2024 ASRJC H2 Chem P2 (Ans)

ASRJC JC2 PRELIM 2024 9729/02/H2 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2024 JC 2 PRELIMINARY EXAMINATION CHEMISTRY 9729/02 Paper 2 Structured Questions SUGGESTED SOLUTIONS Answer all the questions. 1 (a) Phosphorus, sulfur and chlorine are Period 3 elements of the Periodic Table. Table 1.1 shows some properties of the three elements. Table 1.1 P S Cl number of electrons in 3p subshell number of unpaired electrons (i) Complete Table 1.1 to show the number of electrons in the 3p subshell and the number of unpaired electrons in an atom of P, S and Cl. [2] P [Ne]3s23p3 S [Ne]3s23p4 Cl [Ne]3s23p5 number of electrons in 3p subshell 3 4 5 number of unpaired electrons 3 2 1 (ii) With reference to the Data Booklet, state and explain the trend of the ionic radius of P3–, S2– and Cl–. [2] • Ionic radius decreases from P3– (0.212 nm), S 2– (0.184 nm) and C l– (0.181 nm) • Nuclear charge increases from P3– to Cl–. • Number of filled quantum (electron) shells and shielding effect remains the same since the anions (P3– to Cl–) are isoelectronic. • Therefore, the stronger (electrostatic) forces of attraction between the nucleus and the outer electrons results in the decreasing ionic radius.

(b) Phosphoryl chloride, POCl3, is a colourless liquid that is used to make phosphate esters. P O Cl Cl Cl POCl3 has similar chemical properties as PC l5. It has a melting point of 1 °C and a boiling point of 106 °C. It also reacts vigorously with water, forming misty fumes and an acidic solution of H3PO4. (i) Explain how the information in (b) suggests that the structure and bonding of POCl3 is simple covalent. [2] POCl3 has low melting and boiling point suggests that it has weak instantaneous dipole-induced dipole attraction / permanent dipole-permanent dipole attraction between molecules that required low amount of energy to overcome. [1] The vigorous reaction with water suggested that hydrolysis had taken place. [1] Thus, the structure and bonding of POCl3 is likely to be simple covalent. (ii) Write a balanced equation for the reaction of POCl3 with water. [1] POCl3 + 3H2O H3PO4 + 3HCl [1] (iii) In H3PO4, there is no hydrogen atom directly bonded to the phosphorus atom. Draw the ‘dot and cross’ diagram of H 3PO4 and state the shape of the molecule with respect to P. [2] P O O O O xx xx xx x x x xx x x x xx xx x x x x x x H H H [1] Shape: tetrahedral [1] (c) Phosphoryl chloride, POC l3, is manufactured industrially from phosphorus trichloride and oxygen as shown in equation 1.1. equation 1.1 2PCl3(g) + O2(g) 2POCl3(g) The standard enthalpy changes of formation for these species are shown in Table 1.2. Table 1.2 Enthalpy change of formation of PCl3(g) −289 kJ mol–1 Enthalpy change of formation of POCl3(g) −592 kJ mol–1

ASRJC JC2 PRELIM 2024 9729/02/H2 [Turn over (i) Define the term standard enthalpy change of formation. [1] The amount of heat absorbed or evolved when one mole of a substance is formed from its constituent elements, all in their standard states at 298 K and 1 bar. [1] (ii) Using the data from Table 1.2 and relevant data from the Data Booklet, calculate the bond energy of P=O in POCl3. [2] ∆Hr = ∆Hf (products) – ∆Hf (reactants) = 2(−592) – 2(−289) = −606 kJ mol–1 [1] Hr = BE(bonds broken) − BE(bonds formed) {[2×3BE(P−Cl)] + BE(O=O)} – {[2×3BE(P−Cl) + 2×BE(P=O)} = −606 {(6 × 330) + 496} – {(6 × 330) + 2BE(P=O)} = −606 496 – 2BE(P=O) = −606 BE(P=O) = +551 kJ mol–1 [1] (iii) Predict and explain the sign of the entropy change for the reaction in equation 1.1. [1] S is negative because there is a decrease in the number of gaseous particles. There are less ways to distribute the particles and the energies among these particles , resulting in less disorder in the system. Hence entropy of the system decreases. [1] (iv) Comment on the effect of increasing temperature on the spontaneity of the reaction in equation 1.1. [2] Since Hr < 0 (negative) and –TSr > 0 (positive), –TSr becomes more positive with increasing temperature. [1] Gr becomes more positive and the reaction will become less spontaneous as the temperature of the reaction increases. [1] [Total: 15]

2 The structure of a tetrapeptide T is shown below. H2N O N H O H N O N H OH O OH HO O OH HO O T (a) Name the type of reaction to break T into its constituent amino acids. [1] Hydrolysis [1] (b) The four amino acids formed from the reaction in (a) are glutamic acid, tyrosine, U and V. The structures of glutamic acid and tyrosine are as shown. glutamic acid tyrosine OH H2N O HO O OH H2N O OH Table 2.1 lists the p Ka values of the different functional groups present on each of the four amino acids. Table 2.1 Amino acid pKa -carboxyl group -amino group side chain glutamic acid 2.1 9.5 4.1 tyrosine 2.2 9.2 10.5 U 2.0 9.9 3.9 V 2.2 9.2 -

ASRJC JC2 PRELIM 2024 9729/02/H2 [Turn over (i) Explain why the pKa value of the side chain of glutamic acid is lower than that in tyrosine. You may represent glutamic acid as R−CO2H and tyrosine as R OH . [3] Tyrosine R OH R O + H+ In phenoxide ion, the lone pair of electrons on O atom can delocalise into the  electron cloud of the benzene ring, thus dispersing the negative charge, stabilising the phenoxide through resonance. [1] Glutamic acid RCO2H R C O O + H+ In RCO2–, the p-orbital on the C atom overlaps with the p-orbitals of the two neighbouring O atoms. Hence, the negative charge is more effectively dispersed between the two O atoms, [1] resulting in a more (resonance -) stabilised RCO2– as compared to the phenoxide ion. Since the dissociation of RCOOH to release H + is more favoured , the side chain of glutamic acid is a stronger acid and has a lower pKa value than that of tyrosine. [1] (ii) In the space below, draw the structures of the predominant species of U and V at pH 3.0. U V O H3N O OH O O H3N O OH [2] (c) Solutions containing the zwitterions of V can act as buffers. (i) State what is meant by the term zwitterion. [1] A zwitterion is a species that carries both a positive charge and a negative charge but is electrically neutral. [1]

(ii) With the aid of appropriate equations, explain how a solution containing the zwitterions of V can resist pH changes. You may use H 2NCHRCOOH to represent the structure of V. [2] When a small amount of acid is added: H3N+CHRCOO− + H+ H3N+CHRCOOH When a small amount of base is added: H3N+CHRCOO− + OH− H2NCHRCOO− + H2O Since the acid or base added is removed, the pH is kept relatively constant. (d) (i) Calculate the pH of 0.10 mol dm−3 solution of protonated V. Ignore the effect of p Ka of the -amino group on the pH. [2] H3N+CHRCOOH H3N+CHRCOO− + H+ + - + 3 a + 3 +2 + 3i +2 -2.2 + [H N CHRCOO ][H ] K = [H N CHRCOOH] [H ] [H N CHRCOOH] [H ]10 = 0.10 [H ] = 0.02512  [1] pH = 1.6 [1]

ASRJC JC2 PRELIM 2024 9729/02/H2 [Turn over (ii) A student records the pH of th