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?

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kobenhavn kobenhavn · Answer 1 · 2021-03-03T06:27:09+01:00

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

Explanation:

Moles of [tex]N_2[/tex] = 1.45 mole

Moles of [tex]O_2[/tex] = 1.45 mole

Volume of solution = 6.00 L

Initial concentration of [tex]N_2[/tex] = [tex]\frac{moles}{Volume}=\frac{1.45mol}{6.00L}=0.242M[/tex]

Initial concentration of [tex]O_2[/tex] = [tex]\frac{moles}{Volume}=\frac{1.45mol}{6.00L}=0.242M[/tex]

The given balanced equilibrium reaction is,

[tex]2NO(g)\rightleftharpoons N_2(g)+O_2(g)[/tex]

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,

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

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

[tex]0.185=\frac{(0.242-x)^2}{(2x)^2}[/tex]

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 = 2x M = = 0.260 M