chemistry 12

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sry, a misprint. its:

O2+2F2 <--> 2OF2

The question is looking for Kc.

2.50 mo of oxygen and 2.50 mol of fluorine gas are placed in a 2.00L glass container at room temperature. The container is heated to 400K and the following equilbirum is established:

O2+2F2 <--> 2OF

If 37.2% of the fluorine reacts, what is the value of the equilibrium constant?

Check your equation. It isn't balanced. And do you want Kc or Kp?

Initially.
(O2) = 2.50 mols/2.00L = 1.25 M
(F2) = 2.50 mols/2.00L = 1.25 M
(OF2) = 0

1 mol F2 produces 1 mol OF2 in the reaction; therefore, 37.2% of the 1.25 mols F2 reacted and at equilibrium we have (OF2)= 1.25 x 0.372 = ??
(F2) = 1.25 - what reacted.
(O2) = 1.25 - 1/2 OF2.

Then solve for Kc.
I hope this helps. Check my thinking. Check my arithmetic.

Yup, that seems just about right. Thnk you.

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