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 0.119 moles/liter in a sealed container and no product was present initially?

Answer 1

Br₂(g) + F₂(g) ⇔ 2 BrF(g)

Initial concentration 0.119 mole/L    0.119 mole/L 0

Equilibrium 0.119(1-x)mole/L   0.119(1-x)mole/L 2x

Equilibrium constant K_c = [BrF]²/ [ Br₂​][F₂]

54.7 = 4x²/ 0.119(1-x)mole/L  0.119(1-x)mole/L =  54.7 = 4x²/ (0.119(1-x)mole/L )²

54.7 x(0.119)²(1-x)² = 4x²

0.775(1-2x+x²) = 4x²

4x² - 0.775x² +1.55 x - 0.775 = 0

3.225 x² +1.55 x - 0.775 = 0

So equilibrium concentrations are

BrF = 0.612 mole/L

Br₂ = F₂ = 0.119 - 0.036 = 0.083 mole/L

We can solve this quadratic equation

129/40x²+31/20 x−31/40=0

The solutions are:

x1=−31/129−(4/129)√310 = - ve value so neglect it

x2=−31/129+(4/129)√310 = 0.306