Answer:
The given reaction is:
\[\underset{\begin{smallmatrix}
t=0 \\
At\,equilibrium
\end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\underset{\underset{0.0033-2x}{\mathop{0.0033}}\,}{\mathop{2BrCl(g)}}\,\rightleftharpoons
\underset{\begin{smallmatrix}
0 \\
x
\end{smallmatrix}}{\mathop{B{{r}_{2}}(g)}}\,+\underset{\begin{smallmatrix}
0 \\
x
\end{smallmatrix}}{\mathop{C{{l}_{2}}(g)}}\,\]
\[{{K}_{c}}=\frac{[B{{r}_{2}}][C{{l}_{2}}]}{{{[BrCl]}^{2}}}\]
\[32=\frac{{{x}^{2}}}{{{(0.0033-2x)}^{2}}}\]
\[5.656=\frac{x}{0.0033-2x}\]
\[x=1.5\times {{10}^{-3}}\]
\[[BrCl]=0.0033-2x\]
\[=0.0033-2\times
1.5\times {{10}^{-3}}=3\times {{10}^{-4}}M\]
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