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
