Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IIFree Energy and EquilibriumMedium2 minQB
ChemistryMediumnumerical
The diamonds are formed from graphite under very high pressure. Calculate the equilibrium pressure (in 10⁴ bar) at which graphite is converted into diamond at 25°C. The densities of graphite and diamond are 2.4 and 3.6 g/cm³, respectively, and are independent of pressure. ΔG° for the conversion of graphite into diamond is 5.0 kJ/mol.
Answer:
3.00
Solution:
At equilibrium under pressure P, ΔG(P) = ΔG° + ΔV·(P − P°) ≈ ΔG° + ΔV·P = 0. ΔV = Vdiamond − Vgraphite = 12/3.6 − 12/2.4 = 3.33 − 5.0 = −1.667 cm³ mol⁻¹ = −1.667×10⁻⁶ m³ mol⁻¹. P = −ΔG°/ΔV = 5000/1.667×10⁻⁶ = 3.0×10⁹ Pa = 3.0×10⁴ bar. Hence the equilibrium pressure is 3.0 × 10⁴ bar, i.e. 3 in units of 10⁴ bar.
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Free Energy and Equilibrium
⏱ 2mℹ️ Source: QB
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