\(\Delta G=\Delta H - T \Delta S\)
[Here, \(\Delta G=\) change in Gibbs free energy]
For adsorption of a gas, \(\Delta S\) is negative because randomness decreases.
Thus, in order to make \(\Delta G\) negative [for spontaneous reaction], \(\Delta G\) must be highly negative because reaction is
exothermic. Hence, for the adsorption of a gas, if \(\Delta S\) is negative, therefore, \(\Delta H\) should be highly negative.
$\left( i \right)\,2F{e_2}{O_3}\left( s \right) \to 4Fe\left( s \right) + 3{O_2}\left( g \right)$
${\Delta _r}{G^o} = + 1487.0\,kJ\,mo{l^{ - 1}}$
$\left( {ii} \right)\,2CO\left( g \right) + {O_2}(g) \to 2C{O_2}\left( g \right)$
${\Delta _r}{G^o} = - 514.4\,kJ\,mo{l^{ - 1}}$
તો નીચેની પ્રક્રિયા માટે મુક્ત ઊર્જા ફેરફાર $\Delta_rG^o$ .....$kJ\, mol^{-1}$
$\,2F{e_2}{O_3}\left( s \right) + 6CO\left( g \right) \to 4Fe\left( s \right) + 6C{O_2}\left( g \right)$