IB · CHEMISTRY HL

Chemistry HL

Reactivity 1: entropy and spontaneity (HL) — Reactivity 1 HL

Name: ____________________Date: October 10, 2026

1.
Define lattice enthalpy, and state whether its value is always exothermic or endothermic when defined as the enthalpy change for forming a solid ionic lattice from its gaseous ions.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: Defines lattice enthalpy as the enthalpy change when one mole of a solid ionic compound is formed from its constituent ions in the gaseous state. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that this process is always exothermic (negative), since oppositely charged gaseous ions attract and release energy as the lattice forms. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Watch the sign convention: IB defines lattice enthalpy as formation from gaseous ions (exothermic), the reverse of lattice dissociation enthalpy (endothermic) — check which definition a question uses.
Marking points
- Defines lattice enthalpy as the enthalpy change when one mole of a solid ionic compound is formed from its constituent ions in the gaseous state.
- States that this process is always exothermic (negative), since oppositely charged gaseous ions attract and release energy as the lattice forms.
Examiner tip: Watch the sign convention: IB defines lattice enthalpy as formation from gaseous ions (exothermic), the reverse of lattice dissociation enthalpy (endothermic) — check which definition a question uses.
2.
Marking analysis: A learner attempts the following task: “Define lattice enthalpy, and state whether its value is always exothermic or endothermic when defined as the enthalpy change for forming a solid ionic lattice from its gaseous ions.” Their response addresses only this point: “Defines lattice enthalpy as the enthalpy change when one mole of a solid ionic compound is formed from its constituent ions in the gaseous state.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Defines lattice enthalpy as the enthalpy change when one mole of a solid ionic compound is formed from its constituent ions in the gaseous state. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that this process is always exothermic (negative), since oppositely charged gaseous ions attract and release energy as the lattice forms. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Defines lattice enthalpy as the enthalpy change when one mole of a solid ionic compound is formed from its constituent ions in the gaseous state.
- Identifies the missing requirement: States that this process is always exothermic (negative), since oppositely charged gaseous ions attract and release energy as the lattice forms.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
3.
Construct a Born-Haber cycle for the formation of sodium chloride, NaCl(s), from its elements, listing the enthalpy terms needed: atomization of sodium, atomization of chlorine, first ionization energy of sodium, first electron affinity of chlorine, and lattice enthalpy.
[5 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: Includes enthalpy of atomization of sodium: Na(s) → Na(g). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: Includes enthalpy of atomization (half of bond dissociation) of chlorine: ½Cl₂(g) → Cl(g). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: Includes first ionization energy of sodium: Na(g) → Na⁺(g) + e⁻. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Develop this part of the answer: Includes first electron affinity of chlorine: Cl(g) + e⁻ → Cl⁻(g). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
6. Develop this part of the answer: Includes lattice enthalpy of formation: Na⁺(g) + Cl⁻(g) → NaCl(s), closing the cycle to the enthalpy of formation of NaCl(s). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A Born-Haber cycle is Hess's law applied to ionic compound formation; the enthalpy of formation always equals the sum of every other step around the closed cycle.
Marking points
- Includes enthalpy of atomization of sodium: Na(s) → Na(g).
- Includes enthalpy of atomization (half of bond dissociation) of chlorine: ½Cl₂(g) → Cl(g).
- Includes first ionization energy of sodium: Na(g) → Na⁺(g) + e⁻.
- Includes first electron affinity of chlorine: Cl(g) + e⁻ → Cl⁻(g).
- Includes lattice enthalpy of formation: Na⁺(g) + Cl⁻(g) → NaCl(s), closing the cycle to the enthalpy of formation of NaCl(s).
Examiner tip: A Born-Haber cycle is Hess's law applied to ionic compound formation; the enthalpy of formation always equals the sum of every other step around the closed cycle.
4.
