\(\Delta H=\Delta U+\Delta n R T\)
or \(\quad \Delta H-\Delta U=\Delta n R T\)
\(\Delta n=1-\frac{1}{2}=\frac{1}{2}\)
\(\Delta H-\Delta U=\frac{1}{2} \times 8.314 \times 298\)
\(=1238.78 \,J \,m o l^{-1}\)
આપેલ : $\Delta H ^{\circ}=-54.07\,kJ\,mol ^{-1}$
$\Delta S ^{\circ}=10\,J\,K ^{-1}\,mol ^{-1}$
$(2.303 \times 8.314 \times 298=5705$ લો.)
કયા ન્યૂન્નતમ તાપમાને તે સ્વયંભૂ (આપ મેળે) થશે તે ............ $K$ માં છે. (પૂર્ણાક)
(ઉપયોગ કરો : $\Delta_{{c}} {H}($ ગ્લુકોઝ $)=-2700\, {~kJ}\, {~mol}^{-1}$ )
$(i)\,\,{C_{12}}{H_{22}}{O_{11}}\,\, + \,\,12{O_2}\,\, \to \,\,12\,\,C{O_2}\, + \,\,11{H_2}O,\,\,\,\,\,\,\,\,\,\,\,\,\Delta H\,\, = \,\, - 5200.7\,kJ\,mo{l^{ - 1}} $
$(ii)\,\,C\,\, + \,\,{O_2}\, \to \,\,C{O_2},\,\,\,\,\,\,\,\,\,\,\,\,\Delta H\,\, = \,\, - \,394.5\,\,kJ\,\,mo{l^{ - 1}}$
$(iii)\,\,{H_2}\,\, + \,\frac{1}{2}{O_2}\,\, \to \,\,\,{H_2}O,\,\,\,\,\,\,\,\,\,\Delta H\,\, = \,\, - \,285.8\,kJ\,\,mo{l^{ - 1}}$