Question

the following concentration were obtained for the formation of Nh3 from N2 and H2 at equilibrium at 500k. n2= 1.5×10^-2M. H2 = 3.0×10^-2.and NH3=1.2×10-2.caculate equilibrium constant

Answer 1

Equilibrium constant Kc=((1.2×10^(-2))^(2))÷((1.5×10^(-2))×(3.0×10^(-2))^(3)))=355.56

Kp=Kc(RT)^(dN)=0.211

Answer 2

Answer: Equilibrium constant will be $0.035\times 10^4$

Explanation:

$[NH_3]=1.2\times 10^{-2}$

$[H_2]=3.0\times 10^{-2}$

$[N_2]= 1.5\times 10^{-2}$

$N_2+3H_2\rightleftharpoons 2NH_3$

So eq'm constant will be:

$K_{c}=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

$K_c=\frac{[1.2\times 10^{-2}]^2}{[3.0\times 10^{-2}]^3[1.5\times 10^{-2}]}$

$K_c=0.035\times 10^4$