PCl5(g) ↔ PCl3(g) + Cl2(g)
What is the value of the equilibrium constant K(T) at T = 400 K? At 25 ºC:
ΔfHº(PCl5,g) = -374.9 kJ / mol ; Smº(PCl5,g) = 364.6 J / (K mol)
ΔfHº(PCl3,g) = -287.0 kJ / mol ; Smº(PCl3,g) = 311.78 J / (K mol)
Smº(Cl2,g) = 223.07 J / (K mol)
2. For the reaction in problem (2) calculate the degree of dissociation
alpha = (number of moles of PCl5 dissociated at equilibrium) / (original number of moles of PCl5 in the system)
at T = 400 K and P = 1 atm. Assume that in the conditions of the problem the equilibrium gas mixture behaves as an ideal gas mixture.
3. Repeat problem P3, but in the case that the temperature is 400 K and the pressure is 15 atm. Continue assuming that the equilibrium gas mixture behaves as an ideal gas mixture.
4. Equimolar amounts of H2(g) and CO(g) are mixed in a container at 25 ºC. The reaction
H2(g) + CO(g) ↔ HCHO(g)
takes place. What is the mol fraction of formaldehyde, HCHO, at equilibrium when the total pressure is 10 atm? Assume that you can treat the system at equilibrium as an ideal gas mixture.
ΔfGº(CO,g) = -137.17 kJ / mol
ΔfGº(HCHO,g) = -102.53 kJ / mol
These solutions may offer step-by-step problem-solving explanations or good writing examples that include modern styles of formatting and construction of bibliographies out of text citations and references. Students may use these solutions for personal skill-building and practice. Unethical use is strictly forbidden.1. Answer
The change of Gibbs free energy of reaction is defined as: ΔrGº=ΔrHº – T*ΔrSº.
ΔrHº= ΔfHº(PCl3,g) + ΔfHº(Cl2,g) – ΔfHº(PCl5,g) = – 287.0 kJ/mol + 0 – (– 374.9 kJ/mol) = 87.9 kJ/mol
ΔrSº= Smº(PCl3,g) + Smº(Cl2,g) – Smº(PCl5,g) = 311.78 J / (K mol) + 223.07 J / (K mol) – 364.6 J / (K mol) = 170.25 J / (K mol) = 0.17025 kJ / (K mol)
T=25 ºC = 298 K
ΔrGº = 87.9 kJ/mol – 298 K *0.17025 kJ / (K mol) = 37.17 kJ/mol
Equilibrium constant K(T) [T=400 K, R=8.314 J / (K mol)]
K= e–ΔrGº/RT= e(– 37170 J/mol / (8.314 J / (K mol) *400 K)) = e-11.177 = 0.000014....
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