DHS H1 CHEM P2 Answer Scheme

© DHS 2015 8872/02 [Turn over Section A Answer all questions in the spaces provided. 1 (a) Calcium ethanedioate, CaC 2O4, is a white needle–like crystalline solid. When a pure sample of anhydrous CaC 2O4 was heated strongly at 400 C until no further change in mass is observed, a white solid B and 0.028 g of carbon monoxide gas were obtained as the only products. (i) Given that 1 mole of CaC 2O4 decomposes to give 1 mole of CO(g), write a balanced equation for the thermal decomposition of CaC2O4 and identify solid B. equation : CaC2O4 CaCO3 + CO Solid B : CaCO3 (ii) Hence, determine the mass of B obtained in the reaction. Moles of of CaCO3 = moles of CO = 028 0280 . . = 0.0010 mol Mass of CaCO3 obtained = 0.0010  100.0 = 0.10 g [3] (b) The formula of potassium hydrogen ethanedioate can be written as K xHy(C2O4)z. In an experiment to determine the values of x, y and z, 4.50 g of this compound was dissolved in water and the solution made up to 1 dm3. 20.0 cm 3 of the solution was pipetted into a conical flask and then titrated with 0.0200 mol dm ––3 KMnO4 in an acidic medium. It was found that 16.50 cm 3 of KMnO 4(aq) was needed for the complete reaction with ethanedioate ions, C 2O4 2–, present. During the titration, effervescence of carbon dioxide is produced. Given the reaction of manganate(VII) in acidic medium as: MnO4 – + 8H+ + 5e– Mn2+ +4H2O (i) Write a balanced half–equation for the reaction of C2O4 2 during titration. half–equation : C2O4 2 2CO2 + 2e– (ii) Calculate the mass of C2O4 2– present in 1 dm3 of the solution. E.c.f. if half–equation from (i) is wrong. Overall equation: 5C2O4 2 + 2MnO4  + 16H+ 10CO2 + 2Mn2+ + 8H2O Moles of C2O4 2 in 20.0 cm3 = 2 5  (0.0200  1000 50.16 ) = 8.25  10–4 mol

© DHS 2015 8872/02 [Turn over Mass of C2O4 2 in 1000 cm3 = (8.25  104 )  88.0  20 1000 = 3.63 g (iii) Given that 4.50 g of potassium hydrogen ethanedioate contains 0.060 g of hydrogen, calculate the mass of potassium present in the sample. Mass of potassium present = 4.50 – 0.060 –3.63 = 0.81 g (iv) Hence, determine the values x, y and z. K H C2O4 Mass / g 0.81 0060 3.63 Moles / mol 0.81/ 39.1 = 0.021 0.060/1.0 = 0.060 3.63/88.0 = 0.041 Ratio 1 3 2 Working to calculate moles of all species: x = 1, y = 3 and z = 2 Formula is KH3(C2O4)2. [6] [Total: 9]

© DHS 2015 8872/02 [Turn over 2 The following table lists the standard enthalpy changes of combustion, Hc , of some monohydric alcohols. alcohol Hc  / kJ mol–1 methanol(l) –715 ethanol(l) –1367 propan–1–ol(l) ? butan–1–ol(l) –2671 . (a) Write an equation to represent the standard enthalpy change of combustion of propan–1– ol. CH3CH2CH2OH(l) + 2 9 O2(g)  3CO2(g) + 4H2O(l) [1] (b) (i) The difference in the standard enthalpy change of combustion of methanol and ethanol is –652 kJ mol –1. By considering the structures of the two alcohols, suggest the significance of this difference. The difference corresponds to the enthalpy change of combustion of a –CH 2– group. (ii) Hence suggest a value of Hc  for propan–1–ol. Suggest a value of Hc  for propan–1–ol : –2019 kJ mol–1 (iii) Given that Hf [H2O(l)] = –286 kJ mol 1 and Hf [CO2(g)] = –394 kJ mol 1, and using your answer to (b)(ii), calculate the standard enthalpy change of formation of propan–1–ol, Hf [CH3CH2CH2OH(l)]. Hc [CH3CH2CH2OH(l)] = 3 Hf [CO2(g)] + 4 Hf [H2O(l)] – Hf [CH3CH2CH2OH(l)] 2019 = [3(394) + 4(286)] – Hf [CH3CH2CH2OH(l) Hf [CH3CH2CH2OH(l) = –307 kJ mol–1 . [4] (c) Ethanol is metabolised in the body by an enzyme called alcohol dehydrogenase. Define enzyme and describe how alcohol dehydrogenase affects the rate of ethanol metabolism in the body. An enzyme is a biological catalyst that speeds up / alters the rate of chemical reactions and remained chemically unchanged at the end of reaction. An enzyme provides an alternate pathway with lower Ea for reactions to take place. Through the formation of an enzyme–substrate complex, an enzyme increases the likelihood of correct geometry for reactions to occur.

