Question

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

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,

$K_c=\frac{[NOCl]^2}{[NO]^2[Cl_2]}$

The $K_c$ for reverse reaction = $\frac{1}{1.6\times 10^{-5}}$

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

$\frac{1}{1.6\times 10^{-5}}=\frac{(2x)^2}{(2-2x)^2\times (1-x)}$

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.