Respuesta :
Answer : The equilibrium concentrations of all species [tex]NO,Cl_2\text{ and }NOCl[/tex] are, 0.05 M, 0.043 M and 0.975 M respectively.
Explanation : Given,
Moles of [tex]NO[/tex] = 2 mole
Moles of [tex]Cl_2[/tex] = 1 mole
Volume of solution = 1 L
Initial concentration of [tex]NO[/tex] = 2 M
Initial concentration of [tex]Cl_2[/tex] = 1 M
The given balanced equilibrium reaction is,
[tex]2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)[/tex]
Initial conc. 2 M 1 M 0
At eqm. conc. (2-2x) M (1-x) M (2x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[NOCl]^2}{[NO]^2[Cl_2]}[/tex]
The [tex]K_c[/tex] for reverse reaction = [tex]\frac{1}{1.6\times 10^{-5}}[/tex]
Now put all the given values in this expression, we get :
[tex]\frac{1}{1.6\times 10^{-5}}=\frac{(2x)^2}{(2-2x)^2\times (1-x)}[/tex]
By solving the term 'x', we get :
x = 0.975
Thus, the concentrations of [tex]NO,Cl_2\text{ and }NOCl[/tex] at equilibrium are :
Concentration of [tex]NO[/tex] = (2-2x) M = (2 - 2 × 0.975) M = 0.05 M
Concentration of [tex]Cl_2[/tex] = (1-x) M = 1 - 0.975 = 0.043 M
Concentration of [tex]NOCl[/tex] = x M = 0.975 M
Therefore, the equilibrium concentrations of all species [tex]NO,Cl_2\text{ and }NOCl[/tex] are, 0.05 M, 0.043 M and 0.975 M respectively.
