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. (ii) BeO is an amphoteric oxide. Write equations for the reaction between BeO with an acid and with a base. BeO + 2 HCl + H2O → BeCl2 + 2 H2O BeO + 2 NaOH + H2O → Na2Be(OH)4 (iii) B2O3 is weakly acidic. The oxide acidity of the Group III elements decreases down the group. Suggest a possible reason for this trend. Charge density decreases down the group OR Size of atoms/ions increase down the group while charge is constant. [6] (d) Halogens are also commonly found in many organic compounds, such as an aromatic compound X with the molecular formula of C8H6Cl2O. Given that one mole of X reacts with one mole of dimethylamine to form a neutral product Y and Y does not react with hot ethanolic ammonia, suggest the structures for compounds X and Y, explaining your reasoning. Hence, discuss the reactivities of the two chlorine atoms in compound X towards substitution. One mole of X undergoes nucleophilic substitution with one mole of dimethylamine to form an amide in Y. One acid chloride functional group is present in compound X. Absence of alkyl chloride group in both X and Y as amine (basic) functional group is not formed in Y and/or the latter does not undergo nucleophilic substitution with hot ethanolic ammonia. Aryl chloride must be present in both compounds. [6]

4 © ACJC 2014 9647/03/Prelim/14 [Turn over Compound X Compound Y or Compound X Com pound Y Acyl chloride is reactive towards nucleophilic substitution as the C atom is highly partially positive ( +) as it is bonded to 2 electronegative atoms. Hence it reacts with dimethylamine to form amide at room temperature. Whereas the C-C l of the aryl chloride is strengthened by overlapping of the p-orbital of C l with the π orbitals of the benzene ring which results in the delocalisation of electrons (partial double bond character), hence making aryl chloride very much less reactive and the substitution of the halogen atom very difficult. Thus there is no reaction with both dimethylamine and hot dilute NaOH. Or Aryl chloride is very much less reactive as the high electron density of the aromatic ring hinders the nucleophilic attack, due to electrostatic repulsion between like charges. [Total: 20]

5 © ACJC 2014 9647/03/Prelim/14 [Turn over 2 (a) Sodium chloride and silver chloride are two simple salts and their solubilities in water are being considered in this question. Salt H soln / kJ mol-1 Ssoln / J mol-1K-1 NaCl +3.6 +43.2 AgCl +65.7 +34.3 (i) Use the values given in the table to calculate Gsoln  for each of the salts and hence deduce its solubility in water. Gsoln (NaCl) = Hsoln  -TSsoln  = +3.6-298(43.2/1000) = -9.27 kJ mol-1<0 (feasible reaction, NaCl is soluble in water) Gsoln (AgCl) = Hsoln  -TSsoln  = +65.7-298(34.3/1000) = +55.5 kJ mol-1>0 (reaction is not feasible, AgCl is not soluble in water) (ii) The solubility product, K sp, of AgC l is related to Gsoln (AgCl) by the following equation, Gsoln  = -2.303RT lg K sp where R is 8.31 J mol -1K-1 and T is the temperature in K. Use the equation given above to calculate the value of Ksp of AgCl at 298K. lg Ksp = -(55.5 x 103)/(2.303x8.31x298) =9.73 Ksp = 10-9.73 = 1.86x10-10 mol2 dm-6 (iii) Explain how solubility of AgCl will change with increasing temperature?  As T increases, -TSsoln  gets more negative  It is assumed that Hsoln  and Ssoln  do not change much with T  an increase in T will cause Gsoln  to be more negative Hence solubility of AgCl will increase with T. OR AgCl(s) +aq Ag+(aq) + Cl-(aq) Hsoln = +65.7 kJ mol-1 As T increases, position of equilibrium shifts to the right in favour of the endothermic reaction, hence solubility of AgCl increases with T [5] (b) (i) Draw a fully labelled diagram of the electrochemical cell you would use to determine the standard electrode potential of the Ag +(aq) l Ag(s) electrode system and show the direction of electron flow .

6 © ACJC 2014 9647/03/Prelim/14 [Turn over (ii) When aqueous sodium chloride is added to the Ag +(aq)lAg(s) electrode system in the above electrochemical cell in (b)(i), explain qualitatively how the Ecell will change as a result. On addition of NaCl,  Ag+ + Cl-  AgCl (s)  Ag+(aq) + e Ag(s) E = +0.80V  POE shifts left to increase [Ag+]  Electrode potential becomes less positive  Ecell = EAg+/Ag- EH2/H+ will decrease (i