Question

The equilibrium constant for the reaction below, at a given temperature is 45.6. If the equilibrium concentrations of F2 and BrF3 are 1.24 x 10-1 M and 1.99 x 10-1 M respectively, calculate the equilibrium concentration of Br2. (4)

Answer 1

This is an incomplete question, here is a complete question.

The given chemical reaction is:

$Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)$

The equilibrium constant for the reaction below, at a given temperature is 45.6. If the equilibrium concentrations of F₂ and BrF₃ are 1.24 × 10⁻¹ M and 1.99 × 10⁻¹ M respectively, calculate the equilibrium concentration of Br₂.

Answer : The equilibrium concentration of Br₂ is, 0.0428 M

Explanation :  Given,

Concentration of $F_2$ at equilibrium = $1.99\times 10^{-1}$

Concentration of $BrF_3$ at equilibrium = $1.24\times 10^{-1}$

Equilibrium constant = 45.6

The given chemical reaction is:

$Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)$

The expression for equilibrium constant is:

$K_c=\frac{[BrF_3]^2}{[Br_2][F_2]^3}$

Now put all the given values in this expression, we get:

$45.6=\frac{(1.24\times 10^{-1})^2}{[Br_2]\times (1.99\times 10^{-1})^3}$

$[Br_2]=0.0428M$

Thus, the equilibrium concentration of Br₂ is, 0.0428 M