The given question is incomplete. The complete question is
If 1.0 M HI is placed into a closed container and the reaction is allowed to reach equilibrium at 25∘C∘C, what is the equilibrium concentration of H2 (g). Given the equilibrium constant is 62.
Answer: The equilibrium concentration of [tex]H_2[/tex] is 0.498 M
Explanation:
Initial concentration of [tex]HI[/tex] = 1.0 M
The given balanced equilibrium reaction is,
[tex]2HI(g)\rightarrow H_2(g)+I_2(g)[/tex]
initial (1.0) M 0 0
At eqm (1.0-2x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[H_2]\times [I_2]}{[HI]^2}[/tex]
Now put all the given values in this expression, we get :
[tex]62=\frac{(x)^2}{1.0-2x}[/tex]
By solving we get :
[tex]x=0.498M[/tex]
Thus the equilibrium concentration of [tex]H_2[/tex] is 0.498 M
