Quick answer: energy change is not the same as the barrier to reaction
An exothermic reaction has products at a lower enthalpy level than reactants; its reaction enthalpy change is negative. An endothermic reaction has products higher; its change is positive. Activation energy is the barrier from the reactants to the peak of an energy profile, not the difference between reactants and products. A catalyst provides a pathway with a lower activation energy but does not change the starting and ending enthalpy levels. These four ideas are easy to learn separately. The examination challenge is to keep all four distinct when a diagram, a number and a written claim appear together.
For a labelled hypothetical energy profile, let R be reactants’ level, P be products’ level, U be the uncatalysed peak and C be the catalysed peak. Then change in enthalpy is P minus R, activation energy forward is U minus R and the reverse activation energy is U minus P. These calculations describe positions on an invented teaching diagram. They are not measured enthalpy changes for real substances. If the vertical axis is in arbitrary units, the answers must stay in those units. If a question supplies kJ/mol, use those units instead.
What the 2027 Chemistry assessment actually tests
The 2027 G3 Chemistry K324 syllabus lists chemical energetics, energy profiles, reaction enthalpy and activation energy. It also requires candidates to explain bond breaking as endothermic and bond formation as exothermic, and to relate catalysts to lower activation energy. Paper 1 is 40 multiple-choice items in 1 hour, worth 30% of the subject assessment. Paper 2 is a 1-hour-45-minute structured and free-response paper worth 50%, and Paper 3 is a 1-hour-50-minute practical worth 20%. The formal practical tests planning, measurements, presentation and evaluation; this chapter’s fictitious profile puzzles are not actual practical instructions or real past-paper questions.
Read this after Volume 0629 on electrical input and heating and Volume 0636 on evidence-safe thermal-response explanations. Those chapters teach heat accounting and respecting a measurement boundary. This chapter moves to chemical reaction energetics: a hot-looking result is not sufficient to identify a catalyst, and a falling energy profile is not the same as a falling activation barrier.
Build the explanation in a safe sequence
First identify the labels on the vertical and horizontal axes. A reaction-coordinate axis describes progress along an imagined pathway; it is not automatically a time axis. Second find reactants and products. Their relative heights settle the sign of reaction enthalpy. Third find the top of each pathway and subtract the reactant height to obtain each forward activation energy. Fourth check whether a catalyst’s pathway has the same endpoints. Fifth use a sentence that connects the numbers to their meanings rather than simply listing a minus sign.
A negative sign does not mean the reacting mixture must be cold. An exothermic process releases energy to the surroundings and may increase the surroundings’ temperature under suitable conditions. An endothermic process takes energy from the surroundings and may cool them. Measurements depend on insulation, quantities, starting conditions and transfer routes. We should not derive an exact enthalpy value from one temperature reading without the information needed for an energy calculation. If the diagram is schematic, keep it schematic.
For bond-energy reasoning, energy is absorbed when bonds break and released when bonds form. A net exothermic reaction releases more in bond formation than it takes to break bonds. A net endothermic reaction needs more for bond breaking than is released on formation. This qualitative distinction is independent of the peak-versus-endpoints distinction. It is also safer than the common shortcut ‘breaking bonds releases energy’, which reverses the sign.
