$(I)$ ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l);$
$\Delta {H^o_{298\,K}} = - 285.9\,kJ\,mo{l^{ - 1}}$
$(II)$ ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);$
$\Delta {H^o_{298\,K}} = - 241.8\,kJ\,mo{l^{ - 1}}$
તો પાણીની મોલર બાષ્પાયન એન્થાલ્પી .....$kJ\,mol^{-1}$
\({H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l);\)
\(\Delta {H^o} = - 285.9\,kJ\,mo{l^{ - 1}}........(1)\)
\({H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);\)
\(\Delta {H^o}= - 241.8\,kJ\,mo{l^{ - 1}}........(2)\)
We have to calculate
\({H_2}O(l) \to {H_2}O(g);\,\Delta {H^o} = ?\)
On substracting eqn. \((2)\) from eqn. \((1)\) we get
\({H_2}O(l) \to {H_2}O(g);\)
\(\Delta {H^o} = - 241.8 - ( - 285.9)\)
\( = 44.1\,kJ\,mo{l^{ - 1}}\)
$Mg^{2+}(aq) = -456.0; OH-(aq) = -1 57.3; Mg(OH)_2 (s) = -833.9$
$\mathrm{Cd}_{(s)}+\mathrm{Hg}_{2} \mathrm{SO}_{4(s)}+\frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CdSO}_{4} \cdot \frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(s)}+2 \mathrm{Hg}_{(l)}$
$25^{\circ} {C}$ પર ${E}_{\text {cell }}^{0}$નું મૂલ્ય $4.315\, {~V}$ છે.
જો $\Delta {H}^{\circ}=-825.2\, {~kJ} \,{~mol}^{-1}$, પ્રમાણિત એન્ટ્રોપી ફેરફાર $\Delta {S}^{\circ}$ ${J} \,{K}^{-1}$માં $........$ છે. (નજીકના પૂર્ણાંકમાં) [આપેલ: ફેરાડે અચળાંક $ = 96487 \, {C} \, {mol}^{-1} $]