RI 2025 Tut 5b Energetics Part 2 Ans

RAFFLES INSTITUTION Year 5 H2 CHEMISTRY 2025 Tutorial 5b – Chemical Energetics 2 Suggested Solutions to Tutorial 5b: Chemical Energetics 2 6. (a) 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. (b) 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. (c) 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) + O 2(g) → CO2(g) The reaction does 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 ∆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.] Convert from kJ to J

-2- 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 rea