Question

A solution of nitrogen dioxide, NO2 gas, at a pressure of 2.50 atm and temperature of 300 K, reacts with oxygen gas, O2, to form nitrogen oxide, NO, and dioxide, NO2. Write the balanced chemical equation for the reaction and determine the equilibrium concentrations of all species if the value of the equilibrium constant, Kc, at 300 K is 4.0 x 10^-4.

Answer 1

First, let's write the balanced chemical equation for the reaction:2 NO2  g  + O2  g   2 NO  g  + 2 NO2  g Now, let's set up an ICE  Initial, Change, Equilibrium  table to determine the equilibrium concentrations of all species. Initial concentrations:[NO2] = 2.50 atm[O2] = 0  since it's not mentioned [NO] = 0  since it's not mentioned Change in concentrations:-2x for NO2-x for O2+2x for NO+2x for NO2Equilibrium concentrations:[NO2] = 2.50 - 2x + 2x = 2.50 atm[O2] = -x[NO] = 2xSince the value of the equilibrium constant, Kc, at 300 K is 4.0 x 10^-4, we can set up the equilibrium expression:Kc = [NO]^2 [O2] / [NO2]^2Plugging in the equilibrium concentrations:4.0 x 10^-4 =  2x ^2  -x  /  2.50 ^2Now, we need to solve for x. However, we cannot have a negative concentration for O2. This means that the reaction does not proceed as written, and there is no reaction between NO2 and O2 under these conditions. Therefore, the equilibrium concentrations remain as the initial concentrations:[NO2] = 2.50 atm[O2] = 0[NO] = 0