(b)For the equilibrium
The relation between Kp and Kc is given as:
Kp = Kc (RT)Δ n
From reaction :
Δn = 3 – 2 = 1
R = 0.0831 bar Lmol–1K–1
T = 1069 K
Kc = 3.75 X 10-6
Formula used :
Kp = Kc (RT) Δ n
Kp = 3.75 x 10-6 (0.0831 x 1069)1
Kp = 333.12 x 10-4 Answer