$\therefore \quad-\Delta G = RTlnK _{ c }$
| $H - H$ બંધઊર્જા | $:\, 431.37 \,kJ\, mol^{-1}$ |
| $C= C$ બંધઊર્જા | $:\, 606.10\, kJ \,mol^{-1}$ |
| $C - C$ બંધઊર્જા | $:\, 336.49\, kJ\, mol^{-1}$ |
| $C - H$ બંધઊર્જા | $:\, 410.50\, kJ\, mol^{-1}$ |
પ્રક્રિયા : $\begin{array}{*{20}{c}}
{H\,\,\,\,H} \\
{|\,\,\,\,\,\,\,\,|} \\
{C = C} \\
{|\,\,\,\,\,\,\,\,\,|} \\
{H\,\,\,\,H}
\end{array}\, + \,H - H\, \to \,\begin{array}{*{20}{c}}
{H\,\,\,\,H} \\
{|\,\,\,\,\,\,\,\,|} \\
{H - C - C - H} \\
{|\,\,\,\,\,\,\,\,\,|} \\
{H\,\,\,\,H}
\end{array}\,$
$\frac{1}{2}C{l_2}_{(g)}\,\xrightarrow{{\frac{1}{2}{\Delta _{diss}}{H^\Theta }}}\,Cl_{(g)}\,\,\xrightarrow{{{\Delta _{eg}}{H^\Theta }}}\,\,C{l^ - }_{(g)}\,\xrightarrow{{{\Delta _{hyd}}{H^\Theta }}}\,C{l^ - }_{(aq)}$
$({\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} $
${\Delta _{hyd}}H_{C{l^ - }}^\Theta {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} - {\mkern 1mu} 381{\mkern 1mu} kJ{\mkern 1mu} {\mkern 1mu} mo{l^{ - 1}})$
$A$. પ્રવાહીનું બાષ્પમાં બાષ્પીભવન થાય છે.
$B$. સ્ફટિકમય ધનનું તાપમાન $130 \mathrm{~K}$ માંથી $0 \mathrm{~K}$ નીચું (ધટાડવામાં આવે છે) લઈ જવામાં આવે છે.
$C$. $2 \mathrm{NaHCO}_{3(\mathrm{~s})} \rightarrow \mathrm{Na}_2 \mathrm{CO}_{3(\mathrm{~s})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$
$D$. $\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{Cl}_{(\mathrm{g})}$
નીચે આપેલા વિકલ્પોમાંથી સાચો જવાબ શોધો.
${H_2}{O_{(l)}} \to \,\,H_{(aq)}^ + + \,\,OH_{(aq)}^ - \,;\,\,\,\Delta H\,\, = \,\,57.32\,\,KJ\,;$
${H_2}_{(g)} + \,\,\frac{1}{2}\,\,{O_2}_{(g)} \to \,\,{H_2}{O_{(1)}}\,;\,\,\Delta H\,\, = \,\, - 286.20\,\,KJ$