At 800 ?, the equilibrium constant Kp for the synthesis of ammonia (below) is 4.91 x 10 -5. What is the value of Kc? N2(g) + 3 H2(g) = 2 NH3(g)

Question

Hi please help on this problem. Thank you.

 At 800 ?, the equilibrium constant Kp for the synthesis of ammonia (below) is 4.91 x 10^-5. What is the value of Kc?

N₂(g) + 3 H₂(g) = 2 NH₃(g)

Answer 1

Based on mass action law:

 

K_p = (P_NH3)²/(P_N2x(P_H2)³)

Assuming that all gases behave as an ideal gas:

P_ixV = n_ixRxT ⇒ Pi = (ni/V)xRxT = [i]xRxT 

 

K_p = [NH3]²x(RxT)²/(([N₂]xRxT)x([H₂]xRxT)³)

 

K_p = {[NH₃]²/([N₂]x([H₂]³}x(RxT)^-2 = K_C x(RxT)^-2

 

K_C = K_px(RxT)² = 4.91x10^-5 x(8.314x(800+273)² = 3907.5