Marking analysis: A learner attempts the following task: “Construct a Born-Haber cycle for the formation of sodium chloride, NaCl(s), from its elements, listing the enthalpy terms needed: atomization of sodium, atomization of chlorine, first ionization energy of sodium, first electron affinity of chlorine, and lattice enthalpy.” Their response addresses only this point: “Includes enthalpy of atomization of sodium: Na(s) → Na(g).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Includes enthalpy of atomization of sodium: Na(s) → Na(g). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Includes enthalpy of atomization (half of bond dissociation) of chlorine: ½Cl₂(g) → Cl(g). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Includes first ionization energy of sodium: Na(g) → Na⁺(g) + e⁻. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Includes first electron affinity of chlorine: Cl(g) + e⁻ → Cl⁻(g). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
6. Requirement 5: Identifies the missing requirement: Includes lattice enthalpy of formation: Na⁺(g) + Cl⁻(g) → NaCl(s), closing the cycle to the enthalpy of formation of NaCl(s). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Includes enthalpy of atomization of sodium: Na(s) → Na(g).
- Identifies the missing requirement: Includes enthalpy of atomization (half of bond dissociation) of chlorine: ½Cl₂(g) → Cl(g).
- Identifies the missing requirement: Includes first ionization energy of sodium: Na(g) → Na⁺(g) + e⁻.
- Identifies the missing requirement: Includes first electron affinity of chlorine: Cl(g) + e⁻ → Cl⁻(g).
- Identifies the missing requirement: Includes lattice enthalpy of formation: Na⁺(g) + Cl⁻(g) → NaCl(s), closing the cycle to the enthalpy of formation of NaCl(s).
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
5.
Using the Born-Haber cycle for NaCl (ΔHf = −411, atomization Na = +107, atomization Cl = +122, IE1 Na = +496, EA1 Cl = −349, all in kJ mol⁻¹), calculate the lattice enthalpy of NaCl.
[4 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: States Hess's law relation: ΔHf = ΔHatom(Na) + ΔHatom(Cl) + IE1(Na) + EA1(Cl) + ΔHlattice. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Sums the known terms: 107 + 122 + 496 − 349 = 376 kJ mol⁻¹. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Rearranges: ΔHlattice = ΔHf − 376 = −411 − 376. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Work through this mathematical step: Obtains ΔHlattice = −787 kJ mol⁻¹. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Set up the algebraic sum of all known steps equal to the enthalpy of formation first, then solve for the one unknown — this avoids sign errors from trying to trace arrows visually.
Marking points
- States Hess's law relation: ΔHf = ΔHatom(Na) + ΔHatom(Cl) + IE1(Na) + EA1(Cl) + ΔHlattice.
- Sums the known terms: 107 + 122 + 496 − 349 = 376 kJ mol⁻¹.
- Rearranges: ΔHlattice = ΔHf − 376 = −411 − 376.
- Obtains ΔHlattice = −787 kJ mol⁻¹.
Examiner tip: Set up the algebraic sum of all known steps equal to the enthalpy of formation first, then solve for the one unknown — this avoids sign errors from trying to trace arrows visually.
6.
Marking analysis: A learner attempts the following task: “Using the Born-Haber cycle for NaCl (ΔHf = −411, atomization Na = +107, atomization Cl = +122, IE1 Na = +496, EA1 Cl = −349, all in kJ mol⁻¹), calculate the lattice enthalpy of NaCl.” Their response addresses only this point: “States Hess's law relation: ΔHf = ΔHatom(Na) + ΔHatom(Cl) + IE1(Na) + EA1(Cl) + ΔHlattice.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States Hess's law relation: ΔHf = ΔHatom(Na) + ΔHatom(Cl) + IE1(Na) + EA1(Cl) + ΔHlattice. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Sums the known terms: 107 + 122 + 496 − 349 = 376 kJ mol⁻¹. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Rearranges: ΔHlattice = ΔHf − 376 = −411 − 376. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains ΔHlattice = −787 kJ mol⁻¹. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States Hess's law relation: ΔHf = ΔHatom(Na) + ΔHatom(Cl) + IE1(Na) + EA1(Cl) + ΔHlattice.
- Identifies the missing requirement: Sums the known terms: 107 + 122 + 496 − 349 = 376 kJ mol⁻¹.
- Identifies the missing requirement: Rearranges: ΔHlattice = ΔHf − 376 = −411 − 376.
- Identifies the missing requirement: Obtains ΔHlattice = −787 kJ mol⁻¹.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
7.
