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Chapter 17: Problem 20

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

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

Step by step solution

01

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 \(^{\circ} \mathrm{C}\).
02

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.
03

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: \[ \Delta G^{\circ} = - (8.314) \times (298) \times \ln(1.0 \times 10^{-14}) \]
04

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 \( \Delta G^{\circ} \approx 38.07 \ \text{kJ/mol} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

dissociation of water
The dissociation of water is a chemical process in which water ( H鈧侽) molecules split into hydronium ions ( H鈧僌鈦�) and hydroxide ions ( OH鈦�). This is a reversible reaction that can be expressed by the equation:
  • H鈧侽(l) 鈬� H鈦�(aq) + OH鈦�(aq)
In pure water at 25掳C, the concentration of both H鈦� and OH鈦� ions is very low, approximately 1.0 x 10鈦烩伔 M. Consequently, the equilibrium constant Kw for this reaction is also very small, calculated as the product of the ion concentrations:
  • Kw = [H鈦篯[OH鈦籡 = 1.0 x 10鈦宦光伌
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 H鈧侽 molecules and their dissociation products, H鈦� (or H鈧僌鈦�) and OH鈦� ions:
  • Kw = [H鈦篯[OH鈦籡
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 ( 鈭咷掳), 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 = 掳C + 273.15
In this exercise, water's dissociation is analyzed at 25掳C, equivalent to 298K. Using Kelvin ensures consistency when applying the free energy equation ( 鈭咷掳 = -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路K)
It ensures consistency in calculations involving temperature and energy, like the 鈭咷掳 calculation for water dissociation. By integrating R into the equation 鈭咷掳 = -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.

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Most popular questions from this chapter

For the melting of sodium chloride, \(\Delta H_{\text {fusion }}=28.16 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta S_{\text {fusion }}=26.22 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol}) .\) Calculate \(\Delta S_{\text {surr }}\) and \(\Delta S_{\text {total }}\) at: (a) \(1050 \mathrm{~K}\) (b) \(1074 \mathrm{~K}\) (c) \(1100 \mathrm{~K}\) (d) Does \(\mathrm{NaCl}\) melt at \(1100 \mathrm{~K}\) ? Calculate the melting point of \(\mathrm{NaCl}\)

Give an equation that relates the entropy change in the surroundings to the enthalpy change in the system. What is the sign of \(\Delta S_{\text {surr }}\) for the following? (a) An exothermic reaction (b) An endothermic reaction

What is the entropy change when the volume of \(1.6 \mathrm{~g}\) of \(\mathrm{O}_{2}\) increases from \(2.5 \mathrm{~L}\) to \(3.5 \mathrm{~L}\) at a constant temperature of \(75^{\circ} \mathrm{C} ?\) Assume that \(\mathrm{O}_{2}\) behaves as an ideal gas.

Tell whether the following processes are spontaneous or nonspontaneous: (a) Dissolving sugar in hot coffee (b) Decomposition of \(\mathrm{NaCl}\) to solid sodium and gaseous chlorine at \(25^{\circ} \mathrm{C}\) and 1 atm pressure (c) Uniform mixing of bromine vapor and nitrogen gas (d) Boiling of gasoline at \(25^{\circ} \mathrm{C}\) and 1 atm pressure

(a) Diffusion of perfume molecules from one side of a room to the other side (b) Heat flow from a cold object to a hot object (c) Decomposition of rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot \mathrm{H}_{2} \mathrm{O}\right)\) to iron metal, oxygen, and water (d) A mixture of gaseous \(\mathrm{N}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\), with the partial pressures \(\left(P_{\mathrm{N}_{2}}=1 \mathrm{~atm}, P_{\mathrm{H}_{2}}=0.25 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=10 \mathrm{~atm}\right)\), converts some of the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) to \(\mathrm{NH}_{3}\). $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \stackrel{\text { Catalyst }}{\rightleftharpoons} 2 \mathrm{NH}_{3}(g) \text { at } 300 \mathrm{~K} \quad K_{\mathrm{p}}=4.4 \times 10^{5} $$
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