IB · CHEMISTRY SL

Chemistry SL

Reactivity 2: how much, how fast, how far — Reactivity 2

Name: ____________________Date: October 10, 2026

1.
25.0 cm³ of 0.250 mol dm⁻³ hydrochloric acid exactly neutralises 20.0 cm³ of sodium hydroxide solution. Calculate the concentration of sodium hydroxide, using HCl + NaOH → NaCl + H₂O.
[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: Calculates moles of HCl: 0.250 × 0.0250 = 0.00625 mol. 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: Uses the 1:1 equation ratio to state moles of NaOH = 0.00625 mol. 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: Uses concentration = moles ÷ volume(dm³) = 0.00625/0.0200. 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 concentration = 0.313 mol dm⁻³. 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: Always convert volumes from cm³ to dm³ (divide by 1000) before using concentration in mol dm⁻³.
Marking points
- Calculates moles of HCl: 0.250 × 0.0250 = 0.00625 mol.
- Uses the 1:1 equation ratio to state moles of NaOH = 0.00625 mol.
- Uses concentration = moles ÷ volume(dm³) = 0.00625/0.0200.
- Obtains concentration = 0.313 mol dm⁻³.
Examiner tip: Always convert volumes from cm³ to dm³ (divide by 1000) before using concentration in mol dm⁻³.
2.
Marking analysis: A learner attempts the following task: “25.0 cm³ of 0.250 mol dm⁻³ hydrochloric acid exactly neutralises 20.0 cm³ of sodium hydroxide solution. Calculate the concentration of sodium hydroxide, using HCl + NaOH → NaCl + H₂O.” Their response addresses only this point: “Calculates moles of HCl: 0.250 × 0.0250 = 0.00625 mol.” 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: Calculates moles of HCl: 0.250 × 0.0250 = 0.00625 mol. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Uses the 1:1 equation ratio to state moles of NaOH = 0.00625 mol. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Uses concentration = moles ÷ volume(dm³) = 0.00625/0.0200. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains concentration = 0.313 mol dm⁻³. 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: Calculates moles of HCl: 0.250 × 0.0250 = 0.00625 mol.
- Identifies the missing requirement: Uses the 1:1 equation ratio to state moles of NaOH = 0.00625 mol.
- Identifies the missing requirement: Uses concentration = moles ÷ volume(dm³) = 0.00625/0.0200.
- Identifies the missing requirement: Obtains concentration = 0.313 mol dm⁻³.
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.
Magnesium reacts with hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. When 0.10 mol Mg reacts with 0.15 mol HCl, identify the limiting reactant and calculate the amount of hydrogen formed.
[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: Uses the equation ratio 1 mol Mg : 2 mol HCl. 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: Recognises that 0.10 mol Mg would require 0.20 mol HCl, more than is available. 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: Identifies HCl as the limiting reactant. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Work through this mathematical step: Obtains n(H₂) = 0.15 ÷ 2 = 0.075 mol. 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: Compare available amounts only after applying the equation coefficients — the limiting reactant is not necessarily the one present in the smallest raw amount.
Marking points
- Uses the equation ratio 1 mol Mg : 2 mol HCl.
- Recognises that 0.10 mol Mg would require 0.20 mol HCl, more than is available.
- Identifies HCl as the limiting reactant.
- Obtains n(H₂) = 0.15 ÷ 2 = 0.075 mol.
Examiner tip: Compare available amounts only after applying the equation coefficients — the limiting reactant is not necessarily the one present in the smallest raw amount.
4.
Marking analysis: A learner attempts the following task: “Magnesium reacts with hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. When 0.10 mol Mg reacts with 0.15 mol HCl, identify the limiting reactant and calculate the amount of hydrogen formed.” Their response addresses only this point: “Uses the equation ratio 1 mol Mg : 2 mol HCl.” 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: Uses the equation ratio 1 mol Mg : 2 mol HCl. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Recognises that 0.10 mol Mg would require 0.20 mol HCl, more than is available. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Identifies HCl as the limiting reactant. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains n(H₂) = 0.15 ÷ 2 = 0.075 mol. 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: Uses the equation ratio 1 mol Mg : 2 mol HCl.
