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 m