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23 May, 01:37

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?

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  1. 23 May, 01:58
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    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
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