Question

For the equilibrium: 2 NO​ (g) ​ ⇌ N​ 2(g)​+ O​ 2(g)​at 300 K, the equilibrium constant, Kc, is 0.185. If 1.45 moles each of N​ 2(g)​and O​ 2(g)​are introduced in a container that has a volume of 6.00 liters and allowed to reach equilibrium at 300 K, what are the concentrations of N​ 2(g )​ , O​ 2(g)​ ,and NO​ (g)​at equilibrium?

Answer 1

Answer: The concentrations of $N_2,O_2\text{ and }NO$ at equilibrium are 0.112 M, 0.112 M  and 0.260 M

Explanation:

Moles of $N_2$ = 1.45 mole

Moles of  $O_2$ = 1.45 mole

Volume of solution = 6.00 L

Initial concentration of $N_2$ = $\frac{moles}{Volume}=\frac{1.45mol}{6.00L}=0.242M$

Initial concentration of $O_2$ = $\frac{moles}{Volume}=\frac{1.45mol}{6.00L}=0.242M$

The given balanced equilibrium reaction is,

           $2NO(g)\rightleftharpoons N_2(g)+O_2(g)$

Initial conc.          0 M             0.242 M              0.242 M  

At eqm. conc.     (2x) M      (0.242-x) M        (0.242-x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[N_2]\times [O_2]}{[NO]^2}$  

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

$0.185=\frac{(0.242-x)^2}{(2x)^2}$

By solving the term 'x', we get :

x = 0.130

Thus, the concentrations of at equilibrium are :

Concentration of = (0.242-x) M  = (0.242-0.130) M = 0.112 M

Concentration of = (0.242-x) M  = (0.242-0.130) M = 0.112 M

Concentration of = 2x M = = 0.260 M