JPJC Prelim 2025 Paper 2 Answers H2 Chem

© Jurong Pioneer Junior College [Turn Over Answers for JPJC 2025 H2 Chem Prelim Paper 2 1 Nitrogen is found in inorganic compounds such as the oxides of nitrogen, NO 2 and NO. (a) NO2 can be produced from the thermal decomposition of gaseous N2O5. 2N2O5(g) → 4NO2(g) + O2(g) H1 Table 1.1 gives some data relevant to this question. Table 1.1 process H / kJ mol−1 standard enthalpy change of formation of N2O5(g) +11.3 standard enthalpy change of formation of NO(g) +89.0 NO(g) + 1 2 O2(g) → NO2(g) −58.1 (i) Explain what is meant by standard enthalpy change of formation. It is the heat change when 1 mole of substance is formed from its constituent elements in their standard states under the standard conditions of 298 K and 1 bar. [1] (ii) Use data from Table 1.1 to calculate H1. You may find it helpful to draw an energy cycle. H1 = −2(+11.3) + 4(+89.0) + 4(−58.1) = +101 kJ mol−1 [2] 2N2O5(g) 4NO2(g) + O2(g) 2N2(g) + 5O2(g) 2(+11.3) 4NO(g) + 3O2(g) 4(+89.0) 4(−58.1) H1

2 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2025 (iii) In the solid state, N 2O5 has an ionic structure and consists of the ions, NO2+ and NO3−. Draw and name the shapes of NO2+ and NO3−. ion NO2+ NO3− diagram of shape name of shape linear trigonal planar [3] (b) Nitrogen dioxide, NO 2, and dinitrogen tetraoxide, N 2O4, exist in dynamic equilibrium with each other. N2O4(g) = 2NO2(g) H = +58 kJ mol−1 At 50 C and a pressure of 1.68  105 Pa, 4.60 g of the equilibrium gaseous mixture occupies 1.00 dm3. (i) Assuming the gaseous mixture behaves ideally, calculate the average relative molecular mass, Mr, of the gaseous mixture. Mr = 53 4.60 8.31 (273 50) 1.68 10 1.00 10 −   +    = 73.5 [1] (ii) Using the following relationships, calculate the mole fraction of N 2O4, m, and the mole fraction of NO2, n, in the mixture. m + n = 1 Average Mr = 92m + 46n 73.5 = 92m + 46n 73.5 = 92(1 – n) + 46n 73.5 = 92 – 92n + 46n 46n = 18.5 n = 0.402 m = 1 – 0.402 = 0.598 [1] (iii) Hence calculate the partial pressures of N2O4 and NO2 in the mixture. 24NOP = 0.598  1.68  105 = 1.00  105 Pa 2NOP = 0.402  1.68  105 = 6.75  104 Pa [1] N O O + N O O − O • •

3 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2025 [Turn Over (iv) Write an expression for equilibrium constant, Kp, for the reaction, and calculate its value. Include units in your answer. Kp = ( )2 24 2 NO NO p p Kp = ( ) 24 5 6.75 10 1.00 10   = 4.54  104 Pa (or 45.4 kPa) [3] (v) State and explain the effect of increasing the temperature on the average Mr of the equilibrium mixture. Increasing the temperature will shift the position of equilibrium in N2O4(g) = 2NO2(g) to the right to favour the endothermic reaction so as to absorb some heat. Hence, there is a higher mole fraction of NO 2, so the average Mr will decrease. [2] (c) With the aid of suitable equations, describe and explain the role of NO 2 in the oxidation of atmospheric sulfur dioxide. NO2 acts as a homogeneous catalyst in the oxidation of atmospheric SO2 since both NO2 and SO2 have the same gaseous phase. Step 1 : NO2 + SO2 → NO + SO3 Step 2 : NO + ½O2 → NO2 [2] [Total: 16] 2 Copper(I) salts in aqueous solution are unstable as shown by equation 1. equation 1 2Cu+(aq) → Cu(s) + Cu2+(aq) (a) (i) Using relevant data from the Data Booklet, calculate G, in kJ mol −1, for the above reaction. Cu2+ + e− = Cu+ +0.15 Cu+ + e− = Cu +0.52 Ecell = +0.52 – (+0.15) = +0.37 V G = -nFEcell = -(1)(96500)(0.37) = -35705 = -35.7 kJ mol−1 [2] (ii) Deduce the sign of S for the reaction and explain your answer. S is negative (or S < 0) as there is a decrease in disorderliness since the more disordered aqueous ions react to form a more ordered solid Cu. OR number / amount of aqueous ions decreases from 2 to 1 as the reaction proceeds. [1]

4 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2025 (iii) Hence, determine if the reaction in equation 1 is exothermic or endothermic. Explain your answer. G = H − TS S < 0, -TS > 0 , G < 0, so H < 0. Forward reaction is exothermic. [1] (b) Some copper(I) compounds are used as reagents in organic reactions where new carbon-carbon bonds are formed. Larger alkanes can also be formed through the reaction of alkanes with some halogens. However, the yield of the larger alkanes obtained through such a reaction is low. (i) Trace amount of alkane A is obtained when 2-methylpropane reacts with bromine in the presence of ultraviolet light. A Outline the mechanism of this reaction, clearly showing how A is formed in the above reaction. (Free radical substitution) Initiation: Propagation: Termination: [3] uv

5 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2025 [Turn Over (ii) Chloroalkanes can be formed by the above mechanism but not iodoalkanes. Use relevant data from the Data Booklet to explain why iodoalkanes cannot be formed. Iodoalkanes cannot be made by free –radical substitution as the first propagation step is MORE endothermic (H = 410 – 299 = +111 kJ mol–1) OR LESS energy is evolved from the formation of the WEAK H–I bond (299 kJ mol–1) compared to that of H–Br (366 kJ mol–1). [1] (iii) Iodoalkanes can be made by warming a bromoalkane with a solution of sodium iodide in dry propanone, in which sodium bromide is almost insoluble. CH3CH2CH2Br + NaI = CH3CH2CH2I + NaBr Suggest why the above reaction produces a high yield of CH 3CH2CH2I despite the C−I bond being weaker than the C−Br bond. Sodium bromide being insoluble in propanone can be precipitated, causing the [NaBr] to be low. This will cause the position of equilibrium to shift to the right to form more NaBr, allowing the reaction to go to almost to completion. [1] (iv) A student wanted to distinguish 2–bromo–2–methylpropane, (CH 3)3CBr, from compound B shown below. The student suggested the following method: Step 1: To 2 cm 3 of each compound, add an equal volume of NaOH(aq). Step 2: Then add 1 cm3 of AgNO3(aq). Step 3: Then add excess of dilute HCl(aq). Identify and explain two improvements to the student’s proposed method. Improvement 1: Heating is required in step 1. Heating is necessary to break the C‒X bond via nucleophilic substitution to give X− for precipitation with silver nitrate. Improvement 2: Mixture should be acidified before adding AgNO3. Any excess NaOH will react with Ag + to form a brown Ag 2O precipitate, making it difficult to observe the white AgCl ppt. Improvement 3: HNO3 should be used to acidify the mixture. Cl‒ in HCl gives white ppt with AgNO3, giving wrong conclusion. [2] [Total: 11]

6 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2025 3 Amino acids are the fundamental building blocks of proteins and play crucial roles in various biological processes. Amino acids are crystalline solids with high melting points, are water-soluble, and exist as zwitterions. The general structure of an −amino acid is given below. where R represents the side−chain on the −carbon of amino acid. (a) Explain why amino acids exist as crystalline solids at room temperature. A lot of energy is required to overcome the strong ionic bo