possible to minimise electrostatic repulsion. The arrangement of all electron domains determines the electron geometry, while the arrangement of only the atoms (ignoring lone pairs) determines the molecular shape. Lone pairs repel more strongly than bonding pairs, which can compress bond angles slightly, but the basic shape is dictated by the number of domains.

Understanding the Question

The question provides the Lewis structure of sulfur dioxide (SO₂) and asks for its molecular shape. The image shows a central sulfur atom double-bonded to two oxygen atoms, with one lone pair of electrons on the sulfur atom and two lone pairs on each oxygen atom. We need to determine the 3D shape of the molecule based on the electron domains around the central sulfur atom.

Approach

1. Identify the central atom (sulfur).
2. Count the number of electron domains around the central atom (bonding regions + lone pairs). Note that double and triple bonds count as a single electron domain.
3. Determine the electron geometry based on the total number of domains.
4. Deduce the molecular shape by considering only the positions of the atoms, accounting for the lone pair(s).
5. Match the deduced shape to the given options.

Step-by-Step Reasoning

• Central atom: Sulfur (S) is the central atom.
• Bonding domains: Sulfur is double-bonded to two oxygen atoms. In VSEPR theory, each double bond counts as one bonding domain. So, there are 2 bonding domains.
• Lone pairs: The Lewis structure shows one lone pair of electrons on the sulfur atom. This counts as 1 non-bonding domain.
• Total electron domains: 2 (bonding) + 1 (lone pair) = 3 domains.
• Electron geometry: Three electron domains arrange themselves in a trigonal planar geometry with ideal bond angles of 120°.
• Molecular shape: The molecular shape only considers the positions of the atoms. With 2 bonding domains and 1 lone pair, the atoms form a bent or V-shaped arrangement. This is described as non-linear.
• Evaluating options:
  • A (linear): Requires 2 electron domains (e.g., CO₂, which has no lone pairs on carbon). Incorrect.
  • B (non-linear): Matches our deduction (bent/V-shaped due to 3 domains with 1 lone pair). Correct.
  • C (pyramidal): Requires 4 electron domains (3 bonding, 1 lone pair, e.g., NH₃). Incorrect.
  • D (tetrahedral): Requires 4 bonding domains and no lone pairs on the central atom (e.g., CH₄). Incorrect.

Key Takeaways

• VSEPR theory predicts molecular shape based on the repulsion between electron domains around a central atom.
• Multiple bonds (double or triple) count as a single electron domain when determining the basic geometry.
• Lone pairs occupy space and affect the molecular shape, making it different from the electron geometry (e.g., trigonal planar electron geometry with 1 lone pair gives a bent/non-linear molecular shape).

Common Mistakes

• Forgetting the lone pair on sulfur: If a student misses the lone pair on sulfur, they will count only 2 domains (the two double bonds) and incorrectly predict a linear shape (like CO₂).
• Counting double bonds as two domains: If a student counts each bond in a double bond as a separate domain, they will get 4 bonding domains + 1 lone pair = 5 domains, leading to an incorrect geometry. VSEPR treats multiple bonds as a single region of electron density.
• Confusing electron geometry with molecular shape: A student might correctly identify the electron geometry as trigonal planar but fail to recognise that the molecular shape (the arrangement of atoms) is non-linear due to the lone pair.

Things to Be Careful About

• Always count the lone pairs on the central atom carefully from the provided Lewis structure.
• Remember that in VSEPR, a double or triple bond is treated as one electron domain for the purpose of determining the basic shape and electron geometry.
• The term "non-linear" is used in the options to describe a bent or V-shaped molecule. Ensure you match the correct terminology to the given choices.
• The bond angle in SO₂ is slightly less than 120° (approximately 119°) because the lone pair repels the bonding pairs more strongly than they repel each other, but the question only asks for the general shape.

Background Concept

An ideal gas obeys the equation of state

$$pV = nRT$$

where $p$ is pressure, $V$ is volume, $n$ is amount of gas in moles, $R$ is the gas constant and $T$ is temperature in kelvin. Density is mass per unit volume, $\rho = m/V$. Since $n = m/M$, where $M$ is the molar mass, substituting gives

$$\rho = \frac{pM}{RT}$$

This is the key relation: the density of an ideal gas depends only on its molar mass, pressure and temperature, not on the amount present.

For fluorine, the gas is $\text{F}_2$, so $M = 2 \times 19 = 38\text{ g mol}^{-1}$. The pressure is $100\,000\text{ Pa}$ and the temperature is $32^\circ\text{C}$, which must be converted to kelvin: $T = 32 + 273 = 305\text{ K}$.

