2022 EJC Prelims Paper 1 Worked Solutions

2022 JC2 Preliminary Examination H2 Chemistry 9729 Paper 1 Worked Solution 1 24 H OH 25 1.0 10 2.50 10 mol1000nn+− −−= =   =  1 Ca2+ ≡ 2 H+ 2 4 Ca H 1 1.25 10 mol2nn ++ −= =    2 4 2 Ca 4 33 1.25 10CaSO Ca 50 1000 2.50 10 mol dm n V + − + −− = = = =  A 2 Large jump between 7th and 8th I.E.  7 electrons in valence shell  Group 17  C 3  C 4 pV nRT= mpV RTM= Plot of pV against T will be a straight line that passes through the origin (0 K) with gradient mR M . Since M(E) < M(F), the gradient for E will be steeper than the gradient for F.  A 5 G = H – TS 1 : H > 0, S < 0, G > 0 (non- spontaneous) at all T 2 : H < 0, S < 0, G > 0 (non- spontaneous) when HT S   3 ✓: H < 0, S > 0, G < 0 (spontaneous) at all T  D 6 A ✓: 1 mol of H 2O is formed from the reaction of an acid and a base B : lattice energyH ( )( )3NaHCO s is for the reaction Na+(g) + HCO – 3(g) → NaHCO3(s) C ✓: Being a strong acid, the HSO– 4 ion will fully dissociate in water to give H+(aq) and SO2– 4 (aq) D ✓: formationH involves forming (NH4)2SO4 from the constituent elements, S, H 2, N 2 and O 2 in their standard states  B 7 1 ✓: From the slow step, rate = k[(CH3)2C+(OH)][H2O] = k′[(CH3)2C(=O)][H3O+], k′ = kKc At low pH, [H 3O+] is high and is essentially a constant. rate = k″[(CH3)2C(=O)] 2 : rate constant, aE RTk Ae − = depends only on the Ea and T. 3 ✓: I2 is only involved in a fast step after the slow (rate determining) step. Hence rate of reaction is independent of [I2].  D 8 A : Since there is no change in the number of gaseous particles upon reaction, the total pressure in the vessel remains constant. Hence the amount of reactants and products cannot be determined. B : Slow cooling allows the equilibrium to readjust to the new t emperature. Hence, the Kc determined will be that at 20 ºC and not 500 ºC. C ✓: Rapid cooling will slow down the reaction, allowing the amount of I2 in the equilibrium mixture at 500 ºC, and hence Kc, to be determined. D : If the HI present is decomposed back to H2 and I2, the amount of reactants and product in the equilibrium mixture cannot be determined.  C 9 A : Since H < 0, position of equilibrium lies more to the left with increasing T. Hence p (g) 1K p= G es with ing T. B : npV nRT p RT CRTV=  = =   c p (g) 11 (g) KK p RT RT= = = G G C :  depends only on k1, k2 and pG. It is independent of the total surface area of the catalyst. D ✓: At high pG, 11 22 1 kk ppkk+ GG . Hence, 1 , which is independent of T.  D 10 As HCl is titrated against Na2CO3, Na2CO3 is added from the burette into HCl in the conical flask, the pH increases with the volume of titrant. HCl will completely react with any Na2CO3 added Na2CO3 + 2HCl → 2NaCl + H2O + CO2 Hence there will be only one end-point.  A 11 Considering the species present: 1.9 8.1 5.02p +=I (zwitterion is neu