Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIGibb's Free EnergyHard2 minQB
ChemistryHardnumerical

At 298 K, the standard enthalpy of a combustion reaction is −2000 kJ mol⁻¹ and the standard Gibb's energy of the reaction is −2298 kJ mol⁻¹. The additional (non-expansion) work (in kJ) that may be obtained by raising the temperature to 30°C is

Answer:
5.00
Solution:

ΔS = (ΔH − ΔG)/T = (−2000 − (−2298))/298 = 1 kJ K⁻¹ mol⁻¹. Assuming ΔH and ΔS unchanged, at 303 K (30°C): ΔG₃₀₃ = ΔH − TΔS = −2000 − 303 × 1 = −2303 kJ mol⁻¹. Additional non-expansion work = (−ΔG₃₀₃) − (−ΔG₂₉₈) = 2303 − 2298 = 5 kJ. (The trap: work tracks ΔG, not ΔH, and ΔG must be re-evaluated at the new temperature through ΔS.)

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

Doubts & Discussion

Loading discussions...