A non-equilibrium mixture of 1.00 mol of A(g) (partial pressure 1.00 atm) and 1.00 mol of B(g) (partial pressure 1.00 atm) is allowed to equilibrate at 25°C.
Calculate the partial pressure of A(g) at equilibrium.
From a previous problem we were given the rxn.... A <--> B, with GA = 8748 J/mol and GB = 11024 J/mol
We calculated the Gibbs free energy, G = GB - GA = 2276 J/mol
and we calculated k = 0.399, where k = PB/PA (PB, PA: pressures of B and A respectively).
The equilibrium shifts towards A, so we can write...
k = (PB - x) / (PA + x), where k=.399, PA = 1.00 atm and PB = 1.00 atm
solving for x...
x = .601/1.3999 = 0.4295 atm ( x is the partial pressure change)
So the equilibirum partial pressure of A will be PA + 0.4295 atm = 1.00 + 0.4295
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