Question

Consider the following reaction, equilibrium concentrations, and equilibrium constant at a particular temperature. Determine the equilibrium concentration of NO2(g). N2O4(g) ↔ 2 NO2(g) Kc = 0.21 [N2O4]eq = 0.033 M

Answer 1

Answer: Thus, the equilibrium concentration of
`NO_2` is 0.083 M

Step-by-step explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
`K_c`

Molarity of
`N_2O_4` at equilibrium = 0.033 M

The given balanced equilibrium reaction is,

`N_2O_4(g)\rightleftharpoons 2NO_2(g)`

At eqm. conc. (0.033) M x M

The expression for equilibrium constant for this reaction will be,

`K_c=([NO_2]^2)/([N_2O_4])`

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

`0.21=((x)^2)/((0.033))`

By solving the term 'x', we get :

x = 0.083 M

Thus, the concentrations of
`NO_2` at equilibrium is 0.083 M