( $373\, K$ તાપમાને પાણી નું $\Delta H _{\text {vap }}$ $K =41$ કિલોજૂલ/મોલ $\left. R =8.314\, JK ^{-1} mol ^{-1}\right)$)
$\frac{1}{2}C{l_2}(g)\xrightarrow{{\frac{1}{2}{\Delta _{diss}}{H^\Theta }}}Cl(g)\xrightarrow{{{\Delta _{eg}}{H^\Theta }}}$ $C{l^ - }(g)\xrightarrow{{{\Delta _{Hyd}}{H^\Theta }}}C{l^ - }(aq)$
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})$
$(i)$ $NH_3$ $_{(g)} + aq$ $\rightarrow$ $NH_3$ $_{(aq)}$, $\Delta H$ $= -8.4 \,Kcal.$
$(ii)$ $HCl_{(g)} + aq$ $\rightarrow$ $HCl{(aq)}$, $\Delta H =$ ${-1}7.3\, Kcal.$
$(iii)$ $NH_3$ $_{(aq)} + HCl_{(aq)}$ $\rightarrow$ $NH_4Cl $ $_{(aq)}$, $\Delta H = -12.5\, Kcal$.
$(iv)$ $NH_4Cl$ $_{(s)} + aq$ $\rightarrow$ $NH_4Cl$ $_{(aq)}$, $\Delta H = +3.9 \,Kcal.$