State and explain the second law of thermodynamics in terms of the entropy of the universe.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that the entropy of the universe (system plus surroundings) increases in any spontaneous process. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: Explains that this is because spontaneous processes move towards states of greater disorder or greater number of possible microstates overall. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Entropy of the system alone can decrease in a spontaneous process, as long as the entropy of the surroundings increases by more — always consider the total.
Marking points
- States that the entropy of the universe (system plus surroundings) increases in any spontaneous process.
- Explains that this is because spontaneous processes move towards states of greater disorder or greater number of possible microstates overall.
Examiner tip: Entropy of the system alone can decrease in a spontaneous process, as long as the entropy of the surroundings increases by more — always consider the total.
8.
Marking analysis: A learner attempts the following task: “State and explain the second law of thermodynamics in terms of the entropy of the universe.” Their response addresses only this point: “States that the entropy of the universe (system plus surroundings) increases in any spontaneous process.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that the entropy of the universe (system plus surroundings) increases in any spontaneous process. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Explains that this is because spontaneous processes move towards states of greater disorder or greater number of possible microstates overall. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that the entropy of the universe (system plus surroundings) increases in any spontaneous process.
- Identifies the missing requirement: Explains that this is because spontaneous processes move towards states of greater disorder or greater number of possible microstates overall.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
9.
Predict, with a reason, whether the entropy change is positive or negative for the reaction: CaCO₃(s) → CaO(s) + CO₂(g).
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that entropy increases (positive ΔS). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: Explains that a gas (CO₂) is produced from a solid reactant, and gases have far greater disorder/more accessible microstates than solids, so the total number of ways to arrange the system increases. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A quick entropy check: count moles of gas on each side. An increase in moles of gas almost always means a positive entropy change.
Marking points
- States that entropy increases (positive ΔS).
- Explains that a gas (CO₂) is produced from a solid reactant, and gases have far greater disorder/more accessible microstates than solids, so the total number of ways to arrange the system increases.
Examiner tip: A quick entropy check: count moles of gas on each side. An increase in moles of gas almost always means a positive entropy change.
10.
Marking analysis: A learner attempts the following task: “Predict, with a reason, whether the entropy change is positive or negative for the reaction: CaCO₃(s) → CaO(s) + CO₂(g).” Their response addresses only this point: “States that entropy increases (positive ΔS).” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that entropy increases (positive ΔS). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Explains that a gas (CO₂) is produced from a solid reactant, and gases have far greater disorder/more accessible microstates than solids, so the total number of ways to arrange the system increases. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that entropy increases (positive ΔS).
- Identifies the missing requirement: Explains that a gas (CO₂) is produced from a solid reactant, and gases have far greater disorder/more accessible microstates than solids, so the total number of ways to arrange the system increases.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
11.
State the Gibbs free energy equation and explain, using this equation, why a reaction with a positive ΔH and a positive ΔS can become spontaneous at high temperature.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Work through this mathematical step: States ΔG = ΔH − TΔS. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Develop this part of the answer: States that a reaction is spontaneous when ΔG is negative. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: Explains that as T increases, the TΔS term (positive, since ΔS is positive) grows large enough to outweigh the positive ΔH, making ΔG negative overall. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Use the sign combinations of ΔH and ΔS to predict spontaneity: (−,+) always spontaneous, (+,−) never spontaneous, (+,+) spontaneous only at high T, (−,−) spontaneous only at low T.
Marking points
- States ΔG = ΔH − TΔS.
- States that a reaction is spontaneous when ΔG is negative.
- Explains that as T increases, the TΔS term (positive, since ΔS is positive) grows large enough to outweigh the positive ΔH, making ΔG negative overall.
Examiner tip: Use the sign combinations of ΔH and ΔS to predict spontaneity: (−,+) always spontaneous, (+,−) never spontaneous, (+,+) spontaneous only at high T, (−,−) spontaneous only at low T.
12.
Marking analysis: A learner attempts the following task: “State the Gibbs free energy equation and explain, using this equation, why a reaction with a positive ΔH and a positive ΔS can become spontaneous at high temperature.” Their response addresses only this point: “States ΔG = ΔH − TΔS.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States ΔG = ΔH − TΔS. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that a reaction is spontaneous when ΔG is negative. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Explains that as T increases, the TΔS term (positive, since ΔS is positive) grows large enough to outweigh the positive ΔH, making ΔG negative overall. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States ΔG = ΔH − TΔS.
