Question

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?

Answer 1

Answer:

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