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The molar enthalpy of vaporization of benzene at its normal boiling point \(\left(80.09^{\circ} \mathrm{C}\right)\) is \(30.72 \mathrm{kJ} \cdot \mathrm{mol}^{-1}\). Assuming that \(\Delta_{\mathrm{rap}} \bar{H}\) and \(\Delta_{\mathrm{vap}} \bar{S}\) stay constant at their values at \(80.09^{\circ} \mathrm{C},\) calculate the value of \(\Delta_{\mathrm{vap}} \bar{G}\) at \(75.0^{\circ} \mathrm{C}, 80.09^{\circ} \mathrm{C},\) and \(85.0^{\circ} \mathrm{C} .\) Interpret these results physically.

Short Answer

Expert verified
\( \Delta_{\mathrm{vap}} \bar{G} = 0.426 \mathrm{kJ/mol} \) at \(75.0^\circ C \), \(0 \) at \(80.09^\circ C \), and \(-0.420 \mathrm{kJ/mol} \) at \(85.0^\circ C \).

Step by step solution

01

Understanding the Relationship between \/ \( \Delta_{\mathrm{vap}} \bar{G} \)\ and the Gibbs Free Energy Equation

The Gibbs free energy change (\( \Delta_{\mathrm{vap}} \bar{G} \)) for vaporization can be related to enthalpy (\( \Delta_{\mathrm{vap}} \bar{H} \)) and entropy (\( \Delta_{\mathrm{vap}} \bar{S} \)) by the equation \( \Delta_{\mathrm{vap}} \bar{G} = \Delta_{\mathrm{vap}} \bar{H} - T \Delta_{\mathrm{vap}} \bar{S} \). To solve for \( \Delta_{\mathrm{vap}} \bar{G} \), we need to determine \( \Delta_{\mathrm{vap}} \bar{S} \) using the boiling point conditions.
02

Calculating \( \Delta_{\mathrm{vap}} \bar{S} \)

At the boiling point \((80.09^\circ\mathrm{C})\), vaporization is at equilibrium, so \( \Delta_{\mathrm{vap}} \bar{G} = 0 \). Therefore, \( \Delta_{\mathrm{vap}} \bar{S} = \frac{\Delta_{\mathrm{vap}} \bar{H}}{T} \).Convert temperature to Kelvin: \( 80.09 + 273.15 = 353.24 \mathrm{K} \).Thus, \( \Delta_{\mathrm{vap}} \bar{S} = \frac{30.72 \mathrm{kJ/mol}}{353.24 \mathrm{K}} = 0.08696 \mathrm{kJ} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \).
03

Calculating \( \Delta_{\mathrm{vap}} \bar{G} \) at \( 75.0^\circ C \)

Convert 75.0°C to Kelvin: \( 75.0 + 273.15 = 348.15 \mathrm{K} \). Substitute into the Gibbs free energy equation: \( \Delta_{\mathrm{vap}} \bar{G} = 30.72 - 348.15 \times 0.08696 = 0.426 \mathrm{kJ/mol} \).
04

Calculating \( \Delta_{\mathrm{vap}} \bar{G} \) at \( 80.09^\circ C \)

At 80.09°C, which is the boiling point, the system is at equilibrium:\( \Delta_{\mathrm{vap}} \bar{G} = 30.72 - 353.24 \times 0.08696 = 0 \mathrm{kJ/mol} \).
05

Calculating \( \Delta_{\mathrm{vap}} \bar{G} \) at \( 85.0^\circ C \)

Convert 85.0°C to Kelvin: \( 85.0 + 273.15 = 358.15 \mathrm{K} \). Substitute into the Gibbs free energy equation:\( \Delta_{\mathrm{vap}} \bar{G} = 30.72 - 358.15 \times 0.08696 = -0.420 \mathrm{kJ/mol} \).
06

Physical Interpretation of the Results

At \( 75.0^\circ C \), \( \Delta_{\mathrm{vap}} \bar{G} \) is positive, indicating vaporization is not spontaneous. At \( 80.09^\circ C \), \( \Delta_{\mathrm{vap}} \bar{G} \) is zero, indicating the system is at equilibrium (boiling point). At \( 85.0^\circ C \), \( \Delta_{\mathrm{vap}} \bar{G} \) is negative, indicating vaporization is spontaneous.

