The reaction is written as:
2 NOCl (g) <=> 2 NO (g) + Cl2 (g)
Initial concentration of NOCl = 1 mol/ 1 L = 1 M
9.0 % is decomposed => 0.09 x 1 M = 0.09 M
It remains (equilibirum) = 1 M - 0.09 M= 0.91 M
---------------------------
Inicial concentration of NO and Cl2 is equal to 0
In equilibrium:
For NO) +2x = 0.09 M (is formed) => x = 0.045 M
For Cl2) +x = 0.045 M
---------------------------
Equilibrium concentrations:
For NOCl = 0.91 M
For NO = 0.09 M
For Cl2 = 0.045 M
---------------------------
K is calculated as (Kc):
Kc =
[tex]Kc\text{ = }\frac{\lbrack NO\rbrack^2\lbrack Cl2\rbrack}{\lbrack NOCl\rbrack^2}=\frac{(0.09M)^2(0.045M)}{(0.91)^2}=4.40x10^{-4}[/tex]
Remember: for the next reaction: a A + b B <=> c C + d D:
[tex]Kc\text{ = }\frac{\lbrack C\rbrack^c\lbrack D\rbrack^d}{\lbrack A\rbrack^a\lbrack B\rbrack^b}[/tex]
Answer: K = Kc = 4.40x10^-4