Understanding the Question

This is a one-mark multiple-choice question asking for the density of fluorine gas under stated conditions. The word 'density' tells us we need $\rho = m/V$; the phrase 'ideal gas' tells us we may use $pV = nRT$. We are given pressure, temperature and the identity of the gas, so we can calculate the density directly.

The four options are close together (1.4 to 1.7 g dm^-3), so the calculation must be done carefully, especially the temperature conversion and the molar mass.

Approach

1. Convert the Celsius temperature to kelvin.
2. Find the molar mass of $\text{F}_2$.
3. Substitute $p$, $M$, $R$ and $T$ into $\rho = pM/RT$.
4. Check units: with $p$ in Pa and $M$ in kg mol^-1, $\rho$ comes out in kg m^-3, which equals g dm^-3. Alternatively use $p$ in kPa, $V$ in dm^3 and $M$ in g mol^-1 with $R = 8.314\text{ kPa dm}^3\text{ K}^{-1}\text{ mol}^{-1}$.

Step-by-Step Reasoning

• Convert temperature:

$$T = 32 + 273 = 305\text{ K}$$

• Molar mass of fluorine gas:

$$M(\text{F}_2) = 2 \times 19 = 38\text{ g mol}^{-1} = 0.038\text{ kg mol}^{-1}$$

• Substitute into the density expression:

$$\rho = \frac{100\,000 \times 0.038}{8.314 \times 305}$$ $$\rho = \frac{3800}{2535.8} = 1.498\text{ kg m}^{-3}$$

Since $1\text{ kg m}^{-3} = 1\text{ g dm}^{-3}$, this is $1.5\text{ g dm}^{-3}$ to two significant figures.

This matches option B.

If you prefer to keep $M$ in g mol^-1, use $p = 100\text{ kPa}$ and $R = 8.314\text{ kPa dm}^3\text{ K}^{-1}\text{ mol}^{-1}$:

$$\rho = \frac{100 \times 38}{8.314 \times 305} = 1.5\text{ g dm}^{-3}$$

Either route gives the same answer.

Key Takeaways

• The density of an ideal gas is $\rho = pM/RT$.
• Always convert temperature to kelvin in gas calculations.
• Use consistent units: either SI (Pa, kg m^-3, kg mol^-1) or kPa/dm^3/g mol^-1.
• $1\text{ kg m}^{-3} = 1\text{ g dm}^{-3}$.

Common Mistakes

• Using the Celsius temperature directly. If $T = 32\text{ K}$ were used, the density would be much too large.
• Using the molar mass of atomic fluorine (19 g mol^-1) instead of molecular fluorine (38 g mol^-1).
• Mixing units, e.g. using $p$ in Pa but $M$ in g mol^-1; this gives a value off by a factor of 1000.
• Rounding too early; the options differ by only 0.1 g dm^-3.

Things to Be Careful About

• The gas is fluorine, $\text{F}_2$, not $\text{F}$.
• The ideal gas equation uses kelvin, so $32^\circ\text{C}$ must become 305 K.
• $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$ equals $8.314\text{ kPa dm}^3\text{ K}^{-1}\text{ mol}^{-1}$; choose units to match $M$.
• The answer should be quoted to two significant figures, consistent with the data and options.

Background Concept

The physical properties of simple molecular substances, such as boiling point, melting point, and volatility, are determined by the strength of the intermolecular forces between the molecules, not by the strength of the covalent bonds within the molecules. Boiling is a physical change where molecules are separated from each other; the covalent bonds holding the atoms together within each molecule remain intact.

The main types of intermolecular forces relevant here are:

1. Van der Waals' (London dispersion) forces: Present in all molecules. Their strength increases with the number of electrons (and thus the size/polarizability of the electron cloud). For simple molecules down a group (like the Group 16 hydrides), this is the dominant force.
2. Hydrogen bonding: A particularly strong type of dipole-dipole interaction that occurs when hydrogen is bonded to a highly electronegative atom with a small atomic radius (specifically N, O, or F). It requires significantly more energy to break than van der Waals' forces.

Understanding the Question

The question provides a graph of the boiling points of Group 16 hydrides: H$_2$O, H$_2$S, H$_2$Se, and H$_2$Te. The graph shows a steady increase in boiling point from H$_2$S to H$_2$Te, but H$_2$O is plotted far above this trend line (100 °C). The task is to identify the correct chemical explanation for this anomaly.

Approach

To solve this, we must:

1. Recognize that boiling point depends on intermolecular forces.
2. Explain the general trend for H$_2$S, H$_2$Se, H$_2$Te using van der Waals' forces.
3. Explain the anomaly for H$_2$O using hydrogen bonding.
4. Systematically eliminate the incorrect options by identifying the conceptual errors in each statement.

