please help within five minutes

Chemistry
Tutor: None Selected Time limit: 1 Day

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.

Mar 12th, 2015

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


Mar 12th, 2015

Are you studying on the go? Check out our FREE app and post questions on the fly!
Download on the
App Store
...
Mar 12th, 2015
...
Mar 12th, 2015
Dec 11th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer