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
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