$\frac{1}{2}C{l_2}(g)\xrightarrow{{\frac{1}{2}{\Delta _{diss}}{H^\Theta }}}Cl(g)\xrightarrow{{{\Delta _{eg}}{H^\Theta }}}$ $C{l^ - }(g)\xrightarrow{{{\Delta _{Hyd}}{H^\Theta }}}C{l^ - }(aq)$
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})$
\(\frac{1}{2} C_{2}(g)\) to \(C l^{-1}(a q)\) is given by
\(\Delta H=\frac{1}{2} \Delta_{d i s s} H_{C l_{2}}^{(-)}+\Delta_{e g} H_{C l}^{(-)}+\Delta_{h y l} H_{C l}^{(-)}\)
Substituting various values from given data, we get
\(\Delta H=\left(\frac{1}{2} \times 240\right)+(-349)+(-381)\, k J \,m o l^{-1}\)
\(=(120-349-381)\, k J\, m o l^{-1}\)
\(=-610 \,k J \,m o l^{-1}\)

| $\Delta H \,(kJ/mol)$ | |
| $\frac 12 A \rightarrow B$ | $+150$ |
| $3B \rightarrow 2C + D$ | $-125$ |
| $E + A \rightarrow 2D$ | $+350$ |