- Identifies the missing requirement: Recognises that 0.10 mol Mg would require 0.20 mol HCl, more than is available.
- Identifies the missing requirement: Identifies HCl as the limiting reactant.
- Identifies the missing requirement: Obtains n(H₂) = 0.15 ÷ 2 = 0.075 mol.
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.
State the collision theory requirements for a chemical reaction to occur between two particles.
[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 particles must collide with each other. 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 the collision must have sufficient energy (at least the activation energy) and correct (favourable) orientation. 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 collision with enough energy but the wrong orientation still fails to produce a reaction — both conditions must be met simultaneously.
Marking points
- States that particles must collide with each other.
- States that the collision must have sufficient energy (at least the activation energy) and correct (favourable) orientation.
Examiner tip: A collision with enough energy but the wrong orientation still fails to produce a reaction — both conditions must be met simultaneously.
6.
Marking analysis: A learner attempts the following task: “State the collision theory requirements for a chemical reaction to occur between two particles.” Their response addresses only this point: “States that particles must collide with each other.” 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 particles must collide with each other. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that the collision must have sufficient energy (at least the activation energy) and correct (favourable) orientation. 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 particles must collide with each other.
- Identifies the missing requirement: States that the collision must have sufficient energy (at least the activation energy) and correct (favourable) orientation.
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.
Explain, using collision theory, why increasing temperature increases the rate of a chemical reaction.
[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 increasing temperature increases the average kinetic energy of the particles. 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 increases both the frequency of collisions and, more significantly, the proportion of collisions with energy exceeding the activation energy. 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 a greater proportion of successful (effective) collisions per second results in a faster reaction rate. 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: The increase in the proportion of particles with energy above the activation energy is the dominant effect of temperature on rate, far more significant than the modest increase in collision frequency alone.
Marking points
- States that increasing temperature increases the average kinetic energy of the particles.
- States that this increases both the frequency of collisions and, more significantly, the proportion of collisions with energy exceeding the activation energy.
- States that a greater proportion of successful (effective) collisions per second results in a faster reaction rate.
Examiner tip: The increase in the proportion of particles with energy above the activation energy is the dominant effect of temperature on rate, far more significant than the modest increase in collision frequency alone.
8.
Marking analysis: A learner attempts the following task: “Explain, using collision theory, why increasing temperature increases the rate of a chemical reaction.” Their response addresses only this point: “States that increasing temperature increases the average kinetic energy of the particles.” 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 increasing temperature increases the average kinetic energy of the particles. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that this increases both the frequency of collisions and, more significantly, the proportion of collisions with energy exceeding the activation energy. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that a greater proportion of successful (effective) collisions per second results in a faster reaction rate. 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 increasing temperature increases the average kinetic energy of the particles.
- Identifies the missing requirement: States that this increases both the frequency of collisions and, more significantly, the proportion of collisions with energy exceeding the activation energy.
- Identifies the missing requirement: States that a greater proportion of successful (effective) collisions per second results in a faster reaction rate.
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.
Sketch (in words) how a catalyst affects the activation energy of a reaction, and explain why a catalyst increases reaction rate without being consumed.
[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 a catalyst provides an alternative reaction pathway (mechanism) with a lower activation energy. 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 means a greater proportion of collisions now have sufficient energy to react, increasing the rate. 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 the catalyst participates in the mechanism but is regenerated at the end, so it is not consumed 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: A catalyst lowers activation energy, it does not change the overall enthalpy change of the reaction or the position of equilibrium.
Marking points
- States that a catalyst provides an alternative reaction pathway (mechanism) with a lower activation energy.
- States that this means a greater proportion of collisions now have sufficient energy to react, increasing the rate.
- States that the catalyst participates in the mechanism but is regenerated at the end, so it is not consumed overall.
