The equilibrium constant for the reaction of fluorine gas with bromine gas at 300 K is 54.7 and the reaction is: Br2 (g) + F2 (g) ⇔ 2 BrF (g) What is the equilibrium concentration of fluorine if the initial concentrations of bromine and fluorine were 0.127 moles/liter in a sealed container and no product was present initially?

The equilibrium concentration of fluorine 0.027 mol/L

Explanation:

This is the equilibrium reaction:

Br₂ (g) + F₂ (g) ⇄ 2 BrF (g)

Initially 0.127 0.127 -

Intially we have 0.127 moles in both reactants.

React x x 2x

Some amount (x) has reacted. As ratio is 1:2, we have double x in BrF.

So in equilibrium we have 2x of BrF (initially we don't have anything), and 0.127 - x of reactants.

Eq 0.127-x 0.127-x 2x

Let's make K expression:

K = [BrF]² / [Br₂]. [F₂]

54.7 = (2x) ² / (0.127-x). (0.127-x)

54.7 = 4x² / (0.127-x) ²

54.7 = 4x² / (0.127² - 2. 0.127x + x²)

54.7 (0.127² - 0.254x + x²) = 4x²

0.882 - 13.89x + 54.7x² = 4x²

0.882 - 13.89x + 54.7x² - 4x² = 0

0.882 - 13.89x + 50.7x² = 0

a = 50.7

b = - 13.89

c = 0.882

Let's replace the values in the quadratic formula

(-b + -√ (b²-4ac)) / (2a)

- (-13.89) + - √ ((-13.89) ² - 4. 50.7. 0.882)) / 2. 50.7

x₁ = 0.173

x₂ = 0.1

We addopt 0.1 as the result, because 0.127 - 0.173 is a negative number, it can't be possible a negative concentration

In equilirium

[Br₂]; [F₂] = 0.127 - 0.1 = 0.027

[BrF] = 0.1.2 = 0.2