Answer :
Answer:
[tex]\% Decomposition=47.4\%[/tex]
Explanation:
Hello,
In this case, for the given decomposition of phosphorous pentachloride:
[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+ Cl_2(g)[/tex]
As the equilibrium constant is [tex]1.1x10^{-2}[/tex] and the initial concentration of phosphorous pentachloride is:
[tex][PCl_5]_0=\frac{1.0gPCl_5*\frac{1molPCl_5}{208.24gPCl_5} }{250mL*\frac{1L}{1000mL} } =0.019M[/tex]
Hence, by writing the law of mass action equation:
[tex]Kc=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]
We must introduce the change [tex]x[/tex] occurring due to the reaction extent and the concentrations at equilibrium (ICE table methodology):
[tex]Kc=\frac{(x)(x)}{[PCl_5]_0-x}=\frac{x^2}{0.019-x}=1.1x10^{-2}[/tex]
Thus, solving for [tex]x[/tex] we obtain:
[tex]x=0.01M[/tex]
In such a way, the equilibrium concentration of phosphorous pentachloride results:
[tex][PCl_5]_{eq}=[PCl_5]_0-x=0.019M-0.01M\\[/tex]
[tex][PCl_5]_{eq}=0.009M[/tex]
Finally, the percent decomposition is computed by:
[tex]\% Decomposition=\frac{[PCl_5]_0}{[PCl_5]_{eq}}*100\%=\frac{0.009M}{0.019M} *100\%\\\\\% Decomposition=47.4\%[/tex]
Best regards.