The reaction O2 ⇄ 2 O has a K=1.2x10-10. If a 1.7 L container holds 3.8 moles
of O atoms (and nothing else), what will the concentration of O be when
equilibrium is reached?
So far, I have 1.2x10^-10 = 2.2^2 / 0
but I don't know how to do the rest. When it says "when equilibrium is reached" does it mean that the left side and the right side has to match? Help!!
To determine the concentration of O at equilibrium, we will use the given equilibrium constant and the stoichiometry of the reaction.
The equilibrium constant expression for the reaction O2 ⇄ 2 O is written as:
K = [O]^2 / [O2]
Given that K = 1.2x10^-10, we can substitute this value into the equation:
1.2x10^-10 = [O]^2 / [O2]
Now, let's find the concentration of O by solving for [O].
First, we need to determine the equilibrium concentration of O2. We know that the total moles of O atoms present in the container is 3.8 moles. Since the reaction O2 ⇄ 2 O produces twice the number of O atoms compared to O2 molecules, the total moles of O2 will be half of the total moles of O atoms:
[O2] = 3.8 moles / 2 = 1.9 moles
Substituting this value back into the equilibrium constant expression:
1.2x10^-10 = [O]^2 / 1.9
Rearranging the equation to solve for [O]:
[O]^2 = 1.2x10^-10 * 1.9
[O]^2 = 2.28x10^-10
Taking the square root of both sides:
[O] = √(2.28x10^-10)
[O] ≈ 1.511x10^-5 M
Therefore, the concentration of O at equilibrium is approximately 1.511x10^-5 M.