Answer:
The equilibrium concentrations of all species :
[tex][H_2]= 0.314 M[/tex]
[tex][F_2]=0.314 M[/tex]
[tex][HF]=3.372 M[/tex]
Explanation:
Moles of hydrogen gas = 3.00 mole
Concentration of hydrogen gas = [tex][H_2]=\frac{3.00 mol}{1.5 L}=2.0 M[/tex]
Moles of fluorine gas = 3.00 mole
Concentration of fluorine gas = [tex][F_2]=\frac{3.00 mol}{1.5 L}=2.0 M[/tex]
Given the equilibrium constant of the reaction = [tex]K_c=1.15\times 10^2[/tex]
[tex]H_2(g)+F_2(g)\rightleftharpoons 2HF(g)[/tex]
Initial
2.0 M 2.0 M 0
At equilibrium
(2.0-x)M (2.0 -x)M 2x
An expression of [tex]K_c[/tex] is given by :
[tex]K_c=\frac{[HF]^2}{[H_2][F_2]}[/tex]
[tex]1.15\times 10^2=\frac{(2x)^2}{(2.0-x)(2.0 -x)}[/tex]
On solving for x , we get :
x = 1.686 M
The equilibrium concentrations of all species :
[tex][H_2]=(2.0 -1.686)M = 0.314 M[/tex]
[tex][F_2]=(2.0 -1.686)M = 0.314 M[/tex]
[tex][HF]=(2\times 1.686)M = 3.372 M[/tex]
