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

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be