Answer :
Step 1 - Understanding the reaction
The reaction between NO2 (a brown gas) and N2O4 (a colourless gas) reaches equilibrium according to the reaction bellow:
[tex]2NO_{2(g)}\rightleftarrows N_2O_{4(g)}[/tex]The original concentration of NO2 is 0.9 M. Note that any variation in this concentration is due to the formation of N2O4.
Step 2 - Finding the equilibrium concentrations
The exercise already tells us that the equilibrium concentration for NO2 is 0.26 M. To find the equilibrium concentration of N2O4, we must first subtract the equilibrium concentration of NO2 from its original concentration:
[tex]0.9-0.26=0.64M[/tex]This is the concentration of NO2 that reacted. Note that each 2 moles of NO2 produce only one mole of N2O4. The equilibrium concentration of N2O4 will be thus half the concentration of NO2 that reacted:
[tex]\lbrack N_2O_4\rbrack_{eq}=0.32M[/tex]Step 3 - Finding the equilibrium constant
The expression for this equilibrion constant is:
[tex]K=\frac{\lbrack N_2O_4\rbrack_{eq}}{\lbrack NO_2\rbrack^2_{eq}}[/tex]Substituting the values [N2O4]=0.32M and [NO2]=0.26M, we have:
[tex]K=\frac{0.32}{0.26^2}=\frac{0.32}{0.06}=5.33[/tex]The equilibrium constant for this reaction is thus 5.33
