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

Step-by-step 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` =
`(moles)/(Volume)=(1.45mol)/(6.00L)=0.242M`

Initial concentration of
`O_2` =
`(moles)/(Volume)=(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=([N_2]* [O_2])/([NO]^2)`

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

`0.185=((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