Answer:
[tex][HCl]_{eq}=0.05M[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction, the law of mass action at equilibrium results:
[tex]Kc=\frac{[H_2]_{eq}[Cl_2]_{eq}}{[HCl]^2_{eq}}[/tex]
Next, in terms of the change [tex]x[/tex] due to reaction extent, it is rewritten, considering an initial concentration of HCl of 0.25M (1mol/4L), as:
[tex]4.00=\frac{(x)(x)}{(0.25-2x)^2}[/tex]
Thus, solving for [tex]x[/tex] via quadratic equation or solver, the following results are obtained:
[tex]x_1=0.1M\\x_2=0.17M[/tex]
Clearly, the solution is [tex]x=0.1M[/tex] as the other result will provide a negative concentration for the hydrochloric acid at equilibrium, thereby, its equilibrium concentration turns out:
[tex][HCl]_{eq}=0.25M-2*0.1M[/tex]
[tex][HCl]_{eq}=0.05M[/tex]
Best regards.
