Question

Consider the following reaction: C2H4(g) + H2O(g) ⇌ C2H5OH(g) ΔHo = -45.8 kJ/mol At 298K the equilibrium constant, KP, for this reaction is 28.5. What is ΔG for this reaction at 298K under the following conditions: p(C2H4) = p(H2O) = p(C2H5OH) = 3.0 atm?

Answer 1

ΔG = 2721.891 J/mol

Step-by-step explanation:

• C2H4(g) + H2O(g) ↔ C2H5OH(g)

∴ ΔH°rxn = - 45.8 KJ/mol

∴ Kp = 28.5

If p(C2H4) = p(H2O) = p(C2H5OH) = 3.0 atm ⇒ ΔG°rxn = ?

⇒ Kp = (PC2H5OH/P*) / (PC2H4/P*)(PH2O/P*)

∴ P* = 1 atm

⇒ Kp = (3 atm) / (3 atm)(3 atm) = 0.33

• Kp = e∧( - ΔG/RT)

⇒ Ln Kp = - ΔG/RT

⇒ ΔG = - RT LnKp

∴ T = 298 K

∴ R = 8.314 J/mol.K

⇒ ΔG = - ((8.314 J/mol.K)(298 K)) Ln(0.33)

⇒ ΔG = 2721.891 J/mol