AJC_H2_CHEM_P3_Solutions

©2014AndersonJC/CHEM 1 H2 Chemistry 9647 2014 JC2 Prelim P3 Suggested Solutions 1 (a) (i) C: Mg(OH)2 D: [Al(OH)4]– E: Al(OH)3 F: Al2O3 [1] [1] [1] [1] (ii) acid–base reaction [1] (iii) No. of moles of Mg(OH) 2 = 2(17.0)24.3 0.18  = 3.087 x 10–3 mol No. of moles of Mg = 3.087 x 10–3 mol Mass of Mg in the alloy = 3.087 x 10–3 x 24.3 = 0.07501 g % composition of Mg in the alloy = 1001.75 0.07501 = 4.29% [1] (b) (i) BeCl2 + H2O BeO + 2HCl accept BeCl2 + 2H2O Be(OH)2 + 2HCl [1] (ii) Al atom in A lCl3 molecule is electron–deficient as it has only 6 valence electrons around it. Hence, the A l atom can accept another 2 electrons from the lone pair of N atom in one molecule of NH 3 to achieve the (stable) octet configuration. [1] (iii) Al Cl Cl Cl N H H H [1] (iv) NH3 NH3 BeCl Cl accept condensed formula: BeCl2(NH3)2 [1]

©2014AndersonJC/CHEM 2 (c) (i) Electrophilic substitution Generation of electrophile: CH3Cl + AlCl3 CH3 + + AlCl4 – CH3 + HC H 3 CH3 + H++ slow fast Regeneration of catalyst: AlCl4 – + H+ AlCl3 + HCl [1m] for correct equations for generation of electrophile & regeneration of catalyst [1m] for curly arrows & charges + slow & fast [1m] for correct structure of intermediate [3] (ii) H : ClCH 2CH2Cl OH CH2CH2Cl J: O CH2CH2OH K: OH CH2COOH L: OH CH2COCl M: Step 2: NaOH(aq), heat under reflux Step 3: K 2Cr2O7(aq), dil. H2SO4, heat under reflux Step 4: PCl5 or SOCl2 [1m] for correct structure of M and reagents/conditions for step 4 [1] [1] [1] [1] [1] [1] [1]

©2014AndersonJC/CHEM 3 2 (a) (i) When the temperature is increased, the forward endothermic reaction is favoured. The position of equilibrium will shift to the right in order to reduce temperature by absorbing extra heat. Partial pressure of hydrogen will increase. When the pressure is increased, the reverse reaction is favoured. The position of equilibrium shifts to the left in order to decrease pressure by producing less gaseous particles. Partial pressure of hydrogen will decrease. [1m] for each change [2] (ii) The two assumptions are: There are negligible intermolecular forces between the gaseous particles. The volume of the gaseous particles is negligible compared to the volume it occupies. H 2O(g) has stronger hydrogen bonding between molecules compared to the weaker van der Waals’ forces between H 2 molecules, and the volume of H 2O molecules is greater than that of H 2, hence H2O(g) deviates more from ideality than H2. [1m] for each assumption [1m] for correct explanation [3] (iii) Kp = 2 2 O 2 CO 2 CO pp p 2CO(g) + O 2(g) 2CO2(g) Initial amt / mol 5.00 2.50 0 change in amt / mol – 4.95 – 2 95.4 + 4.95 amt at eqm / mol 0.05 0.025 4.95 total amt at eqm = 4.95 + 0.05 + 0.025 = 5.025 mol p(CO2) at eqm = 025.5 95.4 (101) = 99.5 kPa p