Fluorine reacts with oxygen to yield oxygen difluoride.

2 F2(g) + O2(g) 2 OF2(g)
What is the value of K if the following concentrations are found at equilibrium: [O2] = 0.200 mol/L, [F2] = 0.0100 mol/L, and [OF2] = 0.0633 mol/L

See my response above about arrows.

K = (OF2)^2/(O2)(F2)^2
Substitute and solve for K.

To find the value of K, we can use the equilibrium expression:

K = ([OF2]^2) / ([F2]^2 * [O2])

Given that [O2] = 0.200 mol/L, [F2] = 0.0100 mol/L, and [OF2] = 0.0633 mol/L, we can substitute these values into the equation:

K = (0.0633^2) / (0.0100^2 * 0.200)
K ≈ 20.07

Therefore, the value of K is approximately 20.07.

To find the value of K (the equilibrium constant), we need to use the given concentrations of the reactants and products at equilibrium and plug them into the equation:

K = [OF2]^2 / ([F2]^2 * [O2])

Given:
[O2] = 0.200 mol/L
[F2] = 0.0100 mol/L
[OF2] = 0.0633 mol/L

Substitute these values into the equation:

K = (0.0633)^2 / ((0.0100)^2 * 0.200)

Now, solve the equation:

K = 0.0040089 / 0.00002

K ≈ 200.445

Therefore, the value of K is approximately 200.445.