Answer:
The given reaction is:
\[\underset{\begin{smallmatrix}
Mass \\
Moles
\end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\underset{-}{\mathop{C(s)}}\,+\underset{\begin{smallmatrix}
9.45 \\
\frac{9.45}{44}=0.215
\end{smallmatrix}}{\mathop{C{{O}_{2}}(g)}}\,\rightleftharpoons
\underset{\begin{smallmatrix}
90.55 \\
\frac{90.55}{28}=3.234
\end{smallmatrix}}{\mathop{2CO(g)}}\,\,\,\Delta
{{n}_{g}}=2-1=1\]
\[\Sigma n=0.215+3.234\]
\[=3.449=3.45\]
\[{{K}_{p}}=\frac{{{({{n}_{CO}})}^{2}}}{{{n}_{C{{O}_{2}}}}}\times
{{\left( \frac{p}{\Sigma n} \right)}^{\Delta {{n}_{g}}}}\]
\[=\frac{{{(3.234)}^{2}}}{0.215}\times {{\left( \frac{1}{3.45}
\right)}^{1}}=14.1\]
\[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\]
\[{{K}_{c}}=\frac{{{K}_{p}}}{{{(RT)}^{\Delta
{{n}_{g}}}}}=\frac{14.1}{{{(0.0821\times 1127)}^{1}}}\]
\[=0.152mol\,{{L}^{-1}}\]
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