Question

Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 X 102 at a certain temperature. In a particular experiment, 3.00 mole of each component was added to a 1.50 L flask. Calculate the equilibrium concentrations of all species.

Answer 1

Answer:

The equilibrium concentrations of all species :

$[H_2]= 0.314 M$

$[F_2]=0.314 M$

$[HF]=3.372 M$

Explanation:

Moles of hydrogen gas = 3.00 mole

Concentration of hydrogen gas = $[H_2]=\frac{3.00 mol}{1.5 L}=2.0 M$

Moles of fluorine gas = 3.00 mole

Concentration of fluorine gas = $[F_2]=\frac{3.00 mol}{1.5 L}=2.0 M$

Given the equilibrium constant of the reaction = $K_c=1.15\times 10^2$

$H_2(g)+F_2(g)\rightleftharpoons 2HF(g)$

Initial

2.0 M  2.0 M     0

At equilibrium

(2.0-x)M  (2.0 -x)M         2x

An expression of $K_c$ is given by :

$K_c=\frac{[HF]^2}{[H_2][F_2]}$

$1.15\times 10^2=\frac{(2x)^2}{(2.0-x)(2.0 -x)}$

On solving for x , we get :

x = 1.686 M

The equilibrium concentrations of all species :

$[H_2]=(2.0 -1.686)M = 0.314 M$

$[F_2]=(2.0 -1.686)M = 0.314 M$

$[HF]=(2\times 1.686)M = 3.372 M$