Hence, $\Delta G=\Delta H-T_{e} \Delta S=0$
$\therefore \quad \Delta H=T_{e} \Delta S$
or $\quad T_{e}=\frac{\Delta H}{\Delta S}$
For a spontaneous reaction
$\Delta G$ must be negative
which is possible only if $\Delta H < T \Delta S$
or $\quad T > \frac{\Delta H}{\Delta S} ; T_{e} < T$
$\Delta {U_{BC}} = - 5\,kJ\,mo{l^{ - 1}},{q_{AB}} = 2\,kJ\,mo{l^{ - 1}}$
$\Delta {W_{AB}} = - 5\,kJ\,mo{l^{ - 1}},{W_{CA}} = 3\,kJ\,mo{l^{ - 1}}$
$CA$ પ્રક્રમ દરમિયાન પ્રણાલી દ્વારા શોષાતી ઉષ્મા ......$kJ\,mo{l^{ - 1}}$
$Mg^{2+}(aq) = -456.0; OH-(aq) = -1 57.3; Mg(OH)_2 (s) = -833.9$