SAJC Prelim P1 ANS

1 [Turn Over SAJC Prelim 2015 H2 Chemistry Paper 1 Worked Solutions 1 2 3 4 5 6 7 8 9 10 C B D A B C A B B B 11 12 13 14 15 16 17 18 19 20 D B C D B A D C C C 21 22 23 24 25 26 27 28 29 30 C B C D D B D B C D 31 32 33 34 35 36 37 38 39 40 C A A B B B B A D D 1 The mass percentage of magnesium in a mixture of magnesium chloride (Mr = 95.3) and magnesium nitrate (Mr = 148.3) was found to be 21.25%. What mass of magnesium chloride is present in 323 g of the mixture? A 151 g B 165 g C 172 g D 181 g (24.3/95.3)(x) + (24.3/148.3)(1 – x) = 0.2125 Solving for x, x = 0.5338 Mass of magnesium chloride = 0.5338 x 323 = 172g 2 Which conversion results in a loss of electrons for the underlined species? A NO3 − to NO2 (+5 to +4) B NO2 − to NO3 − (+3 to +5) A loss of electrons results in an increase in oxidation state. C S2O8 2− to SO4 2− (+7 to +6) D S4O6 2− to S2O3 2− (+2.5 to +2)

2 [Turn Over 3 Which of the following pairs shows Gas Y deviating more from ideal gas behaviour than Gas X at the same temperature and pressure? Gas X Gas Y A HF HCl The pd-pd interactions between HCl molecules are weaker than the hydrogen bonds between HF molecules, so HCl possesses less significant IMF and is more ideal. B H2O H2 The id-id interactions between H 2 molecules are weaker than the hydrogen bonds between H2O molecules, so H2 possesses less significant IMF and is more ideal. C I2 Br2 The id -id interactions between Br 2 molecules are weaker than t he id -id interactions between I 2 molecules, as Br 2 has a smaller electron cloud , so Br2 possesses less significant IMF and is more ideal. D O2 NO2 The pd-pd interactions between NO2 molecules are stronger than the id-id interactions between O2 molecules, so NO2 possesses more significant IMF and is less ideal. 4 Use of the Data Booklet is relevant to this question. Which of the following shows the correct increasing trend for the angle of deflection when placed in an electric field? A Cl − < Ru3+ < Zn2+ B Ru3+ < Cl − < Zn2+ C Cl − < Zn2+ < Ru3+ D Zn2+ < Ru3+ < Cl − Angle of deflection 𝛼 𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 Cl − < Ru3+ < Zn2+ 1/35.5 < 3/101 < 2/65.4 0.0282 < 0.0297 < 0.0306

3 [Turn Over 5 Which of the following cannot form dimers? A CH3COOH dimerization via hydrogen bonds B CH3CHO Incapable of forming intermolecular hydrogen bonding between CH3CHO molecules C AlCl3 dimerization via dative bonds D NO2 dimerization via formation of covalent bond

4 [Turn Over 6 Which of the following species is planar and does not have dative bonding? A PO4 3− tetrahedral (non-planar), no dative bonds B SO4 2− tetrahedral (non-planar), no dative bonds C CO3 2− trigonal planar (planar), no dative bonds D NO3 − trigonal planar (planar), 1 dative bond

5 [Turn Over 7 The behavior of Group II sulfates and hydroxides show different trends down the group. Group II sulfates become less soluble down the group, while Group II hydroxides become more soluble down the group. Which of the following statements helps to explain this trend? A Down the group, the magnitude of the lattice energy of Group II hydroxides has a more significant decrease than the magnitude of the sum of hydration energies of the ions. B Down the group, the magnitude of the lattice energy of Group II sulfates has a more significant decrease than the magnitude of the sum of hydration energies of the ions. C The hydration energy of the sulfate anion is larger in magnitude than the hydroxide anion. D The magnitude of the lattice energy of Group II ionic compounds increases down the group. ∆Hsoln = ∆Hhyd – LE For ∆Hhyd, it is also always negative, but a decrease in the magnitude of ∆Hhyd leads to a more positive / less negative ∆Hsoln according to the above expression. For LE, as it is always negative (exothermic), a decrease in the magnitude of LE leads to a less positive / more negative ∆Hsoln according to the above expression. So if hydroxides become more soluble down the group, their ∆Hhyd has to become more negative, and that can only be explained by answer A. Statement B leads to a more negative ∆Hhyd and hence increased solubility of the sulfates as we go down the group, which contradicts the information in the question. Statement C is wrong as hydration energy is proportionate to charge density. Sulfate ion despite having a -2 charge, is also much larger than the hydroxide ion, and has a lower charge density as a result. Statement D is wrong as the cation size increases down group II, and since lattice energy is proportionate to q+q-/(r++r-), lattice energy should decrease in magnitude.

6 [Turn Over 8 Phosphorus exists in several allotropes. White phosphorus has the structure P 4, while black phosphorus has a structure similar to graphite. While the transformation of white phosphorus to black phosphorus is spontaneous, the rate is very slow, and usually involves the use of a metal salt catalyst to increase its rate. Which of the following is true? A Black phosphorus has a simple molecular structure. Untrue as the question stated black phosphorus has a structure similar to graphite, and hence it must have a giant covalent structure. B H for the transformation of white phosphorus to black phosphorus is less than zero. True because ∆G = ∆H - T∆S, and since ∆S is negative (see statement C), ∆H must be negative, otherwise the reaction will never be spontaneous. C S for the transformation of white phosphorus to black phosphorus is greater than zero. Untrue as the conversion from white phosphorus to black involves merging simple covalent molecules to form a giant covalent molecule. There is a decrease in the number of molecules, hence entropy change is negative. D The uncatalysed reaction of transforming white phosphorus to black phosphorus has a small activation energy. Untrue as the information stated that the rate is very slow without a catalyst, hence it must have a large activation energy. 9 What is the half-life of a first-order reaction where the concentration of the reactant drops to 1 5 of its initial concentration after 51 minutes? A 20 min B 22 min

7 [Turn Over C 24 min D 26 min [A]t = [A]0(1/2)n where n is the no. of half-life 1/5 = 1(1/2)n lg(1/5) = n lg(1/2) n = 2.322 t1/2 = 51 / 2.322 = 22.0 min 10 The graph below shows the change in rate for an enzyme-catalysed reaction. Which of the following statements is incorrect? A At low [substrate], the rate is first order with respect to [substrate]. B At low [substrate], the reaction slows down due to increasing concentration of the products. It should be: The reaction slows down as the enzyme active sites are gradually occupied by the substrate molecules. C At high [substrate], the rate is constant as all the enzyme active sites are used up. D At high [substrate], increasing [enzyme] increases the rate.

8 [Turn Over 11 A ketone such as propanone can undergo a rearrangement of its atoms to form an enol when dissolved in a non-polar solvent. Enols exist together with their ketone forms in equilibrium. ∆H = +48 kJ mol-1 Which of the following statements can be determined from the information above? A Increasing the temperature results in the shift of the position of equilibrium to the left. Wrong, as the enthalpy change is positive, an increase in temperature would shift the equilibrium to the right. B The entropy change of the reaction is positive under aqueous conditions. Wrong, as there is no change in the no. of molecules, entropy change is negligible. C Adding of warm alkaline iodine results in the shift of the position of equilibrium to the right. Wrong, as the iodoform reaction consumes pro