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
Chemistry - DrBob222, Thursday, March 1, 2012 at 11:23pm
Kc = (OF2)^2/(F2)^2(O2)
Substitute and solve.
Chemistry - Sally, Friday, March 2, 2012 at 12:15am
I get 8.01 for an answer, but webassign says it's wrong.
[0.0633]^2/[o.0100]^2[o.200]=8.01378
Sally, look at your original post. You must be punching the wrong buttons.
To find the value of K (the equilibrium constant) for the given reaction, we need to substitute the concentrations of the chemicals at equilibrium into the equilibrium expression and solve for K.
The equilibrium expression for the reaction is:
Kc = (OF2)^2 / (F2)^2 * (O2)
Substituting the given concentrations into the equilibrium expression:
Kc = (0.0633 mol/L)^2 / (0.0100 mol/L)^2 * (0.200 mol/L)
Calculating this expression leads to:
Kc = 8.01378
Therefore, the value of K for the given reaction is approximately 8.01378.
It seems that the previous answer you obtained is correct, so there may be a mistake with the answer provided by webassign. Double-check the calculations and make sure all values are accurate.