- Identifies the missing requirement: States that a reaction is spontaneous when ΔG is negative.
- Identifies the missing requirement: Explains that as T increases, the TΔS term (positive, since ΔS is positive) grows large enough to outweigh the positive ΔH, making ΔG negative overall.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
13.
A reaction has ΔH = +58 kJ mol⁻¹ and ΔS = +176 J K⁻¹ mol⁻¹. Calculate the minimum temperature (in K) at which the reaction becomes spontaneous.
[4 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: States that the reaction becomes spontaneous when ΔG = 0, i.e. the boundary condition ΔH = TΔS. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Develop this part of the answer: Converts ΔH to consistent units: 58 000 J mol⁻¹. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Work through this mathematical step: Rearranges to T = ΔH / ΔS = 58 000 / 176. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Develop this part of the answer: Obtains T ≈ 330 K. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Always convert ΔH (kJ) and ΔS (J) to the same energy unit before dividing — a very common error is forgetting the factor of 1000.
Marking points
- States that the reaction becomes spontaneous when ΔG = 0, i.e. the boundary condition ΔH = TΔS.
- Converts ΔH to consistent units: 58 000 J mol⁻¹.
- Rearranges to T = ΔH / ΔS = 58 000 / 176.
- Obtains T ≈ 330 K.
Examiner tip: Always convert ΔH (kJ) and ΔS (J) to the same energy unit before dividing — a very common error is forgetting the factor of 1000.
14.
Marking analysis: A learner attempts the following task: “A reaction has ΔH = +58 kJ mol⁻¹ and ΔS = +176 J K⁻¹ mol⁻¹. Calculate the minimum temperature (in K) at which the reaction becomes spontaneous.” Their response addresses only this point: “States that the reaction becomes spontaneous when ΔG = 0, i.e. the boundary condition ΔH = TΔS.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that the reaction becomes spontaneous when ΔG = 0, i.e. the boundary condition ΔH = TΔS. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Converts ΔH to consistent units: 58 000 J mol⁻¹. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Rearranges to T = ΔH / ΔS = 58 000 / 176. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains T ≈ 330 K. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that the reaction becomes spontaneous when ΔG = 0, i.e. the boundary condition ΔH = TΔS.
- Identifies the missing requirement: Converts ΔH to consistent units: 58 000 J mol⁻¹.
- Identifies the missing requirement: Rearranges to T = ΔH / ΔS = 58 000 / 176.
- Identifies the missing requirement: Obtains T ≈ 330 K.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
15.
Explain why lattice enthalpies calculated from a Born-Haber cycle (experimental) sometimes differ from those calculated theoretically assuming a purely ionic model.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that the theoretical (purely ionic) model assumes the ions are perfect spheres with no distortion of electron clouds. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that in reality, the cation can polarize (distort) the anion's electron cloud, introducing a degree of covalent character into the bonding. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: States that this extra covalent character makes the experimental (Born-Haber) lattice enthalpy more exothermic (larger magnitude) than the purely ionic theoretical value. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A large discrepancy between experimental and theoretical lattice enthalpy is itself evidence of significant covalent character in an otherwise 'ionic' compound.
Marking points
- States that the theoretical (purely ionic) model assumes the ions are perfect spheres with no distortion of electron clouds.
- States that in reality, the cation can polarize (distort) the anion's electron cloud, introducing a degree of covalent character into the bonding.
- States that this extra covalent character makes the experimental (Born-Haber) lattice enthalpy more exothermic (larger magnitude) than the purely ionic theoretical value.
Examiner tip: A large discrepancy between experimental and theoretical lattice enthalpy is itself evidence of significant covalent character in an otherwise 'ionic' compound.
16.
Marking analysis: A learner attempts the following task: “Explain why lattice enthalpies calculated from a Born-Haber cycle (experimental) sometimes differ from those calculated theoretically assuming a purely ionic model.” Their response addresses only this point: “States that the theoretical (purely ionic) model assumes the ions are perfect spheres with no distortion of electron clouds.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that the theoretical (purely ionic) model assumes the ions are perfect spheres with no distortion of electron clouds. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that in reality, the cation can polarize (distort) the anion's electron cloud, introducing a degree of covalent character into the bonding. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that this extra covalent character makes the experimental (Born-Haber) lattice enthalpy more exothermic (larger magnitude) than the purely ionic theoretical value. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that the theoretical (purely ionic) model assumes the ions are perfect spheres with no distortion of electron clouds.
