Question

For the autoionization of water at \(25^{\circ} \mathrm{C}\), $$ \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{w}}\) is \(1.0 \times 10^{-14}\). What is \(\Delta G^{\circ}\) for the process?

Short answer

\(\Delta G^{\circ} = 77.99 \, \mathrm{kJ} \, \mathrm{mol}^{-1}\)

Step-by-step solution

Convert the temperature from Celsius to Kelvin

The problem specifies a temperature of \(25^{\circ} \mathrm{C}\). To convert this to Kelvin, add 273.15, therefore, \( T = 25 + 273.15 = 298.15 \, \mathrm{K}\).

Substitute the given values into the equation

The equation for \(\Delta G^{\circ}\) is \(\Delta G^{\circ} = -RT \ln K\). Substituting \(\ln K_{\mathrm{w}} = \ln (1.0 \times 10^{-14})\), \(R = 8.314 \, \mathrm{J} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}\) (as is standard in such calculations) and \(T = 298.15 \, \mathrm{K}\), the equation becomes \(\Delta G^{\circ} = -8.314 \times 298.15 \times \ln (1.0 \times 10^{-14})\).

Solve for \(\Delta G^{\circ}\)

Perform the multiplication and the natural logarithm operation to find the numerical value: \(\Delta G^{\circ} = -8.314 \times 298.15 \times (-31.42) = 77.99 \, \mathrm{kJ} \, \mathrm{mol}^{-1}\).

Key concepts

Autoionization of Water

Water is a fascinating molecule that can undergo a process known as autoionization. This process involves water molecules reacting with each other to form hydrogen ions (\( \text{H}^{+} \)) and hydroxide ions (\( \text{OH}^{-} \)). The balanced chemical equation is given by: \[ \text{H}_2\text{O}(l) \rightleftharpoons \text{H}^{+}(aq) + \text{OH}^{-}(aq) \] This means that even in pure water, a very small number of molecules dissociate to form these ions. At room temperature (around 25°C), the concentration of these ions in water is low, approximately \(1.0 \times 10^{-7} \text{ mol/L} \) each. This balance is described by the autoionization constant, \(K_w\).

• The value of \(K_w\) at 25°C is \(1.0 \times 10^{-14}\).
• This value represents the product of the concentrations of \( \text{H}^{+} \) and \( \text{OH}^{-} \).

Knowing \(K_w\) is essential because it helps us understand the behavior of acidic and basic solutions. For neutral water, \( [\text{H}^{+}] = [\text{OH}^{-}]= 1.0 \times 10^{-7} \text{ mol/L} \). This gives us a pH of 7, which is considered neutral.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), plays a critical role in understanding chemical reactions in equilibrium. For a given reaction at equilibrium, the constant \(K\) quantifies the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients from the balanced equation.For the autoionization of water:\[ \text{H}_2\text{O}(l) \rightleftharpoons \text{H}^{+}(aq) + \text{OH}^{-}(aq) \]The equilibrium constant \(K_w\) is expressed as:\[ K_w = [\text{H}^{+}][\text{OH}^{-}] = 1.0 \times 10^{-14}\] This constant is temperature-dependent and varies with changes in temperature.Some key points about equilibrium constants include:

• If \(K\) is large, the reaction favors the formation of products.
• If \(K\) is small, the reaction favors the formation of reactants.

For water, the small value of \(K_w\) indicates that there are very few ions in pure water, making it largely neutral unless other substances are added.

Thermodynamics

Thermodynamics fundamentally deals with energy transformations in chemical processes. For the autoionization of water, the changes in free energy are of particular interest. Gibbs free energy (\(\Delta G^{\circ}\)) is a measure that indicates the spontaneity of a process.The relationship between the Gibbs free energy change and the equilibrium constant is expressed by the equation:\[ \Delta G^{\circ} = -RT \ln K \] where:

• \(R\) is the universal gas constant, equal to 8.314 J/mol·K
• \(T\) is the temperature in Kelvin
• \(K\) is the equilibrium constant for the reaction

For the autoionization of water:- You use \(K_w = 1.0 \times 10^{-14}\) and \(T = 298.15 \text{ K}\) (as 25°C converted to Kelvin).- Plugging these into the formula gives a \(\Delta G^{\circ}\) value of about 77.99 kJ/mol, indicating this process is non-spontaneous under standard conditions.Thus, spontaneous reactions have \(\Delta G^{\circ} < 0\), while non-spontaneous reactions have \(\Delta G^{\circ} > 0\). The value calculated suggests energy is required for the autoionization of water, confirming its low tendency to occur.