Chemistry 9701/12 — May/June 2025

Background Concept

Melting and boiling points depend on the amount of energy needed to overcome the forces holding particles together.

• Silicon dioxide (SiO2) is a giant covalent network solid. Each silicon atom is bonded to four oxygen atoms and each oxygen atom is bonded to two silicon atoms. Breaking this structure requires breaking many strong covalent bonds, so its melting point is very high.
• Sulfur dioxide (SO2) is a simple molecular substance. Individual SO2 molecules are held together only by weak van der Waals forces, so little energy is needed to separate them. Its melting point is low.

Ammonia and phosphine are both molecular substances, but the intermolecular forces differ:

• Ammonia (NH3) has hydrogen bonding because hydrogen is bonded to nitrogen, a highly electronegative atom with a lone pair. Hydrogen bonding is a relatively strong intermolecular force.
• Phosphine (PH3) has only weak van der Waals forces because phosphorus is not electronegative enough to create a sufficiently polar P–H bond for hydrogen bonding.

Understanding the Question

The question gives two independent statements and asks which are correct. We need to evaluate each statement separately using structure and bonding principles.

Approach

1. Compare the structure of SiO2 with SO2 to judge statement 1.
2. Compare the intermolecular forces in NH3 with PH3 to judge statement 2.
3. Choose the option that matches the correct combination.

Step-by-Step Reasoning

Statement 1:

• SiO2 is a giant covalent network, not a simple molecule.
• Strong covalent bonds must be broken to melt it.
• SO2 is a simple molecular substance with weak intermolecular forces.
• Therefore SiO2 has a higher melting point than SO2.
• Statement 1 is correct.

Statement 2:

• NH3 has hydrogen bonding because N–H bonds are polar and N has a lone pair.
• PH3 has only van der Waals forces because P–H bonds are much less polar.
• Hydrogen bonding is stronger than van der Waals forces.
• Therefore NH3 has a higher boiling point than PH3.
• Statement 2 is correct.

Since both statements are correct, the answer is A.

Key Takeaways

• A giant covalent structure gives a very high melting point because many strong covalent bonds must be broken.
• Simple molecular substances have low melting and boiling points because only weak intermolecular forces are involved.
• Hydrogen bonding occurs when hydrogen is bonded to nitrogen, oxygen, or fluorine.
• A molecule with hydrogen bonding has a higher boiling point than a similar molecule without it.

Common Mistakes

• Assuming all oxides are simple molecular. SiO2 is a giant covalent network solid.
• Thinking PH3 can hydrogen bond just because phosphorus has a lone pair. Hydrogen bonding requires H bonded to N, O, or F, and a sufficiently electronegative atom.

Things to Be Careful About

• Distinguish between giant covalent structures and simple molecular structures.
• Remember that hydrogen bonding is an intermolecular force, not an intramolecular bond.
• Check both statements independently before choosing an option.

Background Concept

An atom is described by two key numbers:

• Proton number (atomic number, $Z$) — the number of protons in the nucleus. This identifies the element.
• Mass number (nucleon number, $A$) — the total number of protons and neutrons in the nucleus.

Since an atom is neutral, the number of electrons equals the number of protons, but the neutrons are neutral particles that contribute almost all the extra mass beyond the protons.

The relationship is:

$$\text{number of neutrons} = A - Z$$

For iron, the proton number is 26. This is a fact worth memorising: iron is element 26, with symbol $\text{Fe}$.

Understanding the Question

The question gives the mass number of an atom of iron as 60 and asks how many neutrons it contains. It does not give the proton number explicitly, so we must recall that iron has atomic number 26. The answer must be a whole number of neutrons, since neutrons are discrete particles.

Four options are given, and only one matches the correct calculation.

Approach

1. Identify the element from its name: iron = $\text{Fe}$, atomic number 26.
2. Use the formula: neutrons = mass number $-$ proton number.
3. Subtract 26 from 60.
4. Select the matching option.

Step-by-Step Reasoning

• Iron is element 26, so every atom of iron has 26 protons.
• The mass number is given as 60. The mass number counts all nucleons: protons + neutrons.
• Therefore:

$$\text{neutrons} = 60 - 26 = 34$$

• This matches option C.

Option A (26) is the proton number, not the neutron number.
Option B (30) would be obtained by incorrectly halving the mass number.
Option D (60) confuses the mass number with the number of neutrons.

Key Takeaways

• Neutrons are found using $A - Z$.
• The mass number is the total of protons and neutrons.
• The atomic number identifies the element and is constant for all atoms of that element.
• Knowing the atomic numbers of common elements such as iron is essential for quick calculation.

Common Mistakes

• Choosing 26: this is the proton number, not the neutron number.
• Choosing 60: this is the mass number, the total of both protons and neutrons.
• Forgetting that neutrons are neutral and do not affect the charge of the atom.

Things to Be Careful About

• Always use the mass number, not the relative atomic mass, when asked for neutrons in a specific isotope.
• The mass number is always a whole number in such questions.
• If the atom is an ion, the number of electrons changes but the number of protons and neutrons does not; here the atom is neutral, so no correction is needed.

