$\Delta H = {(\Delta {H_{f}})_{CaO}} + {(\Delta H_{f})_{C{o_2}}} - (\Delta {H_{f}})CaC{O_3}$
$42 = - 152 - 94 - {(\Delta {H_{f}})_{CaC{O_3}}}$
$ \Rightarrow \,{(\Delta {H_{f}})_{CaC{O_3}}} = - 246 - 42 = - 288\,kJ.$
$Pt ( s )\left| H _{2}( g )\right| H ^{+}( aq ) \| Ag ^{+}( aq ) \mid Ag ( s )$
$E _{\text {Cell }}^{0}=+0.5332 \,V$.
$\Delta_{ f } G ^{0}$ નું મૂલ્ય..........$k\,J\, mol ^{-1}$ છે. (નજીકનો પૂર્ણાંક)
[અચળ કદે મોલર ઉષ્માક્ષમતા $\bar{c}_{\mathrm{v}}$ છે]