Consider the reaction N2(g) + 3H2(g)2NH3(g) Using the standard thermodynamic data in the tables linked above...

Question

Consider the reaction

N₂(g) + 3H₂(g)2NH₃(g)

Using the standard thermodynamic data in the tables linked above, calculate G_rxn for this reaction at 298.15K if the pressure of each gas is 29.86 mm Hg.

Table: https://cxp.cengage.com/contentservice/assets/owms01h/references/chemtables/gen_chem/thermodynamic.html

ANSWER: _____ kJ/mol

Answer 1

N₂(g) + 3H₂(g) 2NH₃(g)

(1) Find ∆Hº by ∑products - ∑reactants = (2x-46.1) - (0 + 0) = -92.2 kJ

(2) FInd ∆Sº same way = (2x192.5) - (191.6 + 3x130.7) = 385 - 584 = -199 J = -0.199 kJ

(3) Find ∆Gº from ∆Gº = ∆Hº - T∆Sº = -92.2 kJ - (298.15 x -0.199) = -92.2 kJ + 59.33 kJ = -32.87 = -33 kJ

NOTE: you can also get this same value by ∆Gº products - ∆Gºreactants = 2x -16.5 - 0 = -33 kJ

Since the reaction is at non-standard conditions, find Q, the reaction quotient:

Q = (PNH₃)²/(PN₂)(PH₂)³ = (29.86)²/(29.86)(29.86)³ = 0.001122 (assuming that ALL gases are at 29.86 mm)

Now, we can apply ∆G = ∆Gº + RT ln Q = -33 + (8.314)(298.15) ln 0.001122

∆G = -33 + -16,838 = 16,871 kJ/mol