RI H2 CHEM P3 ANS Prelim

Question 1 (a) Finely powdered Ca metal has a large surface area which increases its rate of reaction with water vapour in the air to produce hydrogen gas. Ca + H2O  CaO + H2 The heat produced causes hydrogen gas to combust spontaneously with oxygen gas in the air, causing explosions. 2H2 + O2  2H2O This generates more water which spark off further explosive reactions with finely powdered Ca metal. (b)(i) Q = I x t = ne x F ne = 1×105×3×60×60 96500 =1.12×104mol Ca2+ (l) + 2e  Ca(s) Mass of Ca formed = 1.12×104 2 × 40.1 = 2.24 × 105𝑔 = 0.224 tonnes (b)(ii) From Data Booklet, 2H2O + 2e  2H2 + 2OH E=0.83V Ca2+ + 2e  Ca E=2.87 V Since E(H2O/H2) is less negative than E(Ca2+/Ca), H 2O will be preferentially reduced at the cathode instead of Ca2+. No Ca metal will be obtained. (c)(i) Ca(CH3CO2)2  CaCO3 + CH3COCH3 solid M (c)(ii) White residue contains CaCO3 and CaO. Method 1 Amount of CaCO3 formed at 450 oC = Amount of Ca(CH3CO2)2 = 2.0 158.1 = 0.01265 mol Let the amount of CO2 evolved = y mol. Amount/mol CaCO3  CaO + CO2 Initial 0.01265 - - Change -y +y +y Final 0.01265 - y y y Mass of CaCO3 in the residue = (0.01265 – y)(100.1) = (1.266 – 100.1y) g Mass of CaO in the residue = 56.1y g mass of CaCO3in the residue mass of CaO in the residue = 70 30  1.266−100.1𝑦 56.1 y = 7 3 1.266 – 100.1y = 130.9y y = 0.005483 Amount of CO2 evolved = 0.005483 mol Mass of CO2 evolved = (0.005483)(44.0) = 0.241 g

Method 2 Mass of CaCO3 formed at 450°C = 2 /158.1 x 100.1 = 1.266 g CaCO3  CaO + CO2 mass after partial decomp / g 1.266 – x – y x y Since mass of white residue = (1.266 – y) g and %(CaO in white residue) = 30% ⇒ x 0.3 1.266 y  Since amt of CaO formed = amt of CO2 formed, ⇒ xy 56.1 44 Solving simultaneous equations, y = mass of CO2 lost = 0.241 g Method 3 Let mass of white residue be x g and mass of CO2 evolved by y g. Mass of CaO = (0.3x) g Since CaCO3  CaO + CO2, amt of CaO = amt of CO2 0.3x y 56.1 44 By conservation of mass, x + y = 1.266 g Solving simultaneous equations, y = 0.241 g.) (c)(iii) Ba(CH3CO2)2 will decompose at a higher temperature. Ba 2+ ion has a larger ionic radius as compared to Ca 2+ ion resulting in a lower charge density. Ba 2+ ion therefore has weaker polarizing power, hence weakening the covalent bonds within CH 3CO2 - ions to a smaller extent as compared to Ca2+ ion, resulting in higher thermal decomposition temperature. (d)(i) Nucleophilic addition

(d)(ii) (e)(i) Compare experiments 1 and 2. When [Q] is increased 1.25 times, rate is increased 1.25 times.  rate  [Q]  Order of reaction with respect to Q = 1 Compare experiments 1 and 3. When [propanone] is doubled, rate is doubled.  rate  [propanone]  Order of reaction with respect to propanone = 1 Comparing experiments 2 and 4: n n )150.0)(100.0( )100.0)(200.0( = 5 5 1096.3 1028.5     ⇒ n)3 2(2 = 3 4 Hence, n = 1  Order of reaction with respect to OH = 1 The rate equation is: rate = k [Q] [propanone] [OH-] (e)(ii) Step B is the slow step i.e. the rate-determining step. From the slow step: rate = k’ [ ] [H2O] [propanone] = k [Q] [OH–] [propanone] The rate equation obtained agrees with that in (e)(i). (e)(iii) The hydroxide ion acts as a base. OR The hydroxide ion acts as a catalyst.

