Question

Dinitrogen tetroxide partially decomposes according to the following equilibrium: N2O4(g) → 2NO2(g) A 1.000-L flask is charged with 3.00 × 10-2 mol of N2O4. At equilibrium, 2.36 × 10-2 mol of N2O4 remains. Keq for this reaction is ________.

I am wondering why you do not have to convert the moles to molality to solve this problem?

Answer 1

For calculating K_e, we will use molar concentration in terms of mol/L, hence we dont require molality. Further, there is no solvent as such.

At equilibrium, moles of N₂O₄ remaining = 2.36 x 10^-2 mol.

N₂O₄ reacted = ( 3 - 2.36) x 10^-2 mol = 0.64 x 10^-2

Thus, from stoichiometry, moles of NO₂ formed at equilibrium = 2 x 0.64 x 10^-2 = 1.28 x 10^-2 moles

Since volume is 1 L, no. of moles = molar concentration in mol/L

K_eq = [NO₂]²/[N₂O₄] = (1.28)²/(2.36) = 0.6942 x 10^-2 mol/L ( This is K_c)