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Key Concepts

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

enthalpy of vaporization
The enthalpy of vaporization is a crucial concept when discussing phase changes between liquid and gas for substances. It represents the amount of energy required to convert one mole of a liquid to its gaseous form at a constant pressure, usually at the liquid's boiling point. For benzene, at its normal boiling point of 80.09°C, the molar enthalpy of vaporization is given as 30.72 kJ/mol.

This value indicates the strength of intermolecular forces that need to be overcome for the phase change to occur. When these forces are strong, a higher enthalpy of vaporization is required, meaning more energy is needed for vaporization. In calculations related to Gibbs free energy, this enthalpy value plays a significant role because it is directly used in the equation to determine the spontaneity of the vaporization process.

Enthalpy of vaporization is temperature dependent, though in many calculations, it is often assumed to be constant unless stated otherwise. This assumption simplifies calculations while still providing an accurate enough depiction of the thermodynamic behavior of the studied system.
entropy change
Entropy, symbolized as \(\Delta S\), is a measure of the disorder or randomness in a system. The entropy change associated with vaporization, \(\Delta_{\mathrm{vap}} \bar{S}\), is significant because moving from the ordered structure of a liquid to the disordered state of a gas results in an increase in entropy.

To calculate entropy change during vaporization, we use the equation:
  • \(\Delta_{\mathrm{vap}} \bar{S} = \frac{\Delta_{\mathrm{vap}} \bar{H}}{T}\)
where \(T\) is the absolute temperature in Kelvin at which the phase change occurs. This relationship stems from the understanding that at equilibrium (the boiling point), \(\Delta_{\mathrm{vap}} \bar{G} = 0\).

For benzene, the calculated entropy change at its boiling point is approximately 0.08696 kJ/molâ‹…K, indicating an increase in randomness as the liquid benzene evaporates. This measure of entropy is crucial in assessing the energy distribution in a system and is integral in determining the directionality of a process under different temperatures.
spontaneity of vaporization
The spontaneity of vaporization is an essential factor in understanding whether a liquid will naturally convert to vapor at a given temperature. This spontaneity is determined using the Gibbs free energy change equation for vaporization, \(\Delta_{\mathrm{vap}} \bar{G} = \Delta_{\mathrm{vap}} \bar{H} - T \Delta_{\mathrm{vap}} \bar{S}\).

The sign of \(\Delta_{\mathrm{vap}} \bar{G}\) indicates the spontaneity:
  • If \(\Delta_{\mathrm{vap}} \bar{G} > 0\), vaporization is non-spontaneous at that temperature.
  • If \(\Delta_{\mathrm{vap}} \bar{G} = 0\), the system is at equilibrium, meaning no net change occurs. This typically happens at the boiling point.
  • If \(\Delta_{\mathrm{vap}} \bar{G} < 0\), vaporization is spontaneous, indicating that the process will occur naturally without any additional input of energy.
Calculations show that for benzene above 80.09°C, vaporization becomes spontaneous, while below this temperature, it is non-spontaneous. This knowledge is vital in many practical applications, such as designing industrial processes where controlled phase changes are required.

