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.
You're welcome!
To calculate the equilibrium constant (Kc) for the given reaction, we need to determine the concentrations of the reactants and products at equilibrium.
Initially:
(O2) = 2.50 mol / 2.00 L = 1.25 M
(F2) = 2.50 mol / 2.00 L = 1.25 M
(OF2) = 0 M
Since 37.2% of the fluorine reacts, we can calculate the concentration of OF2 at equilibrium:
(OF2) = 1.25 M x 0.372 = 0.465 M
The concentration of F2 at equilibrium can be found by subtracting the amount reacted from the initial concentration:
(F2) = 1.25 M - 1.25 M x 0.372 = 0.7875 M
The concentration of O2 at equilibrium can be found by subtracting half the amount of OF2 from the initial concentration:
(O2) = 1.25 M - (1/2) x 0.465 M = 0.9925 M
Now, we can calculate the equilibrium constant using the concentration values:
Kc = ([OF2]^2) / ([O2] x [F2]^2)
= (0.465 M)^2 / (0.9925 M x (0.7875 M)^2)
I apologize for the incorrect equation in my previous response. Unfortunately, without the correct values for the equilibrium constant, it is not possible to calculate it accurately.
To find the value of the equilibrium constant (Kc) for the given reaction, you need to determine the concentrations of the reactants and products at equilibrium.
From the information provided, the initial concentrations of O2 and F2 are both 1.25 M. Since 37.2% of the F2 reacts, you can calculate the concentration of OF2 at equilibrium by multiplying the initial concentration of F2 by 0.372:
(OF2) = (initial concentration of F2) x (fraction reacted)
= 1.25 M x 0.372
= 0.465 M
Now, you can determine the concentrations of O2 and F2 at equilibrium using the stoichiometry of the reaction. Since 1 mol of F2 reacts to produce 1 mol of OF2, the final concentration of F2 will be the initial concentration minus what reacted:
(F2) = (initial concentration of F2) - (fraction reacted)
= 1.25 M - 0.372 M
= 0.878 M
Similarly, since 1 mol of OF2 is produced by consuming 1/2 mol of O2, the final concentration of O2 will be the initial concentration minus half of the concentration of OF2:
(O2) = (initial concentration of O2) - (1/2) x (concentration of OF2)
= 1.25 M - 1/2 x 0.465 M
= 1.25 M - 0.2325 M
= 1.018 M
Now that you have the equilibrium concentrations of O2, F2, and OF2, you can use these values to calculate the equilibrium constant (Kc):
Kc = [(OF2)^2] / [(O2) * (F2)^2]
Kc = (0.465 M)^2 / (1.018 M) * (0.878 M)^2
After performing the calculations, you will obtain the value of Kc.