Answer: For the given chemical reaction, value of [tex]k_{eq}[/tex] is [tex]1.616\times 10^{-5}[/tex]
Explanation: The chemical reaction according to the question is:
[tex]2NOCl\rightarrow 2NO+Cl_2[/tex]
Equilibrium constant, [tex]k_{eq}[/tex] for the following reaction is given by:
[tex]k_{eq}=\frac{[NO]^2[Cl_2]}{[NOCl]^2}[/tex]
Given:
[tex][NO]=1.2\times 10^{-3}M[/tex]
[tex][Cl_2]=2.2\times 10^{-3}M[/tex]
[tex][NOCl]=1.4\times 10^{-2}M[/tex]
Putting values in above equation we get:
[tex]k_{eq}=\frac{(1.2\times 10^{-3})^2(2.2\times 10^{-3})}{(1.4\times 10^{-2})^2}[/tex]
[tex]k_{eq}=1.616\times 10^{-5}[/tex]
