$\Delta \mathrm{U}=2.1 \mathrm{Kcal}, \Delta \mathrm{S}=20 \mathrm{cal} \mathrm{K}^{-1}$ at $300 \mathrm{K}$
$\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$
$\Delta \mathrm{G}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}-\mathrm{T} \Delta \mathrm{S}$
$=2.1+\frac{2 \times 2 \times 300}{1000}-\frac{300 \times 20}{1000}$
$\left(\mathrm{R}=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right)$
$=2.1+1.2-6=-2.70 \mathrm{Kcal} / \mathrm{mol}$
(તટસ્થીકરણ એન્થાલ્પી $=57\, kJ \,mol ^{-1}$ અને પાણીની વિશિષ્ટ ઉષ્મા= $4.2 \,J\,K ^{-1} \,g ^{-1}$ )
$(i)\,\,{C_{12}}{H_{22}}{O_{11}}\,\, + \,\,12{O_2}\,\, \to \,\,12\,\,C{O_2}\, + \,\,11{H_2}O,\,\,\,\,\,\,\,\,\,\,\,\,\Delta H\,\, = \,\, - 5200.7\,kJ\,mo{l^{ - 1}} $
$(ii)\,\,C\,\, + \,\,{O_2}\, \to \,\,C{O_2},\,\,\,\,\,\,\,\,\,\,\,\,\Delta H\,\, = \,\, - \,394.5\,\,kJ\,\,mo{l^{ - 1}}$
$(iii)\,\,{H_2}\,\, + \,\frac{1}{2}{O_2}\,\, \to \,\,\,{H_2}O,\,\,\,\,\,\,\,\,\,\Delta H\,\, = \,\, - \,285.8\,kJ\,\,mo{l^{ - 1}}$
તો પ્રક્રિયા $C(s) + 2{H_2}(g)\, \to \,C{H_4}(g)$ માટે $(\Delta {H^o})$નું મૂલ્ય ........$kcal$ થશે.
$\frac{1}{2}C{l_2}_{(g)}\,\xrightarrow{{\frac{1}{2}{\Delta _{diss}}{H^\Theta }}}\,Cl_{(g)}\,\,\xrightarrow{{{\Delta _{eg}}{H^\Theta }}}\,\,C{l^ - }_{(g)}\,\xrightarrow{{{\Delta _{hyd}}{H^\Theta }}}\,C{l^ - }_{(aq)}$
$({\mkern 1mu} {\Delta _{diss}}{\mkern 1mu} H_{C{l_2}}^\Theta {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} 240{\mkern 1mu} {\mkern 1mu} kJ{\mkern 1mu} {\mkern 1mu} mo{l^{ - 1}},{\mkern 1mu} {\mkern 1mu} {\Delta _{eg}}{\mkern 1mu} H_{Cl}^\Theta {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} - 349{\mkern 1mu} {\mkern 1mu} kJ{\mkern 1mu} {\mkern 1mu} mo{l^{ - 1}},{\mkern 1mu} {\mkern 1mu} $
${\Delta _{hyd}}H_{C{l^ - }}^\Theta {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} - {\mkern 1mu} 381{\mkern 1mu} kJ{\mkern 1mu} {\mkern 1mu} mo{l^{ - 1}})$
${\Delta _H}\, = \, - 57.2\,kJ\,mo{l^{ - 1}}$ અને ${K_C} = 1.7\, \times \,{10^{16}}$ છે. નીચેના પૈકી ક્યુ વિધાન ખોટુ છે ?