Question

At 35°C, Kc = 1.6 multiplied by10-5 for the following reaction

2 NOCl(g) reverse reaction arrow 2 NO(g)+ Cl2(g)

Calculate the concentrations of all species at equilibrium if

2.0 mol NO and 1.0 mol of Cl2 are placed in a 1.0 L flask

Answer 1

Answer : The equilibrium concentrations of all species
`NO,Cl_2\text{ and }NOCl` are, 0.05 M, 0.043 M and 0.975 M respectively.

Explanation : Given,

Moles of
`NO` = 2 mole

Moles of
`Cl_2` = 1 mole

Volume of solution = 1 L

Initial concentration of
`NO` = 2 M

Initial concentration of
`Cl_2` = 1 M

The given balanced equilibrium reaction is,

`2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)`

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=([NOCl]^2)/([NO]^2[Cl_2])`

The
`K_c` for reverse reaction =
`(1)/(1.6* 10^(-5))`

Now put all the given values in this expression, we get :

`(1)/(1.6* 10^(-5))=((2x)^2)/((2-2x)^2* (1-x))`

By solving the term 'x', we get :

x = 0.975

Thus, the concentrations of
`NO,Cl_2\text{ and }NOCl` at equilibrium are :

Concentration of
`NO` = (2-2x) M = (2 - 2 × 0.975) M = 0.05 M

Concentration of
`Cl_2` = (1-x) M = 1 - 0.975 = 0.043 M

Concentration of
`NOCl` = x M = 0.975 M

Therefore, the equilibrium concentrations of all species
`NO,Cl_2\text{ and }NOCl` are, 0.05 M, 0.043 M and 0.975 M respectively.