Background Concept

Collision theory states that for a reaction to occur, reactant particles must collide with sufficient energy (greater than or equal to the activation energy, $E_{\text{A}}$) and with the correct orientation. The Maxwell-Boltzmann distribution curve shows the distribution of molecular kinetic energies in a gas at a given temperature. The area under the curve to the right of a vertical line representing $E_{\text{A}}$ is proportional to the number (or proportion) of molecules with energy $\ge E_{\text{A}}$, which determines the reaction rate.

Understanding the Question

The question asks to identify which combination of temperature ($T_1$ or $T_2$) and activation energy ($E_{\text{A}(1)}$ or $E_{\text{A}(2)}$) will yield the highest reaction rate, based on the provided Boltzmann distribution graph. We need to maximize the area under the curve to the right of the chosen $E_{\text{A}}$ line.

Approach

We evaluate the two variables independently:

1. Temperature: Compare $T_1$ and $T_2$. The curve for the higher temperature is shifted to the right and has a lower peak (to maintain the same total area, representing the total number of molecules). Thus, $T_2 > T_1$. Higher temperature increases the proportion of molecules with energy $\ge E_{\text{A}}$.
2. Activation energy: Compare $E_{\text{A}(1)}$ and $E_{\text{A}(2)}$. Energy increases from left to right on the x-axis. Thus, $E_{\text{A}(1)} < E_{\text{A}(2)}$. A lower activation energy means a larger proportion of molecules have energy $\ge E_{\text{A}}$.
   The combination that maximizes the successful collision proportion is the highest temperature and the lowest activation energy.

Step-by-Step Reasoning

• Looking at the graph, the curve labeled $T_2$ is flatter and its peak is shifted to the right compared to $T_1$. This indicates that $T_2$ is the higher temperature. At $T_2$, a greater proportion of molecules have higher kinetic energies.
• The vertical line for $E_{\text{A}(1)}$ is to the left of $E_{\text{A}(2)}$, meaning $E_{\text{A}(1)}$ represents a lower activation energy. A lower $E_{\text{A}}$ threshold is easier to overcome, so a larger fraction of molecules will have energy $\ge E_{\text{A}(1)}$ than $\ge E_{\text{A}(2)}$.
• To get the highest reaction rate, we need the maximum proportion of molecules with energy $\ge E_{\text{A}}$. This occurs at the highest temperature ($T_2$) and the lowest activation energy ($E_{\text{A}(1)}$).
• Therefore, the correct combination is $T_2$ and $E_{\text{A}(1)}$, which corresponds to option D.

Key Takeaways

• Higher temperature shifts the Maxwell-Boltzmann curve to the right and lowers the peak, increasing the number of molecules with energy $\ge E_{\text{A}}$.
• Lowering the activation energy (e.g., by using a catalyst) increases the area under the curve to the right of the $E_{\text{A}}$ line, increasing the reaction rate.
• Both higher temperature and lower activation energy independently increase the reaction rate.

Common Mistakes

• Assuming the curve with the higher peak is at a higher temperature: The peak is actually lower at higher temperatures because the curve spreads out to the right, and the total area (total number of molecules) must remain constant.
• Confusing the activation energy values: Remember that energy increases from left to right on the x-axis, so $E_{\text{A}(1)}$ is lower than $E_{\text{A}(2)}$.

Things to Be Careful About

• Always check the x-axis direction: energy increases to the right.
• Remember that the area under the curve is constant for a given amount of gas, so a rightward shift must result in a lower peak.
• The question asks for the highest reaction rate, so choose the combination that maximizes the area to the right of the activation energy line (highest $T$, lowest $E_{\text{A}}$).

Background Concept

This question tests three connected ideas from chemical equilibria:

• Mole fraction and partial pressure: for an ideal gas mixture, the partial pressure of a component is its mole fraction times the total pressure. Comparing partial pressures is therefore equivalent to comparing mole fractions.
• Units of $K_p$: $K_p$ is written in terms of partial pressures, and its units depend on the change in the number of moles of gas, $\Delta n$.
• Le Chatelier's principle: a change in temperature shifts the equilibrium position in the direction that opposes the change; for an exothermic forward reaction, lowering the temperature favours the forward direction.

Understanding the Question

We start with $1.0$ mol of $\text{SO}_2$ and $1.0$ mol of $\text{O}_2$ in a closed container. At equilibrium, $0.8$ mol of $\text{SO}_3$ is present. We must decide which of four statements is correct. The reaction is exothermic ($\Delta H = -197 \text{ kJ mol}^{-1}$), which matters for statement C.

Approach

1. Build an ICE table to find the equilibrium amounts of all three gases.
2. Compare mole fractions (hence partial pressures) of $\text{SO}_2$ and $\text{O}_2$ for statement A.
3. Determine the units of $K_p$ from the stoichiometry for statement B.
4. Apply Le Chatelier's principle to the temperature change for statement C.
5. Calculate the mole fraction of $\text{SO}_3$ for statement D.

Step-by-Step Reasoning

Step 1 — ICE table

Stoichiometry: $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$.

Since $0.8$ mol of $\text{SO}_3$ forms, the change in $\text{SO}_2$ is $-0.8$ mol (2:2 ratio) and the change in $\text{O}_2$ is $-0.4$ mol (2:1 ratio).

Total moles at equilibrium $= 0.2 + 0.6 + 0.8 = 1.6$ mol.

