For the reaction
NO2(g) + NO(g) N2O(g) + O2(g)
Kc = 0.914. Equal amounts of NO and NO2 are to be placed into a 5.00 L container until the N2O concentration at equilibrium is 0.0446 M. How many moles each of NO and NO2 must be placed into the container?
......NO2 + NO ==> N2O + O2
I......x.....x.....0......0
C.....-y.....-y....y......y
E.....x-y...x-y....y......y
So you want (N2O) = (O2) = 0.0446M
mols = 0.0446*5L = 0.223 = y
Kc = (N2O)(O2)/(NO2)(NO)
Kc = (x-0.223)(x-0.223)/(0.223)(0.223)
Solve for x = mols NO2 = mols NO
Then (NO2) = (NO) = mols/L = mols from above/5L = ?
To answer this question, we need to use the equation for the reaction and the given equilibrium constant (Kc). From the balanced equation, we can see that the stoichiometry of the reaction is for every 2 moles of NO2 and 2 moles of NO, we get 1 mole of N2O and 1 mole of O2.
Let's assume that x moles of NO2 and x moles of NO are initially placed into the container. At equilibrium, the concentration of N2O will be given as 0.0446 M. Therefore, the molar concentration of N2O can be calculated as x/(5.00 L) since the total volume is 5.00 L.
Now, we can express the equilibrium constant (Kc) in terms of the molar concentrations of the substances:
Kc = [N2O][O2] / [NO2][NO]
Since the stoichiometry of the reaction is 1:1 for both NO2 and NO with N2O and O2, we can say that:
[N2O] = [O2] = x/(5.00 L)
[NO2] = x/(5.00 L)
[NO] = x/(5.00 L)
Now let's plug these values into the equation for Kc:
0.914 = (x/(5.00 L) * x/(5.00 L)) / ((x/(5.00 L)) * (x/(5.00 L)))
Simplifying this equation, we get:
0.914 = x^2 / (5.00 L * 5.00 L)
Rearranging and solving for x^2, we get:
x^2 = 0.914 * (5.00 L * 5.00 L)
x^2 = 22.85 L^2
Taking the square root of both sides, we find:
x = √22.85 L ≈ 4.78 moles
Since equal amounts of NO and NO2 are initially placed into the container, the moles of NO and NO2 must each be approximately 4.78 moles.