Examiner tip: A catalyst lowers activation energy, it does not change the overall enthalpy change of the reaction or the position of equilibrium.
10.
Marking analysis: A learner attempts the following task: “Sketch (in words) how a catalyst affects the activation energy of a reaction, and explain why a catalyst increases reaction rate without being consumed.” Their response addresses only this point: “States that a catalyst provides an alternative reaction pathway (mechanism) with a lower activation energy.” 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 a catalyst provides an alternative reaction pathway (mechanism) with a lower activation energy. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that this means a greater proportion of collisions now have sufficient energy to react, increasing the rate. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that the catalyst participates in the mechanism but is regenerated at the end, so it is not consumed 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 that a catalyst provides an alternative reaction pathway (mechanism) with a lower activation energy.
- Identifies the missing requirement: States that this means a greater proportion of collisions now have sufficient energy to react, increasing the rate.
- Identifies the missing requirement: States that the catalyst participates in the mechanism but is regenerated at the end, so it is not consumed 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.
11.
Write the equilibrium constant expression, Kc, for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(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: Places products (raised to their coefficients) in the numerator and reactants (raised to their coefficients) in the denominator. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Work through this mathematical step: Writes Kc = [NH₃]²/([N₂][H₂]³). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Only gases and aqueous species appear in the Kc expression; pure solids and pure liquids are omitted since their 'concentration' is effectively constant.
Marking points
- Places products (raised to their coefficients) in the numerator and reactants (raised to their coefficients) in the denominator.
- Writes Kc = [NH₃]²/([N₂][H₂]³).
Examiner tip: Only gases and aqueous species appear in the Kc expression; pure solids and pure liquids are omitted since their 'concentration' is effectively constant.
12.
Marking analysis: A learner attempts the following task: “Write the equilibrium constant expression, Kc, for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g).” Their response addresses only this point: “Places products (raised to their coefficients) in the numerator and reactants (raised to their coefficients) in the denominator.” 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: Places products (raised to their coefficients) in the numerator and reactants (raised to their coefficients) in the denominator. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Writes Kc = [NH₃]²/([N₂][H₂]³). 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: Places products (raised to their coefficients) in the numerator and reactants (raised to their coefficients) in the denominator.
- Identifies the missing requirement: Writes Kc = [NH₃]²/([N₂][H₂]³).
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.
State Le Chatelier's principle, and use it to predict the effect of increasing pressure on the position of equilibrium for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
[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 Le Chatelier's principle: if a system at equilibrium is subjected to a change, the equilibrium shifts to partially counteract that change. 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 increasing pressure shifts the equilibrium towards the side with fewer moles of gas. 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 since there are 4 moles of gas on the left and 2 on the right, the equilibrium shifts to the right (towards NH₃), partially counteracting the pressure increase. 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: For pressure changes, count total moles of gas on each side of the equation — the equilibrium always shifts towards the side with fewer gas moles when pressure increases.
Marking points
- States Le Chatelier's principle: if a system at equilibrium is subjected to a change, the equilibrium shifts to partially counteract that change.
- States that increasing pressure shifts the equilibrium towards the side with fewer moles of gas.
- States that since there are 4 moles of gas on the left and 2 on the right, the equilibrium shifts to the right (towards NH₃), partially counteracting the pressure increase.
Examiner tip: For pressure changes, count total moles of gas on each side of the equation — the equilibrium always shifts towards the side with fewer gas moles when pressure increases.
14.
Marking analysis: A learner attempts the following task: “State Le Chatelier's principle, and use it to predict the effect of increasing pressure on the position of equilibrium for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).” Their response addresses only this point: “States Le Chatelier's principle: if a system at equilibrium is subjected to a change, the equilibrium shifts to partially counteract that change.” 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 Le Chatelier's principle: if a system at equilibrium is subjected to a change, the equilibrium shifts to partially counteract that change. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that increasing pressure shifts the equilibrium towards the side with fewer moles of gas. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that since there are 4 moles of gas on the left and 2 on the right, the equilibrium shifts to the right (towards NH₃), partially counteracting the pressure increase. 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 Le Chatelier's principle: if a system at equilibrium is subjected to a change, the equilibrium shifts to partially counteract that change.