Twelve worked profiles: one data model, five different skills each
Profile 1: display A — read endpoints before discussing the pathway
The source is a labelled exothermic profile. Its invented vertical levels are R = 80, P = 50, uncatalysed peak U = 150 and catalysed peak C = 118, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 50 − 80 = -30. This is exothermic; its magnitude is 30 model units and its enthalpy sign is negative. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 150 − 80 = 70. Reverse activation energy on the same pathway is U − P = 150 − 50 = 100. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 118 − 80 = 38, a reduction of 32 units, or 45.714% of the original forward barrier. Reverse barrier becomes C − P = 68. The catalyst does not change P − R = -30; it changes the available pathway. It is wrong to claim that a 45.714% lower barrier guarantees a 45.714% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The exothermic sign says the overall reaction releases more energy in forming bonds than is needed to break the relevant bonds, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The draft caption says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains -30 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 2: display B — read endpoints before discussing the pathway
The source is an endothermic diagram. Its invented vertical levels are R = 65, P = 94, uncatalysed peak U = 158 and catalysed peak C = 125, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 94 − 65 = 29. This is endothermic; its magnitude is 29 model units and its enthalpy sign is positive. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 158 − 65 = 93. Reverse activation energy on the same pathway is U − P = 158 − 94 = 64. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 125 − 65 = 60, a reduction of 33 units, or 35.484% of the original forward barrier. Reverse barrier becomes C − P = 31. The catalyst does not change P − R = 29; it changes the available pathway. It is wrong to claim that a 35.484% lower barrier guarantees a 35.484% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The endothermic sign says the overall reaction requires more energy to break relevant bonds than is released when bonds form, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The visitor leaflet says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains 29 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 3: display C — read endpoints before discussing the pathway
The source is a revised data sheet. Its invented vertical levels are R = 117, P = 76, uncatalysed peak U = 184 and catalysed peak C = 148, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 76 − 117 = -41. This is exothermic; its magnitude is 41 model units and its enthalpy sign is negative. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 184 − 117 = 67. Reverse activation energy on the same pathway is U − P = 184 − 76 = 108. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 148 − 117 = 31, a reduction of 36 units, or 53.731% of the original forward barrier. Reverse barrier becomes C − P = 72. The catalyst does not change P − R = -41; it changes the available pathway. It is wrong to claim that a 53.731% lower barrier guarantees a 53.731% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The exothermic sign says the overall reaction releases more energy in forming bonds than is needed to break the relevant bonds, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The school presentation says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains -41 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 4: display D — read endpoints before discussing the pathway
The source is a sketch shown by two speakers. Its invented vertical levels are R = 52, P = 89, uncatalysed peak U = 139 and catalysed peak C = 109, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 89 − 52 = 37. This is endothermic; its magnitude is 37 model units and its enthalpy sign is positive. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 139 − 52 = 87. Reverse activation energy on the same pathway is U − P = 139 − 89 = 50. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 109 − 52 = 57, a reduction of 30 units, or 34.483% of the original forward barrier. Reverse barrier becomes C − P = 20. The catalyst does not change P − R = 37; it changes the available pathway. It is wrong to claim that a 34.483% lower barrier guarantees a 34.483% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The endothermic sign says the overall reaction requires more energy to break relevant bonds than is released when bonds form, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The planning notes says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains 37 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 5: display E — read endpoints before discussing the pathway
The source is a model energy chart. Its invented vertical levels are R = 96, P = 41, uncatalysed peak U = 173 and catalysed peak C = 137, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 41 − 96 = -55. This is exothermic; its magnitude is 55 model units and its enthalpy sign is negative. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 173 − 96 = 77. Reverse activation energy on the same pathway is U − P = 173 − 41 = 132. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 137 − 96 = 41, a reduction of 36 units, or 46.753% of the original forward barrier. Reverse barrier becomes C − P = 96. The catalyst does not change P − R = -55; it changes the available pathway. It is wrong to claim that a 46.753% lower barrier guarantees a 46.753% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The exothermic sign says the overall reaction releases more energy in forming bonds than is needed to break the relevant bonds, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The science exhibition panel says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains -55 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 6: display F — read endpoints before discussing the pathway
The source is a teacher’s hypothetical exercise. Its invented vertical levels are R = 78, P = 105, uncatalysed peak U = 166 and catalysed peak C = 142, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 105 − 78 = 27. This is endothermic; its magnitude is 27 model units and its enthalpy sign is positive. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 166 − 78 = 88. Reverse activation energy on the same pathway is U − P = 166 − 105 = 61. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 142 − 78 = 64, a reduction of 24 units, or 27.273% of the original forward barrier. Reverse barrier becomes C − P = 37. The catalyst does not change P − R = 27; it changes the available pathway. It is wrong to claim that a 27.273% lower barrier guarantees a 27.273% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The endothermic sign says the overall reaction requires more energy to break relevant bonds than is released when bonds form, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The committee minutes says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains 27 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 7: display G — read endpoints before discussing the pathway
