The initial concentration of NO2 in a closed container is 0.750M. At equilibrium the concentration of O2 is 0.125M. What are the concentrations of NO2 and NO at equilibrium?
2NO2 (aq) ==== 2NO (aq) + O2 (aq)
......2NO2 (aq) ==== 2NO (aq) + O2(aq)
I....0.750............0..........0
C......-2x.............2x.........x
E....0.750-2x..........2x.........x
So the problem tells you x [which is (O2)] = 1.25M which makes 2x (the concn of NO) just twice that. NO2 then is 0.75-2x
To find the concentrations of NO2 and NO at equilibrium, we can use the stoichiometry and the given information.
First, let's define the initial concentration of NO2 as [NO2]₀ = 0.750 M.
At equilibrium, the balanced chemical equation shows that 2 moles of NO2 is converted to 2 moles of NO and 1 mole of O2. Therefore, the change in concentration for NO2 is -2x, the change for NO is +2x, and the change for O2 is +x.
Since the initial concentration of NO2 is 0.750 M, the equilibrium concentration of NO2 is [NO2] = [NO2]₀ - 2x.
We are given that the concentration of O2 at equilibrium is 0.125 M, which means [O2] = x.
Since 2 moles of NO2 gives 2 moles of NO, the equilibrium concentration of NO is [NO] = 2x.
We can now set up an equation using the given information:
[O2] = 0.125 M // Concentration of O2 at equilibrium
Since [O2] = x, we have x = 0.125 M.
Substituting the value of x into the equation [NO] = 2x, we get:
[NO] = 2(0.125 M) = 0.250 M. Therefore, the concentration of NO at equilibrium is 0.250 M.
Finally, substituting the value of x into the equation [NO2] = [NO2]₀ - 2x, we get:
[NO2] = 0.750 M - 2(0.125 M) = 0.500 M. Therefore, the concentration of NO2 at equilibrium is 0.500 M.
In summary, at equilibrium, the concentration of NO2 is 0.500 M, and the concentration of NO is 0.250 M.