- Identifies the missing requirement: States that in reality, the cation can polarize (distort) the anion's electron cloud, introducing a degree of covalent character into the bonding.
- Identifies the missing requirement: States that this extra covalent character makes the experimental (Born-Haber) lattice enthalpy more exothermic (larger magnitude) than the purely ionic theoretical value.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
17.
Explain, in terms of entropy, why ice melts spontaneously above 0°C but not below 0°C, given that ΔH for melting is positive (endothermic) at all temperatures.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that melting has a positive ΔS, since the liquid state is more disordered than the solid state. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Work through this mathematical step: States that ΔG = ΔH − TΔS, and since ΔH and ΔS are both positive, ΔG becomes negative only when T is large enough for TΔS to exceed ΔH. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: States that above 0°C (273 K), T is large enough for melting to be spontaneous (ΔG < 0); below 0°C it is not (ΔG > 0), consistent with 0°C being the equilibrium melting point where ΔG = 0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The melting/boiling point of a substance is exactly the temperature at which ΔG = 0 for that phase change, i.e. T = ΔH/ΔS.
Marking points
- States that melting has a positive ΔS, since the liquid state is more disordered than the solid state.
- States that ΔG = ΔH − TΔS, and since ΔH and ΔS are both positive, ΔG becomes negative only when T is large enough for TΔS to exceed ΔH.
- States that above 0°C (273 K), T is large enough for melting to be spontaneous (ΔG < 0); below 0°C it is not (ΔG > 0), consistent with 0°C being the equilibrium melting point where ΔG = 0.
Examiner tip: The melting/boiling point of a substance is exactly the temperature at which ΔG = 0 for that phase change, i.e. T = ΔH/ΔS.
18.
Marking analysis: A learner attempts the following task: “Explain, in terms of entropy, why ice melts spontaneously above 0°C but not below 0°C, given that ΔH for melting is positive (endothermic) at all temperatures.” Their response addresses only this point: “States that melting has a positive ΔS, since the liquid state is more disordered than the solid state.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that melting has a positive ΔS, since the liquid state is more disordered than the solid state. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that ΔG = ΔH − TΔS, and since ΔH and ΔS are both positive, ΔG becomes negative only when T is large enough for TΔS to exceed ΔH. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that above 0°C (273 K), T is large enough for melting to be spontaneous (ΔG < 0); below 0°C it is not (ΔG > 0), consistent with 0°C being the equilibrium melting point where ΔG = 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that melting has a positive ΔS, since the liquid state is more disordered than the solid state.
- Identifies the missing requirement: States that ΔG = ΔH − TΔS, and since ΔH and ΔS are both positive, ΔG becomes negative only when T is large enough for TΔS to exceed ΔH.
- Identifies the missing requirement: States that above 0°C (273 K), T is large enough for melting to be spontaneous (ΔG < 0); below 0°C it is not (ΔG > 0), consistent with 0°C being the equilibrium melting point where ΔG = 0.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
19.
Outline one limitation of using standard enthalpy of formation values alone (without considering entropy) to predict whether a reaction will occur spontaneously.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that enthalpy alone does not account for the entropy change of the reaction. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that some reactions with unfavourable (positive) ΔH can still be spontaneous if ΔS is sufficiently positive and TΔS outweighs ΔH, meaning enthalpy alone gives an incomplete or incorrect prediction of spontaneity. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Spontaneity is always determined by ΔG, not ΔH alone; ΔH is only a reliable spontaneity predictor when ΔS is close to zero or its effect is negligible at the temperature considered.
Marking points
- States that enthalpy alone does not account for the entropy change of the reaction.
- States that some reactions with unfavourable (positive) ΔH can still be spontaneous if ΔS is sufficiently positive and TΔS outweighs ΔH, meaning enthalpy alone gives an incomplete or incorrect prediction of spontaneity.
Examiner tip: Spontaneity is always determined by ΔG, not ΔH alone; ΔH is only a reliable spontaneity predictor when ΔS is close to zero or its effect is negligible at the temperature considered.
