NJC H2 Chem 2012 Prelim P3 Soln

NJC SH2 Preliminary Examinations Paper 3 Solutions 1(a) Pressure is approximately zero at A Volume occupied by gas particles is small compared to the volume of the container \NH3 is almost an ideal gas. Low pressure at B, intermolecular attractions are dominant, thus NH3 molecules are pulled closer together. This reduces the impact on wall collisions and makes them softer. P NH3 < P ideal , hence the negative deviation. Volume occupied by gas particles is still relatively small compared to container. High pressure at D , mainly repulsion between particles. The volume of the molecules becomes a significant fraction of the volume of the container, and thus is no longer negligible. VNH3> Videal, hence the positive deviation. At C, NH3 appears to follow the ideal gas equation, because the two deviations balance out. 1(b)(i) A relatively low temperature of 400 o C is used to increase the yield of the exothermic reaction. Iron catalyst is used to compensate for any reduction in rate due to the low temperature used. A high pressure of 200 atm caters to the forward reaction, which is accompanied by a reduction in volume. Pressure exceeding 200 atm would increase the cost of production. 1(b)(iii) PNH3 = 0.363 x 200 = 72.6 atm PH2 = 0.47775 x 200 = 95.55 atm PN2 = 0.15925 x 200 = 31.85 atm Kp = = 1.90 x 10-4 atm-2 1(c) The - NH2 in hydrazine has a stronger electron-withdrawing effect than the – H in ammonia \the lone pair of electrons in ammonia is more available for donation to a proton \ the stronger base & the bigger Kb value 1(d)(i) Hydrazine and water form a miscible mixture Free intermingling of the molecules of hydrazine & water through mutual H- bonding \ Exists as a homogeneous phase 1

2 dd dd N N H H H H dd dd N N H H H H d d d d NN H H H H d d d d N N H H H H H O H d+ d+ H O H d+ d+ H O H d+ d+ H O Hd+ d+ H-bond Hydrazine and trichloromethane are immiscible Segregation between the molecules of hydrazine & trichloromethane into 2 distinct layers • NH2NH2 have strong H-bond with one another • CHCl3 have van der Waals attraction for one another Each cannot establish any significant attraction for the other and will be squeezed out from one another’s layer N N H H H H N N H H H H N N H H H H N N H H H HN N H H H H _________________________________________Phase boundary C H Cl Cl Cl d+ d- Van der Waals forces H-bond

3 1(d)(ii) -2 oxidation state in N of hydrazine is oxidized to 0 in N2 1(e)(i) C CH 3H2NN CHI3 CH(OH)CH3 Compound E Compound F Compound G 1(e)(ii) Reagent X is I2(aq) with NaOH(aq); and the reaction mixture is warmed. 1(f) CH2CHO Compound H 2 a (i) Distillation and collect the distillate that boils over at 150 OC. (ii) Solid BaSO4 will be formed and it can be removed by filtration. (iii) b (i) Ba + 2H2O ® Ba(OH)2 + H2 (ii) Reaction in b(i) involves the reduction of water by the group II metals. Reaction is more vigorous with barium as barium is a stronger reducing agent than magnesium as evidenced by the following data: Mg ® Mg 2+ + 2e Eooxid = +2.38V Ba ® Ba2+ + 2e Eooxid = +2.90V c (i) Standard enthalpy of formation of barium chloride is the heat change or enthalpy change when 1 mole of solid barium chloride is formed from its elements in their standard states ( 25 oC and 1 atmosphere) i,e, formed from barium solid and chlorine gas. Ba 2 + ·´ ·· ·· O ´´ ´´ ´o o· O 2-