Step 2 — Statement A

Partial pressure is proportional to mole fraction at constant temperature and volume.

Mole fraction of $\text{SO}_2 = \frac{0.2}{1.6} = 0.125$

Mole fraction of $\text{O}_2 = \frac{0.6}{1.6} = 0.375$

So $p(\text{SO}_2) < p(\text{O}_2)$. Statement A is false.

Step 3 — Statement B

$K_p = \frac{(p_{\text{SO}_3})^2}{(p_{\text{SO}_2})^2 \, (p_{\text{O}_2})}$

Units: $\frac{(\text{pressure})^2}{(\text{pressure})^2 \times \text{pressure}} = \text{pressure}^{-1}$ (e.g. $\text{Pa}^{-1}$ or $\text{atm}^{-1}$).

This is not $\text{mol dm}^{-3}$. Statement B is false.

Step 4 — Statement C

The forward reaction is exothermic ($\Delta H < 0$). Lowering the temperature shifts the equilibrium in the exothermic direction — the forward direction — producing more $\text{SO}_3$. So the equilibrium amount of $\text{SO}_3$ would be greater than $0.8$ mol, not lower. Statement C is false.

Step 5 — Statement D

Mole fraction of $\text{SO}_3 = \frac{0.8}{1.6} = 0.5$.

Since $0.5 < 0.8$, statement D is true.

Key Takeaways

• Use an ICE table to track equilibrium amounts; the stoichiometric ratios determine the changes.
• For ideal gases, partial pressure is proportional to mole fraction.
• The units of $K_p$ depend on $\Delta n$ (moles of gaseous products minus moles of gaseous reactants). Here $\Delta n = 2 - 3 = -1$, so $K_p$ has units of $\text{pressure}^{-1}$.
• Le Chatelier's principle: an exothermic forward reaction is favoured by a decrease in temperature.

Common Mistakes

• Wrong O2 change: forgetting that $\text{O}_2$ changes by half the amount of $\text{SO}_3$ formed ($-0.4$ mol, not $-0.8$ mol).
• Confusing mole fraction with moles: the mole fraction of $\text{SO}_3$ is $0.5$, not $0.8$.
• Direction of temperature shift: for an exothermic reaction, lowering temperature favours products, not reactants.

Things to Be Careful About

• Mole fractions are always between 0 and 1, so a mole fraction of $0.5$ is definitely less than $0.8$ — but the calculation is needed to confirm it.
• When determining $K_p$ units, count only gaseous species in $\Delta n$.
• The total pressure is not given, but it is not needed — mole fractions alone suffice for statements A and D.

Background Concept

Oxidation state (oxidation number) is the charge an atom would have if all bonding electrons were assigned to the more electronegative atom. In ionic compounds, the oxidation state equals the actual ionic charge. For a neutral compound, the sum of all oxidation states multiplied by their atom counts must equal zero.

Understanding the Question

The question tells us that 'red lead' has the formula $\text{Pb}_3\text{O}_4$ and that each formula unit contains three lead ions and four oxide ions. It also states that lead exists in two different oxidation states within the compound. We need to determine which two oxidation states those are.

Approach

1. Recognise that oxygen in oxides normally has oxidation state $-2$.
2. Calculate the total negative charge from the four oxide ions.
3. Since the compound is neutral, the three lead ions must balance that charge exactly.
4. Test which pair of oxidation states from the options can sum to the required total with three lead ions.

Step-by-Step Reasoning

1. Charge from oxide ions:
   Each $\text{O}^{2-}$ contributes $-2$. With four oxide ions:
   $4 \times (-2) = -8$

2. Required positive charge:
   For neutrality, the three lead ions must sum to $+8$.

3. Test the options:

   • A (+1 and +2): Maximum possible total with three ions is $2 + 2 + 2 = +6$, which is less than $+8$. Impossible.
   • B (+1 and +3): Maximum total is $3 + 3 + 3 = +9$. Could we reach $+8$? Try $3 + 3 + 1 = 7$, $3 + 3 + 3 = 9$. No combination of $+1$ and $+3$ sums to $+8$ (possible sums: 3, 5, 7, 9). Impossible.
   • C (+2 and +4): Try $4 + 2 + 2 = 8$. This works! Two $\text{Pb}^{2+}$ ions and one $\text{Pb}^{4+}$ ion.
   • D (+2 and +6): Try $6 + 2 + 2 = 10$, $6 + 6 + 2 = 14$, $2 + 2 + 2 = 6$. No combination sums to $+8$ (possible sums: 6, 10, 14). Impossible.

4. Confirm with algebra:
   Let $x$ = number of $\text{Pb}^{2+}$ and $y$ = number of $\text{Pb}^{4+}$.
   $x + y = 3$
   $2x + 4y = 8$
   Substituting $x = 3 - y$:
   $2(3 - y) + 4y = 8 \Rightarrow 6 + 2y = 8 \Rightarrow y = 1$
   So $x = 2$. This confirms two $\text{Pb}^{2+}$ and one $\text{Pb}^{4+}$.

Key Takeaways

• In a neutral compound, the sum of oxidation states (each multiplied by its atom count) equals zero.
• Oxygen almost always has oxidation state $-2$ in oxides.
• When a compound contains the same element in two oxidation states, set up charge-balance equations to find the ratio of ions.

