Liquid Nitrogen tetroxide, N2O4(l), was used as a fuel in Apollo missions to the moon. In a closed container the gas N2O4(g) decomposes to nitrogen dioxide, NO2(g). The equilibrium constant,k, for this reaction is 0.87 at 55 degrees Celsius. A vessel filled with N2O4(g) at this temperature is analyzed twice during the course of the reaction and found to contain the following concentrations:

Question

(a) [N2O4(g)] = 5.30 mol/L; [NO2(g)] = 2.15 mol/L

(b) [N2O4(g)] = 0.80 mol/L;[NO2(g)]] = 1.55 mol/L

In each case determine wheter the system. If not, predict the direction in which the reaction will proceed to acheive equilibrium.

Answer 1

You didn't finish your question sentence. I assume you meant to write, "In each case determine whether the system 'is in equilibrium.'". I fnot......
determine the equilibrium quotient for each case.
(a) K_quot = (NO2)^2/(N2O4)= (2.15)^2/(5.30) = 0.872 which is essentially the same as 0.87 so the system is in equilibrium. I suppose, if we wish to get picky, that 0.872 is SLIGHTLY larger than Keq, so the system will shift ever so slightly to the left.

(b) K_quot = 1.55^2/0.80 = 3.00. Therefore, the system is not in equilibrium. We get a numbre larger than Keq because the products are too large and the reactants too small. That means the reaction must shift to the left to achieve equilibrium.