Question

For the reaction N2O4 <--> 2 NO2 (a) What is the equilibrium expression? (b) If [N2O4] initial = 0.25 M and K = 4.5 what are the concentration of reactants and products at equilibrium?

Answer 1

We gots...`N_2O_4(g) rightleftharpoons2NO_2(g)`

Explanation:

And `K_"eq"=([NO_2]^2)/([N_2O_4])=4.5`

Now initially...`[N_2O_4]=0.25*mol*L^-1`, and WE ASSUME that `x*mol*L^-1` dissociates then....

`K_"eq"=4.5=(4x^2)/(0.25-x)`

And so `x=sqrt((4.5xx(0.25-x)/4)`...

And so `x^2=4.5xx(0.25-x)/4=0.28125-1.125x`

And so `x^2+1.125x-0.28125=0`...and we solve this directly using the quadratic equation..`x=0.211`

...and so `[N_2O_4]_"eq"=0.039*mol*L^-1`; `[NO_2]=0.422*mol*L^-1`