Question

${K_P}$ for the reaction $MgC{O_3}\left( g \right)\, \to \,MgO\left( g \right) + C{O_2}\left( g \right)$ is $9 \times {10^{ - 10}}$ . Calculate $\Delta {G^ \circ }$ for the reaction at ${25^ \circ }C$ .

Answer 1

Hint : ${K_P}$ is the equilibrium constant which is used to express the relationship between partial pressure of products and partial pressure of reactants. $\Delta {G^ \circ }$ is the Gibbs Free Energy of a chemical reaction under standard conditions. Use the formula $\Delta {G^ \circ }\, = \, - RT\ln {K_P}$ to calculate the value of $\Delta {G^ \circ }$ at ${25^ \circ }C$ .

Complete Step By Step Answer:
The equation given in the question is: $MgC{O_3}\left( g \right)\, \to \,MgO\left( g \right) + C{O_2}\left( g \right)..........\left( 1 \right)$ .
Now, at equilibrium $\sum\limits_i {{\nu _i}{\mu _i}\, = \,0}$ , where ${\nu _i}$ is the stoichiometric coefficient of component $i$ with proper sign and ${\mu _i}$ is the chemical potential of the component $i$ .
Applying this to equation $\left( 1 \right)$ , ${\mu _{MgO}} + {\mu _{C{O_2}}} - {\mu _{MgC{O_3}}}\, = \,0........\left( 2 \right)$ .
Now we know, ${\mu _i}\, = \,{\mu _i}^ \circ \left( T \right)\, + \,RT\ln \left( {\dfrac{{{P_i}}}{{{P_i}^ \circ }}} \right)$ .
Applying this to equation $\left( 2 \right)$ we get,
${\mu _{MgO}}^ \circ \, + \,{\mu _{C{O_2}}}^ \circ \, - \,{\mu _{MgC{O_3}}}^ \circ \, + \,RT\ln \left( {\dfrac{{{P_{MgO}}}}{{{P_{MgO}}^ \circ }}} \right)\, + \,RT\ln \left( {\dfrac{{{P_{C{O_2}}}}}{{{P_{C{O_2}}}^ \circ }}} \right)\, - \,RT\ln \left( {\dfrac{{{P_{MgC{O_3}}}}}{{{P_{MgC{O_3}}}^ \circ }}} \right)\, = \,0$
$\Rightarrow \,\left( {{\mu _{MgO}}^ \circ \, + \,{\mu _{C{O_2}}}^ \circ \, - \,{\mu _{MgC{O_3}}}^ \circ } \right)\, = \, - RT\ln \left( {\dfrac{{{P_{MgO}}}}{{{P_{MgO}}^ \circ }}} \right)\, - \,RT\ln \left( {\dfrac{{{P_{C{O_2}}}}}{{{P_{C{O_2}}}^ \circ }}} \right)\, + \,RT\ln \left( {\dfrac{{{P_{MgC{O_3}}}}}{{{P_{MgC{O_3}}}^ \circ }}} \right)$
\Rightarrow \,\left( {{\mu _{MgO}}^ \circ \, + \,{\mu _{C{O_2}}}^ \circ \, - \,{\mu _{MgC{O_3}}}^ \circ } \right)\, = \, - RT\ln \left\\{ {\dfrac{{\left( {\dfrac{{{P_{MgO}}}}{{{P_{MgO}}^ \circ }}} \right)\left( {\dfrac{{{P_{C{O_2}}}}}{{{P_{C{O_2}}}^ \circ }}} \right)}}{{\left( {\dfrac{{{P_{MgC{O_3}}}}}{{{P_{MgC{O_3}}}^ \circ }}} \right)}}} \right\\}............\left( 3 \right)
Now, the left hand side of the above equation is the standard Gibbs free energy change i.e. $\Delta {G_P}^ \circ$ of the reaction and is given by $\Delta {G_P}^ \circ \, = \,{\mu _{MgO}} + {\mu _{C{O_2}}} - {\mu _{MgC{O_3}}}$ .
So, the equation $\left( 3 \right)$ becomes,
\Delta {G_P}^ \circ \, = \, - RT\ln \left\\{ {\dfrac{{\left( {\dfrac{{{P_{MgO}}}}{{{P_{MgO}}^ \circ }}} \right)\left( {\dfrac{{{P_{C{O_2}}}}}{{{P_{C{O_2}}}^ \circ }}} \right)}}{{\left( {\dfrac{{{P_{MgC{O_3}}}}}{{{P_{MgC{O_3}}}^ \circ }}} \right)}}} \right\\}..........\left( 4 \right) .
Now the thermodynamic equilibrium constant of the reaction, ${K_P}\,($ taking ${P^ \circ }\, = \,1\,bar$ as the standard state pressure $)$ is given by,
${K_P}\, = \,\dfrac{{\left( {\dfrac{{{P_{MgO}}}}{{{P_{MgO}}^ \circ }}} \right)\left( {\dfrac{{{P_{C{O_2}}}}}{{{P_{C{O_2}}}^ \circ }}} \right)}}{{\left( {\dfrac{{{P_{MgC{O_3}}}}}{{{P_{MgC{O_3}}}^ \circ }}} \right)}}$ .
So equation $\left( 4 \right)$ changes to
$\Delta {G_P}^ \circ \, = \, - RT\ln {K_P}..........\left( 5 \right)$ .
Now, it is given that the thermodynamic equilibrium constant ${K_P}\, = \,9 \times {10^{ - 10}}$ .
Therefore, $\ln {K_P}\, = \,\ln \left( {9 \times {{10}^{ - 10}}} \right)\, = \, - 20.83$ .
The temperature given is ${25^ \circ }C$ .
So $T\, = \,\left( {25 + 273} \right)\,K\, = \,298\,K$ .
Now the universal gas constant $R\, = \,8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$ .
Putting the values in equation $\left( 5 \right)$ we get,
$\Delta {G_P}^ \circ \, = \, - 8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}} \times \,298\,K\, \times \,\left( { - 20.83} \right)$
$\Rightarrow \,\Delta {G_P}^ \circ \, = \,51607.8\,J\,mo{l^{ - 1}}$
$\Rightarrow \,\Delta {G_P}^ \circ \, = \,51.6\,KJ\,mo{l^{ - 1}}$ .

Additional Information:
Now the thermodynamic equilibrium constant can also be represented in terms of concentration and is given by, ${K_C}\, = \,\dfrac{{\left( {\dfrac{{{C_{MgO}}}}{{{C_{MgO}}^ \circ }}} \right)\left( {\dfrac{{{C_{C{O_2}}}}}{{{C_{C{O_2}}}^ \circ }}} \right)}}{{\left( {\dfrac{{{C_{MgC{O_3}}}}}{{{C_{MgC{O_3}}}^ \circ }}} \right)}}$ , where ${C^ \circ }\, = \,1\,mol\,d{m^{ - 3}}$ . Therefore change in Gibbs Free Energy is given by, $\Delta {G_C}^ \circ \, = \, - RT\ln {K_C}$ . The relation between ${K_P}$ and ${K_C}$ is: ${K_P}\, = \,{K_C}{\left( {\dfrac{{{C^ \circ }RT}}{{{P^ \circ }}}} \right)^{\Delta n}}$ .

Note :
In this question units play an important role so convert all the quantities given to either SI or CGS units. Proceed in a stepwise manner in order to avoid mistakes. Do remember that ${K_p}$ is a dimensionless quantity as it is the ratio of the partial pressures of products to the partial pressure of reactants.