Question

A tabulation of more precise thermodynamic data than are presented in Appendix D lists the following values for \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(298.15 \mathrm{K},\) at a standard state pressure of 1 bar. $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ}, & \Delta G_{f,}^{\circ} & S_{\prime}^{\circ} \\\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{H}_{2} \mathrm{O}(1) & -285.830 & -237.129 & 69.91 \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & -241.818 & -228.572 & 188.825 \\ \hline \end{array}$$ (a) Use these data to determine, in two different ways, \(\Delta G^{\circ}\) at \(298.15 \mathrm{K}\) for the vaporization: \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{bar}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 1 \mathrm{bar}) .\) The value you obtain will differ slightly from that on page 838 because here, the standard state pressure is 1 bar, and there, it is 1 atm. (b) Use the result of part (a) to obtain the value of \(K\) for this vaporization and, hence, the vapor pressure of water at \(298.15 \mathrm{K}\) (c) The vapor pressure in part (b) is in the unit bar. Convert the pressure to millimeters of mercury. (d) Start with the value \(\Delta G^{\circ}=8.590 \mathrm{kJ}\), given on page 838 and calculate the vapor pressure of water at 298.15 K in a fashion similar to that in parts (b) and (c). In this way, demonstrate that the results obtained in a thermodynamic calculation do not depend on the convention we choose for the standard state pressure, as long as we use standard state thermodynamic data consistent with that choice.

Short answer

The vapor pressure of water at standard state is 23.57 mmHg when \(\Delta G^{\circ} = 8.557 kJ/mol\) as calculated, and 23.18 mmHg when \(\Delta G^{\circ} = 8.590 kJ/mol\) as given. The minor discrepancy can be attributed to rounding errors and the slightly different standard state pressures. These values verify that thermodynamic results remain consistent regardless of standard state pressure conventions, provided consistent standard state thermodynamic data are used.

Step-by-step solution

- Calculate \( \Delta G^{\circ} \)

Calculate the Gibbs free energy change, \( \Delta G^{\circ} \), for the vaporization process using the Gibbs free energy of formation equation: \( \Delta G^{\circ} = \Delta G_f^{\circ} (products) - \Delta G_f^{\circ} (reactants) = -228.572 kJ/mol - (-237.129 kJ/mol) = 8.557 kJ/mol .

- Calculate \( K \)

Calculate the equilibrium constant (\( K \)) with the Gibbs free energy equation: \( \Delta G^{\circ} = -RT \ln K \). Solve it for \( K \) so that it becomes: \( K = e^{(-\Delta G^{\circ} / (RT))} \). Substituting all the values, and take \( R = 8.314 J/ (mol.K) \) we arrive to \( K = e^{(-8.557 kJ / (8.314 J/mol.K * 298.15 K))} = 0.0314 \). The value of \( K \) obtained here is equal to the vapor pressure of water in the standard state.

- Convert Pressure Units

Convert the vapor pressure from bars to millimeters of mercury (mmHg), knowing that 1 bar = 750.062 mmHg, we find that the vapor pressure = 0.0314 * 750.062 mmHg = 23.57 mmHg.

- Repeat Calculation with Different \( \Delta G^{\circ} \)

Repeat the steps using \( \Delta G^{\circ} = 8.590 \, kJ \) given in the question. Accordingly for Step 2: \( K = e^{(-8.590 kJ / (8.314 J/mol.K * 298.15 K))} = 0.0309 \). Multiplying by the conversion factor, we obtain: vapor pressure = 0.0309 * 750.062 mmHg = 23.18 mmHg. It shows results are consistent regardless of standard state pressure.

Key concepts

Gibbs Free Energy

Gibbs free energy, often denoted as \( G \), is a thermodynamic quantity that measures the maximum amount of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. Named after the scientist Josiah Willard Gibbs, it's a central concept in chemical thermodynamics and plays a vital role in predicting the direction of chemical reactions.In a chemical reaction, the change in Gibbs free energy \( \Delta G \) is given by the difference in the Gibbs energy of the products and the reactants. The \( \Delta G \)-criterion for spontaneous processes states that a reaction is spontaneous if \( \Delta G \) is negative, non-spontaneous if it is positive, and at equilibrium if it is zero. The \( \Delta G^\circ \) value is calculated using standard Gibbs free energies of formation, which are determined under standard conditions (1 bar pressure, and typically at 298.15 K).When evaluating vaporization processes, \( \Delta G^\circ \) is particularly useful because it can directly correlate to the vapor pressure and equilibrium constant for that process. Understanding how to calculate \( \Delta G^\circ \) for these reactions and interpret its value is crucial for students of chemistry education trying to determine the spontaneity and feasibility of reactions.

Vapor Pressure Calculation

Vapor pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. It is a measure of a substance's volatility and is directly related to its tendency to evaporate.For students grappling with vapor pressure calculations, it is important to realize that it can be derived from the equilibrium constant of a phase transition reaction, such as vaporization. In the provided exercise, the use of the equilibrium constant, \( K \) for the vaporization process enables the calculation of the vapor pressure of water at a specific temperature.Application in Chemical Equilibrium:

• The equilibrium constant \( K \) calculated from Gibbs free energy reflects the ratio of the concentrations of the products to reactants at equilibrium.
• In the case of vaporization, the equilibrium constant is equal to the vapor pressure of the liquid.
• The unit of vapor pressure may vary, and it is often necessary to convert between different units, such as bars and millimeters of mercury (mmHg), as demonstrated in the exercise.

The ability to perform these conversions accurately is an essential skill for students, as it underpins further calculations in thermodynamics and plays a role in the interpretation of experimental data.

Chemical Equilibrium Constant

The chemical equilibrium constant \( K \) is a dimensionless value that quantifies the ratio of the concentrations of products to reactants for a reversible reaction at equilibrium. The magnitude of \( K \) provides valuable information regarding the relative amounts of products and reactants present when the reaction has reached a state of balance.Understanding how to calculate \( K \) using Gibbs free energy (\( \Delta G^\circ \) and the gas constant \( R \) allows students to predict not only the direction in which a reaction will proceed but also the extent of the reaction under standard conditions. This calculation can be performed through the equation \( \Delta G^\circ = -RT \ln K \), where \( T \) is the temperature in Kelvin, and \( R \) is the gas constant, an understanding of which is essential for students studying chemical reactions.Connection to Vapor Pressure:

• In cases like vaporization, the equilibrium constant \( K \) is directly proportional to the vapor pressure of the substance.
• A high \( K \) value at a given temperature suggests a high vapor pressure indicating a substance that easily transitions to the gas phase.
• This proportional relationship allows the use of \( K \) in the indirect determination of vapor pressure, as has been elaborated upon in the exercise.

Through the exercise provided, students can learn not only the theoretical aspect but also a practical application of these fundamental concepts, crucial for their understanding of the behavior of substances in response to changes in their surroundings.