Question

The equilibrium constant for the reaction N2O4 ⇌ 2 NO2 is 8.75 x 10-2. At equilibrium, a vessel contains N2O4 at a concentration of 1.72 x 10-2 M. What is the concentration of NO2?

Answer 1

K_eq = C(NO)²/C(N₂O₄),

So:

C(NO) = $\sqrt{\smash[b]{C(N_2O_4)*K_{eq}}}$ = $\sqrt{\smash[b]{8.75*10^{-2}*1.72*10^{-2}}}$ = 3.88 * 10^-2 M

Answer: concentration of NO2 is 3.88 * 10^-2 M.