Đáp án:

${K_C} = {K_p} = {10^{5,3}}$

Giải thích các bước giải:

$\begin{gathered}
  {K_p} = \dfrac{{p_{_{C{O_2}}}^3}}{{p_{CO}^3}} = {\left( {\dfrac{{{p_{C{O_2}}}}}{{{p_{CO}}}}} \right)^3};{K_p} = {e^{\dfrac{{ - \Delta G}}{{RT}}}} \hfill \\
  {K_C} = \dfrac{{{{\left[ {C{O_2}} \right]}^3}}}{{{{\left[ {CO} \right]}^3}}};{K_p} = {K_C}.{\left( {RT} \right)^{\Delta n}} \hfill \\ 
\end{gathered} $

Δn = tổng số mol khí sản phẩm - tổng số mol khí ban đầu

$\begin{gathered}
  \Delta H_{298pu}^o = 3\Delta H_{298C{O_2}}^o + \Delta H_{298Fe}^o - \Delta H_{298F{e_2}{O_3}}^o - 3\Delta H_{298CO}^o \hfill \\
   =  - 104,05.3 - ( - 196,5) - 3.( - 36,42) \hfill \\
   =  - 6,39\left( {kcal/mol} \right) \hfill \\
   =  - 26747\left( {J/mol} \right) \hfill \\ 
\end{gathered} $

$\begin{gathered}
  \Delta S_{298pu}^o = 3S_{298C{O_2}}^o + 2S_{298Fe}^o - S_{298F{e_2}{O_3}}^o - 3S_{298CO}^o \hfill \\
   = 61,06.3 + 6,5.2 - 21,5 - 3.57,3 \hfill \\
   = 2,78\left( {cal/mol.K} \right) \hfill \\
   = 11,64\left( {J/mol} \right) \hfill \\ 
\end{gathered} $

$\begin{gathered}
  \Delta G_{298}^o = \Delta H_{298}^o - T\Delta S_{298}^o =  - 26747 - 298.11,64 =  - 30215,72\left( J \right) \hfill \\
  \Delta G_{298}^o =  - RT\ln {K_p} \hfill \\
   \Rightarrow {K_p} = {e^{\dfrac{{ - \Delta G_{298}^o}}{{RT}}}} = {e^{\dfrac{{30215,72}}{{8,314.298}}}} = {10^{5,3}} \hfill \\
  \Delta n = 3 - 3 = 0 \hfill \\
   \Rightarrow {K_C} = \dfrac{{{K_p}}}{{{{(RT)}^{\Delta n}}}} = {K_p} = {10^{5,3}} \hfill \\ 
\end{gathered} $