$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
માટે નીચેનામાંથી કઈ શરત સાચી છે ?
\(\mathrm{PCl}_{5(\mathrm{g})}=\mathrm{PCl}_{3(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})}\)
\(\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}\)
\(\Delta n_{g}=\) Change in number of moles of product and reactant species.
since \(\Delta n_{g}=+v e,\) hence \(\Delta H=+v e\)
also one mole of \(\mathrm{PCl}_{5}\) is dissociated into two moles of \(\mathrm{PCl}_{5}\) and \(\mathrm{Cl}_{2}\) in the same phase.
Therefore, \(\Delta S=S_{\text {product }}-S_{\text {reactant }}\)
\(\Delta S=+v e\)
$\left( i \right)\,2F{e_2}{O_3}\left( s \right) \to 4Fe\left( s \right) + 3{O_2}\left( g \right)$
${\Delta _r}{G^o} = + 1487.0\,kJ\,mo{l^{ - 1}}$
$\left( {ii} \right)\,2CO\left( g \right) + {O_2}(g) \to 2C{O_2}\left( g \right)$
${\Delta _r}{G^o} = - 514.4\,kJ\,mo{l^{ - 1}}$
તો નીચેની પ્રક્રિયા માટે મુક્ત ઊર્જા ફેરફાર $\Delta_rG^o$ .....$kJ\, mol^{-1}$
$\,2F{e_2}{O_3}\left( s \right) + 6CO\left( g \right) \to 4Fe\left( s \right) + 6C{O_2}\left( g \right)$
${C_{\left( {graphite} \right)}} + {O_{2\left( g \right)}} \to C{O_{2\left( g \right)}}\,;\Delta H = - 393.5\,kJ$
${H_{2\left( g \right)}} + 1/2{O_{2\left( g \right)}} \to {H_2}{O_{\left( l \right)}}\,;\,\Delta H = - 286.2\,kJ$
${C_2}{H_{4\left( g \right)}} + 3{O_{2\left( g \right)}} \to 2C{O_{2\left( g \right)}} + 2{H_2}{O_{\left( l \right)}}\,;\,\Delta H = - 1410.8\,kJ$
મોલર ઉષ્મા ક્ષમતા $N _{2} O\,100\,J\,K ^{-1}\,mol\,^{-1}$ )