Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIGibb's Free EnergyMedium2 minQB
ChemistryMediumnumerical

At 298 K, the standard enthalpy of combustion of sucrose is −5737 kJ mol⁻¹ and the standard Gibb's energy of the reaction is −6333 kJ mol⁻¹. The additional (non-expansion) work (in kJ) that may be obtained by raising the temperature to 29°C is

Answer:
8.00
Solution:

Maximum non-expansion work = −ΔG. ΔS = (ΔH − ΔG)/T = (−5737 − (−6333))/298 = 596/298 = 2 kJ K⁻¹ mol⁻¹. Assuming ΔH and ΔS unchanged, at 302 K (29°C): ΔG₃₀₂ = ΔH − TΔS = −5737 − 302 × 2 = −6341 kJ mol⁻¹. Additional work = (−ΔG₃₀₂) − (−ΔG₂₉₈) = 6341 − 6333 = 8 kJ.

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Gibb's Free Energy
⏱ 2mℹ️ Source: QB

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