$\Delta_{f} {H}^{\ominus}$ ${KCl}=-436.7 \,{~kJ}\, {~mol}^{-1}$
$\Delta_{\text {sub }} {H}^{\ominus}$ ${K}=89.2 \,{~kJ}\, {~mol}^{-1}$
$\Delta_{\text {ionization }} \,{H}^{-}$ ${K}=419.0\, {~kJ}\, {~mol}^{-1}$
$\Delta_{\text {electron gain }} {H}^{\ominus}$ ${Cl}_{(\text {e) }}=-348.6 \,{~kJ} \,{~mol}^{-1}$
$\Delta_{{bond}} {H}^{-}$ ${Cl}_{2}=243.0 \,{~kJ} \,{~mol}^{-1}$
${KCl}$ની લેટિસ એન્થાલ્પીની તીવ્રતા $.....$ ${kJ} {mol}^{-1}$ છે.
\(+\Delta_{\text {electron gain }} {H}_{({Cl})}^{\ominus}+\Delta_{\text {lattice }} {H}_{({KCl})}^{\ominus}\)
\(\Rightarrow-436.7=89.2+419.0+\frac{1}{2}(243.0)+\{-348.6\}\)
\(+\Delta_{\text {lattice }} {H}_{({KCl})}^{\ominus}\)
\(\Rightarrow \Delta_{\text {lattice }} {H}_{({KCl})}^{\ominus}=-717.8 {~kJ}\, {~mol}^{-1}\)
The magnitude of lattice enthalpy of \({KCl}\) in \({kJ}\,{mol}^{-1}\) is \(718\) (Nearest integer).
${\Delta _r}{G^o}$ (in $kJ\,mol^{-1}$) $=120-\frac {3}{8}\,T$
તો $T$ તાપમાને પ્રક્રિયા મિશ્રણનો મુખ્ય ઘટક કયો?
($1\,L\,atm\, = 101.32\,J$)