\(\Delta G = \Delta H - T \Delta S \Delta H = T \Delta S\)
$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O ; \Delta H = -2900 \,KJ/mole $
$ C(s)\,\, + \,\,{O_2}(g)\,\, \to \,\,C{O_2}\,(g)$ $\Delta H = \,\, - \,94\,\,kcal$
${H_2}\,(g)\,\, + \,\,\frac{1}{2}\,{O_2}\,(g)\,\, \to \,\,{H_2}O\,(g),$ $\Delta H\,\, = \,\, - \,68\,\,kcal$
${C_2}{H_5}OH\,(\ell )\,\, + \,\,3{O_2}\,(g)\,\, \to \,\,2C{O_2}\,(g)\,\, + \,\,3{H_2}O\,(\ell ),$$\Delta H\,\, = \,\,\, - \,327\,\,kcal$
$\frac{1}{2}C{l_2}(g)\xrightarrow{{\frac{1}{2}{\Delta _{diss}}{H^\Theta }}}Cl(g)\xrightarrow{{{\Delta _{eg}}{H^\Theta }}}$ $C{l^ - }(g)\xrightarrow{{{\Delta _{Hyd}}{H^\Theta }}}C{l^ - }(aq)$
તો $\frac{1}{2}C{l_2}(g)$ ના $Cl^-_{(aq)}$ માં રૂપાંતમાં ઊર્જાનો ફેરફાર ............. $\mathrm{kJ\,mol}^{-1}$ જણાવો.
$({{\Delta _{diss}}H_{C{l_2}}^\Theta } = 240\,kJ\,mol^{-1}, {{\Delta _{eg}}H_{C{l}}^\Theta }= -349 \,kJ\,mol^{-1},$${{\Delta _{Hyd}}H_{C{l}}^\Theta }= -381 \,kJ\,mol^{-1})$