Question

Consider the dissociation of a weak acid $\mathbf{HA}\mathbf{}\left(K_a=4.5\times {10}^{-3}\right)$in water:
$\mathbf{HA}\left(\mathrm{aq}\right)\mathbf{\rightleftharpoons }{\mathbf{H}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{A}}^{\mathbf{-}}\left(\mathrm{aq}\right)$
Calculate $\mathbf{∆}{\mathbf{G}}^{\mathbf{o}}$ for this reaction at 25°C.

Short answer

The value of $\mathbf{∆}{\mathbf{G}}^{\mathbf{o}}$for the given reaction at 25°C is $\mathbf{13}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$

Step-by-step solution

Step 1: Introduction to the Concept

The Gibbs free energy for an equilibrium reaction is related to the equilibrium constant (K) by the Gibbs-Helmholtz isotherm:

$\mathbf{∆}{\mathbf{G}}_{\mathbf{R}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{RTlnK}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(1\right)$

Where $\mathbf{R}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{molK}$and T is the temperature.

Step 2: Solution Explanation

The given equation,

$\mathrm{HA}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right)$ ${\mathrm{K}}_{\mathrm{a}}=4.5\times {10}^{-3}\mathrm{at}{25}^{\mathrm{o}}\mathrm{C}$

From equation (1),

$\mathbf{∆}{\mathbf{G}}_{\mathbf{R}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{R}\mathbf{\times }\mathbf{T}\mathbf{\times }\mathbf{lnK}$

$$\begin{gathered} ∆{\mathrm{G}}_{\mathrm{R}}^{\mathrm{o}}=-0.008314\mathrm{kJ}/\mathrm{molK}\times \left(25+273\mathrm{K}\right)\times \ln \left(4.5\times {10}^{-3}\right) \\ ∆{\mathrm{G}}_{\mathrm{R}}^{\mathrm{o}}=13.4\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Therefore$\mathbf{∆}{\mathbf{G}}^{\mathbf{o}}$for given reaction is $\mathbf{13}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$.