Calculate the standard reaction entropy at 298 K of C 12 H 12 O 11(s) + 12 O 2(g) ā 12 CO 2(g) + 11H 2 O (l) . ļ f S o C 12 H 12 O 11(s) 360.2 O 2(g) 205.1 CO 2(g) 213.74 H 2 O (l) 109.6
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The formula for calculating the standard reaction entropy is: \[ \Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}} \] Where \( S^\circ \) is the standard molar entropy. Show moreā¦
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A Standard entropy changes cannot be measured directly in the laboratory. They are calculated from experimentally obtained values of $\Delta G^{0}$ and $\Delta H^{0} .$ From the data given here, calculate $\Delta S^{0}$ at $298 \mathrm{~K}$ for each of the following reactions. $$ \text { (a) } \mathrm{OF}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) $$ $$ \begin{array}{lr} \Delta H^{0}=-323.2 \mathrm{~kJ} / \mathrm{mol} & \Delta G^{0}=-358.4 \mathrm{~kJ} / \mathrm{mol} \\ \text { (b) } \mathrm{CaC}_{2}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s})+\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \\ \Delta H^{0}=-125.4 \mathrm{~kJ} / \mathrm{mol} & \Delta G^{0}=-145.4 \mathrm{~kJ} / \mathrm{mol} \\ \text { (c) } \mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \\ \Delta H^{0}=81.5 \mathrm{~kJ} / \mathrm{mol} & \Delta G^{0}=-26.20 \mathrm{~kJ} / \mathrm{mol} \end{array} $$
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