RI Prelim P3 (Ans)

1 2020 RI H2 Chemistry Prelim Paper 3 – Suggested Solutions Section A 1(a)(i) OR 1(a)(ii) CH3OCH3 has 2 bond pairs and 2 lone pairs around O , hence it has a bent shape. Since lone pairs exert a greater repulsion than bond pairs, it has a bond angle of 105o 1(a)(iii) The permanent dipole-permanent dipole (pd-pd) intermolecular forces in CH3OCH3 is stronger than the pd−pd intermolecular forces in CF2Cl2. This is because CH3OCH3 has a greater net dipole moment than CF2Cl2. 1(b)(i) The two assumptions are:  The gas particles exert negligible attractive forces/intermolecular forces of attraction on each other.  The volume of gas particles is negligible. Other assumptions:  The gas particles are in constant random motion.  Collision between gas particles are perfectly elastic.  The average kinetic energy of gas particles is proportional to absolute temperature. 1(b)(ii) 1 p ideal gas 293 K

2 1(b)(iii) At higher temperature of 500 K, the N2O gas molecules have higher kinetic energy and are able to overcome the intermolecular attractive forces. Therefore, the intermolecular attractive forces between them are less significant and the N2O gas molecules deviate from ideal ity to a smaller extent. 1(c)(i) Kp = PCH3OCH3PCO2 (PCO)3 (PH 2 )3 units: atm−4 OR Pa−4 1(c)(ii) pV = nRT p = nRT/V Partial pressure of CH3OCH3 at equilibrium = 732  8.31  500 101325  10 = 3 atm (shown) 1(c)(iii) Let the initial partial pressure of CO be x. 3CO(g) + 3H2(g) ⇌ CH3OCH3(g) CO2(g) initial / atm x 40 – x 0 0 change / atm –0.6x = –9 –0.6x +3 +3 equilibrium/ atm 0.4 x 40 – 1.6x 3 3 Change in partial pressure of CO: 0.60 x = (3)(3) = 9 atm Equilibrium pressure of CO = 0.4x = ( 0.4 0.6) (9) = 6.0 atm 1(c)(iv) Equilibrium pressure of H2 = 28 – 6 – 3 – 3 = 16.0 atm Kp = P(CH3OCH3)P(CO2) P(CO)3P(H2)3 = (3)(3) (6)3(16)3 = 1.02  10−5 atm−4 OR 9.65  10−26 Pa−4 1 p 500 K ideal gas 293 K

3 1(c)(v) When Ar(g) is added at constant volume, although the total pressure in the vessel increases, the partial pressure of the reactants and products remains the same. Hence there is no shift in position of equilibrium (or Q still equals to K) and the partial pressure of CH3OCH3 will remain the same. 1(c)(vi) When temperature increases, position of equilibrium (1) will shift left to favour the backwards endothermic reaction. As a result less products and more reactants will be formed and the Kp value