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
Kc = (OF2)^2/(F2)^2(O2)
Substitute and solve.
I get 8.01 for an answer, but webassign says it's wrong.
[0.0633]^2/[o.0100]^2[o.200]=8.01378
You have it set up right but your must be punching the wrong buttons on the calculator. I get 200.344 which must be rounded, of course.
To find the value of K (the equilibrium constant), we need to use the given concentrations at equilibrium and plug them into the equilibrium expression.
The equilibrium expression for this reaction is:
K = ([OF2]^2) / ([F2]^2 * [O2])
Substituting the given values:
K = (0.0633^2) / ((0.0100)^2 * 0.200)
K = 0.00402 / 0.000020
K = 201
Therefore, the value of K for this reaction is 201.