Common Mistakes

• Forgetting that the total charge must be divided among three lead ions, not one.
• Assuming the oxidation states must be consecutive integers; they do not have to be, but here the charge balance forces $+2$ and $+4$.
• Miscounting: some students incorrectly think $\text{Pb}_3\text{O}_4$ means one lead is $+4$ and the other two are $+2$ without checking the charge balance.

Things to Be Careful About

• Always multiply each oxidation state by the number of atoms of that element.
• Check that the chosen pair actually sums to the required total with exactly three ions.
• Remember that this is a neutral compound, so positive and negative charges must cancel exactly.

Background Concept

Hess's law states that the enthalpy change of a reaction is independent of the route taken, provided the initial and final states are the same. This works because enthalpy is a state function — its value depends only on the starting and ending conditions, not on the path between them. In practice, this means we can add, subtract, and scale enthalpy changes of known reactions to obtain the enthalpy change of a reaction that is difficult or impossible to measure directly.

The standard enthalpy change of combustion, $\Delta H_c^\ominus$, is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. The standard enthalpy change of formation, $\Delta H_f^\ominus$, is the enthalpy change when one mole of a compound is formed from its elements in their standard states. Here we are given three combustion enthalpies and asked to find a formation enthalpy — a classic Hess's law application.

Understanding the Question

The table gives:

• $\Delta H_c(\text{C}) = -394 \text{ kJ mol}^{-1}$
• $\Delta H_c(\text{H}_2) = -286 \text{ kJ mol}^{-1}$
• $\Delta H_c(\text{CH}_3\text{OH}) = -726 \text{ kJ mol}^{-1}$

We must find $\Delta H_f$ of methanol, i.e. the enthalpy change for:

$$\text{C(s)} + 2\text{H}_2(\text{g)} + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OH(l)}$$

The four options are $-240$, $-46$, $46$, and $240$ kJ mol$^{-1}$. The sign and magnitude of the answer depend entirely on correctly applying Hess's law with the right stoichiometric coefficients.

Approach

The key insight is that both routes below start from the same reactants and end at the same products:

Route 1: Elements (C, H$_2$, O$_2$) → combustion products (CO$_2$ + H$_2$O)
Route 2: Elements → methanol (formation) → combustion products (CO$_2$ + H$_2$O)

Since both routes connect the same initial and final states, their total enthalpy changes are equal:

$$\Delta H_c(\text{C}) + 2\Delta H_c(\text{H}_2) = \Delta H_f(\text{CH}_3\text{OH}) + \Delta H_c(\text{CH}_3\text{OH})$$

Rearranging gives:

$$\Delta H_f(\text{CH}_3\text{OH}) = \Delta H_c(\text{C}) + 2\Delta H_c(\text{H}_2) - \Delta H_c(\text{CH}_3\text{OH})$$

The factor of 2 on $\Delta H_c(\text{H}_2)$ arises because two moles of H$_2$ are needed to form one mole of CH$_3$OH.

Step-by-Step Reasoning

1. Write the three combustion equations explicitly:

   $$\text{C(s)} + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g)} \quad \Delta H = -394 \text{ kJ mol}^{-1}$$ $$\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O(l)} \quad \Delta H = -286 \text{ kJ mol}^{-1}$$ $$\text{CH}_3\text{OH(l)} + \frac{3}{2}\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g)} + 2\text{H}_2\text{O(l)} \quad \Delta H = -726 \text{ kJ mol}^{-1}$$

2. Combine the combustions of the elements to match the combustion products of methanol:

   Combustion of C gives one CO$_2$; combustion of 2 H$_2$ gives two H$_2$O. Together:

   $$\text{C(s)} + 2\text{H}_2(\text{g)} + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g)} + 2\text{H}_2\text{O(l)}$$

   with $\Delta H = -394 + 2(-286) = -966 \text{ kJ mol}^{-1}$.

3. Reverse the combustion of methanol (flipping the sign):

   $$\text{CO}_2(\text{g)} + 2\text{H}_2\text{O(l)} \rightarrow \text{CH}_3\text{OH(l)} + \frac{3}{2}\text{O}_2(\text{g)} \quad \Delta H = +726 \text{ kJ mol}^{-1}$$

4. Add steps 2 and 3 and cancel common species:

   CO$_2$ and 2H$_2$O cancel on both sides; O$_2$ cancels as $2 - \frac{3}{2} = \frac{1}{2}$:

   $$\text{C(s)} + 2\text{H}_2(\text{g)} + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OH(l)}$$

   with $\Delta H = -966 + 726 = -240 \text{ kJ mol}^{-1}$.

5. Select the option: The value $-240 \text{ kJ mol}^{-1}$ matches option A.

Key Takeaways

• Hess's law lets us calculate formation enthalpies from combustion data without performing the formation reaction directly.

• The general relationship is:

  $$\Delta H_f(\text{compound}) = \sum \Delta H_c(\text{elements}) - \Delta H_c(\text{compound})$$

  with each term scaled by its stoichiometric coefficient.

• Reversing a reaction flips the sign of $\Delta H$; multiplying a reaction by a factor multiplies $\Delta H$ by the same factor.