© DHS 2015 8872/02 [Turn over [3] [Total: 8] 3 (a) Below is the incomplete sketch of the plot of first ionisation energies against atomic (proton) numbers for ten elements. (i) On the diagram above, use crosses () to mark the first ionisation energies of Si and S. (ii) Explain why the first ionisation energy of potassium is lower than that of sodium. The valence electron of potassium is located in a shell with a higher principal quantum number. Hence the valence electrons are further away from the nucleus and less strongly attracted to the nucleus. Smaller amount of energy is needed to remove the valence electron from K. (iii) State which of the ten elements is likely to have the lowest second ionisation energy and write an equation to represent the second ionisation energy of this element. Calcium Ca+(g) + e → Ca2+(g) [4] (b) X, Y and Z are elements in Period 3 of the Periodic Table. The oxide of X dissolves sparingly in water to give a weak ly alkaline solution. The oxide of Y does not react with water, but reacts with both acids and bases. Z reacts with oxygen to form a covalent oxide with a very high melting point. 0 200 400 600 800 1000 1200 1400 1600 10 11 12 13 14 15 16 17 18 19 20 11Na 12Mg 13Al 14Si 15P 16S 17Cl 18Ar  19K  20Ca first ionisation energy / kJ mol–1 11Na 12Mg 13Al 14Si 15P 16S 17Cl 18Ar 19K 20Ca atomic number  

© DHS 2015 8872/02 [Turn over (i) Fill in the table below by identifying the elements X, Y and Z and suggest the pH of their chlorides in water. . identity of element pH of the chloride in water X Mg 6.5 Y Al 3 Z Si 2 (ii) Write balanced chemical equations to show how oxide of Y reacts with both hydrochloric acid and calcium hydroxide. Al2O3 + 6HCl → 2AlCl3 + 3H2O Al2O3 + 2NaOH + 3H2O → 2NaAl(OH)4 [4] [Total: 8] 4 Data concerning methanol and two of its analogues are given in the table below. compound Mr atomic radius of atom Z in CH3ZH / nm dipole moment / D boiling point / C pKa (pKa = –log10Ka) CH3OH 32.0 0.073 1.7 65 15.5 CH3SH 48.1 0.102 1.5 6 10.4 CH3SeH 95.0 0.116 1.3 25  5 No calculation is required. (a) Explain the differences in boiling points of the three compounds.  All three substances are simple molecular.  A larger amount of energy is needed overcome strong hydrogen bonding between CH 3OH molecules than the weaker van der Waals' interactions between CH3SH or CH3SeH molecules.  Thus CH 3OH has the highest boiling point.  Number of electrons in CH 3SeH is larger than that in CH3SH. OR Size of electron cloud of CH3SeH is larger than that of CH3SH.  CH 3SeH is more easily polarized than CH3SH.  A larger amount of energy is needed to overcome stronger and more significant van der Waals’ interactions OR instantaneous – induced dipole interactions between CH3SeH molecules.  Thus CH 3SeH has a higher boiling point than CH3SH. [3]

© DHS 2015 8872/02 [Turn over (b) Arrange the three compounds in order of decreasing acidity and suggest a reason for this trend.  Strongest CH3SeH > CH3SH > CH3OH Weakest (Z = O, S or Se)  Size of atom decreases from Se to O.  Z–H bond length decreases from SeH to OH.  Bond strength increases from Se H to OH.  Ease of breaking Z–H bond decreases from Se to O. OR  Size of atom decreases from Se to O.  Charge density of conjugate bas