Question

At elevated temperatures, BrF5 establishes the following equilibrium. 2 BrF5(g) ⇋ Br2(g) + 5 F2(g) The equilibrium concentrations of the gases at 1200°C are 0.011 M for BrF5, 0.635 M for Br2, and 0.854 M for F2. Calculate the value of Kc. Enter a number to 2 decimal places.

Answer 1

Answer:

Kc = 2383.84

Explanation:

Let's consider the following reaction at equilibrium.

2 BrF₅(g) ⇋ Br₂(g) + 5 F₂(g)

The concentration equilibrium constant (Kc) is the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

Kc = [Br₂].[F₂]⁵/[BrF₅]²

Kc = (0.635).(0.854)⁵/(0.011)²

Kc = 2383.84

Answer 2

Answer:

Kc = 0.02 M^4

Explanation:

For reaction involving gas, gas equilibrium constant, Kc is used.

Kc = [product]/[reactant]

aA(g) + bB(g) --> cC(g) + dD(g)

Kc = ([C]^c * [D]^d)/([A]^a * [B]^b)

2BrF5(g) ⇋ Br2(g) + 5 F2(g)

Concentrations at equilibrium:

BrF5 = 0.011M

Br2 = 0.635M

F2 = 0.0854M

Kc = ([Br2]^1 * [F2]^5)/([BrF5]^2)

= ([0.635]^1 * [0.085]^5)/([0.011]^2)

= 0.0233 M^4