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Most popular questions from this chapter

We can use the equation \((\partial S / \partial U)_{v}=1 / T\) to illustrate the consequence of the fact that entropy always increases during an irreversible adiabatic process. Consider a twocompartment system enclosed by rigid adiabatic walls, and let the two compartments be separated by a rigid heat-conducting wall. We assume that each compartment is at equilibrium but that they are not in equilibrium with each other. Because no work can be done by this two-compartment system (rigid walls) and no energy as heat can be exchanged with the surroundings (adiabatic walls), \\[U=U_{1}+U_{2}=\text { constant }\\] Show that \\[d S=\left(\frac{\partial S_{1}}{\partial U_{1}}\right) d U_{1}+\left(\frac{\partial S_{2}}{\partial U_{2}}\right) d U_{2}\\] because the entropy of each compartment can change only as a result of a change in energy. Now show that \\[d S=d U_{1}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) \geq 0\\] Use this result to discuss the direction of the flow of energy as heat from one temperature to another.

Use the following data for propene to plot \(\bar{G}(T)-\bar{H}(0)\) versus \(T\). [In this case we will ignore the (usually small) corrections due to nonideality of the gas phase.] $$ \bar{C}_{P}^{\mathrm{s}}(T) / R=\frac{12 \pi^{4}}{5}\left(\frac{T}{\Theta_{\mathrm{D}}}\right)^{3} \quad \Theta_{\mathrm{D}}=100 \mathrm{~K} \quad 0 \mathrm{~K}

The coefficient of thermal expansion of water at \(25^{\circ} \mathrm{C}\) is \(2.572 \times 10^{-4} \mathrm{K}^{-1}\), and its isothermal compressibility is \(4.525 \times 10^{-5} \mathrm{bar}^{-1} .\) Calculate the value of \(C_{p}-C_{v}\) for one mole of water at \(25^{\circ} \mathrm{C}\). The density of water at \(25^{\circ} \mathrm{C}\) is \(0.99705 \mathrm{g} \cdot \mathrm{mL}^{-1}\).

Show that the molar enthalpy of a substance at pressure \(P\) relative to its value at one bar is given by \\[\bar{H}(T, P)=\bar{H}(T, P=1 \mathrm{bar})+\int_{1}^{P}\left[\bar{V}-T\left(\frac{\partial \bar{V}}{\partial T}\right)_{P}\right] d P^{\prime}\\] Calculate the value of \(\bar{H}(T, P)-\bar{H}(T, P=1 \text { bar) at } 0^{\circ} \mathrm{C} \text { and } 100\) bar for mercury given that the molar volume of mercury varies with temperature according to \\[\bar{V}(t)=\left(14.75 \mathrm{mL} \cdot \mathrm{mol}^{-1}\right)\left(1+0.182 \times 10^{-3} t+2.95 \times 10^{-9} t^{2}+1.15 \times 10^{-10} t^{3}\right)\\] where \(t\) is the Celsius temperature. Assume that \(\bar{V}(0)\) does not vary with pressure over this range and express your answer in units of \(\mathrm{kJ} \cdot \mathrm{mol}^{-1}\).

In this problem, we will derive the equation $$ \bar{H}(T, P)-H^{\circ}(T)=R T(Z-1)+\int_{\bar{V}^{\mathrm{id}}}^{\bar{V}}\left[T\left(\frac{\partial P}{\partial T}\right)_{V}-P\right] d \bar{V}^{\prime} $$ where \(\bar{V}^{\text {id }}\) is a very large (molar) volume, where the gas is sure to behave ideally. Start with \(d H=T d S+V d P\) to derive $$ \left(\frac{\partial H}{\partial V}\right)_{T}=T\left(\frac{\partial S}{\partial V}\right)_{T}+V\left(\frac{\partial P}{\partial V}\right)_{T} $$ and use one of the Maxwell relations for \((\partial S / \partial V)_{T}\) to obtain $$ \left(\frac{\partial H}{\partial V}\right)_{T}=T\left(\frac{\partial P}{\partial T}\right)_{V}+V\left(\frac{\partial P}{\partial V}\right)_{T} $$ Now integrate by parts from an ideal-gas limit to an arbitrary limit to obtain the desired equation.

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