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OCR A AS Jun 2016 Paper 1 Q24
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![24 Aqueous lead(II) nitrate, Pb(NO3)2(aq), and aqueous potassium iodide, KI(aq), react together. The equation is shown below. 18 Pb(NO3)2(aq) + 2KI(aq)PbI2(s) + 2KNO3(aq) A student carries out an experiment to determine the enthalpy change of reaction, rH, of this reaction. The student follows the method outlined below. Add 50.0 cm3 of 1.50 mol dm3 Pb(NO3)2(aq) to a polystyrene cup. (cid:129) (cid:129) Measure out 50.0 cm3 of a solution of KI(aq), which is in excess. (cid:129) Measure the temperature of both solutions. (cid:129) Add the KI(aq) to the polystyrene cup, stir the mixture and record the maximum temperature. Temperature readings Initial temperature of both solutions = 19.5 C Maximum temperature of mixture = 30.0 C (a) Calculate rH, in kJ mol1, for the reaction shown in the equation above. Give your answer to an appropriate number of significant figures. Assume that the density of all solutions and specific heat capacity, c, of the reaction mixture is the same as for water. rH =kJ mol1 [4]OCR 2016<br />
(b) Write an ionic equation for the reaction that the student carries out. 19 Include state symbols.[1] (c) The 50.0 cm3 of KI(aq) used in the experiment contains 10% more KI than is needed to react with 50.0 cm3 of 1.50 mol dm3 Pb(NO3)2(aq). Pb(NO3)2(aq) + 2KI(aq)PbI2(s) + 2KNO3(aq) Calculate the concentration, in mol dm3, of KI that the student used. concentration of KI =mol dm3 [2]OCR 2016 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/June_2016_QP_-_Paper_1_OCR_Chemistry_A_AS-Level_24-0.jpg)
![H032/01 Mark Scheme June 2016 Question (a) 24 Marks Answer 1000 = 0.075(0) (mol)FIRST, CHECK THE ANSWER ON ANSWER LINE IF rH = 58.5 (kJ mol1) award 4 marks --------------------------------------- Energy released in J OR kJ = 100.04.1810.5 = 4389 (J) OR 4.389 (kJ)Correctly calculates n(Pb(NO3)2) = 1.5050 H value in J OR kJ Answer MUST divide energy by n(Pb(NO3)2) () 4389 OR () (Sign ignored and/or more than 3 SF) Correct rH in kJ ANDsign AND 3 SF = 58.5 (kJ mol1)0.0750 OR ()58520 (J) 4.389 0.0750 ()58.52 (kJ) OR 13 AO element AO2.4 AO2.4 AO2.8 AO2.8 Guidance FULL ANNOTATIONS MUST BE USED ----------------------------------- ALLOW 4390 J; 4.39 kJ DO NOT ALLOW less than 3 SF IGNORE units i.e. ALLOW correctly calculated number in J OR kJ ALLOW ECF from n(Pb(NO3)2) AND/OR Energy ALLOW 58500 (from 4390) IGNORE absence ofsign and 3 SF requirement Final mark requiressign, kJ AND 3 SF Note: From 4390 J, rH = 58.5 (kJ mol1) (SAME) ---------------------------------------------------- Common error 29.3 3 marks (50 g instead of 100 g in mcT)<br />
H032/01 Mark Scheme June 2016 Question (b) Answer Pb2+(aq) + 2I(aq)PbI2(s)State symbols required Marks AO element AO2.7 AO2.8 2 50 = 3 (mol dm3)FIRST, CHECK ANSWER ON ANSWER LINE IF [KI(aq)] rounds to 3.3 mol dm3 e.g. 3.30, 3.33, 3.3 recurring -------------------------------------------------------------------- Method 1 [KI(aq)] for complete reaction = 20.0750 = 0.150 mol1000 10% greater gives 31.1 = 3.3(0)OR-------------------------------------------------------------- Method 2 n(KI(aq)) required = 2.20.0750 = 0.165 mol[KI(aq)] = 0.1651000 50 = 3.3(0) (mol dm3)(c) Total 14 Guidance (aq) on ALLOW Pb+2(aq) IGNORE spectator ions, K+(aq) and 2NO3 both sides ALLOW ECF from incorrect n(Pb(NO3)2) from 24(a) BUT if (a) is incorrect but 0.0750 used here, treat as a fresh start and IGNORE response from 24(a) ALLOW 2 marks for 3.3/3.3 recurring Attempt at increasing concentration by 10% = 20.0750 = 0.150 mol1000 ALLOW ECF from incorrect n(KI) --------------------------------------------------------------------- Common errors 3 1.65 1 mark (no factor of 2 used for KI) 2.7 1 mark (10% less rather than 10% more) 2.73/2.72 1 mark (10% increase in volume: 55 cm3) 1 mark (Correct for KI with no extra 10%) 45 = 3.33 (mol dm3)<br />](https://learnah.org/wp-content/uploads/2018/08/June_2016_MS_-_Paper_1_OCR_Chemistry_A_AS-Level_24-0.jpg)
