question_answer 51)

At 1127 K and 1 atm pressure, a gaseous mixture of CO and\[C{{O}_{2}}\]in equilibrium with solid carbon has 90.55% CO by mass. \[C(s)+C{{O}_{2}}(g)\rightleftharpoons 2CO(g)\] Calculate \[{{K}_{c}}\] for this reaction.

Answer:

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