Question

For the decomposition of phosphorous pentachloride to phosphorous trichloride and chlorine at 400K the KC is 1.1x10-2. Given that 1.0g of phosphorous pentachloride is added to a 250mL reaction flask, find the percent decomposition after the system has reached equilibrium. PCl_5(g) PCl_3(g) Cl_2(g) K_C

Answer 1

Answer:

$\% Decomposition=47.4\%$

Explanation:

Hello,

In this case, for the given decomposition of phosphorous pentachloride:

$PCl_5(g)\rightleftharpoons PCl_3(g)+ Cl_2(g)$

As the equilibrium constant is $1.1x10^{-2}$ and the initial concentration of phosphorous pentachloride is:

$[PCl_5]_0=\frac{1.0gPCl_5*\frac{1molPCl_5}{208.24gPCl_5} }{250mL*\frac{1L}{1000mL} } =0.019M$

Hence, by writing the law of mass action equation:

$Kc=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

We must introduce the change $x$ occurring due to the reaction extent and the concentrations at equilibrium (ICE table methodology):

$Kc=\frac{(x)(x)}{[PCl_5]_0-x}=\frac{x^2}{0.019-x}=1.1x10^{-2}$

Thus, solving for $x$ we obtain:

$x=0.01M$

In such a way, the equilibrium concentration of phosphorous pentachloride results:

$[PCl_5]_{eq}=[PCl_5]_0-x=0.019M-0.01M$

$[PCl_5]_{eq}=0.009M$

Finally, the percent decomposition is computed by:

$\% Decomposition=\frac{[PCl_5]_0}{[PCl_5]_{eq}}*100\%=\frac{0.009M}{0.019M} *100\%\\\\\% Decomposition=47.4\%$

Best regards.