20.
Marking analysis: A learner attempts the following task: “Outline one limitation of using standard enthalpy of formation values alone (without considering entropy) to predict whether a reaction will occur spontaneously.” Their response addresses only this point: “States that enthalpy alone does not account for the entropy change of the reaction.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that enthalpy alone does not account for the entropy change of the reaction. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that some reactions with unfavourable (positive) ΔH can still be spontaneous if ΔS is sufficiently positive and TΔS outweighs ΔH, meaning enthalpy alone gives an incomplete or incorrect prediction of spontaneity. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that enthalpy alone does not account for the entropy change of the reaction.
- Identifies the missing requirement: States that some reactions with unfavourable (positive) ΔH can still be spontaneous if ΔS is sufficiently positive and TΔS outweighs ΔH, meaning enthalpy alone gives an incomplete or incorrect prediction of spontaneity.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
21.
Using standard entropy values S°(CaCO₃(s)) = 93 J K⁻¹ mol⁻¹, S°(CaO(s)) = 40 J K⁻¹ mol⁻¹ and S°(CO₂(g)) = 214 J K⁻¹ mol⁻¹, calculate ΔS° for the reaction CaCO₃(s) → CaO(s) + CO₂(g).
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Uses ΔS° = ΣS°(products) − ΣS°(reactants). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes (40 + 214) − 93. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Obtains ΔS° = +161 J K⁻¹ mol⁻¹. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: This calculated positive ΔS° confirms the qualitative prediction that entropy increases when a gas is produced from a solid reactant — the two approaches (qualitative reasoning and standard entropy data) should always agree.
Marking points
- Uses ΔS° = ΣS°(products) − ΣS°(reactants).
- Substitutes (40 + 214) − 93.
- Obtains ΔS° = +161 J K⁻¹ mol⁻¹.
Examiner tip: This calculated positive ΔS° confirms the qualitative prediction that entropy increases when a gas is produced from a solid reactant — the two approaches (qualitative reasoning and standard entropy data) should always agree.
22.
Marking analysis: A learner attempts the following task: “Using standard entropy values S°(CaCO₃(s)) = 93 J K⁻¹ mol⁻¹, S°(CaO(s)) = 40 J K⁻¹ mol⁻¹ and S°(CO₂(g)) = 214 J K⁻¹ mol⁻¹, calculate ΔS° for the reaction CaCO₃(s) → CaO(s) + CO₂(g).” Their response addresses only this point: “Uses ΔS° = ΣS°(products) − ΣS°(reactants).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Uses ΔS° = ΣS°(products) − ΣS°(reactants). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes (40 + 214) − 93. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains ΔS° = +161 J K⁻¹ mol⁻¹. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Uses ΔS° = ΣS°(products) − ΣS°(reactants).
- Identifies the missing requirement: Substitutes (40 + 214) − 93.
- Identifies the missing requirement: Obtains ΔS° = +161 J K⁻¹ mol⁻¹.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
23.
A reaction (the Haber process) has ΔH° = −92 kJ mol⁻¹ and ΔS° = −199 J K⁻¹ mol⁻¹ at 298 K. Calculate ΔG° for this reaction at 298 K, and state whether it is spontaneous under these conditions.
[4 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Develop this part of the answer: Converts ΔH° to consistent units: −92 000 J mol⁻¹. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Work through this mathematical step: Uses ΔG° = ΔH° − TΔS°. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Substitutes −92 000 − (298 × −199) = −92 000 + 59 302. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Develop this part of the answer: Obtains ΔG° ≈ −32.7 kJ mol⁻¹, which is negative, so the reaction is spontaneous at 298 K. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A negative ΔG° means the reaction is thermodynamically spontaneous, but this says nothing about its rate — the Haber process still needs a catalyst and elevated temperature/pressure industrially because it is spontaneous yet extremely slow at 298 K.
Marking points
- Converts ΔH° to consistent units: −92 000 J mol⁻¹.
- Uses ΔG° = ΔH° − TΔS°.
- Substitutes −92 000 − (298 × −199) = −92 000 + 59 302.
- Obtains ΔG° ≈ −32.7 kJ mol⁻¹, which is negative, so the reaction is spontaneous at 298 K.
