Chemistry 2 - Equilibrium Constant

Question

The equilibrium constant for the chemical equation

N2(g)+3H2(g)↽−−⇀2NH3(g)

is 𝐾p=0.0328 at 201 ∘C. Calculate the value of 𝐾c for the reaction at 201 ∘C.

𝐾c= ?

Answer 1

Kp = Kc(RT)^∆n

Kp = 0.0328

Kc = ?

R = 0.0821

T = 201C + 273 = 474K

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

Solving for Kc, we have...

0.0328 = Kc (0.0821*474)^-2

0.0328 = Kc * 0.160

Kc = 0.205

(be sure to check the math)