Question

The reaction below is carried out at a different temperature at which Kc=0.055. This time, however, the reaction mixture starts with only the product, [NO]=0.0100M, and no reactants. Find the equilibrium concentrations of N2, O2, and NO at equilibrium.

Answer 1

Thee given question is incomplete. the complete question is:

The reaction below is carried out at a different temperature at which Kc = 0.055. This time, however, the reaction mixture starts with only the product, [NO] = 0.0100 M, and no reactants. Find the equilibrium concentrations of N2, O2, and NO at equilibrium. The equation is N2(g) + O2(g) <--> 2NO(g)

Answer: Concentration of
`NO` at equilibrium = 0.001 M

Concentration of
`N_2` = 0.0045 M

Concentration of
`O_2` = 0.0045 M

Step-by-step explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stoichiometric coefficients.

Initial concentration of
`NO` = 0.0100 M

The given balanced equilibrium reaction is,

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

Initial conc. 0 M 0 M 0.0100 M

At eqm. conc. (x) M (x) M (0.0100-2x) M

The expression for equilibrium constant for this reaction will be,

`K_c=([NO]^2)/([N_2][O_2])`

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

`0.055=((0.0100-2x)^2)/((x)* (x))`

By solving the term 'x', we get :

x = 0.0045

Thus, the concentrations of
`N_2,O_2\text{ and }NO` at equilibrium are :

Concentration of
`NO` at equilibrium = (0.0100-2x) M =
`(0.0100-2* 0.0045)=0.001M`

Concentration of
`N_2` = (x) M = 0.0045 M

Concentration of
`O_2` = (x) M = 0.0045 M