Question

A mixture of 0.468 M Cl2, 0.450 M F2, and 0.862 M ClF is enclosed in a vessel and heated to 2500 K.

Cl2(g)+F2(g)↽−−⇀2ClF(g)Kc=20.0 at 2500
Calculate the equilibrium concentration of each gas at 2500 K
Cl2 = ?
F2 = ?
ClF = ?

Answer 1

To calculate the equilibrium concentrations of Cl2, F2, and ClF at 2500 K, use the ICE table and equilibrium expression. The equilibrium concentrations are approximately 0.357 M for Cl2, 0.339 M for F2, and 1.084 M for ClF.

Step-by-step explanation:

To calculate the equilibrium concentrations of each gas at 2500 K, we can use the equilibrium expression and stoichiometry of the reaction:

Cl2(g) + F2(g) ↔ 2ClF(g)

Given that Kc = 20.0 at 2500 K, we can set up an ICE table:

1. Initial concentrations: Cl2 = 0.468 M, F2 = 0.450 M, ClF = 0.862 M
2. Change in concentrations: -x, -x, +2x (since 2 moles of ClF are formed for every mole of Cl2 and F2 reacted)
3. Equilibrium concentrations: Cl2 = 0.468 - x M, F2 = 0.450 - x M, ClF = 0.862 + 2x M

Equating the Kc expression to the equilibrium concentrations:

Kc = [ClF]^2 / ([Cl2] * [F2])

Substituting the equilibrium concentrations and given Kc value:

20.0 = (0.862 + 2x)^2 / ((0.468 - x) * (0.450 - x))

Simplifying the equation and solving for x, we find that x ≈ 0.111 M. Therefore, the equilibrium concentrations are:

Cl2 ≈ 0.468 - 0.111 ≈ 0.357 M

F2 ≈ 0.450 - 0.111 ≈ 0.339 M

ClF ≈ 0.862 + 2(0.111) ≈ 1.084 M

Answer 2

The equilibrium concentrations of Cl2, F2, and ClF at 2500 K are 0.126 M, -0.144 M, and 1.159 M, respectively.

Step-by-step explanation:

To calculate the equilibrium concentrations of Cl2, F2, and ClF, we can use the equilibrium constant Kc and the initial concentrations of the reactants. Given that Kc = 20.0 at 2500 K, we can set up an ICE (Initial, Change, Equilibrium) table to solve for the equilibrium concentrations.

Initial concentrations: [Cl2] = 0.468 M, [F2] = 0.450 M, [ClF] = 0.862 M

Using the stoichiometry of the balanced equation, the change in concentration of Cl2 and F2 will be -2x (since they react to form 2 moles of ClF), and the change in concentration of ClF will be +x (since it is formed from the reaction).

After applying the equilibrium expression, we get: Kc = [ClF]^2 / ([Cl2] * [F2]) = 20.0

Substituting the given initial concentrations and solving for x, we find that x = 0.297 M. Therefore, the equilibrium concentrations are: [Cl2] = 0.468 M - 2x = 0.468 M - 2(0.297 M) = 0.468 M - 0.594 M = 0.126 M, [F2] = 0.450 M - 2x = 0.450 M - 2(0.297 M) = 0.450 M - 0.594 M = -0.144 M (negative value indicates that F2 is completely consumed), and [ClF] = 0.862 M + x = 0.862 M + 0.297 M = 1.159 M.