Common Mistakes

• Forgetting the stoichiometric factor of 2 on $\Delta H_c(\text{H}_2)$ — this would give $-394 - 286 + 726 = +46$, which is option C (wrong sign and wrong magnitude).
• Getting the sign of $\Delta H_c(\text{CH}_3\text{OH})$ wrong — subtracting $-726$ as if it were $+726$ gives $-394 - 572 - 726 = -1692$, which is not among the options but shows the danger of sign errors.
• Confusing formation with combustion — the formation reaction produces one mole of compound from elements, while combustion consumes one mole of compound with oxygen. These are opposite processes in the cycle.
• Dropping the negative signs when substituting values — every value in the table is negative, and each must be substituted with its sign.

Things to Be Careful About

• Always include the stoichiometric coefficient when combining enthalpy changes — the 2 in $2\Delta H_c(\text{H}_2)$ is essential.
• Check the sign of the final answer: formation of methanol is exothermic (negative), consistent with most stable compounds having negative formation enthalpies.
• The units are kJ mol$^{-1}$, referring to the enthalpy change per mole of the reaction as written (one mole of CH$_3$OH formed).
• In an exam, draw the energy cycle or write the equations explicitly to avoid sign errors — it is easy to misplace a minus sign when working purely algebraically.

Background Concept

VSEPR (Valence Shell Electron Pair Repulsion) theory states that electron pairs around a central atom arrange themselves as far apart as possible to minimise repulsion. Both bonding pairs and lone pairs count as electron domains. However, a lone pair is concentrated closer to the central atom and occupies more space, so it repels other electron pairs more strongly than a bonding pair does. This extra repulsion pushes bonding pairs closer together, reducing the bond angle.

The electron-pair geometry describes the arrangement of all electron pairs around the central atom, while the molecular shape describes only the positions of the atoms. For example, both $\text{NH}_3$ and $\text{H}_2\text{O}$ have four electron domains arranged tetrahedrally, but their molecular shapes are trigonal pyramidal and bent respectively.

Understanding the Question

This multiple-choice question asks for the correct order of bond angles from smallest to largest for three molecules: $\text{H}_2\text{O}$, $\text{NH}_3$ and $\text{BF}_3$. Each row in the table gives a different ordering, and only one is correct. The key is to compare the number of lone pairs on the central atom and how they affect the bond angle.

Approach

For each molecule, determine the number of bonding pairs and lone pairs on the central atom. Then use VSEPR to identify the electron-pair geometry and the resulting molecular shape. Finally, compare the bond angles, remembering that more lone pairs on the central atom means greater lone-pair repulsion and therefore a smaller bond angle.

Step-by-Step Reasoning

1. $\text{BF}_3$

   • Boron is in Group 3 and forms three single bonds to fluorine.
   • Central B has 3 bonding pairs and no lone pairs.
   • Electron-pair geometry: trigonal planar.
   • Bond angle: $120^\circ$.

2. $\text{NH}_3$

   • Nitrogen is in Group 5 and forms three single bonds to hydrogen.
   • Central N has 3 bonding pairs and 1 lone pair.
   • The four electron domains point toward the corners of a tetrahedron, but the molecular shape is trigonal pyramidal.
   • The lone pair repels the N–H bonding pairs more strongly than they repel each other, compressing the H–N–H angle from the ideal $109.5^\circ$ to about $107^\circ$.

3. $\text{H}_2\text{O}$

   • Oxygen is in Group 6 and forms two single bonds to hydrogen.
   • Central O has 2 bonding pairs and 2 lone pairs.
   • Again, four electron domains give a tetrahedral electron-pair arrangement, but the molecular shape is bent.
   • Two lone pairs repel the O–H bonding pairs even more, reducing the H–O–H angle to about $104.5^\circ$.

4. Ordering

   • $\text{H}_2\text{O}$: $104.5^\circ$
   • $\text{NH}_3$: $107^\circ$
   • $\text{BF}_3$: $120^\circ$

   So the correct order from smallest to largest bond angle is:

   $\text{H}_2\text{O} < \text{NH}_3 < \text{BF}_3$

   This matches option A.

Key Takeaways

• Bond angle is determined by the total number of electron pairs around the central atom, not just the number of atoms bonded to it.
• Lone pairs compress bond angles because they repel more strongly than bonding pairs.
• The more lone pairs on the central atom, the smaller the bond angle.
• Distinguish between electron-pair geometry and molecular shape: $\text{NH}_3$ is trigonal pyramidal, not tetrahedral; $\text{H}_2\text{O}$ is bent, not tetrahedral.

Common Mistakes

• Choosing option C, which puts $\text{NH}_3$ before $\text{H}_2\text{O}$. This is wrong because $\text{H}_2\text{O}$ has two lone pairs, so its bond angle is smaller than that of $\text{NH}_3$.
• Confusing molecular shape with electron-pair geometry. For example, saying $\text{NH}_3$ is tetrahedral ignores the lone pair and leads to an incorrect bond angle.
• Forgetting that $\text{BF}_3$ has no lone pairs, so its bond angle is the largest at $120^\circ$.

