Edexcel Jun 2017 Paper 1 Q21 (with explained solutions) |

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Edexcel Jun 2017 Paper 1 Q21

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21 Magnesium is in Group 2 of the Periodic Table. It has a number of naturally occurring isotopes, including 24Mg and 26Mg. (a) (i) Explain, in terms of the subatomic particles in the atoms, why 24Mg and 26Mg are isotopes. (2)(ii) A sample of magnesium, which contains only the isotopes 24Mg and 26Mg, has a relative atomic mass of 24.433. Calculate the percentage abundance of each isotope in this sample of magnesium. (2) 18 *P53491A01832* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />
(1) (2) *(iii) Explain why the first ionization energy of magnesium is higher than the first ionization energy of sodium. A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D (b) (i) Give the electronic structure of a magnesium atom, using the s, p, d notation. (1)(ii) Write the equation for the first ionization energy of magnesium. Include state symbols.(iv) Explain why the first ionization energy of magnesium is higher than the first ionization energy of aluminium. (2)A E R A S I H T N I E T I R W T O N O D*P53491A01932* 19 Turn over<br />
(c) Magnesium carbonate decomposes on heating: MgCO3(s) (cid:111) MgO(s) + CO2(g) (cid:168)H1 The enthalpy change can only be determined indirectly, by applying Hesss Law. (i) Explain why the enthalpy change of this reaction cannot be determined directly. (1)D O N O T W R I T E I N T H I S A R E A (ii) State Hesss Law. (1)(d) A class of students carried out an experiment to measure (cid:168)H1 indirectly by determining the enthalpy changes of two reactions: MgCO3(s) + 2HCl(aq) (cid:111) MgCl2(aq) + H2O(l) + CO2(g) (cid:168)H2 MgO(s) + 2HCl(aq) (cid:111) MgCl2(aq) + H2O(l) (cid:168)H3 One group used the following method to measure (cid:168)H2. 20Pipette 50.0 cm3 of 2 mol dm3 hydrochloric acid (a large excess) into a polystyrene cup and note the temperature of the acid.Weigh accurately 2.50 g of magnesium carbonate powder.Add the magnesium carbonate to the acid, stir continuously and note the highest temperature. *P53491A02032* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />
(i) Why is excess hydrochloric acid used? (1)(ii) The students were told that using a polystyrene cup gives better results than using a glass beaker because of its good thermal insulation and its low heat capacity. Explain why these properties improve experimental results. (2) Good thermal insulationA E R A S I H T N I E T I R W T O N O DLow heat capacity(iii) One student using this method measured a temperature increase of 18.5C. Calculate the energy change, in joules, for this reaction. Assume the specific heat capacity of the solution is 4.18 J g1 C1 and use the expression energy change (J) = 50.0specific heat capacitytemperature change (1) A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D *P53491A02132* 21 Turn over<br />
*(iv) Use your answer to (d)(iii) to calculate the molar enthalpy change for the reaction between magnesium carbonate and hydrochloric acid. Give your answer to three significant figures and include a sign and units. (3) (cid:3) (cid:3) (e) The class collected all their results and, after eliminating anomalous results, calculated the mean values of (cid:168)H2 and (cid:168)H3: (cid:168)H2 = 126 kJ mol1 (cid:168)H3 = 231 kJ mol1 (i) Complete the Hess cycle below by adding the missing arrow and species. (1) MgCO3(s) MgO(s) + CO2(g) 2HCl(aq)( ) +( ) +( ) 22 *P53491A02232* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />
(ii) Use your completed Hess cycle and the students mean values for (cid:168)H2 and (cid:168)H3 to calculate the enthalpy change for the thermal decomposition of magnesium carbonate. Include a sign and units. *(f) Data book values for (cid:168)H2 and (cid:168)H3 are (cid:168)H2 = 179.4 kJ mol1 (cid:168)H3 = 296.4 kJ mol1 Most of the values obtained by the students were close to their mean values, and they suggested that the difference between their values and those from the data book was due to the measurement uncertainties in their experiments. Evaluate this suggestion. (2) (2)(Total for Question 21 = 24 marks) A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D *P53491A02332* 23 Turn over<br />

