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.