Question
Les tables de données fournissent les valeurs de référence de l'enthalpie et de l'entropie de différentes substances à $25^{\circ} \mathrm{C}$.Calculer la valeur de l'enthalpie et de l'entropie du dioxygène à $110^{\circ} \mathrm{C}$ sous 10 bar.
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What is the difference between entropies of oxygen at $150 \mathrm{kPa}$ and $39^{\circ} \mathrm{C}$ and oxygen at $150 \mathrm{kPa}$ and $337^{\circ} \mathrm{C}$ on a perunit-mass basis?
The following reaction, sometimes used in the laboratory to generate small quantities of oxygen gas, has $\Delta G^{\circ}=$ $-224.4 \mathrm{kJ} / \mathrm{mol}$ at $25^{\circ} \mathrm{C} :$ $$2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$ Use the following additional data at $25^{\circ} \mathrm{C}$ to calculate the standard molar entropy $S^{\circ}$ of $\mathrm{O}_{2}$ at $25^{\circ} \mathrm{C} : \Delta H^{\circ} \mathrm{AClO}_{3} )=$ $-397.7 \mathrm{kJ} / \mathrm{mol}, \Delta H^{\circ} \mathrm{f}(\mathrm{KCl})=-436.5 \mathrm{kJ} / \mathrm{mol}, S^{\circ}\left(\mathrm{KClO}_{3}\right)=$ $143.1 \mathrm{J} /(\mathrm{K} \cdot \mathrm{mol}),$ and $S^{\circ}(\mathrm{KCl})=82.6 \mathrm{J} /(\mathrm{K} \cdot \mathrm{mol})$.
The standard entropy of $\mathrm{O}_{2}(\mathrm{g})$ at $298.15 \mathrm{K}$ and 1 bar is listed in Table $\mathrm{C}_{2} 2$ as $205.138 \mathrm{JK}^{-1} \mathrm{mol}^{-1},$ and the standard Gibbs energy of formation is listed as $0 \mathrm{kJ} \mathrm{mol}^{-1}$. Assuming that $\mathrm{O}_{2}$ is an ideal gas, what will be the molar entropy and molar Gibbs energy of formation at 100 bar?
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