Chemistry - Thermodynamics - II Question with Solution | TestHub

The reaction for the neutralization of H⁺ and OH⁻ is:

H⁺(aq) + OH⁻(aq) $⟶$ H₂O(l)

The equilibrium constant for this reaction is $K_{eq}=\frac{1}{K_w}$.

Given $K_w=10^{-14}$ at 300 K, $K_{eq}=10^{14}$.

Calculate $\Delta G^{\circ }$ using the formula:

$$\Delta G^{\circ }=-RT\ln K_{eq}$$

Given $R=2cal/(molK)$ and $\ln x=2{\log }_{10}x$ (this is an approximation, usually $\ln x=2.303{\log }_{10}x$, but we will use the given approximation).

$$\Delta G^{\circ }=-(2cal/(molK))\times (300K)\times (2{\log }_{10}10^{14})$$

$$\Delta G^{\circ }=-(2\times 300\times 2\times 14)cal/mol=-16800cal/mol$$

Now, use the Gibbs-Helmholtz equation:

$$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$$

We are given $\Delta H_{neut}=-13.5kcal/mol=-13500cal/mol$.

$$-16800cal/mol=-13500cal/mol-(300K)\Delta S^{\circ }$$

$$-3300cal/mol=-(300K)\Delta S^{\circ }$$

$$\Delta S^{\circ }=\frac{-3300cal/mol}{-300K}=11cal/(molK)$$

The standard entropy change for the reaction is:

$$\Delta S^{\circ }=S^{\circ }(H_{2}O)-S^{\circ }(H^{+})-S^{\circ }(O{H}^{-})$$

It is a convention to set $S^{\circ }(H^{+})=0cal/(molK)$.

$$11cal/(molK)=16cal/(molK)-0cal/(molK)-S^{\circ }(O{H}^{-})$$

$$S^{\circ }(O{H}^{-})=16-11=5cal/(molK)$$