Chemistry - Ionic Equilibrium Question with Solution | TestHub

The question asks for the standard entropy of ${\mathrm{OH}}^{-}$ at 298 K.

We use the following concepts:

1. $\Delta G=\Delta H-T\Delta S$

2. $\Delta G=-RT\ln K$

3. Reaction: ${\mathrm{H}}^{+}$(aq) + ${\mathrm{OH}}^{-}$(aq) $\to$${\mathrm{H}}_2\mathrm{O}$(l)

4. $\Delta S_{rxn}=S^{\circ }({\mathrm{H}}_2\mathrm{O})-S^{\circ }({\mathrm{H}}^{+})-S^{\circ }(\mathrm{O}{\mathrm{H}}^{-})$

Given: $K_w=10^{-14}$, $\Delta H_{neut}=-13.4kcal$, $S^{\circ }({\mathrm{H}}^{+})=0$, $S^{\circ }({\mathrm{H}}_2\mathrm{O})=16.7cal/molK$

$-RT\ln K_w=\Delta H-T\Delta S$

$-R(298)\ln (10^{-14})=-13400-298(16.7-0-S^{\circ }(\mathrm{O}{\mathrm{H}}^{-}))$

$2.303\times 2\times 298\times 14=-13400-298\times 16.7+298S^{\circ }(\mathrm{O}{\mathrm{H}}^{-})$

$19143.44=-13400-4976.6+298S^{\circ }(\mathrm{O}{\mathrm{H}}^{-})$

$S^{\circ }(\mathrm{O}{\mathrm{H}}^{-})=\frac{19143.44+13400+4976.6}{298}=\frac{37520.04}{298}\approx -2.7cal/molK$

Therefore, $S^{\circ }(\mathrm{O}{\mathrm{H}}^{-})\approx -2.7cal/molK$