Step-by-Step Reasoning

• General Trend (H$_2$S, H$_2$Se, H$_2$Te): As we move down Group 16 from S to Te, the central atom gets larger and has more electrons. The resulting molecules (H$_2$S, H$_2$Se, H$_2$Te) have larger, more polarizable electron clouds. This leads to stronger van der Waals' (dispersion) forces between molecules, requiring more thermal energy to overcome, hence the increasing boiling points. S, Se, and Te are not electronegative enough (and are too large) to form hydrogen bonds with hydrogen.

• The Anomaly (H$_2$O): Oxygen is highly electronegative and has a small atomic radius. This creates a strong permanent dipole in the O–H bond and allows the hydrogen atom of one molecule to interact strongly with the lone pair on the oxygen atom of a neighboring molecule. This is a hydrogen bond. Because hydrogen bonds are much stronger than van der Waals' forces, water requires significantly more energy to boil, giving it an anomalously high boiling point compared to what the trend would predict.

• Evaluating Option A: "There are fewer electrons in the oxygen atoms so there is less shielding of the nuclear charge." Shielding and nuclear charge concepts apply to atomic properties like ionisation energy or atomic radius, not to the intermolecular forces that determine boiling points. Incorrect.

• Evaluating Option B: "There are strong hydrogen bonds in water but not in the other compounds." This correctly identifies the presence of hydrogen bonding in water as the reason for its high boiling point, and correctly notes that the other hydrides lack this feature. Correct.

• Evaluating Option C: "The covalent bonds in water are much stronger than in the other compounds." While the O–H bond is indeed strong, boiling does not break covalent bonds (intramolecular forces); it only overcomes intermolecular forces. If covalent bonds had to be broken, water would decompose into hydrogen and oxygen gas rather than boil. Incorrect.

• Evaluating Option D: "The water molecules are smaller and so have stronger van der Waals' forces." This contains a fundamental error in reasoning. Smaller molecules have fewer electrons, which means their electron clouds are less polarizable, resulting in weaker van der Waals' forces, not stronger. Incorrect.

Key Takeaways

• Boiling is a physical process that overcomes intermolecular forces, not intramolecular covalent bonds.
• Hydrogen bonding (occurring in molecules with H–N, H–O, or H–F bonds) causes anomalously high boiling points, melting points, and densities (in the case of ice) compared to the rest of the group.
• For simple molecular substances down a group where hydrogen bonding is absent, boiling points increase due to increasing van der Waals' forces as molecular size and electron count increase.

Common Mistakes

• Confusing intermolecular and intramolecular forces: Students often select options mentioning strong covalent bonds (Option C) because they associate "strong" with "high boiling point". They forget that boiling separates molecules, it does not break them apart.
• Misunderstanding van der Waals' forces: Option D tests the misconception that smaller molecules have stronger van der Waals' forces. In reality, van der Waals' forces increase with molecular size and electron count.
• Applying atomic concepts to physical properties: Option A mixes up atomic structure concepts (shielding, nuclear charge) with physical properties (boiling point) that depend on intermolecular interactions.

Things to Be Careful About

• Always ask yourself: "What is being broken or overcome in this process?" For phase changes (melting, boiling), it is intermolecular forces. For chemical reactions or decomposition, it is intramolecular bonds.
• Remember the specific conditions for hydrogen bonding: hydrogen must be covalently bonded to a small, highly electronegative atom (N, O, or F). Sulfur, selenium, and tellurium do not meet this criterion.
• When evaluating trends down a group for hydrides, always consider both van der Waals' forces (which increase down the group) and the possibility of hydrogen bonding (which causes anomalies for the period 2 elements N, O, F).

Background Concept

Hess's law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same. This allows us to construct energy cycles (enthalpy cycles) to relate different enthalpy changes, such as enthalpies of formation, combustion, and reaction.

Key enthalpy changes to remember:

• Enthalpy of combustion ($\Delta H_c^\ominus$): The enthalpy change when one mole of a substance is completely burned in oxygen. Combustion is always an exothermic process, so $\Delta H_c^\ominus$ is always negative.
• Enthalpy of formation ($\Delta H_f^\ominus$): The enthalpy change when one mole of a compound is formed from its elements in their standard states. Many formation reactions are exothermic (e.g., $\Delta H_f^\ominus$ for $\text{CH}_4$ is $-74.8 \text{ kJ mol}^{-1}$), but not all.

