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 react