Things to Be Careful About

• Use the accepted approximate bond angles: $\text{H}_2\text{O}$ = $104.5^\circ$, $\text{NH}_3$ = $107^\circ$, $\text{BF}_3$ = $120^\circ$.
• Count lone pairs only on the central atom; lone pairs on terminal atoms do not affect the central bond angle in the same way.
• In VSEPR, a double or triple bond is treated as one electron domain, although this is not needed for these three molecules.
• When comparing bond angles, always reason from lone-pair repulsion rather than from the number of bonded atoms alone.

Background Concept

The Haber process is the industrial synthesis of ammonia from nitrogen and hydrogen. It is an equilibrium reaction carried out at high temperature and pressure, and it uses a heterogeneous catalyst. The catalyst is finely divided iron, often promoted with oxides of potassium and aluminium. The balanced equation is:

$\mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightleftharpoons 2\mathrm{NH_3}(g)$

The mole ratio of the two gaseous reactants is therefore 1 : 3 (one mole of nitrogen to three moles of hydrogen).

Understanding the Question

The question asks for two pieces of information:

1. Which transition metal is used as the catalyst in the Haber process?
2. In what mole ratio do the two gaseous reactants (nitrogen and hydrogen) react?

The options pair a metal (Fe or V) with a mole ratio (1 : 2 or 1 : 3).

Approach

Recall the Haber process equation and the identity of its catalyst. The catalyst is iron, and the stoichiometric ratio of N₂ to H₂ is 1 : 3.

Step-by-Step Reasoning

1. The Haber process reaction is:
   $\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$
2. The catalyst is iron (Fe). Vanadium(V) oxide is used in the Contact process for sulfuric acid, not in the Haber process.
3. From the balanced equation, one mole of nitrogen reacts with three moles of hydrogen, so the mole ratio is 1 : 3.
4. This matches option B.

Key Takeaways

The Haber process catalyst is iron. The reactant mole ratio is always N₂ : H₂ = 1 : 3. This is a fundamental fact tested in AS chemistry.

Common Mistakes

• Confusing the Haber process catalyst (Fe) with the Contact process catalyst (V₂O₅).
• Misreading the mole ratio as 1 : 2, which is the ratio of nitrogen to ammonia (1 : 2) rather than nitrogen to hydrogen.

Things to Be Careful About

The question asks for the ratio between the two reactants, not between a reactant and the product. Always read the wording carefully.

Things to Be Careful About

• The catalyst is written as Fe, not Fe₂O₃ or any other compound.
• The mole ratio must be expressed in the order given in the options: metal first, then ratio. Here the ratio is N₂ : H₂ = 1 : 3.

Background Concept

The rate of a reaction is the change in concentration of a reactant or product per unit time. For a reactant being consumed, the rate is:
$\text{rate} = -\frac{\Delta[\text{reactant}]}{\Delta t}$
Here we want the average rate of decomposition of hydrogen peroxide over the 2.0-minute period, so we need the total change in concentration of H₂O₂ divided by the total time.

Understanding the Question

• We have 50 cm³ of 0.50 mol dm⁻³ H₂O₂ solution.
• 120 cm³ of O₂ gas is produced in 2.0 minutes.
• The gas volume is measured at room conditions, so the molar gas volume is 24 dm³ mol⁻¹ (24000 cm³ mol⁻¹).
• We need the average rate of decomposition of H₂O₂ in mol dm⁻³ s⁻¹.

Approach

1. Convert the volume of O₂ to moles using the molar gas volume at room conditions.
2. Use the balanced equation 2H₂O₂ → 2H₂O + O₂ to find the moles of H₂O₂ decomposed (2 mol H₂O₂ per 1 mol O₂).
3. Convert moles of H₂O₂ to a concentration change by dividing by the solution volume in dm³.
4. Divide by the time in seconds to obtain the rate.

Step-by-Step Reasoning

1. Moles of O₂ = 120 cm³ ÷ 24000 cm³ mol⁻¹ = 0.005 mol.
2. From the stoichiometry, moles of H₂O₂ decomposed = 2 × 0.005 = 0.010 mol.
3. Δ[H₂O₂] = 0.010 mol ÷ 0.050 dm³ = 0.20 mol dm⁻³.
4. Time = 2.0 min = 120 s.
5. Average rate = 0.20 mol dm⁻³ ÷ 120 s = 0.00167 mol dm⁻³ s⁻¹ ≈ 0.0017 mol dm⁻³ s⁻¹.

This matches option C.

Key Takeaways

• Always convert gas volumes to moles using the appropriate molar gas volume (24 dm³ mol⁻¹ at room conditions).
• Always apply the stoichiometric ratio from the balanced equation — forgetting it is a classic error.
• Convert all units consistently: cm³ → dm³, minutes → seconds.
• The rate of decomposition of H₂O₂ is twice the rate of formation of O₂ because of the 2:1 mole ratio.

Common Mistakes

• Forgetting the 2:1 stoichiometric ratio and using moles of O₂ directly as moles of H₂O₂ → gives 0.00083 mol dm⁻³ s⁻¹ (option B).
• Using the molar gas volume at STP (22.4 dm³ mol⁻¹) instead of room conditions (24 dm³ mol⁻¹).
• Confusing cm³ and dm³ when converting volume.
• Not converting minutes to seconds.

Things to Be Careful About

• "Room conditions" is a specific term — it means 25 °C and 1 atm, where 1 mol of gas occupies 24 dm³.
• The volume of the solution (50 cm³) is used to convert moles to concentration, not the gas volume.
• The rate is asked in mol dm⁻³ s⁻¹, so time must be in seconds.

Background Concept

Enthalpy changes accompany phase changes. Converting a solid to a liquid (fusion) and a liquid to a gas (vaporisation) both require energy input to overcome intermolecular forces, so both $\Delta H_\text{m}$ and $\Delta H_\text{b}$ are positive (endothermic).

The magnitude of the enthalpy change reflects how much intermolecular force must be overcome. Vaporisation requires breaking essentially all intermolecular attractions, while fusion only needs to loosen the ordered structure enough to allow flow. Hence $\Delta H_\text{b}$ is much larger than $\Delta H_\text{m}$.

Water and ice are both composed of H$_2$O molecules held together by hydrogen bonds (plus weaker van der Waals forces). The type of intermolecular force is the same in both phases.

Understanding the Question

The question gives two equilibria:

• $\text{H}_2\text{O(l)} \rightleftharpoons \text{H}_2\text{O(g)}$ with $\Delta H_\text{b}$
• $\text{H}_2\text{O(s)} \rightleftharpoons \text{H}_2\text{O(l)}$ with $\Delta H_\text{m}$

We must judge which of three statements about these enthalpy changes and the intermolecular forces are correct.

Approach

Evaluate each statement independently using the sign and magnitude of the enthalpy changes, and the nature of the intermolecular forces.

Step-by-Step Reasoning

1. Statement 1: Both $\Delta H_\text{b}$ and $\Delta H_\text{m}$ are negative.

   • Melting and boiling both require energy input. Therefore both enthalpy changes are positive, not negative. Statement 1 is false.

2. Statement 2: $\Delta H_\text{b}$ is greater than $\Delta H_\text{m}$.

   • Boiling breaks all intermolecular forces; melting only weakens them. So $\Delta H_\text{b}$ is considerably larger than $\Delta H_\text{m}$. Statement 2 is true.

3. Statement 3: The intermolecular forces in ice and water are of the same type.

   • Both phases consist of H$_2$O molecules with hydrogen bonding and van der Waals forces. The type is the same. Statement 3 is true.

Thus only statements 2 and 3 are correct, matching option D.

Key Takeaways

• Phase changes that require energy input are endothermic ($\Delta H > 0$).
• Vaporisation enthalpy > fusion enthalpy because more intermolecular force is overcome.
• Intermolecular force type does not change with phase; only the extent of interaction changes.

Common Mistakes

• Assuming that because ice is cold, melting must be exothermic — melting always requires energy.
• Confusing the sign convention: endothermic processes have positive $\Delta H$.
• Thinking that ice has a different type of bonding than water; it is the same hydrogen bonding, just more ordered.

Things to Be Careful About

• Read the arrow direction: the equilibria are written solid → liquid and liquid → gas, which are the endothermic directions.
• Distinguish between the magnitude and the sign of enthalpy changes.
• Remember that hydrogen bonding exists in both ice and liquid water; the difference is the degree of organisation, not the type of force.

Background Concept

A hydrated salt contains water molecules as water of crystallisation. When heated, these water molecules are driven off, leaving the anhydrous salt. In $\text{LiOH}\cdot x\text{H}_2\text{O}$, each formula unit contains one $\text{LiOH}$ and $x$ water molecules. Because the heating causes complete dehydration and no decomposition, the amount of $\text{LiOH}$ is unchanged: all the $\text{LiOH}$ present before heating is still present after heating. The value of $x$ is the mole ratio of water to $\text{LiOH}$:

$x = \frac{\text{moles of }\text{H}_2\text{O}}{\text{moles of }\text{LiOH}}$

The molar mass of $\text{LiOH}$ is approximately $23.9\ \text{g mol}^{-1}$ and that of water is $18\ \text{g mol}^{-1}$.

Understanding the Question

You are given three crucible masses: $Q$ (empty crucible), $R$ (crucible + hydrated salt), and $S$ (crucible + anhydrous salt after heating). The question asks which algebraic expression gives $x$. The key is to identify which mass differences represent the hydrate, the anhydrous salt, and the water lost, then convert those masses to moles.

Approach

1. Find the mass of the hydrated sample from $R - Q$.
2. Find the mass of anhydrous $\text{LiOH}$ from $S - Q$.
3. Find the mass of water lost as the difference between the hydrate and the residue: $R - S$.
4. Convert the masses of $\text{LiOH}$ and $\text{H}_2\text{O}$ to moles using their molar masses.
5. Divide moles of water by moles of $\text{LiOH}$ to obtain $x$.

Step-by-Step Reasoning

Mass of hydrated $\text{LiOH}\cdot x\text{H}_2\text{O}$:

$m_{\text{hydrate}} = R - Q$

Mass of anhydrous $\text{LiOH}$ after heating:

$m_{\text{LiOH}} = S - Q$

Mass of water lost:

$m_{\text{H}_2\text{O}} = (R - Q) - (S - Q) = R - S$

Moles of anhydrous $\text{LiOH}$:

$n_{\text{LiOH}} = \frac{S - Q}{23.9}$

Moles of water lost:

$n_{\text{H}_2\text{O}} = \frac{R - S}{18}$

Therefore:

$x = \frac{n_{\text{H}_2\text{O}}}{n_{\text{LiOH}}} = \frac{\frac{R-S}{18}}{\frac{S-Q}{23.9}} = \frac{23.9(R-S)}{18(S-Q)}$

This matches option C.

