What is the value of Kc?

Question

Consider the synthesis of ammonia:

N₂(g) + 3H₂(g)<=>2NH₃(g)

At 926.5 °C, the equilibrium constant K_p is 4.09  × 10^–5. What is the value of K_c?

Answer 1

Kp = Kc(RT)^∆n

∆n is the change in the number of moles = 2 - 4 = -2

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 926.5 + 273.15 = 1199.7K

4.09x10^-5 = Kc(0.0821 * 1199.7)^-2

4.09x10^-5 = Kc(0.1008)

Kc = 4.06x10^-4

(Please check the math)