Free Energy and Equlibrium?
A(g) <=> B(g) (Note: <=> is an equilibriam sign)
at 25°C. Assume that G°A = 7008 J/mol and G°B = 11336 J/mol.
a. Calculate the value of the equilibrium constant for this reaction.
b. 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.
c. Calculate the partial pressure of B(g) at equilibrium.
Answer:
a. Since G°A = 7008 J/mol and G°B = 11336 J/mol, and the reaction is equimolar, we can calculate
ΔGº = G°B - G°A = (11336 - 7008) J/mol = 4328 J/mol
and we can relate the change in free energy to the equilibrium constant
ΔGº = -RT ln(K)
So, at T = 298ºK (25ºC), and with R = 8.314472 J/(K mol)
4328 = -8.314*298 ln(K)
4328 = -2477,57 ln(K)
-4328/2477,57 = ln(K)
-1.7469 = ln(K)
K = exp(-1.7469) = 0.1743
b. A good thing about equilibriums is that they are reached no matter where you start at. So, the equilibrium constant tells us that
K = (B)/(A)
where (A) and (B) are each gas' partial pressure.
And since the reaction is equimolar, we'll always have 2.00 mol of gas molecules in total, so
(A) + (B) = 2.00 atm => (B) = 2 - (A)
0.1743 = (B)/(A)
0.1743 = [2 - (A)]/(A) = 2/(A) - 1
1.1743 = 2/(A)
(A) = 2/1.1743 = 1.70 atm
c. (B) = 2 - (A)
(B) = 2 - 1.70 = 0.30 atm
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