Examiner tip: A negative ΔG° means the reaction is thermodynamically spontaneous, but this says nothing about its rate — the Haber process still needs a catalyst and elevated temperature/pressure industrially because it is spontaneous yet extremely slow at 298 K.
24.
Marking analysis: A learner attempts the following task: “A reaction (the Haber process) has ΔH° = −92 kJ mol⁻¹ and ΔS° = −199 J K⁻¹ mol⁻¹ at 298 K. Calculate ΔG° for this reaction at 298 K, and state whether it is spontaneous under these conditions.” Their response addresses only this point: “Converts ΔH° to consistent units: −92 000 J mol⁻¹.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Converts ΔH° to consistent units: −92 000 J mol⁻¹. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Uses ΔG° = ΔH° − TΔS°. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Substitutes −92 000 − (298 × −199) = −92 000 + 59 302. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains ΔG° ≈ −32.7 kJ mol⁻¹, which is negative, so the reaction is spontaneous at 298 K. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Converts ΔH° to consistent units: −92 000 J mol⁻¹.
- Identifies the missing requirement: Uses ΔG° = ΔH° − TΔS°.
- Identifies the missing requirement: Substitutes −92 000 − (298 × −199) = −92 000 + 59 302.
- Identifies the missing requirement: Obtains ΔG° ≈ −32.7 kJ mol⁻¹, which is negative, so the reaction is spontaneous at 298 K.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
25.
State the equation relating standard Gibbs free energy change (ΔG°) to the equilibrium constant (K) of a reaction, and explain what a large negative ΔG° indicates about the value of K.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Work through this mathematical step: States ΔG° = −RT ln K. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Develop this part of the answer: Explains that a large negative ΔG° corresponds to a large positive value of ln K, and therefore a very large value of K. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: States that this means the reaction favours products strongly at equilibrium (the position of equilibrium lies far to the right). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: This equation is the formal link between thermodynamics (ΔG°) and equilibrium (K) — a reaction can be thermodynamically very favourable (large K) yet still appear not to proceed if it is kinetically very slow.
Marking points
- States ΔG° = −RT ln K.
- Explains that a large negative ΔG° corresponds to a large positive value of ln K, and therefore a very large value of K.
- States that this means the reaction favours products strongly at equilibrium (the position of equilibrium lies far to the right).
Examiner tip: This equation is the formal link between thermodynamics (ΔG°) and equilibrium (K) — a reaction can be thermodynamically very favourable (large K) yet still appear not to proceed if it is kinetically very slow.
26.
Marking analysis: A learner attempts the following task: “State the equation relating standard Gibbs free energy change (ΔG°) to the equilibrium constant (K) of a reaction, and explain what a large negative ΔG° indicates about the value of K.” Their response addresses only this point: “States ΔG° = −RT ln K.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States ΔG° = −RT ln K. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Explains that a large negative ΔG° corresponds to a large positive value of ln K, and therefore a very large value of K. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that this means the reaction favours products strongly at equilibrium (the position of equilibrium lies far to the right). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States ΔG° = −RT ln K.
- Identifies the missing requirement: Explains that a large negative ΔG° corresponds to a large positive value of ln K, and therefore a very large value of K.
- Identifies the missing requirement: States that this means the reaction favours products strongly at equilibrium (the position of equilibrium lies far to the right).
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
27.
State and explain the general trend in standard entropy (S°) for the same substance in its solid, liquid and gas states.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that S°(solid) < S°(liquid) < S°(gas). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that particles in a solid are highly ordered, with few possible arrangements (microstates). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: States that gas particles are free to move throughout a much larger volume with far more possible positions and arrangements, giving them the greatest entropy of the three states. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: This trend is the reason a change of state from solid to gas (or any step towards a less ordered state) always has a positive ΔS, regardless of which specific substance is involved.
Marking points
- States that S°(solid) < S°(liquid) < S°(gas).
- States that particles in a solid are highly ordered, with few possible arrangements (microstates).
- States that gas particles are free to move throughout a much larger volume with far more possible positions and arrangements, giving them the greatest entropy of the three states.
Examiner tip: This trend is the reason a change of state from solid to gas (or any step towards a less ordered state) always has a positive ΔS, regardless of which specific substance is involved.
