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

`\% Decomposition=47.4\%`

Step-by-step 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=(1.0gPCl_5*(1molPCl_5)/(208.24gPCl_5) )/(250mL*(1L)/(1000mL) ) =0.019M`

Hence, by writing the law of mass action equation:

`Kc=([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=((x)(x))/([PCl_5]_0-x)=(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=([PCl_5]_0)/([PCl_5]_(eq))*100\%=(0.009M)/(0.019M) *100\%\\\\\% Decomposition=47.4\%`

Best regards.