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Chapter 8: Problem 1

The molar enthalpy of vaporization of bencene at its normal boiling point \(\left(80.09^{\circ} \mathrm{C}\right)\) is \(30.72 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\). Assuming that \(\Delta_{\text {rap }} \bar{H}\) and \(\Delta_{\text {vap }} \bar{S}\) stay constant at their values at \(80.09^{\circ} \mathrm{C}\), calculate the value of \(\Delta_{\mathrm{rap}} \overline{\mathrm{G}}\) 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
Vaporization is non-spontaneous at 75掳C, at equilibrium at 80.09掳C, and spontaneous at 85掳C.

Step by step solution

01

Understand the Gibbs Free Energy Change

The change in Gibbs Free Energy (\( \Delta_{ ext{vap}} \overline{G} \)) for the phase transition process can be related to enthalpy and entropy changes using the equation:\[\Delta_{ ext{vap}} \overline{G} = \Delta_{ ext{vap}} \overline{H} - T \Delta_{ ext{vap}} \overline{S}\]To solve the problem, we'll calculate this for three different temperatures.
02

Calculate the Entropy of Vaporization

The entropy of vaporization (\( \Delta_{ ext{vap}} \overline{S} \)) can be determined using the equation:\[\Delta_{ ext{vap}} \overline{S} = \frac{ \Delta_{ ext{vap}} \overline{H} }{ T_{ ext{boiling}} }\]Substitute the given values:\( \Delta_{ ext{vap}} \overline{H} = 30.72 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \)and \( T = 80.09^{\circ} \mathrm{C} = 353.24 \, \mathrm{K} \).Calculate:\[\Delta_{ ext{vap}} \overline{S} = \frac{ 30.72 }{ 353.24 } \, \mathrm{kJ} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1} \approx 0.0869 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}\]
03

Calculate \( \Delta_{ ext{vap}} \overline{G} \) at 75.0掳C

Convert the temperature to Kelvin: \( T = 75.0^{\circ} \mathrm{C} = 348.15 \, \mathrm{K} \).Substitute into the equation:\[\Delta_{ ext{vap}} \overline{G} = 30.72 \, \mathrm{kJ} \, \mathrm{mol}^{-1} - 348.15 \times 0.0869 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}\]Calculate:\[\Delta_{ ext{vap}} \overline{G} \approx 30.72 - 30.26 \, \mathrm{kJ} \, \mathrm{mol}^{-1} = 0.46 \, \mathrm{kJ} \, \mathrm{mol}^{-1}\]
04

Calculate \( \Delta_{ ext{vap}} \overline{G} \) at 80.09掳C

Using the boiling point temperature:\( T = 80.09^{\circ} \mathrm{C} = 353.24 \, \mathrm{K} \).Since we are at the boiling point, \( \Delta_{ ext{vap}} \overline{G} \) is zero:\[\Delta_{ ext{vap}} \overline{G} = 0\]
05

Calculate \( \Delta_{ ext{vap}} \overline{G} \) at 85.0掳C

Convert the temperature to Kelvin: \( T = 85.0^{\circ} \mathrm{C} = 358.15 \, \mathrm{K} \).Substitute into the equation:\[\Delta_{ ext{vap}} \overline{G} = 30.72 \, \mathrm{kJ} \, \mathrm{mol}^{-1} - 358.15 \times 0.0869 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}\]Calculate:\[\Delta_{ ext{vap}} \overline{G} \approx 30.72 - 31.12 \, \mathrm{kJ} \, \mathrm{mol}^{-1} = -0.40 \, \mathrm{kJ} \, \mathrm{mol}^{-1}\]
06

Interpretation of Results

- At 75.0掳C, \( \Delta_{\text{vap}} \overline{G} > 0\) implies the liquid phase is more stable and vaporization is non-spontaneous.- At 80.09掳C, \( \Delta_{\text{vap}} \overline{G} = 0\) signifies equilibrium between the liquid and vapor phases.- At 85.0掳C, \( \Delta_{\text{vap}} \overline{G} < 0\) indicates vaporization is spontaneous as the vapor phase is more stable.

