Question

At \(25^{\circ} \mathrm{C}, K_{\mathrm{w}}\) for the dissociation of water is \(1.0 \times 10^{-14}\). Calculate \(\Delta G^{\circ}\) for the reaction \(2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}(a q) .\)

Short answer

\(\Delta G^{\circ} \approx 38.07 \ \text{kJ/mol}\) for the dissociation of water at 25°C.

Step-by-step solution

Identify the Given Information

We are given the equilibrium constant \( K_{w} = 1.0 \times 10^{-14} \) for the dissociation of water at \(25^{\circ} \mathrm{C}\). We are required to find \( \Delta G^{\circ} \). The temperature \( T = 298 \ \text{K} \) because it is given in \(^{\circ} \mathrm{C}\).

Apply the Formula for Free Energy Change

The standard change in Gibbs free energy \( \Delta G^{\circ} \) can be calculated using the formula: \[ \Delta G^{\circ} = -RT \ln K_{w} \] where \( R \) is the universal gas constant \(8.314 \ J\cdot mol^{-1}\cdot K^{-1}\) and \( T \) is the absolute temperature in Kelvin.

Insert Values into the Formula

Substitute the known values into the formula: \( R = 8.314 \ J\cdot mol^{-1}\cdot K^{-1}, T = 298 \ K, \) and \( K_{w} = 1.0 \times 10^{-14} \). This gives us: \[ \Delta G^{\circ} = - (8.314) \times (298) \times \ln(1.0 \times 10^{-14}) \]

Calculate the Natural Logarithm and Result

Calculate \( \ln(1.0 \times 10^{-14}) \), which is approximately \(-32.236 \). Now, compute \( \Delta G^{\circ} \): \[ \Delta G^{\circ} = - (8.314) \times (298) \times (-32.236) \approx 38,070 \ J\,mol^{-1} \] Convert Joules to kilojoules by dividing by 1000, resulting in \( \Delta G^{\circ} \approx 38.07 \ \text{kJ/mol} \).

Key concepts

dissociation of water

The dissociation of water is a chemical process in which water ( H₂O) molecules split into hydronium ions ( H₃O⁺) and hydroxide ions ( OH⁻). This is a reversible reaction that can be expressed by the equation:

• H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)

In pure water at 25°C, the concentration of both H⁺ and OH⁻ ions is very low, approximately 1.0 x 10⁻⁷ M. Consequently, the equilibrium constant Kw for this reaction is also very small, calculated as the product of the ion concentrations:

• Kw = [H⁺][OH⁻] = 1.0 x 10⁻¹⁴

This small value reflects water's tendency to stay mostly in its undissociated form at neutral pH.

equilibrium constant

The equilibrium constant ( K) provides insights into the position of equilibrium in a chemical reaction. For the dissociation of water, the equilibrium constant ( Kw) quantifies the balance between H₂O molecules and their dissociation products, H⁺ (or H₃O⁺) and OH⁻ ions:

• Kw = [H⁺][OH⁻]

This value depends on temperature and is crucial in determining the direction and extent of the reaction under certain conditions. The logarithm of Kw is used in calculating Gibbs free energy ( ∆G°), which helps predict the spontaneity of the reaction. When K > 1, the products are favored, and when K < 1, reactants are favored.

temperature in Kelvin

Temperature is key in determining the behavior and spontaneity of chemical reactions, including water dissociation. It must be expressed in Kelvin for accurate calculations in thermodynamics. Conversion from Celsius is straightforward:

• K = °C + 273.15

In this exercise, water's dissociation is analyzed at 25°C, equivalent to 298K. Using Kelvin ensures consistency when applying the free energy equation ( ∆G° = -RT ln K), where R is the universal gas constant. Temperature influences the equilibrium constant ( Kw), generally increasing dissociation as temperature rises.

universal gas constant

The universal gas constant ( R) is fundamental in chemical thermodynamics, linking energy, temperature, and moles. In the context of Gibbs free energy calculations, R relates to work and heat transfer:

• R = 8.314 J/(mol·K)

It ensures consistency in calculations involving temperature and energy, like the ∆G° calculation for water dissociation. By integrating R into the equation ∆G° = -RT ln K, we evaluate the spontaneity of reactions at molecular level, reflecting R's role in energetics and equilibrium. Using consistent units with R is critical for achieving accurate and reliable results.