- Identifies the missing requirement: States that increasing pressure shifts the equilibrium towards the side with fewer moles of gas.
- Identifies the missing requirement: States that since there are 4 moles of gas on the left and 2 on the right, the equilibrium shifts to the right (towards NH₃), partially counteracting the pressure increase.
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.
The forward reaction A(g) + B(g) ⇌ C(g) is exothermic. Predict, with a reason, the effect of increasing temperature on the value of Kc.
[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 increasing temperature favours the endothermic direction of the reaction (by Le Chatelier's principle). 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 since the forward reaction is exothermic, the reverse (endothermic) reaction is favoured, shifting equilibrium to the left (towards reactants). 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 Kc therefore decreases with increasing temperature for an exothermic forward reaction. 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: Unlike pressure or concentration changes, a temperature change actually alters the numerical value of Kc itself, not just the position of equilibrium at a fixed Kc.
Marking points
- States that increasing temperature favours the endothermic direction of the reaction (by Le Chatelier's principle).
- States that since the forward reaction is exothermic, the reverse (endothermic) reaction is favoured, shifting equilibrium to the left (towards reactants).
- States that Kc therefore decreases with increasing temperature for an exothermic forward reaction.
Examiner tip: Unlike pressure or concentration changes, a temperature change actually alters the numerical value of Kc itself, not just the position of equilibrium at a fixed Kc.
16.
Marking analysis: A learner attempts the following task: “The forward reaction A(g) + B(g) ⇌ C(g) is exothermic. Predict, with a reason, the effect of increasing temperature on the value of Kc.” Their response addresses only this point: “States that increasing temperature favours the endothermic direction of the reaction (by Le Chatelier's principle).” 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 increasing temperature favours the endothermic direction of the reaction (by Le Chatelier's principle). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that since the forward reaction is exothermic, the reverse (endothermic) reaction is favoured, shifting equilibrium to the left (towards reactants). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that Kc therefore decreases with increasing temperature for an exothermic forward reaction. 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 increasing temperature favours the endothermic direction of the reaction (by Le Chatelier's principle).
- Identifies the missing requirement: States that since the forward reaction is exothermic, the reverse (endothermic) reaction is favoured, shifting equilibrium to the left (towards reactants).
- Identifies the missing requirement: States that Kc therefore decreases with increasing temperature for an exothermic forward reaction.
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.
At a certain temperature, the equilibrium concentrations for the reaction A(g) ⇌ B(g) are [A] = 0.20 mol dm⁻³ and [B] = 0.80 mol dm⁻³. Calculate the value of Kc at this temperature.
[2 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 Kc = [B]/[A]. 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 0.80/0.20 to obtain Kc = 4.0 (no units, since concentration units cancel for this reaction). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Only use equilibrium concentrations (not initial concentrations) when calculating Kc — this is a common source of error.
Marking points
- Uses Kc = [B]/[A].
- Substitutes 0.80/0.20 to obtain Kc = 4.0 (no units, since concentration units cancel for this reaction).
Examiner tip: Only use equilibrium concentrations (not initial concentrations) when calculating Kc — this is a common source of error.
18.
Marking analysis: A learner attempts the following task: “At a certain temperature, the equilibrium concentrations for the reaction A(g) ⇌ B(g) are [A] = 0.20 mol dm⁻³ and [B] = 0.80 mol dm⁻³. Calculate the value of Kc at this temperature.” Their response addresses only this point: “Uses Kc = [B]/[A].” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 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 Kc = [B]/[A]. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes 0.80/0.20 to obtain Kc = 4.0 (no units, since concentration units cancel for this reaction). 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: Uses Kc = [B]/[A].
