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.