question_answer

The initial pressure of \[PC{{l}_{5}}\] present in one litre vessel at \[200\text{ }K\]is 2 atm. At equilibrium the pressure Increases to 3 atm with Temperature increasing to\[250\text{ }K\]. The percentage dissociation of \[PC{{l}_{5}}\]at equilibrium is

A) \[30%\]   

B)                    \[60%\]               

C) \[0.2%\]              

D)        \[20%\]

Correct Answer: D

Solution :

[d] First method Pressure at \[200\text{ }K=2\text{ }atm\] Pressure at \[250\text{ }K=p\text{ }atm\] Using relation; \[\frac{{{P}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}}{{{T}_{2}}}\Rightarrow \frac{P}{250}=\frac{2}{200}\] \[\therefore \,\,\,\,P=2.5\,atm\] \[\frac{a}{a+\alpha }=\frac{2.5}{3}\Rightarrow \,\,\,\alpha =\frac{1}{5}a\] Therefore, % of \[PC{{l}_{5}}\] dissociated  \[=\frac{\alpha }{a}\times 100\] \[=\frac{a}{5\times a}\times 100=20%\] Using relation: \[\alpha =\frac{{{T}_{1}}{{P}_{2}}-{{T}_{2}}{{P}_{1}}}{{{T}_{2}}{{P}_{1}}}\] \[=\frac{200\times 3-250\times 2}{250\times 2}=\frac{600-500}{500}\] \[\alpha =\frac{1}{5}\] % of \[\alpha =\frac{1}{5}\times 100=20%\]