28.
Marking analysis: A learner attempts the following task: “State and explain the general trend in standard entropy (S°) for the same substance in its solid, liquid and gas states.” Their response addresses only this point: “States that S°(solid) < S°(liquid) < S°(gas).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that S°(solid) < S°(liquid) < S°(gas). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that particles in a solid are highly ordered, with few possible arrangements (microstates). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that gas particles are free to move throughout a much larger volume with far more possible positions and arrangements, giving them the greatest entropy of the three states. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that S°(solid) < S°(liquid) < S°(gas).
- Identifies the missing requirement: States that particles in a solid are highly ordered, with few possible arrangements (microstates).
- Identifies the missing requirement: States that gas particles are free to move throughout a much larger volume with far more possible positions and arrangements, giving them the greatest entropy of the three states.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
29.
Construct an enthalpy cycle relating the enthalpy of solution of an ionic compound to its lattice enthalpy (of dissociation) and the enthalpies of hydration of its ions, and hence state the equation linking these three enthalpies.
[4 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that dissolving an ionic solid in water can be considered in two steps: first, breaking apart the lattice into gaseous ions (lattice dissociation enthalpy, endothermic). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that second, the gaseous ions are hydrated (surrounded by water molecules) as they dissolve (enthalpy of hydration, exothermic). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Work through this mathematical step: States the equation: ΔHsolution = ΔHlattice dissociation + ΔHhydration(cation) + ΔHhydration(anion). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Develop this part of the answer: States that whether dissolving is exothermic or endothermic overall depends on the relative magnitudes of the lattice dissociation enthalpy and the combined hydration enthalpies. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: This enthalpy of solution cycle is structurally identical to a Born-Haber cycle — both apply Hess's law by breaking an overall change into a hypothetical sequence of steps through gaseous ions.
Marking points
- States that dissolving an ionic solid in water can be considered in two steps: first, breaking apart the lattice into gaseous ions (lattice dissociation enthalpy, endothermic).
- States that second, the gaseous ions are hydrated (surrounded by water molecules) as they dissolve (enthalpy of hydration, exothermic).
- States the equation: ΔHsolution = ΔHlattice dissociation + ΔHhydration(cation) + ΔHhydration(anion).
- States that whether dissolving is exothermic or endothermic overall depends on the relative magnitudes of the lattice dissociation enthalpy and the combined hydration enthalpies.
Examiner tip: This enthalpy of solution cycle is structurally identical to a Born-Haber cycle — both apply Hess's law by breaking an overall change into a hypothetical sequence of steps through gaseous ions.
30.
Marking analysis: A learner attempts the following task: “Construct an enthalpy cycle relating the enthalpy of solution of an ionic compound to its lattice enthalpy (of dissociation) and the enthalpies of hydration of its ions, and hence state the equation linking these three enthalpies.” Their response addresses only this point: “States that dissolving an ionic solid in water can be considered in two steps: first, breaking apart the lattice into gaseous ions (lattice dissociation enthalpy, endothermic).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that dissolving an ionic solid in water can be considered in two steps: first, breaking apart the lattice into gaseous ions (lattice dissociation enthalpy, endothermic). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that second, the gaseous ions are hydrated (surrounded by water molecules) as they dissolve (enthalpy of hydration, exothermic). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States the equation: ΔHsolution = ΔHlattice dissociation + ΔHhydration(cation) + ΔHhydration(anion). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: States that whether dissolving is exothermic or endothermic overall depends on the relative magnitudes of the lattice dissociation enthalpy and the combined hydration enthalpies. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that dissolving an ionic solid in water can be considered in two steps: first, breaking apart the lattice into gaseous ions (lattice dissociation enthalpy, endothermic).
- Identifies the missing requirement: States that second, the gaseous ions are hydrated (surrounded by water molecules) as they dissolve (enthalpy of hydration, exothermic).
- Identifies the missing requirement: States the equation: ΔHsolution = ΔHlattice dissociation + ΔHhydration(cation) + ΔHhydration(anion).
- Identifies the missing requirement: States that whether dissolving is exothermic or endothermic overall depends on the relative magnitudes of the lattice dissociation enthalpy and the combined hydration enthalpies.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.