question_answer 1)

\[3mol\] of\[PC{{l}_{5}}\] kept in 1 L closed vessel was allowed to attain equilibrium at 380 K. Calculate the composition of the reaction mixture at equilibrium. \[{{K}_{c}}=1.80\]  

Answer:

\[\underset{{{t}_{eq}}}{\overset{{}}{\mathop{{{t}_{0}}}}}\,\underset{\underset{(3-x)}{\mathop{3}}\,}{\mathop{PC{{l}_{5}}(g)}}\,\rightleftharpoons \underset{\underset{x}{\mathop{0}}\,}{\mathop{PC{{l}_{3}}(g)}}\,+\underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{C{{l}_{2}}(g)}}\,\] \[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{x\times x}{(3-x)}=1.80\] \[{{x}^{2}}+1.8x-5.4=0\] \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] \[=\frac{-1.8\pm \sqrt{{{(1.8)}^{2}}+4\times 1\times 5.4}}{2\times 1}\] \[x=1.59\] \[[PC{{l}_{3}}]=[C{{l}_{2}}]=1.59M\] \[[PC{{l}_{5}}]=(3-x)=3-1.59\] =1.41M