Understanding the Question

We are given an energy cycle with three pathways:

• Top: $\text{CH}_4 + 2\text{O}_2$
• Bottom left: $\text{C} + 2\text{H}_2 + 2\text{O}_2$ (elements in standard states, plus excess oxygen)
• Bottom right: $\text{CO}_2 + 2\text{H}_2\text{O}$ (combustion products)

The arrows represent the enthalpy changes:

• X: Between the elements and methane (involving $\text{O}_2$ on both sides, so net change is $\text{CH}_4 \rightleftharpoons \text{C} + 2\text{H}_2$)
• Y: From methane + oxygen to combustion products (combustion of methane)
• Z: From elements + oxygen to combustion products (combustion of elements)

We are told the magnitudes are 890 and 964, and we must determine which of the three values (X, Y, Z) are negative.

Approach

1. Identify the chemical process represented by each arrow (Y and Z are clearly combustion; X is the reverse of formation or decomposition).
2. Recall that combustion is exothermic ($\Delta H < 0$).
3. Use Hess's law or knowledge of formation enthalpies to determine the sign of X.
4. Select the option that matches the negative values.

Step-by-Step Reasoning

Step 1: Analyze Arrow Y
Arrow Y goes from $\text{CH}_4 + 2\text{O}_2$ to $\text{CO}_2 + 2\text{H}_2\text{O}$. This is the complete combustion of methane:
$\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$
Combustion is exothermic, so $\Delta H_Y$ is negative. Given the magnitude is 890, $\Delta H_Y = -890 \text{ kJ mol}^{-1}$.

Step 2: Analyze Arrow Z
Arrow Z goes from $\text{C} + 2\text{H}_2 + 2\text{O}_2$ to $\text{CO}_2 + 2\text{H}_2\text{O}$. This represents the combustion of carbon and hydrogen:
$\text{C}(\text{s}) + 2\text{H}_2(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$
This is also an exothermic process (combustion of elements), so $\Delta H_Z$ is negative. Given the magnitude is 964, $\Delta H_Z = -964 \text{ kJ mol}^{-1}$.

Step 3: Analyze Arrow X using Hess's Law
Following the cycle from bottom left to bottom right:

• Path 1 (direct): $\Delta H = \Delta H_Z = -964 \text{ kJ mol}^{-1}$
• Path 2 (via top): $\Delta H = \Delta H_X + \Delta H_Y = \Delta H_X + (-890)$

By Hess's law:
$\Delta H_X - 890 = -964$
$\Delta H_X = -964 + 890 = +74 \text{ kJ mol}^{-1}$
Since $\Delta H_X$ is positive, arrow X represents an endothermic process. Chemically, looking at the diagram, the arrow X points from $\text{CH}_4 + 2\text{O}_2$ down to $\text{C} + 2\text{H}_2 + 2\text{O}_2$ (decomposition of methane into its elements), which is the reverse of the exothermic formation of methane ($\Delta H_f^\ominus = -74.8 \text{ kJ mol}^{-1}$). Thus, $\Delta H_X = +74 \text{ kJ mol}^{-1}$.

Step 4: Conclusion

• X is positive (+74)
• Y is negative (-890)
• Z is negative (-964)

Therefore, Y and Z are the negative values. This matches option C.

Key Takeaways

• In an enthalpy cycle, combustion pathways (to $\text{CO}_2$ and $\text{H}_2\text{O}$) are always exothermic (negative $\Delta H$).
• Formation pathways from elements to compounds are often exothermic, but the reverse (decomposition) is endothermic (positive $\Delta H$).
• Hess's law can be used to algebraically determine the sign and magnitude of an unknown enthalpy change in a cycle.

Common Mistakes

• Misreading arrow directions: Assuming all arrows point from elements to products. In this diagram, X points from methane to elements (or vice versa depending on exact arrowhead, but the math dictates it must be +74 to balance the cycle). If a student assumes X is formation ($\text{C} + 2\text{H}_2 \rightarrow \text{CH}_4$), they would get -74 (negative), and might incorrectly choose an option like "X, Y and Z" if it existed, or get confused.
• Forgetting combustion is exothermic: Thinking combustion could be endothermic.
• Sign errors in Hess's law: Calculating $\Delta H_X = -890 - (-964) = +74$ correctly, but making an algebra mistake like $-964 - 890 = -1854$.

Things to Be Careful About

• Always check the direction of the arrows in the energy cycle. The arrow for X in the diagram points from $\text{CH}_4 + 2\text{O}_2$ to $\text{C} + 2\text{H}_2 + 2\text{O}_2$ (decomposition), making it endothermic (+ve). If the arrow pointed the other way (formation), it would be exothermic (-ve). The numerical values given (-890 and -964 for Y and Z) force X to be +74 to satisfy Hess's law ($\Delta H_X + \Delta H_Y = \Delta H_Z \Rightarrow \Delta H_X - 890 = -964 \Rightarrow \Delta H_X = +74$).
• Combustion of any substance (including elements like C and H₂) to their standard oxides ($\text{CO}_2$, $\text{H}_2\text{O}$) is always exothermic.