- Identifies the missing requirement: Substitutes 0.80/0.20 to obtain Kc = 4.0 (no units, since concentration units cancel for this reaction).
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.
Explain why a catalyst does not affect the position of equilibrium or the value of Kc for a reversible reaction, even though it increases the rate at which equilibrium is reached.
[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 a catalyst lowers the activation energy of both the forward and reverse reactions by exactly the same amount. 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 since both rates increase by the same factor, the equilibrium is reached faster, but the ratio of forward to reverse rates at equilibrium (and hence Kc and the equilibrium concentrations) remains unchanged. 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 catalyst speeds up how quickly equilibrium is reached but never changes where that equilibrium lies — this is a very commonly tested distinction.
Marking points
- States that a catalyst lowers the activation energy of both the forward and reverse reactions by exactly the same amount.
- States that since both rates increase by the same factor, the equilibrium is reached faster, but the ratio of forward to reverse rates at equilibrium (and hence Kc and the equilibrium concentrations) remains unchanged.
Examiner tip: A catalyst speeds up how quickly equilibrium is reached but never changes where that equilibrium lies — this is a very commonly tested distinction.
20.
Marking analysis: A learner attempts the following task: “Explain why a catalyst does not affect the position of equilibrium or the value of Kc for a reversible reaction, even though it increases the rate at which equilibrium is reached.” Their response addresses only this point: “States that a catalyst lowers the activation energy of both the forward and reverse reactions by exactly the same amount.” 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 a catalyst lowers the activation energy of both the forward and reverse reactions by exactly the same amount. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that since both rates increase by the same factor, the equilibrium is reached faster, but the ratio of forward to reverse rates at equilibrium (and hence Kc and the equilibrium concentrations) remains unchanged. 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 a catalyst lowers the activation energy of both the forward and reverse reactions by exactly the same amount.
- Identifies the missing requirement: States that since both rates increase by the same factor, the equilibrium is reached faster, but the ratio of forward to reverse rates at equilibrium (and hence Kc and the equilibrium concentrations) remains unchanged.
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.
Explain, using collision theory, why increasing the concentration of a reactant in solution increases the rate of a chemical reaction.
[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 increasing concentration increases the number of reactant particles present in a given volume. 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 increases the frequency of collisions between reactant particles per unit time. 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 more frequent collisions result in more successful (effective) collisions per second, increasing the rate of reaction. 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: Unlike temperature, increasing concentration only increases collision frequency — it does not change the proportion of particles with energy above the activation energy.
Marking points
- States that increasing concentration increases the number of reactant particles present in a given volume.
- States that this increases the frequency of collisions between reactant particles per unit time.
- States that more frequent collisions result in more successful (effective) collisions per second, increasing the rate of reaction.
Examiner tip: Unlike temperature, increasing concentration only increases collision frequency — it does not change the proportion of particles with energy above the activation energy.
22.
Marking analysis: A learner attempts the following task: “Explain, using collision theory, why increasing the concentration of a reactant in solution increases the rate of a chemical reaction.” Their response addresses only this point: “States that increasing concentration increases the number of reactant particles present in a given volume.” 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 increasing concentration increases the number of reactant particles present in a given volume. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that this increases the frequency of collisions between reactant particles per unit time. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that more frequent collisions result in more successful (effective) collisions per second, increasing the rate of reaction. 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 increasing concentration increases the number of reactant particles present in a given volume.
- Identifies the missing requirement: States that this increases the frequency of collisions between reactant particles per unit time.
- Identifies the missing requirement: States that more frequent collisions result in more successful (effective) collisions per second, increasing the rate of reaction.
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.
Explain, using collision theory, why increasing the surface area of a solid reactant, e.g. using powder instead of a single lump, increases the rate of reaction.
[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 increasing surface area exposes more reactant particles at the surface to collisions with particles of the other reactant. 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 increases the frequency of collisions between the reacting particles. 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 more frequent collisions result in more successful collisions per second, increasing the rate of reaction. 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: Surface area only matters for reactions involving a solid reacting with a liquid or gas — particles buried inside a solid lump cannot collide with the other reactant until the surface layers above them have already reacted away.