The source is a pair of energy pathways. Its invented vertical levels are R = 134, P = 100, uncatalysed peak U = 210 and catalysed peak C = 174, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 100 − 134 = -34. This is exothermic; its magnitude is 34 model units and its enthalpy sign is negative. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 210 − 134 = 76. Reverse activation energy on the same pathway is U − P = 210 − 100 = 110. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 174 − 134 = 40, a reduction of 36 units, or 47.368% of the original forward barrier. Reverse barrier becomes C − P = 74. The catalyst does not change P − R = -34; it changes the available pathway. It is wrong to claim that a 47.368% lower barrier guarantees a 47.368% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The exothermic sign says the overall reaction releases more energy in forming bonds than is needed to break the relevant bonds, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The draft audio guide says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains -34 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 8: display H — read endpoints before discussing the pathway
The source is an annotated energy graph. Its invented vertical levels are R = 42, P = 83, uncatalysed peak U = 151 and catalysed peak C = 120, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 83 − 42 = 41. This is endothermic; its magnitude is 41 model units and its enthalpy sign is positive. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 151 − 42 = 109. Reverse activation energy on the same pathway is U − P = 151 − 83 = 68. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 120 − 42 = 78, a reduction of 31 units, or 28.44% of the original forward barrier. Reverse barrier becomes C − P = 37. The catalyst does not change P − R = 41; it changes the available pathway. It is wrong to claim that a 28.44% lower barrier guarantees a 28.44% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The endothermic sign says the overall reaction requires more energy to break relevant bonds than is released when bonds form, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The class practice answer says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains 41 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 9: display I — read endpoints before discussing the pathway
The source is a projected slide. Its invented vertical levels are R = 108, P = 62, uncatalysed peak U = 187 and catalysed peak C = 143, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 62 − 108 = -46. This is exothermic; its magnitude is 46 model units and its enthalpy sign is negative. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 187 − 108 = 79. Reverse activation energy on the same pathway is U − P = 187 − 62 = 125. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 143 − 108 = 35, a reduction of 44 units, or 55.696% of the original forward barrier. Reverse barrier becomes C − P = 81. The catalyst does not change P − R = -46; it changes the available pathway. It is wrong to claim that a 55.696% lower barrier guarantees a 55.696% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The exothermic sign says the overall reaction releases more energy in forming bonds than is needed to break the relevant bonds, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The final-check worksheet says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains -46 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 10: display J — read endpoints before discussing the pathway
The source is an interpretation exercise. Its invented vertical levels are R = 91, P = 119, uncatalysed peak U = 178 and catalysed peak C = 148, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 119 − 91 = 28. This is endothermic; its magnitude is 28 model units and its enthalpy sign is positive. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 178 − 91 = 87. Reverse activation energy on the same pathway is U − P = 178 − 119 = 59. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 148 − 91 = 57, a reduction of 30 units, or 34.483% of the original forward barrier. Reverse barrier becomes C − P = 29. The catalyst does not change P − R = 28; it changes the available pathway. It is wrong to claim that a 34.483% lower barrier guarantees a 34.483% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The endothermic sign says the overall reaction requires more energy to break relevant bonds than is released when bonds form, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The rehearsal script says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains 28 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 11: display K — read endpoints before discussing the pathway
The source is a paper-based model. Its invented vertical levels are R = 127, P = 71, uncatalysed peak U = 201 and catalysed peak C = 168, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 71 − 127 = -56. This is exothermic; its magnitude is 56 model units and its enthalpy sign is negative. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 201 − 127 = 74. Reverse activation energy on the same pathway is U − P = 201 − 71 = 130. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 168 − 127 = 41, a reduction of 33 units, or 44.595% of the original forward barrier. Reverse barrier becomes C − P = 97. The catalyst does not change P − R = -56; it changes the available pathway. It is wrong to claim that a 44.595% lower barrier guarantees a 44.595% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The exothermic sign says the overall reaction releases more energy in forming bonds than is needed to break the relevant bonds, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The caption approval sheet says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains -56 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
Profile 12: display L — read endpoints before discussing the pathway
The source is a diagram marked in arbitrary units. Its invented vertical levels are R = 54, P = 88, uncatalysed peak U = 154 and catalysed peak C = 124, all on the same arbitrary scale. Begin with a simple sketch of two pathways sharing a reactant and product. Do not treat the numbers as measurements from a hot pack, a real neutralisation or a physiological response. This model lets us test how carefully a candidate separates location, signed difference and mechanism.
