Question

Calculate kp at 298.15 k for the reactions (a), (b), and (c) using δg°f values.

Answer 1

(a) 2NO(g) + O₂(g) ⇄2NO₂(g)kp
(b)  2N₂O(g)⇄2NO(g) + N₂(g) kp
(c)  N₂(g) + O₂(g)⇄ 2NO(g) kp
Now A is
2NO +O₂⇄2NO₂
ΔG° =ΔG° products - ΔG reactants
=2× 51.3-(256.6)
-70.6kJ/mol.
ΔG° = -RT Inkp
-70.6 = -8.314 ×10⁻³ ˣ 298.15 ˣInkJ
InkJ = 28.48
kp=2.34 ˣ 10¹²

B is 
ΔG° = 2× 86.6 - 2 × 104.2 = -35.2
-35.2 = 8.314 × 10⁻³ ˣ 298.15 ˣInkJ
InkJ = 14.2
kp = 1.47ˣ 10⁶

C is
It is also similar
kp = 4.62 ˣ 10⁻³I