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Acceptable Answer Reject Mark Question Number 21(a)(i) (2) Reject Mark (2) (24Mg and 26Mg atoms) have 12 protons (1) IGNORE (24Mg and 26Mg atoms) have the same number of protons / proton number 24Mg has 12 neutrons but 26Mg has 14 neutrons ALLOW 26Mg has two more neutrons than 24Mg (1) ALLOW for 1 mark Just 24Mg and 26Mg atoms have the same number of protons/proton number and different numbers of neutrons /neutron number IGNORE 24Mg and 26Mg atoms have the same atomic number but different mass numbers References to electrons unless incorrect (24x + 26(100x)100 = 24.433 (1) 2x = 26002443.3 = 156.7 % 24Mg = x = 78.35; % 26Mg = 100x = 21.65 ALLOW 78.4 and 21.6 OR 78.3 and 21.7 (1) Correct answers with no working scores 2 marks. Ignore SF except 1 Acceptable Answer Question Number 21(a)(ii) Percentage 24Mg = x Question Number 21(b)(i) Acceptable Answer 1s2 2s2 2p6 3s2 OR 1s2 2s2 2px 2 2py 2 2pz 2 3s2 Reject Mark (1)<br />
Mark (1) Mark (2) Reject Reject Filled / half-filled orbital has greater stability Magnesium is Mg2+ but sodium is Na+ Mg has higher charge density (than Na) Acceptable Answer Mg(g)Mg+(g) + e()((g)) OR Mg(g)e()((g)Mg+(g) Acceptable Answer Question Number *21(b)(iii) Ionized / outer electrons are in the 3s / same orbital/subshell (for each atom) ALLOW Same shell (for subshell) Atoms have the same inner shell shielding (1) Mg has one more proton (in the nucleus) (so attractive force is greater) ALLOW Higher proton number Greater effective nuclear charge Reverse argument (1) IGNORE References to atomic radius Atomic number Question Number 21(b)(ii)<br />
Question Number 21(b)(iv) Acceptable Answer Reject Filled 3s orbital is stable (The nuclear charge is greater but ) Ionized / outer electron of aluminium is in a (3)p orbital / the (3)p subshell (1) which is further from the nucleus than the (3)s orbital OR is at a higher energy than the (3)s orbital OR is shielded by the (inner) 3s orbital ALLOW Use of 3s subshell for 3s orbital Reverse argument (1) IGNORE Use of 2s and 2p for 3s and 3p if consistent Mark (2) Mark (1) Acceptable Answer Question Number 21(c)(i) When heat is supplied to a system, it is very difficult / impossible to measure the heat absorbed by the reaction OR When heat is supplied to a system, it is very difficult / impossible to measure the temperature change due to the reaction ALLOW When heat is supplied to a system, it is very difficult / impossible to measure the temperature change OR Difficult to measure the temperature of a solid IGNORE Reference to thermicity of the reaction Reject Just difficult / impossible to measure the temperature change<br />
Acceptable Answer Enthalpy / heat change of a reaction is independent of the route. ALLOW Enthalpy / heat change is independent of the route. Reject Acceptable Answer Reject So that all the MgCO3 reacts. ALLOW So that all the solid reacts So that all the solid reacts IGNORE Reference to limiting factors Just to ensure complete reaction Mark (1) Mark (1) Acceptable Answer Reject Mark No heat loss (2) Reject Mark (1) (Good thermal insulation) reduces heat transfer with the surroundings ALLOW Reduces heat loss to the surroundings (1) (Low heat capacity) less / little heat is used to heat / cool the container (1) Acceptable Answer (E) = 50.0 x 4.18 x 18.5 = 3866.5 (J) OR 3.8665 = 3.87 kJ IGNORE SF except 1 SF +/ signs Question Number 21(c)(ii) Question Number 21(d)(i) Question Number 21(d)(ii) Question Number 21(d)(iii)<br />
Acceptable Answer Question Number 21(d)(iv) Molar mass MgCO3 = 84.3 (1) H = ()answer 21(d)(iii)mol MgCO3 = ()3866.5(2.50/84.3 ) OR = ()3866.50.029656 OR = ()3866.5 x 33.72 (1) = ()130378 = 130 000 / 1.30 x 105 J mol1 OR = 130 kJ mol1 (1) TE at each stage Correct answer with no working scores 3 Correct answer with no working and no or incorrect units and / or sign scores 2 Mark (3) Reject Answer not to 3 SF 130 kJ mol1 Question Number 21(e)(i) Acceptable Answer Reject Mark (1) MgCO3(s) MgO(s) + CO2(g) 2HCl(aq) 2HCl(aq) MgCl2(aq) + H2O(l) + CO2(g) IGNORE Omission of 2HCl(aq) (on lhs) Acceptable Answer Question Number 21(e)(ii) H1 = H2H3 (1) Reject Mark (2) = 126(231) = +105 kJ mol1 (1) Correct answer including sign, with no working scores 2 105 kJ mol1 / 105 kJ mol1 / +105 all score 1 mark No TE on incorrect cycle equation<br />
Acceptable Answer The student values were much smaller / smaller magnitude / less negative than the Data Book values which indicates a systematic error (1) Uncertainties will give values scattered about the true value (so cannot explain the discrepancy) OR The results obtained by the students are precise but inaccurate (1) If no other mark is scored allow uncertainties are too small to account for the discrepancy scores 1 IGNORE References to likely sources of error such as heat loss Reject Mark (2) (Total for Question 21 = 24 marks) Question Number 21(f)<br />

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