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OCR A Jun 2017 Paper 1 Q19
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![16 19 Aqueous solutions of hydrogen peroxide, H2O2(aq), decompose as in the equation below. A student investigates the decomposition of H2O2(aq) by measuring the volume of oxygen gas produced over time. All gas volumes are measured at room temperature and pressure. The student uses 25.0 cm3 of 2.30 mol dm3 H2O2. From the results, the student determines the concentration of H2O2(aq) at each time. The student then plots a concentrationtime graph. 2H2O2(aq) 2H2O(l) + O2(g) [H2O2(aq)] / mol dm3 2.5 2.0 1.5 1.0 0.5 0.0 500 1000 1500 time / s 2000 2500 3000 (a) Determine the total volume of oxygen, measured at room temperature and pressure, that the student should be prepared to collect in this investigation. Suggest apparatus that would allow this gas volume to be collected, indicating clearly the scale of working.[3]<br />
17 (b) Suggest a different experimental method that would allow the rate of this reaction to be followed over time.[1] (c)* Determine the initial rate of reaction, the order with respect to H2O2, and the rate constant. Your answer must show full working on the graph and on the lines below.[6]<br />](https://learnah.org/wp-content/uploads/2018/08/June_2017_QP_-_Paper_1_OCR_Chemistry_A_A-Level_19-0.jpg)
![H432/01 Mark Scheme June 2017 Guidance ALLOW 0.69(0) dm3 2nd mark subsumes 1st mark ALLOW 1000 cm3/1 dm3 syringe Needs a name of actual apparatus, not just container measuring cylinder without volume is insufficient DO NOT ALLOW burette For other possible apparatus, contact Team Leader ALLOW volumes from 7001000 cm3 but should be realistic apparatus, e.g. 700, 750, 800, 850, 900, 950. ALLOW weight for mass ALLOW take samples and titrate (remaining H2O2) Question (a) 19 Answer n(H2O2) = 2.30OR = 0.0575 (mol)25.0 1000 Marks vol O2 = 0.057524000 = 690 cm3Collect in 1000 cm3/1 dm3 measuring cylinder(b) Measure mass (loss)13<br />
H432/01 Mark Scheme June 2017 (c)* Question Answer Please refer to the marking instructions on page 5 of mark scheme for guidance on marking this question. Level 3 (56 marks) A comprehensive conclusion using quantitative data from the graph to correctly determine initial rate AND half lives/gradient with 1st order conclusion for H2O2 AND determination of k. There is a well-developed line of reasoning which is clear and logically structured. Clear working for initial rate, half life/gradient and order and k. Units mostly correct throughout. Level 2 (34 marks) Attempts to describe all three scientific points but explanations may be incomplete. OR Explains two scientific points thoroughly with few omissions. There is a line of reasoning with some structure and supported by some evidence. The scientific points are supported by evidence from the graph. Level 1 (12 marks) Reaches a simple conclusion using at least one piece of quantitative data from the graph. Attempts to calculate initial rate OR half life. There is an attempt at a logical structure with a reasoned conclusion from the evidence. 0 marks No response worthy of credit. Marks Guidance Indicative scientific points may include: Initial rateTangent shown on graph as line at t = 0 sGradient determined in range: 1.52.0103 e.g. 2.3 1300 = 1.77103initial rate as gradient value with units: mol dm3 s 1, For other methods contact TL Evidence for 1st order1st order clearly linked to half-life OR 2 gradients: 1. Half life 2 methodsHalf life shown on graphHalf life range 8001000 sTwo constant half lives 50 s 2. Two gradientstwo rates2 tangents shown on graph at c and c/2Gradient at c/2 is half gradient at c e.g. c = 2.3 mol dm3, gradient = 1.6103 AND c = 1.15 mol dm3, gradient = 0.8103For chosen method, conclusion: H2O2 is 1st order Determination of k 2 methodsk clearly linked to rate OR half-life: 1.6103 rate = 7104 k = [H2O2] e.g. k = 2.3 s1 OR k = e.g. k = 0.693 950 = 7.3104 s1 Total 10 14<br />](https://learnah.org/wp-content/uploads/2018/08/June_2017_MS_-_Paper_1_OCR_Chemistry_A_A-Level_19-0.jpg)