4 (ii) BaCl2(s) + aq ® Ba2+(aq) + 2Cl-(aq) DH = -13.2 kJ mol-1 (iii) Ba(s) + Cl2(g) + O2(g) + 2H2(g) BaCl2(s) + O2(g) + 2H2(g) Using Hess’ Law, DHf + 2(-285.9) + (-13.2) = (-1002) + 2(-92.3) + 2(-71.9) + (-114.4) DHf = - 859.8 kJ mol-1 = - 860 kJ mol-1 (iii) BaCl2(s) ® Ba(s) + Cl2(g) Using DS = SSproducts - Sreactants = (63 + 223) - (124) = + 162 JK-1 mol-1 For the reaction to be feasible, DG = DH - TDS must be less than zero. = 5307K Lowest temp for the reaction to be feasible is 5307K. Ba(OH)2(aq) + Cl2(g) + H2(g) Ba(OH)2(aq) + 2HCl(g) DHf + aq (-1002 ) Ba(OH)2 (aq) + 2HCl(aq) BaCl2 (aq) + 2H2O(l) BaCl2(s) + 2H2O(l) 2 (-92.3) +aq 2 (-71.9) (-114.4) +aq (-13.2) 2 (-285.9)

5 d D: CH3COCl E: CH3CHO F: CH3C H CN OH G: CH3C H OH CH2NH2 Reagents and conditions for step (i): LiAlH4 in dry ether or H2 with Pt as catalyst ‘ 3 (a) · W contains chiral carbon. · W is an aldehyde. No of moles of CO2 = 4 . 22 6 . 33 = 1.50 mol nY : n CO2 0.3 : 1.50 1:5 · Y contains 5 carbon atoms. Structure of W: C C C C O H H H H H H H CH3 Structure of Y: C C C C H H H H H H H H CH3 H

6 Structure of X: C C C C H H H H H H CH3 O H (b) Observations Deductions J has molecular formula of C4H7Cl. J is a halogenoalkane. J reacts with acidified potassium manganate(VII), giving effervescence of a colourless gas. J undergoes oxidative cleavage. J contains terminal alkene. J reacts with ethanolic potassium cyanide to produce compound K with molecular formula C5H7N. J undergoes nucleophilic substit ution of Cl with CN. J rotates plane-polarised light. J contains a chiral carbon. Displayed formula of J: C C H HH H C C H H H Cl

7 (c) (i) Comparing Expt 1 & 2, when [ J] triples while [KCN] is kept constant, initial rate of formation of product triples. Hence, reaction is first order wrt J. Comparing Expt 1 & 3, when [J ] doubles and [KCN] doubles, initial rate of formation of product only doubles. Hence, reaction is zero order wrt KCN. Rate = k[J] Mechanism: Nucleophilic Substitution (SN1) Step 1: C CH3 H CHCH2 Cl slow C CH3 H CHCH2 + Cl d+ d- Step 2: C CH3 H CHCH2 + CN fast C CH3 H CHCH2 CN

8 (ii) In[J] In[J] o Time In[J] = In[J] o – kt In[J] = – kt + In[J]o [y = mx + c] Hence, rate constant = – gradient (d) (i) overall order = 2 (ii) Ethanol is more polar than propanone, stabilizes carbocation to a greater extent through formation of stronger ion-dipole interactions . Hence SN1 mechanism is favoured when ethanol is used as the solvent. (e) Fraction of molecules T2 where T2 < T1 T1 Energy Ea Legend: fraction of molecules having energy ≥ Ea at T 1 fraction of molecules having energy ≥ Ea at T 2 At a lower temperature T 2, the fraction of molecules having energy greater than or equal to Ea decreases, from to Hence, the number of effective collisions decreases and rate of reaction decreases.

9 4 (i) NH3 in ethanol, in sealed tube P: CONH2 (ii) Sn, c.HCl ,reflux followed by NaOH(aq) NH2c.HNO3 c. H2SO4 55oC NO2 (iii) Add neutral aqueous iron(III) chloride to both. An intense violet complex will be observed for phenol. No violet complex for phenylamine. Or Add sodium metal to both. Effervescence will be observed for phenol. Gas evolved extinguished burning splinter with a “pop” sound. No gas evolved for ph