What is the value of ΔGrxn? (ΔGºrxn = 41.90 kJ)

Question

The partial pressures of N₂, H₂, and NH₃ were measured as 0.603, 1.00, and 0.0100 atm, respectively, in the reaction shown below. Temperature is 650 K.

What is the value of ΔG_rxn? (ΔGº_rxn = 41.90 kJ)

N2​(g)+3H2​(g)2NH3​(g)

    

Kp​ = 4.3 x 10−4

Answer 1

For this question, we'll want to use the following relationship:

∆G = ∆Gº + RT ln Q

N₂(g) + 3H₂(g) <==> 2NH₃(g)

Q = (NH₃)² / (N₂)(H₂)³

Q = (0.0100)² / (0.603)(1.00)³

Q = 1.66x10^-4 and since this is LESS than Kp, the reaction is NOT at equilibrium and will proceed to the right, i.e. toward the products.

Solving for ∆Grxn, we have...

∆G = 41,900 J + (8.314 J/Kmol)(650K) ln 1.66x10^-4

∆G = -5135 J = -5.13 kJ