problem 11.11 6th edition

[Cynthia Ulloa]

Hi y'all I am having trouble understanding this problem. Can someone break it down for me and explain how to go about each one?

[Ashley Zhu 1A]

You are given that ozone forms oxygen in the equilibrium reaction 2O3 -->/<-- 3O2. It gives you different conditions, one where 0.10 mol of ozone is put in a 1.0L vessel and then another where 0.50 mol is put in a 1.0L vessel, both of which are at the same temperature. This means that the equilibrium constant, K, is the same for both (the only thing that changes K is temperature).

a) since the volume is the same (both in 1.0L vessels), the number of moles of O2 is greater in the second experiment b/c you start out with more ozone
b) likewise, because there is more moles of O2 in the second experiment and both experiments have the same volume, the concentration of O2 in the second experiment will be greater
c) then, by the same logic, the ratio of [O2]/[O3] must be different since the concentrations of each will be different in each experiment due to differing amounts of moles in each
d) [O2]^3/[O3]^2 however, is the same for both experiments because that is what K equals (K = [products]^stoichiometric coefficient/[reactants]^stoichiometric coefficient) and K is the same for both since the temperature is the same
e) [O3]^2/[O2]^3 is also the same for both since it is the reciprocal of the ratio in part d, meaning it is 1/K and K is still the same

Hope this helps!