Question

The following reaction occurs in pure water:
${\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{\to }{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}\left(\mathrm{aq}\right)$
which is often abbreviated as
${\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{\to }{\mathbf{H}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}\left(\mathrm{aq}\right)$
For this reaction, $\mathbf{∆}{\mathbf{G}}^{\mathbf{o}}\mathbf{=}\mathbf{79}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$at 25°C. Calculate the value of $\mathbf{∆}\mathbf{G}$for this reaction at 25°C when $\left[{\mathrm{OH}}^{-}\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{M}$and $\left[H^{+}\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{71}\mathbf{}\mathbf{M}$.

Short answer

The value of $\mathbf{∆}\mathbf{G}$ for the given reaction at 25°C is $\mathbf{74}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$.

Step-by-step solution

Step 1: Introduction to the Concept

For a process at temperatureT, the Gibbs free energy is related to the equilibrium constant (K) and the standard free energy change ($\mathbf{∆}{\mathbf{G}}^{\mathbf{o}}$) in the following way:

$\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{∆}{\mathbf{G}}^{\mathbf{o}}\mathbf{+}\mathbf{RTlnQ}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(1\right)$

$\mathrm{Q}=$Reaction Quotient$\mathbf{=}\frac{\left[\mathrm{Products}\right]}{\left[\mathrm{Reactants}\right]}$

Where $\mathbf{R}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{molK}$

If the change in free energy is negative, i.e. the reaction is said to be spontaneous.

Step 2: Solution Explanation

The given equation,

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

From equation (1),

$\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{∆}{\mathbf{G}}^{\mathbf{o}}\mathbf{+}\mathbf{RTln}\left(\left[{\mathrm{OH}}^{-}\right]\left[H^{+}\right]\right)$

$$\begin{gathered} ∆\mathrm{G}=79.9\mathrm{kJ}/\mathrm{mol}+0.008314\mathrm{kJ}/\mathrm{molK}\times 298\mathrm{K}\times \ln \left(0.15\times 0.71\right) \\ ∆\mathrm{G}=79.9\mathrm{kJ}/\mathrm{mol}-5.548\mathrm{kJ}/\mathrm{mol} \\ ∆\mathrm{G}=74.4\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Therefore $\mathbf{∆}\mathbf{G}$for given reaction is $\mathbf{74}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$.