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

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,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

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

$0.21=\frac{(x)^2}{(0.033)}$

By solving the term 'x', we get :

x = 0.083 M

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