ACJC_H2_CHEM_P3_Answers Prelim

2 © ACJC 2014 9647/03/Prelim/14 [Turn over Answer any four questions. 1 (a) Dinitrogen tetroxide N 2O4 is one of the most important rocket propellants developed. N2O4 forms an equilibrium mixture with nitrogen dioxide NO2. NO2 is favoured at higher temperatures, while at lower temperatures, N 2O4 predominates. (i) Draw dot-and-cross diagrams to show the bonding in the molecules of NO2 and N2O4. (ii) Suggest a value for the bond angle in each of the above two molecules. 120o<Bond angle in NO2 <180o Bond angle in N2O4 is 120o [3] (b) The chemistry of nitrogen oxides is very versatile. (i) Given the following reactions and their standard enthalpy changes Reaction 1 NO(g) + NO2(g) N 2O3(g) Ho r = -39.8 kJ mol-1 Reaction 2 NO(g) + NO2(g) + O2(g) N 2O5(g) Ho r = -112.5 kJ mol-1 Reaction 3 2NO2(g) N2O4(g) Ho r = -57.2 kJ mol-1 Reaction 4 2NO(g) + O2(g) 2NO 2(g) Ho r = -114.2 kJ mol-1 Reaction 5 N2O5(s) N 2O5(g) Ho r = +54.1 kJ mol-1 Calculate the Ho r for Reaction 6 N 2O3(g) + N2O5(s) 2N 2O4(g) By algebraic method, -(Reaction 1)– (Reaction 2) + 2(Reaction 3) + (Reaction 4) +(Reaction 5) = - (-39.8) – (-112.5) + 2 (-57.2) + (-114.2) + (+54.1) Answer : -22.2 kJ mol -1 (ii) By considering the entropy and enthalpy change during reaction 5 and reaction 6, suggest how the standard Gibbs free energy change of the two reactions will compare in sign and in magnitude. Hence predict which reaction will be more spontaneous. Explain your reasoning. S would be +ve in reaction 5 and reaction 6 owing to the increase in the number of moles of gas. H for reaction 5 is +ve, its G would only become negative at high temperatures, whereas since the H for reaction 6 is -ve, the G for this reaction is negative at all temperatures. Thus the reaction 6 is likely to be more spontaneous. [5]

3 © ACJC 2014 9647/03/Prelim/14 [Turn over (c) Compounds of elements in the second and third period of the Periodic Table show similar trends of periodicity. The Period II oxides are given as Li2O BeO B 2O3 CO 2 N 2O3/ N2O5 (i) The melting points of Li2O and CO2 are 1440 oC and -79 oC respectively. Explain for the differences in melting points. Li2O has a giant ionic lattice structure with strong electrostatic forces of attraction between oppositely charged Li+ and O2- ions. CO2 are simple covalent molecules with weak intermolecular Van der Waals forces. Hence more energy is needed to separate the ions in Li2O compared to energy needed to separate the CO2 molecules during melting process. (