Initially a 1 L vessel contains 10 moles of NO2 and 6 moles of O2. At equilibrium, the vessel contains 8.8 moles of NO2. Determine Kc.
Please help
Determine Kc for which reaction?
I'm not sure, this was all that I was given
I will assume that the reaction is as follows and that the concn of NO2 initially is 10 mol/1L = 10 M and 8.8 mols/L or 8.8 M at the end.
......2xNO2 + O2 ==> NO + O3
I.....10....6......0.....0
C.....-x....-x.....x.....x
E....10-x..6-x.....x.....x
Therefore, x must be 10-8.8 = 1.2 which makes 6-1.2 = 4.8 with both NO and O3 at the E line being 1.2
To find Kc = (NO)(O3)/(NO2(O2) substitute the E line values into the Kc expression and evaluate Kc.
If you find that the equation is different than the one I've assumed, the process will be the same process. Remember to balance the equation first.
To determine the equilibrium constant Kc, we can use the balanced chemical equation for the reaction and the molar concentrations of reactants and products at equilibrium.
The balanced chemical equation for the reaction is:
2 NO2(g) + O2(g) ⇌ 2 NO3(g)
Let's define the initial concentrations as follows:
[NO2]₀ = 10 moles/L
[O2]₀ = 6 moles/L
At equilibrium, the concentrations become:
[NO2]eq = 8.8 moles/L
[O2]eq = will be determined
We can assume that the change in the concentration of NO2 is -1.2 moles/L (10 - 8.8), as it decreases from the initial value, while the change in the concentration of O2 is x moles/L.
Using the balanced chemical equation, we can see that the change in the concentration of O2 is (-2)(-1.2) = +2.4 moles/L, as it increases by a factor of 2.
The equilibrium concentrations are then:
[NO2]eq = 8.8 moles/L
[O2]eq = 6 + 2.4 = 8.4 moles/L
Now we can calculate the equilibrium constant Kc using the formula:
Kc = ([NO3]eq^2) / ([NO2]eq^2 * [O2]eq)
= ([NO3]eq^2 / [NO2]eq^2) * (1/[O2]eq)
The stoichiometric coefficients in the balanced chemical equation tell us that [NO3]eq^2 / [NO2]eq^2 = 1 and [O2]eq = 8.4 moles/L.
Therefore, Kc = 1 * (1/8.4) = 0.119 (rounded to three decimal places)
So, the equilibrium constant Kc for this reaction is approximately 0.119.