Marking points
- States that increasing surface area exposes more reactant particles at the surface to collisions with particles of the other reactant.
- States that this increases the frequency of collisions between the reacting particles.
- States that more frequent collisions result in more successful collisions per second, increasing the rate of reaction.
Examiner tip: Surface area only matters for reactions involving a solid reacting with a liquid or gas — particles buried inside a solid lump cannot collide with the other reactant until the surface layers above them have already reacted away.
24.
Marking analysis: A learner attempts the following task: “Explain, using collision theory, why increasing the surface area of a solid reactant, e.g. using powder instead of a single lump, increases the rate of reaction.” Their response addresses only this point: “States that increasing surface area exposes more reactant particles at the surface to collisions with particles of the other reactant.” 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 increasing surface area exposes more reactant particles at the surface to collisions with particles of the other reactant. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that this increases the frequency of collisions between the reacting particles. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that more frequent collisions result in more successful collisions per second, increasing the rate of reaction. 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 increasing surface area exposes more reactant particles at the surface to collisions with particles of the other reactant.
- Identifies the missing requirement: States that this increases the frequency of collisions between the reacting particles.
- Identifies the missing requirement: States that more frequent collisions result in more successful collisions per second, increasing the rate of reaction.
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.
For the equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq), predict and explain, using Le Chatelier's principle, the effect on the position of equilibrium of adding more Fe³⁺(aq) ions.
[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 adding more Fe³⁺(aq) increases the concentration of a reactant. 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, by Le Chatelier's principle, the equilibrium shifts in the direction that reduces this increase, i.e. towards the products (to the right). 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 increases the concentration of FeSCN²⁺ (the product) at the new equilibrium position. 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 iron(III) thiocyanate equilibrium is a classic real demonstration: the deep red FeSCN²⁺ colour visibly intensifies when Fe³⁺ or SCN⁻ is added, and fades when either is removed — a direct visual confirmation of Le Chatelier's principle.
Marking points
- States that adding more Fe³⁺(aq) increases the concentration of a reactant.
- States that, by Le Chatelier's principle, the equilibrium shifts in the direction that reduces this increase, i.e. towards the products (to the right).
- States that this increases the concentration of FeSCN²⁺ (the product) at the new equilibrium position.
Examiner tip: This iron(III) thiocyanate equilibrium is a classic real demonstration: the deep red FeSCN²⁺ colour visibly intensifies when Fe³⁺ or SCN⁻ is added, and fades when either is removed — a direct visual confirmation of Le Chatelier's principle.
26.
Marking analysis: A learner attempts the following task: “For the equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq), predict and explain, using Le Chatelier's principle, the effect on the position of equilibrium of adding more Fe³⁺(aq) ions.” Their response addresses only this point: “States that adding more Fe³⁺(aq) increases the concentration of a reactant.” 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 adding more Fe³⁺(aq) increases the concentration of a reactant. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that, by Le Chatelier's principle, the equilibrium shifts in the direction that reduces this increase, i.e. towards the products (to the right). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that this increases the concentration of FeSCN²⁺ (the product) at the new equilibrium position. 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 adding more Fe³⁺(aq) increases the concentration of a reactant.
- Identifies the missing requirement: States that, by Le Chatelier's principle, the equilibrium shifts in the direction that reduces this increase, i.e. towards the products (to the right).
- Identifies the missing requirement: States that this increases the concentration of FeSCN²⁺ (the product) at the new equilibrium position.
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.
The rate equation for a reaction is rate = k[A][B]². State the order of reaction with respect to A, with respect to B, and the overall order of reaction.
[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 reaction is first order with respect to A. 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 the reaction is second order with respect to B. 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 the overall order of reaction is 3 (1 + 2). 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: The order with respect to a species is simply its exponent in the rate equation, and the overall order is the sum of all the individual exponents — orders must be found experimentally and cannot be deduced from the balanced chemical equation alone.
