Answer : The equilibrium concentrations of all species
are, 0.05 M, 0.043 M and 0.975 M respectively.
Explanation : Given,
Moles of
= 2 mole
Moles of
= 1 mole
Volume of solution = 1 L
Initial concentration of
= 2 M
Initial concentration of
= 1 M
The given balanced equilibrium reaction is,
![2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)](https://tex.z-dn.net/?f=2NO%28g%29%2BCl_2%28g%29%5Crightleftharpoons%202NOCl%28g%29)
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,
![K_c=\frac{[NOCl]^2}{[NO]^2[Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNOCl%5D%5E2%7D%7B%5BNO%5D%5E2%5BCl_2%5D%7D)
The
for reverse reaction = ![\frac{1}{1.6\times 10^{-5}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1.6%5Ctimes%2010%5E%7B-5%7D%7D)
Now put all the given values in this expression, we get :
![\frac{1}{1.6\times 10^{-5}}=\frac{(2x)^2}{(2-2x)^2\times (1-x)}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1.6%5Ctimes%2010%5E%7B-5%7D%7D%3D%5Cfrac%7B%282x%29%5E2%7D%7B%282-2x%29%5E2%5Ctimes%20%281-x%29%7D)
By solving the term 'x', we get :
x = 0.975
Thus, the concentrations of
at equilibrium are :
Concentration of
= (2-2x) M = (2 - 2 × 0.975) M = 0.05 M
Concentration of
= (1-x) M = 1 - 0.975 = 0.043 M
Concentration of
= x M = 0.975 M
Therefore, the equilibrium concentrations of all species
are, 0.05 M, 0.043 M and 0.975 M respectively.