Question 2 (a)(i)  All peptide linkages are involved in intra-chain hydrogen bonding.  It is stabilized by hydrogen bonds between the C=O group of a peptide in one strand and the N-H group of another peptide in the adjacent strand  R groups (side chains) project above and below the sheet and are 90 to the plane of the pleated sheet. (a)(ii) Methionine and tyrosine exists as zwitterions and they can form ion -dipole interactions with water , making them highly soluble in aqueous state. (a)(iii) -NO2 group is electron withdrawing which disperses the negative charge on the O atom of the phenoxide ion , stabilizing the phenoxide ion, making nitro-tyrosine a stronger acid. (a)(iv) At physiological pH of 7.34 -7.45, the phenol group on nitrotyrosines is deprotonated and forms phenoxide ion. With the increase in negatively charged phenoxide ions, it will be bonded to positively charged R groups like ( -NH3 +) via ionic bonding in the tertiary structures, promoting fibril formation. (or disrupts existing hydrogen-bonding) OR At physiological pH of 7.34 -7.45, the phenol group on nitrotyrosines is deprotonated which allows it to form ionic interactions with positively charged R -groups and disrupts the original hydrogen-bonding between the R -groups , causing a conformation chang e in the tertiary structure. OR The intramolecular hydrogen -bond between the –OH and –NO2 of the nitrotyrosine R -group reduces the formation of hydrogen -bonds with R -groups of other amino acid residues. This causes a conformation change in the tertiary structure. (a)(v) Hydrogen bonding

(b)(i) X: COOH (b)(ii) Nucleophilic Substitution Y COOH (b)(iii) Orange ppt formed (c)(i) The second and third K a of protonated tyrosines are too close as they are only about 1 unit apart (i.e. 9.1 vs 10.1). Hence, the titration curve cannot show a distinct sharp rise for the second and third equivalence points. (c)(ii) (c)(iii) H2T ⇌ H+ + HT2 or H2T + H2O ⇌ H3O+ + HT2 pH = 5.7  [H+] = 10-5.7 = 2.00 x 10-6 mol dm-3 (c)(iv) 𝐾𝑎2 = [𝐻+][𝐻𝑇2−] [𝐻2𝑇−] = 10−9.1 𝐾𝑎2 = [10−5.7]2 [𝐻2𝑇−] = 10−9.1 [𝐻2𝑇−] = 5.01 × 10−3 [H3T] = 2 x [H2T-] = 2 x 5.01 x 10-3 = 0.0100 mol dm-3

Question 3 (a)(i) Due to the presence of partially filled d -subshells in transition metals, reactant molecules can form weak interactions with the surface of the catalyst. The H2 and O2 molecules adsorb onto the surface of the Pt catalyst. The adsorption increases the surface concentration of H 2 and O2 on the Pt surface and weakens the covalent bonds in H2 and O2 for reaction. After reaction, the product (H 2O) desorbs from the surface, allowing more reactants to adsorb. (a)(ii) An acid-metal reaction between Zn and H2SO4 produces H2 gas: Zn(s) + H2SO4(g)  ZnSO4(aq) + H2(g) The H2 gas evolved travels out of the outlet and mixes with O 2 from the air on the surface of the Pt catalyst , undergoing an exothermic reaction which ignites the remaining H 2 gas exiting the outlet. (b)(i) Oxidation state of C in C=O in quinone = +2 Oxidation state of C in C–O in hydroquinone = +1 Since there is a decrease in oxidation state from quinone to hydroquinone, reduction has occurred. (b)(ii) Cathode: Ag+(aq) + e–  Ag(s) Anode: (b)(iii) Reading on voltmeter = E cell = 0.80 – (+0.70) = +0.10 V (b)(iv) [Fe(CN)6]3– + e– ⇌ [Fe(CN)6]4– +0.36 V

Since Ecell = +0.70 –(+0.36) = +0.34 V >0, the above reaction occurs. This causes the  [hydroquinone] to increase and [quinone] to decrease.  E(quinone/hydroquinone) to become less positive (become smaller than +0.70V)  Reading on voltmeter to become more positive. (c)(i) i. I2, NaOH, warm ii. HCl(aq) (c)(ii) COCl2 (c)(iii) Test: To separate samples of the compounds, add neutral FeCl3 (aq) Observation: Violet colouration observed in Y. No violet colouration observed in Z. (c)(iv) Products: Question 4 (a)(i) Al: [Ne] 3s23p1 Mg:[Ne] 3s2 The 3p electron to be removed from Al is at a higher energy level than the 3s electron to be removed from Mg. Hence less energy is required to remove the 3