Skill 1 — classify, show the subtraction, name the sign. Compute P − R = 88 − 54 = 34. This is endothermic; its magnitude is 34 model units and its enthalpy sign is positive. A clear explanation does not say ‘the peak is high, so the reaction is endothermic’. The top of a profile describes an activation barrier; the product-versus-reactant difference describes the overall energy change. Students should be able to cover the peak labels and still state the correct category.
Skill 2 — compare forward and reverse barriers. Uncatalysed forward activation energy is U − R = 154 − 54 = 100. Reverse activation energy on the same pathway is U − P = 154 − 88 = 66. These are not identical merely because the transition region is shared. When the products are lower, climbing back from products reaches a larger barrier; when products are higher, it is the other way round. Drawing arrows from the right starting levels prevents most algebra errors.
Skill 3 — introduce the catalyst without moving the endpoints. The catalysed forward barrier is C − R = 124 − 54 = 70, a reduction of 30 units, or 30% of the original forward barrier. Reverse barrier becomes C − P = 36. The catalyst does not change P − R = 34; it changes the available pathway. It is wrong to claim that a 30% lower barrier guarantees a 30% faster reaction. Rate depends on more than this diagram alone.
Skill 4 — explain bond changes correctly. The endothermic sign says the overall reaction requires more energy to break relevant bonds than is released when bonds form, in a qualitative bond-energy account. It does not tell you the names or number of bonds. Without a formula or reaction equation, do not invent the molecules. A reliable structured answer distinguishes the net result from its microscopic explanation and does not confuse bond breaking with an energy-releasing process.
Skill 5 — judge a caption. The unapproved poster says, ‘The catalyst changed the reaction’s enthalpy because its line has a lower maximum.’ The correction is: ‘The catalyst offers a lower activation-energy pathway; the modelled reaction enthalpy remains 34 units.’ If no units or source are supplied on a public figure, state that the levels are illustrative. This is a source-and-meaning check: an accurate calculation becomes poor Science when the sentence attaches it to the wrong physical quantity.
From model diagrams to examination responses
When you meet a multiple-choice question, determine whether it asks about sign, barrier, product level, catalyst or the effect of reversing direction. Do not inspect all four options before forming a prediction. Draw a tiny reactant-to-product arrow, write the relevant subtraction, then choose. In a structured explanation, use the nouns ‘reactants’, ‘products’, ‘activation energy’ and ‘enthalpy change’; replacing them with ‘first line’ and ‘second line’ makes the answer harder to assess and easier to misread.
For a practical evaluation question, separate observed temperature change from inferred energy changes. For example, an uninsulated container can lose energy, and reading the thermometer after the maximum can understate the temperature rise. This is not permission to invent a numerical correction. It is a reason to name the likely direction of error and propose a suitable controlled comparison. Unless a question supplies heat capacities, masses and specific assumptions, do not calculate a reaction enthalpy from a single temperature record.
Final self-test: what changes when a catalyst is added? The activation-energy route. What does not change in the drawn model? Reactant level, product level and hence the sign and magnitude of overall enthalpy change. What changes if the reverse reaction is considered? Starting level, ending level, enthalpy sign and the relevant forward barrier. What remains crucial? Read the question’s direction and its units. These distinctions survive unfamiliar diagrams.
For a foundation-to-secondary bridge, read the eduKateSengkang post-PSLE learning plan and PSLE Science on temperature versus energy transferred. The related subject routes are secondary Mathematics tutorials, English listening improvement and G3 Science lab-report reasoning. These links give background rather than substituting for the examination syllabus.