Background Concept

A reaction pathway (energy profile) diagram plots the energy of the system against the progress of reaction. Key features:

• Reactants start at a certain energy level on the left.
• Products end at a certain energy level on the right.
• The peak of the curve represents the activated complex (transition state). The energy difference between the peak and the reactants is the activation energy ($E_a$).
• The energy difference between the products and the reactants is the enthalpy change ($\Delta H$).

If products are at a higher energy than reactants, $\Delta H > 0$ and the reaction is endothermic. If products are at a lower energy, $\Delta H < 0$ and the reaction is exothermic.

Understanding the Question

We are given two facts:

1. The reaction is endothermic — products must be higher in energy than reactants.
2. The activation energy is numerically equal to twice the enthalpy change: $E_a = 2\Delta H$.

We must choose which of the four diagrams (A, B, C, D) correctly represents both conditions.

Approach

Step 1: Eliminate diagrams that do not show an endothermic reaction (products lower than reactants).
Step 2: For the remaining diagrams, read off the grid values for reactant energy, product energy, and peak energy.
Step 3: Calculate $\Delta H$ and $E_a$ from the grid units.
Step 4: Check whether $E_a = 2\Delta H$.

Step-by-Step Reasoning

Step 1 — Eliminate exothermic diagrams.

• Diagrams C and D show products at a lower energy level than reactants. These represent exothermic reactions ($\Delta H < 0$). Since the question states the reaction is endothermic, C and D are eliminated.

Step 2 — Read diagram B from the grid.

• Reactants are at 2 grid units above the baseline.
• Products are at 3 grid units above the baseline.
• The peak (activated complex) is at 4 grid units above the baseline.

Step 3 — Calculate $\Delta H$ and $E_a$.

• $\Delta H = E_{\text{products}} - E_{\text{reactants}} = 3 - 2 = 1$ grid unit (positive, confirming endothermic).
• $E_a = E_{\text{peak}} - E_{\text{reactants}} = 4 - 2 = 2$ grid units.

Step 4 — Check the condition $E_a = 2\Delta H$.

• $2\Delta H = 2 \times 1 = 2$ grid units.
• $E_a = 2$ grid units.
• Therefore $E_a = 2\Delta H$. ✓

Diagram B satisfies both conditions: it is endothermic and $E_a = 2\Delta H$.

Key Takeaways

• In an endothermic reaction profile, products sit above reactants; in an exothermic profile, they sit below.
• Activation energy is measured from the reactants to the peak, not from the baseline or the products.
• Enthalpy change is measured from the reactants to the products.
• Always read numerical values from grid lines when they are provided.

Common Mistakes

• Confusing $E_a$ with $\Delta H$: Measuring the peak from the baseline instead of from the reactant level gives the wrong $E_a$.
• Choosing an exothermic diagram: C and D show products lower than reactants, which contradicts the endothermic condition.
• Misreading the grid: Carefully count grid squares from reactant level to peak and from reactant level to products.

Things to Be Careful About

• State symbols are not needed for pathway diagrams, but the relative positions of reactants, products, and the peak must be correct.
• The activation energy is always measured from the reactant energy level to the peak, regardless of whether the reaction is endothermic or exothermic.
• For endothermic reactions, $\Delta H$ is positive; for exothermic, it is negative. The numerical relationship $E_a = 2\Delta H$ uses the magnitude (numerical value) of $\Delta H$.

Background Concept

Redox reactions are analysed by assigning oxidation numbers. Oxidation is an increase in oxidation number; reduction is a decrease. In this industrial process, atmospheric oxygen is the oxidising agent.

Understanding the Question

The reaction converts chromite, FeCr2O4, into sodium chromate(VI), Na2CrO4. We need to identify which elements change oxidation state and whether they are oxidised.

Approach

Assign oxidation numbers to every element on both sides of the equation, then compare them. Focus on Fe, Cr, C and O.

Step-by-Step Reasoning

1. In FeCr2O4, four O atoms contribute a total of -8. Since the compound is neutral, Fe + 2Cr = +8. The mineral contains Fe2+ and Cr2O4^2-, so Fe = +2 and each Cr = +3.
2. In Na2CrO4, Na = +1 each and O = -2 each: 2(+1) + Cr + 4(-2) = 0, so Cr = +6.
3. In Fe2O3, O = -2 each, so 2Fe + 3(-2) = 0, giving Fe = +3.
4. In Na2CO3 and CO2, carbon is +4 in both carbonate and carbon dioxide, so carbon is unchanged.
5. O2 is elemental oxygen with oxidation number 0; in oxides it is -2, so oxygen is reduced.

Therefore Cr (+3 to +6) and Fe (+2 to +3) are oxidised; C is not oxidised; O is reduced. The correct option is A.

Key Takeaways

• Always assign oxidation numbers using known rules: O is usually -2, Na is +1, and neutral compounds sum to 0.
• An element is oxidised only if its oxidation number increases.
• In FeCr2O4, chromium is +3, not +6.

Common Mistakes

• Assuming Cr in chromite is already +6 because the product is chromate(VI).
• Forgetting that carbonate carbon stays +4 throughout.
• Counting oxygen as oxidised because it appears in products; O2 goes from 0 to -2, so it is reduced.

Things to Be Careful About

• Use the oxidation number of each atom, not the overall charge of a polyatomic ion.
• The phrase "only elements oxidised" excludes elements that are reduced or unchanged.

Background Concept

This question tests redox stoichiometry — the idea that in a redox reaction the total number of electrons lost by the reducing agent must exactly equal the total number of electrons gained by the oxidising agent. The key skill is to track electrons through oxidation number changes rather than trying to balance the full equation by inspection.

Here, $\text{KMnO}_4$ is the oxidising agent: manganese is reduced from oxidation state $+7$ in $\text{MnO}_4^-$ to $+2$ in $\text{Mn}^{2+}$, a gain of $5$ electrons per manganese ion. Hydrogen peroxide is the reducing agent: each $\text{H}_2\text{O}_2$ molecule is oxidised to $\text{O}_2$, losing $2$ electrons (the oxygen goes from $-1$ in $\text{H}_2\text{O}_2$ to $0$ in $\text{O}_2$, and with two oxygens that is $2$ electrons lost per molecule).

Understanding the Question

The question gives a redox reaction in acidic conditions between $\text{KMnO}_4$ and $\text{H}_2\text{O}_2$, producing $\text{O}_2$, potassium sulfate, manganese(II) sulfate and water. It tells you directly that each $\text{H}_2\text{O}_2$ molecule loses two electrons. You are asked how many moles of $\text{O}_2$ are produced when exactly $1.0\text{ mol}$ of $\text{KMnO}_4$ reacts with excess $\text{H}_2\text{O}_2$.

The word "excess" is important: it means all the $\text{KMnO}_4$ is used up, so the amount of $\text{O}_2$ is limited by the $1.0\text{ mol}$ of $\text{KMnO}_4$, not by the peroxide. The calculation therefore reduces to: how many electrons can $1.0\text{ mol}$ of $\text{KMnO}_4$ accept, and how many $\text{H}_2\text{O}_2$ molecules (and hence $\text{O}_2$ molecules) does that correspond to?

Approach

The strategy is to convert everything into electron moles:

1. Find the oxidation state change of manganese and hence the electrons gained per mole of $\text{KMnO}_4$.
2. Multiply by $1.0\text{ mol}$ to get total electrons accepted.
3. Use the given fact that each $\text{H}_2\text{O}_2$ loses $2$ electrons to find how many moles of $\text{H}_2\text{O}_2$ are oxidised.
4. Recognise that each $\text{H}_2\text{O}_2$ produces exactly one $\text{O}_2$, so the moles of $\text{O}_2$ equal the moles of $\text{H}_2\text{O}_2$ oxidised.

Step-by-Step Reasoning

Step 1 — Electrons gained per $\text{KMnO}_4$.
In $\text{KMnO}_4$, potassium is $+1$ and each oxygen is $-2$ (four oxygens, total $-8$), so manganese must be $+7$. In $\text{MnSO}_4$, the sulfate ion is $-2$, so manganese is $+2$. The change is $+7 \rightarrow +2$, a gain of $5$ electrons per manganese atom.

Step 2 — Total electrons accepted by $1.0\text{ mol KMnO}_4$.
$1.0\text{ mol}$ of $\text{KMnO}_4$ accepts $1.0 \times 5 = 5.0\text{ mol}$ of electrons.

Step 3 — Moles of $\text{H}_2\text{O}_2$ oxidised.
Each $\text{H}_2\text{O}_2$ loses $2$ electrons. To supply $5.0\text{ mol}$ of electrons we need:
$\frac{5.0\text{ mol e}^-}{2\text{ mol e}^- \text{ per mol H}_2\text{O}_2} = 2.5\text{ mol H}_2\text{O}_2$

Step 4 — Moles of $\text{O}_2$ produced.
Each $\text{H}_2\text{O}_2$ molecule loses two electrons and is oxidised to one $\text{O}_2$ molecule:
$\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-$
So $2.5\text{ mol H}_2\text{O}_2$ produces $2.5\text{ mol O}_2$.

This matches option B.

