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

1.430 atm

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up