Question

Free Energy and Equlibrium Consider the system: A(g) = B(g) at 25°C. Assume that G°A = 8038 J/mol and G°B = 12674 J/mol. Calculate the value of the equilibrium constant for this reaction. 1 pts Tries 0/8 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. 1 pts Tries 0/8 Calculate the partial pressure of B(g) at equilibrium

Answer 1

Answer:

G^o(A) = 8038 J/mol

G^o(B) = 12674 J/mol

So G^o (A-->B) = G^oB - G^oA = 12674 J/mol - 8038 J/mol = 4636 J/mol

Equilibrium constant = K = exp (-G^o (A-->B) /RT)

Where R = gas constant = 8.314 J/ml-K

T = Temperature = 25^oC = 298 K

So K = exp (-4636 J/mol / (8.314 J/K-mol x 298 K)) = 0.031

Answer 2:

                            A(g)         ---->          Bb(g)

Initial conc.              1 mol                        1 mol

Final conc. (At equilibrium)   (1 - x) mol                   (1 + x) mol

K = 0.031 = [B(g)]/[A(g)] = (1 + x)/(1 - x)

it gives

x = -0.940

so [A(g)] = 1 - 0.940 = 0.060 mol

[B(g)] = 1 - (- 0.940) = 1.940 mol

So at equilibrium, there will be 0.060 moles of A(g) available and 1.940 moles of B(g) will be available if we start with 1 mole of each A(g) and B(g).