Why the distractors are wrong:

• A (2.0 mol) — would arise from wrongly pairing 5 electrons gained with 2.5 lost per peroxide, or from miscounting the electron ratio.
• C (4.5 mol) and D (5.0 mol) — would come from confusing the number of electrons gained by $\text{KMnO}_4$ (5) with the number of moles of $\text{O}_2$ (5.0), or from adding the electron counts rather than dividing. A common slip is to think "5 electrons per Mn, so 5 mol O2" — forgetting that each O2 needs 2 electrons from one H2O2.

Key Takeaways

• In redox stoichiometry, balance electrons first: electrons lost = electrons gained.
• The oxidising agent's electron acceptance per mole is its oxidation-state drop; the reducing agent's electron donation per mole is its oxidation-state rise.
• "Excess" of one reagent means the other reagent limits the reaction.
• Each $\text{H}_2\text{O}_2 \rightarrow \text{O}_2$ conversion is a clean 1:1 mole ratio with 2 electrons transferred.

Common Mistakes

• Confusing electrons with moles of product: 5 electrons per $\text{KMnO}_4$ does not mean 5 mol of $\text{O}_2$; you must divide by the 2 electrons per peroxide.
• Miscounting the oxidation state of Mn: forgetting that in $\text{MnO}_4^-$ the four oxygens contribute $-8$, so Mn is $+7$, not $+4$.
• Ignoring the "excess" clause and trying to use the $\text{H}_2\text{O}_2$ amount, which is not given and is not limiting.
• Writing the peroxide half-equation incorrectly: each $\text{H}_2\text{O}_2$ gives one $\text{O}_2$ and two electrons, not four.

Things to Be Careful About

• Use the oxidation number change of the element that changes state, not the whole compound.
• Keep the electron bookkeeping in moles: $1.0\text{ mol KMnO}_4$ accepts $5.0\text{ mol e}^-$.
• The final answer is a quantity of moles of gas; the unit "mol" must be stated.
• In acidic conditions the half-equation is balanced with $\text{H}^+$ and water, but for this stoichiometry question you only need the electron counts, not the full balanced equation.

Background Concept

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

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

where each concentration is the equilibrium concentration in mol dm$^{-3}$. The value of $K_c$ is constant at a fixed temperature and indicates the position of equilibrium.

Here the reaction is an esterification — a condensation between an alcohol and a carboxylic acid. The key feature for this calculation is that all four species are liquids in the same reaction mixture, so they all occupy the same volume $V$. When we write $K_c$ in terms of moles $n$ and volume $V$, e.g. $[\text{CH}_3\text{COOR}] = n_{\text{ester}}/V$, every $V$ cancels. This means we can work directly with mole amounts rather than concentrations.

Understanding the Question

We start with 3.0 mol of alcohol ROH, 2.0 mol of ethanoic acid and 1.0 mol of water, with no ester present. At equilibrium 1.5 mol of ester has formed. The question asks for $K_c$.

The crucial information is the amount of ester at equilibrium: because the stoichiometry is 1:1:1:1, the 1.5 mol of ester tells us exactly how much ROH and acid were consumed and how much extra water was produced. We must remember that water was already present at the start, so the equilibrium amount of water is the initial 1.0 mol plus the 1.5 mol produced.

Approach

1. Write the balanced equation and note the 1:1:1:1 stoichiometry.
2. Build an ICE table (Initial, Change, Equilibrium) in moles.
3. Use the equilibrium amount of ester (+1.5 mol) to fix the changes for the other three species.
4. Substitute the equilibrium mole amounts into the $K_c$ expression; since volume cancels, moles are used directly.

Step-by-Step Reasoning

Step 1 — Set up the ICE table.

Step 2 — Why the changes are what they are. The ester is a product, so its change is $+1.5$ mol. Because all coefficients are 1, ROH and CH3COOH each decrease by 1.5 mol, and H2O increases by 1.5 mol.

Step 3 — Equilibrium amounts.

• ROH: $3.0 - 1.5 = 1.5$ mol
• CH3COOH: $2.0 - 1.5 = 0.5$ mol
• CH3COOR: $0 + 1.5 = 1.5$ mol
• H2O: $1.0 + 1.5 = 2.5$ mol

Step 4 — Substitute into $K_c$.

$$K_c = \frac{[\text{CH}_3\text{COOR}][\text{H}_2\text{O}]}{[\text{ROH}][\text{CH}_3\text{COOH}]} = \frac{1.5 \times 2.5}{1.5 \times 0.5} = \frac{3.75}{0.75} = 5.0$$

The answer is D.

Why the distractors are wrong:

• A (0.20) is the reciprocal of the correct value — a candidate who inverted the $K_c$ expression (put reactants over products).
• B (0.25) could come from using the initial amount of water (1.0 mol) instead of the equilibrium amount (2.5 mol): $(1.5 \times 1.0)/(1.5 \times 2.0) = 0.25$, mixing initial and equilibrium values.
• C (2.00) could come from $(1.5 \times 2.0)/(1.5 \times 1.0) = 2.00$, again mixing initial water and initial acid with equilibrium ester and ROH.

Key Takeaways

• Always use an ICE table for equilibrium calculations; the amount of a product formed fixes the changes for all species through stoichiometry.
• When all species share one volume, $K_c$ can be computed from mole amounts because the volume cancels.
• Water produced in the reaction must be added to any water initially present.
• The equilibrium constant uses equilibrium amounts only — never initial amounts.

Common Mistakes

• Forgetting the initial water: using 1.5 mol instead of 2.5 mol for water gives $K_c = (1.5 \times 1.5)/(1.5 \times 0.5) = 3.0$, which is not even an option, but it shows the error.
• Using initial amounts in $K_c$: substituting 3.0 and 2.0 for the reactants instead of the equilibrium 1.5 and 0.5.
• Inverting the expression: writing reactants over products gives 0.20 (option A).
• Ignoring stoichiometry: forgetting that 1.5 mol ester means 1.5 mol of each reactant consumed.

Things to Be Careful About

• All species are liquids in the same mixture, so the volume cancels — you may use moles directly. If volumes differed, you would need concentrations.
• The $K_c$ expression places products in the numerator and reactants in the denominator, each raised to the power of its stoichiometric coefficient (all 1 here).
• The value 5.0 is dimensionless here because the number of product terms equals the number of reactant terms, so the mol units cancel.
• Keep the arithmetic clean: $1.5 \times 2.5 = 3.75$ and $1.5 \times 0.5 = 0.75$; $3.75/0.75 = 5$.

Background Concept

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and re-establish equilibrium. For the Haber process, $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$, the forward reaction is exothermic ($\Delta H = -92 \text{ kJ mol}^{-1}$) and results in a decrease in the number of moles of gas (4 moles $\rightarrow$ 2 moles).

Understanding the Question

The question asks to identify the correct graph showing the percentage of ammonia at equilibrium as a function of pressure (at two different temperatures: 400°C and 500°C). We must determine the direction of the equilibrium shift with increasing pressure and with increasing temperature to select the matching diagram.

Approach

Apply Le Chatelier's principle twice:

1. Consider the effect of pressure on the equilibrium position based on the difference in moles of gas between reactants and products.
2. Consider the effect of temperature on the equilibrium position based on the sign of the enthalpy change (exothermic vs endothermic).

Step-by-Step Reasoning

• Pressure effect: The reaction $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$ has 4 moles of gaseous reactants and 2 moles of gaseous products. Increasing the pressure will cause the equilibrium to shift to the side with fewer moles of gas to reduce the pressure. This is the right-hand side (products). Therefore, as pressure increases, the yield of ammonia increases, meaning the % NH$_3$ at equilibrium increases. The graph must show an upward trend. This rules out graphs A and B.

• Temperature effect: The forward reaction is exothermic (releases heat). Increasing the temperature adds heat to the system. The equilibrium will shift in the endothermic direction (to the left, towards reactants) to absorb the excess heat. Therefore, at a higher temperature (500°C), the equilibrium position is further to the left, resulting in a lower % NH$_3$ compared to a lower temperature (400°C). The curve for 400°C must lie above the curve for 500°C. This rules out graph C, where 500°C is above 400°C.

• Graph D correctly shows % NH$_3$ increasing with pressure and the 400°C curve above the 500°C curve.

Key Takeaways

• Increasing pressure shifts equilibrium towards the side with fewer moles of gas.
• Increasing temperature shifts equilibrium in the endothermic direction.
• For the Haber process, high pressure and low temperature favour ammonia production, though a compromise temperature is used in industry for kinetic reasons (rate of reaction).

Common Mistakes

• Confusing the direction of the shift with temperature: thinking exothermic means higher temperature gives more product. Remember, higher temperature favours the endothermic direction.
• Miscounting moles of gas: forgetting that N$_2$ and H$_2$ are both gases, giving 4 moles on the left, not 1.
• Assuming the graph must start at 0% NH$_3$ at 0 pressure: at very low pressure, the equilibrium lies far to the left, so % NH$_3$ is near 0. The graphs correctly show this starting point.

Things to Be Careful About

• Always count moles of GAS only when applying the pressure rule. Solids and liquids do not affect the pressure shift.
• Remember that "exothermic" means $\Delta H$ is negative, so the forward reaction releases heat, making the reverse reaction endothermic. Higher temperature favours the endothermic (reverse) reaction.
• Ensure the labels on the curves match the logic: lower temperature (400°C) should give higher yield for an exothermic reaction.