Question

2. Hydrogen fluoride can be produced from elemental fluorine and hydrogen according to the reaction H2(g) + F2(g)  2HF(g). The reaction has an equilibrium constant, Kc, of 7.75 x 10 2 at a certain temperature. a. Calculate the equilibrium concentration of HF(g) if 5.750 mol of H2 and F2 are introduced into a 1.500 L flask.

Answer 1

Answer:

The equilibrium concentration of HF is 7.15 M.

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of Hydrogen gas = 5.750 moles

Volume of container = 1.500 L

$[H_2]=\frac{5.750 mol}{1.500 L}=3.833 M$

Moles of fluorine gas = 5.750 moles

Volume of container = 1.500 L

$[F_2]=\frac{5.750 mol}{1.500 L}=3.833 M$

For the given chemical equation:

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

Initially

3.833 M    3.833 M        0

At equilibrium:

(3.833-x) M   (3.833-x) M     2x

The expression of an equilibrium constant will be given as:

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

$7.75\times 10^2=\frac{(2x)^2}{(3.833-x)\times (3.833-x)}$

Solving for x:

x = 3.576 M

The equilibrium concentration of HF :

$[HF]=2x=2\times 3.576 M=7.152 M\approx 7.15 M$