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* 10^(-1)`

Concentration of
`BrF_3` at equilibrium =
`1.24* 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=([BrF_3]^2)/([Br_2][F_2]^3)`

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

`45.6=((1.24* 10^(-1))^2)/([Br_2]* (1.99* 10^(-1))^3)`

`[Br_2]=0.0428M`

Thus, the equilibrium concentration of Br₂ is, 0.0428 M