At 311 K, this reaction has a 𝐾c value of 0.0398 . X(g)+3Y(g)↽−−⇀2Z(g) Calculate 𝐾p K p at 311 K.

Answer 1

X(g) + 3Y(g) <=> 2Z(g)

Kp = Kc(RT)^∆n

Kp = ?

Kc = 0.0398

R = gas constant = 0.0821 Latm/Kmol

T = temp in K = 311K

∆n = change in moles of gas = 2 - 4 = -2

Kp = 0.0398 (0.0821 * 311)^-2

Kp = 0.0398 (25.53)^-2

Kp = 0.0398 (0.0.00153)

Kp = 6.11x10^-5