Marking points
- States that the reaction is first order with respect to A.
- States that the reaction is second order with respect to B.
- States that the overall order of reaction is 3 (1 + 2).
Examiner tip: The order with respect to a species is simply its exponent in the rate equation, and the overall order is the sum of all the individual exponents — orders must be found experimentally and cannot be deduced from the balanced chemical equation alone.
28.
Marking analysis: A learner attempts the following task: “The rate equation for a reaction is rate = k[A][B]². State the order of reaction with respect to A, with respect to B, and the overall order of reaction.” Their response addresses only this point: “States that the reaction is first order with respect to A.” 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 reaction is first order with respect to A. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that the reaction is second order with respect to B. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that the overall order of reaction is 3 (1 + 2). 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 reaction is first order with respect to A.
- Identifies the missing requirement: States that the reaction is second order with respect to B.
- Identifies the missing requirement: States that the overall order of reaction is 3 (1 + 2).
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.
Distinguish between percentage yield and atom economy, and state one reason a reaction might have a high atom economy but a low percentage yield.
[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 that percentage yield = (actual yield ÷ theoretical yield) × 100, comparing the mass of product actually obtained to the maximum possible mass. 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: States that atom economy = (Mr of desired product ÷ sum of Mr of all reactants) × 100, measuring what proportion of the total mass of reactants ends up as the desired product according to the balanced equation itself. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Develop this part of the answer: States a valid reason, e.g. practical losses during the reaction, such as side reactions, incomplete reaction, or losses during purification and filtration, reduce the actual yield obtained even though the reaction's own stoichiometry wastes little mass. 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: Atom economy is a theoretical, equation-based measure fixed by the reaction's stoichiometry, while percentage yield reflects practical losses in a specific experiment — a reaction can score well on one and poorly on the other.
Marking points
- States that percentage yield = (actual yield ÷ theoretical yield) × 100, comparing the mass of product actually obtained to the maximum possible mass.
- States that atom economy = (Mr of desired product ÷ sum of Mr of all reactants) × 100, measuring what proportion of the total mass of reactants ends up as the desired product according to the balanced equation itself.
- States a valid reason, e.g. practical losses during the reaction, such as side reactions, incomplete reaction, or losses during purification and filtration, reduce the actual yield obtained even though the reaction's own stoichiometry wastes little mass.
Examiner tip: Atom economy is a theoretical, equation-based measure fixed by the reaction's stoichiometry, while percentage yield reflects practical losses in a specific experiment — a reaction can score well on one and poorly on the other.
30.
Marking analysis: A learner attempts the following task: “Distinguish between percentage yield and atom economy, and state one reason a reaction might have a high atom economy but a low percentage yield.” Their response addresses only this point: “States that percentage yield = (actual yield ÷ theoretical yield) × 100, comparing the mass of product actually obtained to the maximum possible mass.” 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 percentage yield = (actual yield ÷ theoretical yield) × 100, comparing the mass of product actually obtained to the maximum possible mass. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that atom economy = (Mr of desired product ÷ sum of Mr of all reactants) × 100, measuring what proportion of the total mass of reactants ends up as the desired product according to the balanced equation itself. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States a valid reason, e.g. practical losses during the reaction, such as side reactions, incomplete reaction, or losses during purification and filtration, reduce the actual yield obtained even though the reaction's own stoichiometry wastes little mass. 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 percentage yield = (actual yield ÷ theoretical yield) × 100, comparing the mass of product actually obtained to the maximum possible mass.
- Identifies the missing requirement: States that atom economy = (Mr of desired product ÷ sum of Mr of all reactants) × 100, measuring what proportion of the total mass of reactants ends up as the desired product according to the balanced equation itself.
- Identifies the missing requirement: States a valid reason, e.g. practical losses during the reaction, such as side reactions, incomplete reaction, or losses during purification and filtration, reduce the actual yield obtained even though the reaction's own stoichiometry wastes little mass.
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.