RI 2024 Energetics Part 2 Tut 5b Ans

RAFFLES INSTITUTION Year 5 H2 CHEMISTRY 2023 Tutorial 5b – Chemical Energetics 2 Suggested Solutions to Tutorial 5b: Chemical Energetics 2 6. (a) Entropy increases. There are more ways for the molecules to arrange themselves in the larger volume, resulting in an increase in the disorder and hence entropy of the system. ( b) Entropy increases. This is because t he increase in temperature causes the broadening of the Maxwell -Boltzmann energy distribution of the particles. Thus , there are more possible energy states in which the molecules can adopt at a high temperature, which increases disorder in the system. ( c) Entropy decreases. This is because the reaction results in a decrease in the number of moles of gaseous particles ( from 1 mol e to 0 mol e) in the system , causing a decrease in disorder and hence a decrease in entropy. ( d) Entropy increases. This is because the reaction results in an increase in the number of moles of gaseous particles (from 1 mole to 2 moles). With more particles, there are more ways to arrange the particles and more ways to distribute the energy in the system, and hence creating greater disorder and higher entropy in the system. 7 . (a) (i) C(s) + ½O2(g) → CO(g) ∆Gf = ∆Hf − T∆Sf −137.2 = −110.5 − (298) ∆Sf Hence, ∆Sf[CO(g)] = [(137.2 – 110.5) x 1000]/298 = +89.6 J mol−1 K−1 C(s) + O2(g) → CO2(g) ∆Gf = ∆Hf − T∆Sf −394.4 = −393.5 − (298) ∆Sf Hence, ∆Sf[CO2(g)] = [(394.4 – 393.5) x 1000]/298 = +3.0 J mol−1 K−1 (ii) C(s) + ½O2(g) → CO(g) The reaction results in an increase in the number of moles of gaseous molecules, causing an increase in entropy. Hence the ∆Sf value should be large and positive. C (s) + O2(g) → CO2(g) The reaction doe s not result in any change in the number of moles of gaseous molecules in the system. Hence the ∆Sf value is small and near to zero. (b) C(s) + CO2(g) → 2CO(g) ∆G (298K) = 2∆Gf(CO) − ∆Gf(CO2) = 2( −137.2) −(−394.4) = +120 kJ mol−1 Since ∆G (298K) > 0, the above reaction is not feasible at 298 K. Since rΔH mΔH (products) nΔH (reactants)θθ θ= −∑∑ ff ∆Hr = 2(−110.5) − (−393.5) = +172.5 kJ mol−1 Now ∆G = ∆H − T∆S So ∆S (298K) = (∆H (298K) − ∆G (298K) ) / 298 = (172.5 − 120)(103) / 298 = +176.2 J mol−1 K−1 At 1000 K, Using ∆G = ∆H − T∆S Convert from kJ to J

-2- ∆G (1000K) = (172.5 x 103) − (1000)(176.2) = −3.7 kJ mol−1 Since now ∆G (1000K) < 0, the reaction has now become feasible. [Assumption: ∆H and ∆S remain constant over the temperature range from 298 K to 1000 K.] 8. ( a) ( ) ( ) 1ΔH 8.37 62.43 70.8 kJ molθ −= −− = − ( ) ( ) θθ θ 43.1ΔG ΔH T ΔS 70.8 273 350 1000 −= − = − −+ − = −  143.9 kJ mol (b) Consider the decomposition reaction. WI2(g) → W(s) + I2(g) For the above decomposition reaction, ∆H = +70.8 kJ mol-1 ∆S = +0.0431 kJ mol-1 K-1 ∆G = ∆H –T∆S For the reaction to be spontaneous, ∆G = ∆H –T∆S < 0 70.8 – (T)(0.0431) < 0 T > 70.8/0.0431 T > 1643 K Hence the temperature of the filament above which WI2 will decompose is 1643 K (4 sig. fig.) (NOT 1640K) 9. (a)(i) Br2(l) + 3F2(g) → 2BrF3(l) Br 2(g) + 3F2(g) 2BrF3(g) 2Br(g) + 6F(g) By Hess’ law, 6BE(Br–F) = – (2)(+44) – (2)(–301) + 31 + 193 + (3)(158) BE(Br–F) = +1212/6 = +202 kJ mol -1 Av erage bond energy of the Br –F bond in BF3 = +202 kJ mol-1 (a )(ii) ∆Gf = ∆Hf − T∆Sf −241 = −301 − 298(∆Sf) ∆Sf = −301+241 298 = −0.201 kJ mol−1 K−1 ∆S f is negative which is expected as the reaction results in a decrease in the number of moles of gas particles, which reduces entropy. (2)(–301) kJ mol-1 +31 kJ mol-1 (2)(+44) kJ mol-1 6BE(Br–F) +193 + (3)(158) kJ mol-1

-3- (b)(i) 2 O=Cl=O(g) → Cl2(g) + 2O2(g) 2Cl(g) + 4O(g) By Hess’ law, 4BE(Cl=O) = –204 + 244 + (2)(496) BE(Cl=O) = +1032/4 = +258 kJ mol-1 Hence Cl=O bond energy in ClO2 = +258 kJ mol-1 ( b)(ii) ∆G = ∆H – T∆S ∆H < 0 ∆S > 0 ⇒ –T∆S < 0 Since both ∆H and –T∆S are negative, the sign of ∆G is negative. Since both ∆H and –T∆S are negative, the value of ∆G is more negative than ∆H. 10 (a) ∆G = ∆H – T∆S At 298 K, Reaction (1): ∆G = – 5465 – (298)(–553/1000) = –5300 kJ mol–1 Reaction (2): ∆G = – 3261 – (298)(138/1000) = –3300 kJ mol–1 (b) Since ∆G (1) is more negative, reaction (1) takes places predominantly. (c) (d) At high temperatures, ∆G (2) will be more negative than ∆G (1) or reaction (2) is more exergonic than reaction (1) and CO will be produced in greater proportion. –204 kJ mol-1 4BE(Cl=O) +244 + (2)(496) kJ mol-1

-4- 11. ( a) The magnitude of hydration energy is dependent on the charge density of the cation. Mg2+ has a higher charge and smaller ionic size than Na+ and hence has a higher charge density. (b) ∆Hsoln = –L.E. + Σ∆Hhyd For NaOH: – 44 = 896 + (–390) + ∆Hhyd (OH–) ∆Hhyd(OH–) = –550 kJ mol–1 For Mg(OH)2: x = 2995 – 1890 + 2(–550) = +5 kJ mol–1 (c) Although the disruption of crystal lattice during dissolution in both cases increases disorder, t he hydration process decreases disorder about the Na + and Mg2+ ions because it puts the hydrating water molecules into an orderly arrangement about the ions. So, there is a decrease in entropy in the hydration process. Due to the higher charge and smaller size of Mg 2+ ion, more water molecules are attached to (or ordered around) the Mg2+ ion. Hence, on the average, the entropy change in the hydration process decreases more than that of Na+. Hence the net entropy changes of solution of Mg(OH)2 is likely to be more negative (or less positive) than that of NaOH. (d) Mg(OH)2 dissolves with the formation of more mobile aqueous ions; so the entropy (of the system) increases (or ∆S is positive).