VJC 2021 Prelim H2P2 Ans [Final]

1 Ó VJC 2021 9729/02/PRELIM [Turn over Victoria Junior College 2021 H2 Chemistry Prelim Exam 9729/2 Suggested Answers 1 The table below shows the fifth to eighth ionisation energies of two consecutive elements, X and Y in the second period of the Periodic Table. successive ionisation energies / kJ mol–1 5th 6th 7th 8th X 10990 13330 71330 84080 Y 11020 15160 17870 92040 (a) (i) With the aid of an equation, define the term first ionisation energy with reference to X. [2] First ionisation energy of X is the amount of energy required to remove one mole of electrons from one mole of X gaseous atoms, producing one mole of gaseous X+ ions. X(g) ® X+(g) + e– (ii) State and explain the group number of X. [1] Group 16. There is a big jump from 6th to 7th IE, indicating that the 7th electron is removed from the inner principal quantum shell [OR there are 6 valence electrons]. (b) (i) Write down the full electronic configurations of X+ and Y+. [1] X+: 1s22s22p3 Y+: 1s22s22p4 (ii) Hence, explain which element has a less endothermic second ionisation energy. [2] Y has a less endothermic second ionisation energy. It is easier to remove a paired 2p electron in Y+ due to the presence of interelectronic repulsion between electrons in the same orbital. (c) Boron is another Period 2 element which reacts vigorously with fluorine to form boron trifluoride, BF3, an important reactant in organic syntheses. Boron trifluoride is a very reactive gas and it is hard to handle at room temperature. It can be converted to a liquid compound which is easily stored by reacting it with diethyl ether in the mole ratio of 1:1. Diethyl ether can be represented by the formula ROR, where –R represents the ethyl group. (i) Explain why boron trifluoride can form a compound with diethyl ether. [2] ROR has a lone pair of electrons on O while the B in BF3 has a vacant orbital, hence, B can accept lone pair of electrons to form a stable octet structure via dative bond formation.

2 Ó VJC 2021 9729/02/PRELIM [Turn over (ii) Draw the structure of the compound formed, indicating clearly the shape and bond angle around each central atom. [2] Structure should include • Correct structure with wedge and hash bonds • Dative bond from O to B • Around B: tetrahedral, 109.5o • Around O: trigonal pyramidal, 107o [Total: 10] S2 (a) Use of the Data Booklet is relevant to this part of the question. Xenon hexafluoride, XeF6 was one of the first noble gas compounds synthesised. XeF6 reacts with the silicon dioxide, SiO2 in glass to form liquid xenon oxytetrafluoride, XeOF4 and gaseous silicon tetrafluoride, SiF4 as shown by the equation below: 2XeF6 + SiO2 ® 2XeOF4 + SiF4 (i) An unknown amount of XeF6 was allowed to react with SiO2 in a 2 m3 closed vessel at 25 ºC. When all the XeF6 has reacted, a pressure of 0.505 kPa was measured in the vessel. Assuming that the gas inside the vessel behaves ideally, calculate the amount of XeF6 reacted in the vessel. [2] Only gaseous SiF4 exerts a pressure of 0.505 kPa at the end of the reaction. PV = nRT 0.505 x 103(2) = n(8.31)(25 + 273) n = 0.408 mol Hence, amount of XeF6 reacted = 2 x 0.408 = 0.816 mol (ii) Hence, calculate the mass of XeF6 reacted. [1] Mass of XeF6 reacted = 0.816 (131.3 + 19.0(6)) = 200 g (b) In a given vessel of a fixed volume, N2O5 gas decomposed to NO2 and O2 as shown below at 50 oC. 2N2O5(g) ® 4NO2(g) + O2(g) The rate equation for this thermal decomposition is as follows. Rate = k[N2O5]

3 Ó VJC 2021 9729/02/PRELIM [Turn over Theory shows that under the conditions of the experiment, the following relationship between the concentration of N2O5 and time is as follows. equation 1 ln[N2O5]t = –kt + ln[N2O5]initial where [N2O5]t = concentration of N2O5 present at time t k = rate constant (i) The following graph of ln[N2O5]t against t is plotted with the data obtained from the experiment. With the help of equation 1 and the graph above, determine I. the initial concentration of N2O5 [1] From the graph, ln[N2O5]initial = –4.1 [N2O5]initial = 1.66 x 10–2 mol dm–3 II. rate constant, k, including its units [1] k = –gradient = –(–5.0 + 4.1) / 1800 = 5.00 x 10–4 s–1 (ii) It is known that the rate of the reaction doubles with every 10 oC rise in temperature. Calculate the initial rate of the thermal decomposition of N2O5 when the volume of vessel is halved at 70 oC. [2] Rate = k[N2O5] = 5.00 x 10–4 x 4 (1.66 x 10–2 x 2) = 6.64 x 10–5 mol dm–3 s–1 –4.0 4.04 –4.2 4.04 –4.4 4.04 –4.6 –4.8 –5.0 4.04 –5.2 4.04 0 600 1200 1800 2400 3000 3600 Time / s 4.04.0 ln[N2O5]t

4 Ó VJC 2021 9729/02/PRELIM [Turn over (c) In another experiment, O2 reacts with NO as follows. O2 + 2NO ® 2NO2 (i) Using the following data, determine the order of reaction with respect to O2 and NO. Hence, write down the rate equation for the reaction. [2] Comparing 1 and 2, when [O2] is doubled, initial rate is doubled. Hence, first order wrt O2. Comparing 1 and 3, when [NO] is doubled, initial rate is increased by four times. Hence, second order wrt NO. Rate = k[O2][NO]2 experiment [O2] / mol dm–3 [NO] / mol dm–3 initial rate / mol dm–3 s–1 1 1.10 x 10–2 1.30 x 10–2 3.21 x 10–3 2 2.20 x 10–2 1.30 x 10–2 6.40 x 10–3 3 1.10 x 10–2 2.60 x 10–2 12.8 x 10–3 (ii) The following relationship can be used to calculate the activation energy of the reaction between O2 and NO: ln= –(– ) where k = rate constant at the respective temperature in Kelvins. Calculate the activation energy of the reaction if the values of the rate constants are 7.00 x 10–3 and 9.21 x 10–3 at 25 oC and 60 oC respectively. Include the correct units in your answer. [2] ln (9.21 x 10–3 / 7.00 x 10–3) = (–Ea/8.31)(1/333 – 1/298) Ea = 6460 J mol–1 = 6.46 kJ mol–1 (iii) With an appropriate sketch of the Boltzmann distribution, explain why a rise in temperature increases the value of rate constant, k. [2] At higher temperatures, the proportion of molecules with kinetic ene