Answer: [tex]3.8\times 10^{-3}M[/tex]
Explanation:
Moles of [tex]NOCl[/tex] = 1 mole
Moles of [tex]Cl_2[/tex] = 1 mole
Volume of solution = 1 L
Initial concentration of [tex]NOCl[/tex] = 1 M
Initial concentration of [tex]Cl_2[/tex] = 1 M
The given balanced equilibrium reaction is,
[tex]2NOCl(g)\rightleftharpoons 2NO(g)+Cl_2(g)[/tex]
Initial conc. 1 M 0M 1 M
At eqm. conc. (1-2x) M (2x) M (1+x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[NO]^2[Cl_2]}{[NOCl]^2}[/tex]
The [tex]K_c[/tex]= [tex]1.6\times 10^{-5}}[/tex]
Now put all the given values in this expression, we get :
[tex]{1.6\times 10^{-5}}=\frac{(2x)^2\times (1+x)}{(1-2x)^2}[/tex]
By solving the term 'x', we get :
[tex]x=0.0019[/tex]
Concentration of [tex]NO[/tex] at equilibrium= (2x) M = [tex]2\times 0.0019=3.8\times 10^{-3}M[/tex]
