m $60\\text{ s}$ to $90\\text{ s}$ the curve still rises.","highlights":[{"text":"The rate of reaction stayed at zero","index":2,"location":"eliminate"}]},{"type":"text","order":4,"title":"Select decreased","content":"Choose **“The rate of reaction decreased”**. The gradient becomes less steep between $60\\text{ s}$ and $90\\text{ s}$, so hydrogen is produced more slowly. This clear statement and gradient reason earn the **B1** mark.","highlights":[{"text":"The rate of reaction decreased","index":1,"location":"option"}]}]}],"generatedAt":"2026-07-21T04:08:48.808Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]},{"id":"ad0f5723-0987-4d14-9b4d-f9da2029f179","label":"c","content":"The student repeated the experiment at a lower temperature, keeping all other variables constant. How would the initial line of best fit compare to the original line on the graph?\n- Initially the line of best fit would be less steep\n- Initially the line of best fit would be the same steepness\n- Initially the line of best fit would be steeper","markscheme":"- State that initially the line of best fit would be less steep (since lower temperature reduces the rate of reaction) B1","marks":1,"order":2,"markingRubric":null,"explanation":{"methods":[{"id":"rate_comparison","label":"Use collision theory","steps":[{"type":"text","order":0,"title":"Link temperature to rate","content":"A lower temperature gives the reacting particles less kinetic energy, so there are fewer successful collisions each second. The reaction is therefore slower. This uses the condition **\"keeping all other variables constant\"**: temperature is the factor being changed."},{"type":"text","order":1,"title":"Choose the gradient","content":"On a reaction graph, the initial line of best fit shows the rate at the start: a slower reaction has a smaller gradient. Therefore, the correct answer is **Initially the line of best fit would be less steep**. This earns the B1 mark: state both the correct comparison and the reason that lower temperature reduces the rate.","highlights":[{"text":"Initially the line of best fit would be less steep","index":0,"location":"option"},{"text":"The student repeated the experiment at a lower temperature","location":"specification"}]}]}],"generatedAt":"2026-07-21T04:09:38.734Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"27cb1259-e2a6-43e2-bee8-fa84a81d6883","revisionId":"41d3cb2f-c94d-42a0-b942-affb936d5af2","questionNumber":null,"questionType":"LA","specification":"A student investigated the rate of reaction for the catalytic decomposition of hydrogen peroxide. The table below shows the volume of oxygen gas collected after 90 seconds in four repeat tests:\n\n| Test | Volume of gas collected in \$\\text{cm}^3\$ |\n| :--- | :--- |\n| Test 1 | 92 |\n| Test 2 | 97 |\n| Test 3 | 91 |\n| Test 4 | 48 |","difficulty":3,"difficultyBand":"easy","totalMarks":4,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"747ee116-8b27-47eb-afff-c3b18e020e4c","label":null,"content":"Calculate the mean volume of gas collected at 90 seconds.\n\nDo **not** include the anomalous result in your calculation.\n\nGive your answer to 2 significant figures.","markscheme":"- Identifies $48\\text{ cm}^3$ (Test 4) as the anomalous result B1\n- Calculates the sum of the concordant results: $92 + 97 + 91 = 280$ M1\n- Calculates the mean by dividing by the number of concordant trials: $\\frac{280}{3} \\approx 93.33$ M1\n- Rounds to 2 significant figures to obtain $93\\text{ cm}^3$ A1","marks":4,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"mean_excluding_anomaly","label":"Mean excluding the anomaly","steps":[{"type":"text","order":0,"title":"Identify the anomaly","content":"Test 4, $48\\text{ cm}^3$, is anomalous: it is much lower than the other three volumes, which are all between $91\\text{ cm}^3$ and $97\\text{ cm}^3$. Do not include it in the mean. Identifying this

Rate of reaction

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Rate of reaction

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