ACSI Paper 2 Prelim (Student Version) Answer

1 YEAR 6 PRELIMINARY EXAMINATION 2023 INTERNATIONAL BACCALAUREATE DIPLOMA PROGRAMME CHEMISTRY HIGHER LEVEL PAPER 2 Wednesday 13 th September 2023 2 hours 15 minutes INSTRUCTIONS TO CANDIDATES  Do not open this examination paper until instructed to do so.  Write your candidate session number in the box above.  A calculator is required for this paper.  A copy of the Chemistry Data Booklet is required for this paper.  Write your answers in the boxes provided.  If you use additional sheets of paper for your answer, attach them to the booklet. Indicate the question number clearly on these sheets.  All drawings must be in ink. For examiner’s use Qn 1 /6 Qn 2 /22 Qn 3 /8 Qn 4 /10 Qn 5 /13 Qn 6 /9 Qn 7 /15 Qn 8 /7 Wrong s.f. /units Total /90 ________________________________________________________________________ This question paper consists of 31 printed pages including this cover page. Name Candidate Number

2 Answer all questions. Write your answers in the boxes provided. 1. Carbon consists of 4 allotropes in the form of diamond, graphite, fullerene and graphene. (a) State and explain the difference in electrical conductivity of diamond and graphite. [2] Diamond is a non-conductor Graphite is a good conductor Diamond uses all the valence electrons in bonding/ no free electrons Graphite has delocalized electrons to conduct electricity Marker comments Most students can answer the question. Some students misused words like “lone pairs” or lone electrons (b) Explain why the carbon-carbon bonds in graphite are stronger than the carbon- carbon bonds found in diamond. [2] Resonance or delocalization of pi electrons in graphite/ overlap of pi electrons or overlap sp2-sp2 and sp3-sp3 C Resulting in shorter bonds/ partial double bond character in C-C bonds/ 1.333 Bond order Marker comments A few students related this to Diamond having a weaker London Dispersion Force compared to the dipole-dipole forces in graphite, this is a conceptual misunderstanding. Students are to note that the bond resulting from delocalised electrons is not 1.5 bond order and it is not a double bond in graphite but rather a partial double bond. (c) State the type of hybridization around the carbon atoms in diamond and graphite. [2] Diamond: sp3 Graphite: sp2 Marker comments Generally well done

3 2. The energy cycle shown can be used to calculate the enthalpy change of formation of KCl (s). Each arrow indicates a transformation, W, X, and Y. Each transformation consists of one or more steps. (a) (i) Calculate the value of the following transformations using section 8 of the data booklet and the following information. Enthalpy change of atomisation of K (s) +89 kJ mol─1 Enthalpy change of atomisation of Cl2 (g) +121 kJ mol─1 Enthalpy change of solution of KCl (s) 17 kJ mol─1 Enthalpy change of hydration of Cl─(g) 365 kJ mol─1 Enthalpy change of hydration of K+(g) 340 kJ mol─1 [4] Transformation W: K (s) + 1 2Cl2 (g)  K+ (g) + Cl─ (g) ...................................................................................................................................................... ...................................................................................................................................................... ...................................................................................................................................................... Transformation Y: K+ (g) + Cl─ (g)  KCl (aq) Marker comments For W, many students divided 121 by 2 as they mistook it as bond energy. Some students used the BE in the data booklet but this is not allowed as Q did not require them to refer to section 11 of the data booklet. For Y, some students mistook and included -17 in their calculation K(s)  K+ (g) H = +89 + 419 = +508 kJ mol─1 1 2Cl2 (g)  Cl─ (g) H = +121 + (─349) = ─228 kJ mol─1 H = 508 ─ 228 = +280 kJ mol─1 H = ─365 ─ 340 = ─705 kJ mol─1

4 (ii) Using your answer to (a)(i) and the given data to calculate the enthalpy change of formation of KCl (s). If you did not get answers to (a)(i), use +200 kJ mol─1 and ─650 kJ mol─1 for W and Y respectively, but these are not the correct answers. [2] ...................................................................................................................................................... ...................................................................................................................................................... Marker comments Most students are awarded full ECF. (iii) State and explain the sign for the entropy change for the formation of KCl (s). [1] ...................................................................................................................................................... Marker comments: Many students did not make clear reference to gas but just stated fewer ways to distribute energy or less disorder. Quite a few misread the Q as enthalpy and explain using bond formation and bond breaking. (iv) Deduce, giving reasons, whether altering the temperature would change the spontaneity for the formation of KCl (s). [2] ...................................................................................................................................................... Marker comments: Most students were able to get this Q correct. (b) Explain why the enthalpy change of hydration of K+(g) is less exothermic than that of Ca2+ (g). [1] ...................................................................................................................................................... ...................................................................................................................................................... Marker comments Most students were able to get this Q correct. for substituting enthalpy change of solution H = +280 + (─705) + (17) = ─408 kJ mol─1 Or H = +200 + (─650) + (17) = ─433 kJ mol─1 The sign will be negative as the number of moles of gases decreases from 0.5 mol to 0 mol. Accept change in phase from gas to solid As temperature increases, ─TS becomes more positive,hence G becomes more positive and reaction becomes non-spontaneous. Ca2+ has a higher charge density than K+, hence it forms stronger ion-dipole interaction with water molecules and releases more hydration energy. Accept explanation using charge and ionic size.

5 (c) (i) Explain why the ionic radius of K+ is smaller than that of Cl─. [2] ...................................................................................................................................................... Marker comments Many students mentioned the same number of quantum shells. Some incorrect answers include, K lost an electron and 1 quantum shell but not for Cl, or greater interelectronic repulsion in Cl-. (ii) Explain the decreasing trend in the first ionization energies of group 1 and group 17 elements down the group. [1] ...................................................................................................................................................... Marker comments Most students were able to get this Q correct. (iii) The Periodic Table groups together elements with similar properties. In most Periodic Tables, hydrogen is placed at the top of Group 1, but in some it is placed at the top of Group 17. Using your knowledge of chemistry, state one reason for hydrogen being placed above either Group 1 or Group 17. [2] Reason why hydrogen is placed above Group 1: ....................................................