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Question

At 1120 K, ∆G° = 58.3 kJ/mol for the reaction 3 A (g) + B (g) →2 C (g). If the partial pressures of A, B, and C are 11.5 atm, 8.60 atm, and 0.510 atm respectively, what is the free energy for this reaction?

What is ∆G for a reaction where ∆G° = 4.5 kJ/mol and Q = 7.8 at 295 K? (R = 8.314 J/mol ･ K)

Answer 1

3A(g) + B(g) ==> 2C(g)

11.5.....8.60.........0.510 ..... non-equilibrium partial pressures

Q = (C)² / (A)³(B)

Q = (0.510)² / (11.5)³(8,60)

Q = 1.99x10^-5

∆G = ∆Gº + RT ln Q

∆G = ?

∆Gº = 58.3 kJ/mol

RT = (8.314 J/Kmol)(1120K) = 9312 J/mol = 9.312 kJ/mol (change units to match those of ∆Gº)

ln Q = ln 1.99x10^-5 = -10.8

∆G = 58.3 + (9.312)(-10.8) = 58.3 - 100.6

∆G = -42.3 J/mol

∆G = ∆Gº + RT ln Q

∆G = ?

∆Gº = 4.5 kJ/mol

R = 8.314 J/Kmol

T = 295K

Q = ?

Plug in the values (see solution above) and solve for Q. Remember to keep units of R and ∆Gº the same.