Option A, $\frac{S-R}{18}$, is negative because the residue is lighter than the hydrate; water lost is $R - S$, not $S - R$. Option B gives only the moles of water and ignores the anhydrous $\text{LiOH}$, so it cannot give $x$. Option D uses the wrong mass difference $S - R$ and also has an incorrect numerator sign.

Key Takeaways

• Water of crystallisation is found from the mass loss on heating.
• Always use mole ratios, not mass ratios, to find $x$.
• Subtract the empty crucible mass to obtain the mass of each solid sample.
• The molar masses of the anhydrous compound and water must both be used.

Common Mistakes

• Using $S - R$ instead of $R - S$ for the mass of water lost.
• Using $R - Q$ (the hydrated sample mass) in the denominator instead of $S - Q$ (the anhydrous mass).
• Forgetting to include the molar mass of $\text{LiOH}$.
• Treating $x$ as a mass ratio rather than a mole ratio.

Things to Be Careful About

• Molar mass of $\text{LiOH}$: $6.9 + 16.0 + 1.0 = 23.9\ \text{g mol}^{-1}$; molar mass of water is $18\ \text{g mol}^{-1}$.
• Complete dehydration means no decomposition; if decomposition occurred, the residue would not be pure $\text{LiOH}$ and the calculation would be invalid.
• Units cancel in the ratio, so $x$ has no unit.
• Check the denominator carefully: it is $18(S-Q)$, not $18(R-Q)$.

Background Concept

The ideal gas equation is $pV = nRT$, where $p$ is pressure, $V$ is volume, $n$ is amount in moles, $R$ is the gas constant and $T$ is absolute temperature. For an ideal gas, this equation is exact. In SI units, $R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$, so pressure must be in $\text{Pa}$, volume in $\text{m}^3$ and temperature in $\text{K}$. The equation is usually rearranged as $n = \frac{pV}{RT}$ when the amount is required.

Understanding the Question

The question gives a volume of $240\text{ cm}^3$, a temperature of $37^\circ\text{C}$ and a pressure of $100\text{ kPa}$, and asks for the number of moles of ideal gas. The chemistry is simple; the trap is that the data are not in SI units. You must convert before substituting.

Approach

Convert each quantity to SI units: $100\text{ kPa} = 1.00 \times 10^5\text{ Pa}$; $240\text{ cm}^3 = 240 \times 10^{-6}\text{ m}^3$; $37^\circ\text{C} = 310\text{ K}$. Then substitute into $n = \frac{pV}{RT}$ and compare with the options.

Step-by-Step Reasoning

1. Pressure: $100\text{ kPa} = 100 \times 10^3\text{ Pa} = 1.00 \times 10^5\text{ Pa}$.
2. Volume: $1\text{ cm}^3 = 10^{-6}\text{ m}^3$, so $240\text{ cm}^3 = 240 \times 10^{-6}\text{ m}^3 = 2.40 \times 10^{-4}\text{ m}^3$.
3. Temperature: $T/\text{K} = 37 + 273 = 310$.
4. Rearrange $pV = nRT$: $n = \frac{pV}{RT}$.
5. Substitute:

$$n = \frac{(1.00 \times 10^5)(240 \times 10^{-6})}{8.31 \times 310} = \frac{24}{2576.1} = 9.32 \times 10^{-3}\text{ mol}.$$

This matches option B.

Key Takeaways

The ideal gas equation is only used correctly when all quantities are in SI units. Always convert $^\circ\text{C}$ to $\text{K}$, $\text{kPa}$ to $\text{Pa}$, and $\text{cm}^3$ to $\text{m}^3$ before using $R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$. The same equation can be rearranged to find $p$, $V$ or $T$.

Common Mistakes

• Using the Celsius temperature directly: $T = 37\text{ K}$ gives $n = 78.1\text{ mol}$, which is option D. The Kelvin scale must always be used.
• Leaving the volume in $\text{cm}^3$: if the $10^{-6}$ conversion is missed, the numerical answer is too large by a factor of $10^6$; if the conversion is applied incorrectly as $10^{-9}$, the answer becomes $9.32 \times 10^{-6}\text{ mol}$, option A.
• Forgetting to convert $\text{kPa}$ to $\text{Pa}$ while using $R$ in SI units: this introduces a factor of $1000$.
• Mixing unit systems, such as using $\text{kPa}$ with $\text{cm}^3$ and $R = 8.31$, without adjusting the units of $R$.

Things to Be Careful About

• Use $273$ (or $273.15$) when converting Celsius to Kelvin; either gives the same answer to three significant figures here.
• Keep the answer in standard form with the correct unit: $9.32 \times 10^{-3}\text{ mol}$.
• The gas is described as ideal, so no correction for intermolecular forces or molecular volume is needed; the simple equation applies.
• If a question gives volume in $\text{dm}^3$, remember $1\text{ dm}^3 = 10^{-3}\text{ m}^3 = 1000\text{ cm}^3$; here the volume is in $\text{cm}^3$.