Postby **Megan Lu 3D** » Sun Jan 17, 2021 11:15 am

Hi! For the final part of this question, we can set up an ICE box.

(I) The initial concentration of N2O4 is unchanged, so we can leave it at 0.373 mol/L; the initial concentration of NO2 needs to account for the additional 1.00 mol added, so that would be 2.04 + 1.00 mol/L.

(C) Because the product NO2 has been added, the reaction will shift left as it moves towards equilibrium. Thus, we can say that the reactant N2O4 is changes +x and the product NO2 changes -2x, as there are 2 mols of NO2 per 1 mol of N2O4 in the reaction equation.

(E) Putting it all together, we get [N2O4] = 0.373 + x and [NO2] = 2.04 + 1.00 - 2x

After creating this ICE box, we can set the expression [NO2]^2 / [N2O4] equal to the equilibrium constant value from the beginning of the question, and then solve for x. After solving for x, plug the value in to get the final equilibrium concentrations. Hope this helps!