J.R. S. answered 04/06/23
Ph.D. University Professor with 10+ years Tutoring Experience
N2(g) + 3 H2(g) <==> 2 NH3(g)
1.75.........2.15..............0...........Initial
-x............-3x................+2x........Change
1.75-x....2.15-3x...........2x........Equilibrium
2x = 1.34 M
x = 0.67 M
Equilibrium concentrations are as follows:
[N2] = 1.75 - 0.67 = 1.08 M
[H2] = 2.15 - 2.01 = 0.14 M
[NH3] = 1.34 M
Kc = [NH3]2 / [N2][H2]3 = (1.34)2 / (1.08)(0.14)3 = 606
∆Gº = -RT ln K
∆Gº = -(8.314 J/Kmol)(298K) ln 606
∆Gº = -2478 x 6.41
∆Gº = -15,884 J/mole
∆Gº = -15.9 kJ/mol
The reaction is SPONTANEOUS. This is true because ∆Gº is negative indicating spontaneity. This is also in agreement with the large value of Kc indicating the reaction lies far to the right (product side).