Question

2NOCl(g) ↔ 2NO(g) + Cl2(g) with K = 1.6 x10–5. In an experiment, 1.00 mole of pure NOCl and 1.00 mole of pure Cl2 are placed in a 1.00-L container. If x moles of NOCl react, what is the equilibrium concentration of NO?

Answer 1

Answer: $3.8\times 10^{-3}M$

Explanation:

Moles of  $NOCl$ = 1 mole

Moles of  $Cl_2$ = 1 mole

Volume of solution = 1 L

Initial concentration of $NOCl$ = 1 M

Initial concentration of $Cl_2$ = 1 M

The given balanced equilibrium reaction is,

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

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,

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

The $K_c$= $1.6\times 10^{-5}}$

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

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

By solving the term 'x', we get :

$x=0.0019$

Concentration of $NO$ at equilibrium= (2x) M  =  $2\times 0.0019=3.8\times 10^{-3}M$