Mcpherson

Chemistry

21 May, 14:05

I already posted this question, but im still confused on it.

Dinitrogentetraoxide partially decomposes according to the following equilibrium:

N2O4 (g) - > 2NO2 (g) < - A 1.00-L flask is charged with 0.400 mol of N2O4. At equilibrium at 373 K, 0.0055 mol ofN2O4

remains. Keq for this reaction is?. so my steps: N2O4 (g) - > 2NO2 (g) Initial conc. 0.400 mol 0 mole (no reaction yet) Change - 0.3945 mol + 2 x 0.3945 mole (based on equation above)

Equilibrium 0.400-0.3945 mol 0.7890 mole 0.0055 mol 0.7890 Keq = [NO2]2/[N2O4] Keq = (0.789) ^2 / (0.0055) = 113 but my book says the answer is 0.87?