NYJC 2023 H2 Chemistry 9729 P2 Answer

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 1 H2 Chemistry Prelim Exam Answers Paper 2 Answers 1 (a) (i)     3 3 3 NOC 125 101325 8.31 150 273 3603 mol m 3603 mol dm1000 3 60 mol dm pV nRT np V RT           . l (ii) The pressure of NOC l is very high, causing the volume of NOC l gas particles to become more significant relative to the total volume of the gas. There is significant intermolecular forces of attraction (i.e. permanent dipole - permanent dipole or instantaneous dipole-induced dipole interactions between NOCl molecules) as it is polar/has a relatively large electron cloud size.

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 2 (b) (i) (ii) The graph is straight line passing through origin, rate increases linearly with (ρNOCl)2. The order of reaction with respect to ρNOCl is 2. (iii) Rate = k(ρNOCl)2 Units: N-1 m2 s-1 Initial rate / N m-2 s-1 (ρNOCl)2 / N2 m-4 0.20 0.40 0.60 0.80 1.00 50 100 150 200 x x x x x

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 3 (c) Increasing temperature will lead to  an increase in the average kinetic energy of reactant molecules,  more molecules will have energy greater or equal to Ea.  frequency of effective collisions increases.  rate of reaction increases,  hence the rate constant increases (since rate = k[conc] and [conc] is unchanged). (d) rate = k[Cl2][NO]2 2 (a) (i) Lattice energy is the energy evolved when one mole of an ionic compound is formed

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 4 from its constituent gaseous ions under standard condition. (ii) Ionic radius of Cl─ ion = 0.181 nm Ionic radius of I─ ion = 0.216 nm Ionic radius of Cl─ ion is smaller than the ionic radius of I─ ion. Since Cl─ ion in Ag Cl and I─ ion in AgI have the same ionic charge, the theoretical lattice energy of AgCl is more exothermic than the theoretical lattice energy of AgI. (iii) Ag+ has high charge density and hence high polarising power. It can distort the electron cloud of Cl─ ion to result in sharing of electron density. This results in covalent character in the ionic bonds of AgCl. (iv) AgI has greater covalent character as I─ has a larger and more polarisable electron cloud. (b) (i) ( ( (- (  ─ ( ( ( (  ─ (ii) By Hess’s L w, ∆Hf = (+178) + (+½ x 244) + (+590) + (349) + (687) = 146 kJ mol─1 (iii)

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 5 y 2CaCl(s) Ca(s) + CaCl2(s) From equation 4, 2CaCl(s)  2Ca+(g) + 2Cl─(g)  ∆Heqn4 From equation 5, Ca+(g) + e─  Ca(g) ∆Heqn5 y ∆Heqn2 ∆Heqn3  ∆Heqn4 ∆Heqn5 ∆Heqn6 = 178 + (2310) + (687 x 2) + (590) + 1150 = 554 kJ mol─1 (c) (i) Cl─ ion has a smaller ionic radius hence a higher charge density than I─ ions. Cl─ ion forms a stronger ion-dipole interaction with H2O molecules. The ion-dipole interaction formed between Ag + and Cl ─ ions with H 2O molecules is more exothermic than that formed between Ag + and I─ ions with H 2O molecules. Hence, more energy is released to overcome the stronger ionic bonds in AgC l and hydrogen bonds between H2O molecules. (ii) ∆Go ppt = 2.303 RT log Ksp = 2.303 × 8.31 × 298 × log (2.0 × 10-10) = 55310 J mol-1 = 55.3 kJ mol-1 (iii) ∆Go ppt = ∆Ho ppt  T∆So ppt ∆So ppt =  0.03591 kJ mol-1 K-1 =  35.9 J mol-1 K-1 (iv) ∆So ppt is negative. Entropy of the system decreases as AgCl is precipitated. There are less ways of arranging the ions in the solid than in the aqueous state, resulting in a more disordered system to a less disordered system. (v) ∆Ho ppt is neg tive nd ∆So ppt is negative. Hence T∆So ppt is positive. When temperature increased to 1000 K, T∆So ppt becomes more positive. ∆Go ppt will become more positive / less negative. Precipitation will be less spontaneous, hence solubility of AgCl(s) increases. 3 (a) (i) Concentrations of the different sugar in 100g of ice-cream are: 2Ca+(g) + 2Cl(g) Ca+(g) + e + Ca2+(g) + 2Cl(g) Ca(g) + Ca2+(g) + 2Cl (g) Ca(s) + Ca2+(g) + 2Cl(g) 178 590 +1150 2(+687) 2310

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 6 sucrose = 3g glucose = 7g lactose = 1.4g maltose = 8.5g Taking the relative sweetness values from table sucrose = 1 x 3g = 3.000 units of relative sweetness glucose = 0.75 x 7g = 5.250 units of relative sweetness lactose = 0.2 x 1.4g = 0.2800 units of relative sweetness maltose = 0.5 x 8.5g = 4.250 units of relative sweetness Overall relative sweetness of the 150g serving of ice-cream = 3.000 + 5.250 + 0.2800 + 4.250 = 12.78 = 12.8 units of relative (ii) Sugar relative sweetness price /kg price per unit of relative sweetness Glucose 0.75 $8.63 $11.51 Fructose 1.15 $25.70 $22.35 Lactose 0.2 $15.49 $77.45 Maltose 0.5 $13.87 $27.74 The price per unit of relative sweetness of lactose is the highest at $77.45. To maintain the same level of relative sweetness, the manufacturer using more lactose will incur a much higher cost. Hence they would minimise use of lactose. (iii) Volume of ice-cream after melting = 150 g / 0.5 g ml-1 = 300 ml Based on percentage mass given, 150g of ice -cream contains 4.5g of sucrose, 10.5g of glucose, 2.1g of lactose and 12.75g of maltose. Total amount of sugar in the serving of ice-cream = 4.5g of sucrose + 10.5g of glucose + 2.1g of lactose + 12.75g of maltose = 29.85 g Since sugar content = 29.85 g / 300 ml = 9.95 g / 100 ml The assigned grade would be C. (b) (i) Hydrolysis

Nanyang Junior College 2023 J2 H2 Chemistry Prelim Exam Answers 7 (ii) Glucose and fructose have simple molecular structure with hydrogen bonds between their molecules. A lot of energy is required to overcome the strong hydrogen bonds which results in a high boiling point and low volatility. Hence these molecules do not get v aporise and travel easily to get picked up by smell receptors of the human body. (iii) It is reducing in n ture/ It c used reducti n f T ens’ re gent/ It underg es oxidation. (c) (i) Dynamic equilibrium occurs in a reversible system, where the rates of the forward and reverse reactions are the same. There is no net change in the concentration of the reactants and products. (ii) percentage of the glucose left = 52.2 19.0 33.2 0.3516 35.2%113.4 19.0 94.4     percentage of the glucose formed = 100 – 35.16 = 64.84% Assuming there was 1 mol of glucose at the start, Kc =     0.6484 1 1.844 1.840.3516 1 glucose glucose           (iii) The new equilibrium mixture will contain more glucose and less glucose. When the te per ture is incre sed, by Le h te ier’s Princip e, the reverse endothermic reaction is favoured to reduce the added heat. Hence the position of equilibrium will shift to the left. (iv) The final measured optical rotation remains at +52.2o but it will be achieved faster. The acid acts as catalyst for the conversion, lowering the activation energy of both the forward and reverse reactions to the same extent, therefore, increasing the rate of the forward and reverse reactions to the same extent, allowing dynamic equilibrium to be reached faster. Addition of a catalyst has no effect on position of equilibrium, composition of equ