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Key Concepts

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

Enthalpy of Vaporization
Enthalpy of vaporization is the energy required to change a substance from a liquid to a gas at constant temperature and pressure. For benzene, this process at its boiling point requires 30.72 kJ/mol. This value represents how much energy is needed per mole to overcome the intermolecular forces holding the liquid molecules together. Understanding enthalpy of vaporization helps explain why boiling indicates great energy changes. Energy added increases molecule movement, allowing transition from liquid to gas. The enthalpy of vaporization aids us in knowing when and how this phase change occurs. In our calculations, it's crucial to recognize this value remains constant at the boiling point, simplifying the analysis of Gibbs Free Energy change under varying conditions.
Entropy of Vaporization
Entropy of vaporization measures the disorder increase as a liquid turns into gas. This disorder comes because gas molecules are further apart than liquid, creating more possible arrangements.In our exercise, entropy of vaporization \[\Delta_{\text{vap}} \overline{S} = \frac{30.72{\text{ kJ mol}}^{-1}}{353.24{\text{ K}}}\] approximates 0.0869 kJ/mol路K.This value shows the increased randomness and is key when using the Gibbs Free Energy equation. Helmholtz鈥檚 relation notes that higher entropy increases lessens the energy needed to change states, making vaporization more favorable. It reveals how much of the energy, provided by heat, is convertible into useful work as the substance changes phase.
Boiling Point
The boiling point is the temperature where a liquid's vapor pressure equals the atmospheric pressure, allowing the liquid to become gas. For benzene, this is 80.09掳C. At this temperature, the enthalpy of vaporization guides the conversion of liquid to vapor.At the boiling point, \(\Delta_{\text{vap}} \overline{G}=0\), indicating no net energy gain or loss, and meaning both liquid and vapor phases balance perfectly. This equilibrium showcases how critical the boiling point is, facilitating our understanding of phase changes in chemistry. As conditions shift around this point, Gibbs Free Energy computations help determine the spontaneity and stability of the involved phases, as characterized by signs and values found during our calculations.

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Most popular questions from this chapter

The coefficient of thernal expansion of water at \(25^{\circ} \mathrm{C}\) is \(2.572 \times 10^{-4} \mathrm{~K}^{1}\), and its isuthermal compressibility is \(4.525 \times 10^{5}\) 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 \(0.99705 \mathrm{~g} \cdot \mathrm{mL}^{-1}\).

We introduced the Joule-Thomson effect and the Joule-Thomson coefficient in Problems 5-52 through 5-54.The Joule-Thomson coefficient is defined by $$ \mu_{J T}=\left(\frac{\partial T}{\partial P}\right)_{H}=-\frac{1}{C_{P}}\left(\frac{\partial H}{\partial P}\right)_{T} $$ and is a direct measure of the expected temperature change when a gas is expanded through a throttle. We can use one of the equations derived in this chapter to vbtain a convenicit working equation for \(\mu_{11}\). Show that $$ \mu_{\mathrm{JT}}=\frac{1}{C_{p}}\left[T\left(\frac{\partial V}{\partial T}\right)_{p}-V\right] $$ Use this result to show that \(\mu_{1 T}=0\) for an ideal gas.

Determine \(\bar{C}_{p}-\bar{C}_{V}\) for a gas that obeys the equation of state \(P(\bar{V}-b)=R T\).

Show that the molar enthalpy of a substance at pressure \(P\) relative to its value at one bar is given by $$ \bar{H}(T, P)=\bar{H}(T, P=1 \text { bar })+\int_{1}^{F}\left[\bar{V}-T\left(\frac{\partial \bar{V}}{\partial T}\right)_{P}\right] d P^{\prime} $$ Calculate the value of \(\bar{H}(T, P)-\bar{H}(T, P=1\) bar \()\) at \(0^{\circ} \mathrm{C}\) and 100 bar for mercury given that the molar volume of mercury varies with temperature according to $$ \bar{V}(t)=\left(14.75 \mathrm{~mL} \cdot \mathrm{mol}^{-1}\right)\left(\mathrm{I}+0.182 \times 10^{-3} t+2.95 \times 10^{-9} t^{2}+1.15 \times 10^{-10} t^{3}\right) $$ 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 \(\mathrm{kJ} \cdot \mathrm{mol}^{-1}\).

Show that the enthalpy is a function of only the temperature for a gas that obeys the equation of state \(P(\bar{V}-b T)=R T\), where \(b\) is a constant.
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