Answer:
[tex][PCl_3]_{eq}=0.117M[/tex]
[tex][Cl_2]_{eq}=0.117M[/tex]
[tex][PCl_5]_{eq}=8x10^{-3}M[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction at equilibrium, we can write the law of mass action as shown below:
[tex]Kc=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]
That in terms of the ICE methodology is written by means of the change [tex]x[/tex] due to the reaction extent:
[tex]Kc=\frac{x*x}{[PCl_5]_0-x}[/tex]
Thus, we need to compute the initial concentration of phosphorous pentachloride:
[tex][PCl_5]_0=\frac{13.0g*\frac{1mol}{208.25g} }{0.500L} =0.125M[/tex]
So we write:
[tex]1.80=\frac{x^2}{0.125-x}[/tex]
That we solve via either solver or quadratic equation to obtain the solution:
[tex]x=0.117M[/tex]
Thereby, the equilibrium concentrations are:
[tex][PCl_3]_{eq}=x=0.117M[/tex]
[tex][Cl_2]_{eq}=x=0.117M[/tex]
[tex][PCl_5]_{eq}=0.125M-x=0.125M-0.117M=8x10^{-3}M[/tex]
Best regards.
