Question

the equation for the equilibrium constant for the reaction in terms of k1 k2 and k3

Answer 1

hope it helps!!!!!!!

Answer 2

Answer:

The equilibrium constant in terms of given constants:

$K_4=\frac{K_2(K_3)^3}{K_1}$

Explanation:

$N_2(g)+3H_2\rightaleftharpoons 2NH_3(g),K_1$..[1]

$K_1=\frac{[NH_3]^2}{[N_2][H_2]^3}$

$N_2(g)+O_2\rightaleftharpoons 2NO(g),K_2$..[2]

$K_2=\frac{[NO]^2}{[N_2][O_2]}$

$H_2(g)+\frac{1}{2}O_2\rightaleftharpoons H_2O(g),K_3$...[3]

$K_3=\frac{[H_2O]}{[H_2][O_2]^{\frac{1}{2}}}$

[te]2NH_3(g)+\frac{5}{2}O_2\rightaleftharpoons 2NO+3H_2O(g),K_4[/tex]...[4]

$K_4=\frac{[NO]^2[H_2O]^3}{[NH_3]^2[O_2]^{\frac{5}{2}}}$

Multiply and divide $[N_2]$ and $[H_2]^3$:

$K_4=\frac{[NO]^2[H_2O]^3}{[NH_3]^2[O_2]^{\frac{5}{2}}}\times \frac{[N_2][H_2]^3}{[N_2][H_2]^3}$

$K_4=\frac{K_2\times [H_2O]^3}{K_1\times [O_2]^{\frac{3}{2}}}\times \frac{1}{[H_2]^3}$

$K_4=\frac{K_2}{K_1}\times (\frac{[H_2O]}{[H_2][O_2]^{\frac{1}{2}}})^3$

$K_4=\frac{K_2(K_3)^3}{K_1}$