Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IIGibb's Free EnergyHard2 minQB
ChemistryHardnumerical
An athlete in the weight room lifts a 50 kg mass through a vertical distance of 2.0 m. The mass is allowed to fall through the 2.0 m distance while coupled to an electrical generator. The electrical generator produces an equal amount of electrical work, which is used to produce aluminium by Hall electrolytic process.
Al2O3(solution) + 3C(graphite) → 2 Al(l) + 3CO(g); Δ G° = 600 kJ
How many times must the athlete lift the 50 kg mass to provide sufficient Gibbs energy to produce 27 g Al? (g = 10 m/s2)
Answer:
300.00
Solution:
27 g Al = 1 mol. The reaction gives 2 mol Al for 600 kJ, so 1 mol needs 300 kJ. Work per lift = mgh = 50×10×2.0 = 1000 J. Lifts = 300000/1000 = 300. Ans = 0300.
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Gibb's Free Energy
⏱ 2mℹ️ Source: QB
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