\({A} \underset{{T}\, 300\, {~K}}{\longrightarrow} {B}\)
\([\Delta {G}]_{{P}, {T}}=-49.4\, {~kJ} / {mol}\)
\(\Delta {H}_{{rxn}}=51.4 \,{~kJ} / {mol}\)
\(\Delta {S}_{{rxn}}=?\)
\(\Rightarrow\) From the relation \([\Delta {G}]_{{P}, {T}}=\Delta {H}-{T} \Delta {S}\)
\(\Rightarrow \Delta S_{\text {rxn }} =\frac{\Delta H_{r x n}-[\Delta G]_{p, T}}{T}\)
\(=\frac{[51.4-(-49.4)] \times 100}{300} \,\frac{{J}}{{mol} {K}}\)
\(\Rightarrow \Delta {S}_{{rxn}} =336\, \frac{{J}}{{mol} {K}}\)
(આપેલ : $\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )
${C_{\left( {graphite} \right)}} + {O_{2\left( g \right)}} \to C{O_{2\left( g \right)}}\,;\Delta H = - 393.5\,kJ$
${H_{2\left( g \right)}} + 1/2{O_{2\left( g \right)}} \to {H_2}{O_{\left( l \right)}}\,;\,\Delta H = - 286.2\,kJ$
${C_2}{H_{4\left( g \right)}} + 3{O_{2\left( g \right)}} \to 2C{O_{2\left( g \right)}} + 2{H_2}{O_{\left( l \right)}}\,;\,\Delta H = - 1410.8\,kJ$
$R = 8.314\, J\,K^{-1}\,mol^{-1} \,\,\,2.303 \times 8.314 \times 298 = 5705$
$\Delta {S_{({x_2})}}\,\, = \,\,60\,$ જૂલ/મોલ કેલ્વિન, $\Delta {S_{({y_2})}}\,\, = \,\,40$ જૂલ/મોલ કેલ્વિન $\Delta {S_{(x{y_3})}}\,\, = \,\,50\,$ જૂલ/મોલ કેલ્વિન
હોય, તો સંતુલને તાપમાને ......$K$