At the end of the experiment, the reaction mixtures will need to be treated. a.
ID: 486119 • Letter: A
Question
At the end of the experiment, the reaction mixtures will need to be treated. a. Approximately how many mL of liquid will need to be treated at the end of the experiment? b. What is the name of the red precipitate that is removed from the mixture at the end of the experiment? c. What should you do if the filtration of the red precipitate slows down? A reaction O_2(g) + N_2(g) rightarrow 2NO(g) was prepared with the following initial concentrations: [O_2]_0 = [N_2]_0 = 0.0100 M, [No]_0 = 0 M. At equilibrium [NO] was measured and found to be 2.00 times 10^-4 M. a. Set-up an ICE table and calculate the equilibrium concentration for the two reactants. b. Calculate K_eq. The same reaction is now run with the following initial concentrations: [O_2]_0 = 0.100 M; [N_2]_0 = 1.00 times 10^-4 M, [NO]_0 = 0 M. Because the concentration of oxygen is one thousand times as concentrated as that of nitrogen, the reaction is driven to the right effectively using up all of the N_2. a. What are the equilibrium concentrations for all species in this reaction? [NO] = ________ [NO_2] = ________ [O_2] = ________ Why is this not the best system to use when determining a K_eq?Explanation / Answer
I can answer question 2 and 3, since I do not have information to answer question 1.
2. Write down the equation as given:
O2 (g) + N2 (g) ------> 2 NO (g)
[O2]0 = [N2]0 = 0.0100 M and [NO]0 = 0 M.
(a) Let x be the molar consumption of O2 and N2 at equilibrium. Set up the ICE Chart as below:
O2 (g) + N2 (g) <=====> 2 NO (g)
initial 0.0100 0.0100 0
change -x -x +2x
equilibrium (0.0100 –x) (0.0100 – x) 2x
(b) Given 2x = 2.00*10-4 M, therefore, x = (2.00*10-4 M)/2 = 1.00*10-4 M
Therefore, [O2]eq = [N2]eq = (0.0100 – 1.00*10-4) M = 0.0099 M
Keq = [NO]2/[O2][N2] = (2.00*10-4)2/(0.0099).(0.0099) = (4.00*10-8)/(9.801*10-5) = 4.08*10-4 4.1*10-4 (ans).
3. Here we have [O2]0 = 0.100 M; [N2] = 1.00*10-4 M and [NO]0 = 0 M
(a) Given that [O2]0 = 1000*[N2]0, it is imperative that N2 is completely used up in the reaction. Infact, the reaction is no longer an equilibrium system but rather goes to completion.
As per the balanced stoichiometric reaction for the reaction,
[NO] = 2*[N2] = 2*1.00*10-4 M = 2.00*10-4 M.
Since N2 is completely used up, hence at the end of the reaction, [N2] = 0 and [O2] = (0.100 M) – ½*[NO] = (0.100 M) – ½*(2.00*10-4 M) = 0.0999 M
Ans: [NO] = 2.00*10-4 M; [N2] = 0.00 M and [O2] = 0.0999 M
(b) The above system (in 3a) is not a good system to determine Keq because the concentration of O2 is extremely high as compared to the concentration of the other reactant, N2. This essentially forces the reaction to progress in the forward direction and suppresses the reverse reaction. Therefore, the system essentially goes to completion and the equilibrium cannot be studied.
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