The standard entropy of vaporization of benzene is approximately \(85 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\) at its boiling point. (a) Estimate the standard enthalpy of vaporization of benzene at its normal boiling point of \(80 .^{\circ} \mathrm{C}\). (b) What is the standard entropy change of the surroundings when \(10 . \mathrm{g}\) of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), vaporizes at its normal boiling point?

Short Answer

Expert verified
The standard enthalpy of vaporization of benzene at its normal boiling point is approximately 30.02 kJ\(\cdot\)mol^{-1}, and the standard entropy change of the surroundings when 10 g of benzene vaporizes at its normal boiling point is 0.31 J\(\cdot\)K^{-1}.

Step by step solution

01

Understand the problem and the given data

We are given the standard entropy of vaporization of benzene, which is 85 J\(\cdot\)K^{-1}\(\cdot\)mol^{-1}, and the normal boiling point, which is 80 \(^{\circ}\)C. We are asked to find the enthalpy of vaporization at this temperature for part (a) and the standard entropy change of the surroundings when 10 g of benzene vaporizes for part (b).
02

Calculate the standard enthalpy of vaporization using Gibbs free energy equation

Using the Gibbs free energy equation \(\Delta G = \Delta H - T\Delta S\), at the boiling point, \(\Delta G\) is zero because the liquid and vapor are in equilibrium. Thus, \(0 = \Delta H_{vap} - T\Delta S_{vap}\). Rearrange to calculate the enthalpy of vaporization, \(\Delta H_{vap}\):\[\Delta H_{vap} = T\Delta S_{vap}\]. Before we substitute the values in, convert the boiling point from Celsius to Kelvin by adding 273.15: \(T = 80 \(^{\circ}\)C + 273.15 = 353.15\) K. Now substitute the values into the equation: \(\Delta H_{vap} = 353.15\) K \(\times\) \(85\) J\(\cdot\)K^{-1}\(\cdot\)mol^{-1} = 30018.25 J\(\cdot\)mol^{-1} = 30.02 kJ\(\cdot\)mol^{-1}.
03

Calculate the molar mass of benzene

The molar mass of benzene (C6H6) is calculated by adding the mass of six carbon atoms and six hydrogen atoms: 6\(\times\)12.01 g/mol (carbon) + 6\(\times\)1.008 g/mol (hydrogen) = 72.06 g/mol + 6.048 g/mol = 78.108 g/mol.
04

Calculate the number of moles of benzene vaporizing

Convert the mass of benzene vaporizing from grams to moles using the molar mass calculated in the previous step: \(n = \frac{mass}{molar\ mass} = \frac{10 g}{78.108 g/mol} = 0.1280 mol\).
05

Calculate the standard entropy change of the surroundings

The entropy change of the surroundings (\(\Delta S_{surroundings}\)) during a process is the negative of the entropy change of the system (\(\Delta S_{system}\)) divided by the temperature. If we consider the vaporization of benzene the system, then \(\Delta S_{system} = 85\) J\(\cdot\)K^{-1}\(\cdot\)mol^{-1}. Thus, for the surroundings: \(\Delta S_{surroundings} = -\frac{\Delta S_{system} \times n}{T} = -\frac{85 J\(\cdot\)K^{-1}\(\cdot\)mol^{-1} \times 0.1280 mol}{353.15 K} = -0.31 J\(\cdot\)K^{-1}\). Since we are interested in the total change, we can drop the negative sign, resulting in 0.31 J\(\cdot\)K^{-1}.

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Key Concepts

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

Understanding the Gibbs Free Energy Equation
The Gibbs free energy equation is fundamental in thermodynamics and describes the total amount of usable energy called 'free' energy, within a chemical system. Given by the formula \( \Delta G = \Delta H - T\Delta S \), it serves as a critical tool in predicting the direction of a chemical reaction under constant temperature and pressure.

The equation combines the concepts of enthalpy (\(\Delta H\)), which denotes the heat content or total energy, and entropy (\(\Delta S\)), indicative of the disorder or randomness of the system. Temperature (\(T\)) in Kelvin degrees acts as a pivot to balance between enthalpy and entropy. At equilibrium, especially during phase changes like vaporization at the boiling point, Gibbs free energy change \(\Delta G\) is zero. This is because the process occurs spontaneously and no additional energy is required to drive it forward. By rearranging the equation to solve for \(\Delta H_{vap}\) when \(\Delta G = 0\) at the boiling point, we obtain an elegant approach to estimate the enthalpy of vaporization — a measure of energy required to convert a substance from liquid to gaseous state.
Enthalpy of Vaporization
Enthalpy of vaporization, often denoted as \(\Delta H_{vap}\), quantifies the heat energy needed to vaporize a specific amount of a substance at its boiling point under standard pressure. This physical property is essential because it provides insight into the strength of intermolecular forces within a liquid – the stronger these forces, the higher the enthalpy of vaporization.

The calculation in our textbook problem cleverly employs the condition where Gibbs free energy is zero at the boiling point. The enthalpy of vaporization is then directly proportional to the entropy of vaporization and the absolute temperature. It's noteworthy that enthalpies of vaporization are typically expressed in Joules per mole (J/mol) or kilojoules per mole (kJ/mol), providing a clear understanding of the energy required on a per-mole basis.
Molar Mass Calculation
Molar mass refers to the mass of one mole of a substance and is a crucial concept for converting between mass and number of moles in chemistry. The molar mass of a compound like benzene, \(C_6H_6\), is calculated by summing the molar masses of its constituent atoms, carbon and hydrogen, which are determined from the periodic table.

Specifically, the molar mass aids in quantifying how much of a substance is present. This calculation is fundamental in stoichiometry, the part of chemistry that deals with the quantities of reactants and products in chemical reactions. In practical terms, knowing the molar mass allows a student to determine the number of molecules in a given mass of a compound – a critical step in many quantitative exercises like the one illustrated here.
Entropy Change of Surroundings
The concept of entropy extends beyond the system in question to include the surroundings – essentially, the rest of the universe. In a thermodynamic process, the total entropy change of the universe equals the sum of the changes in entropy of the system and its surroundings. When a substance like benzene vaporizes, the system absorbs heat, increasing its own entropy. The Second Law of Thermodynamics ensures that this is compensated by a greater entropy increase in the surroundings, to favor the spontaneous process.

The entropy change for the surroundings during vaporization is calculated negatively from the system's entropy change (because energy is taken from the surroundings) and normalized by the temperature. This key step highlights a universal balance: while a system may become more ordered by losing entropy, its surroundings must become more disordered, maintaining the natural trend towards increased overall disorder or entropy in the universe.

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Most popular questions from this chapter

Without performing any calculations, predict whether there is an increase or a decrease in entropy for each of the following processes: (a) \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl}(\mathrm{aq})+\mathrm{HClO}(\mathrm{aq})\); (b) \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{~s}) \longrightarrow 3 \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{PO}_{4}^{3-}\) (aq); (c) \(\mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{Br}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \operatorname{HBr}(\mathrm{aq})\)

According to current theories of biological evolution, complex amino and nucleic acids were produced from randomly occurring reactions of compounds thought to be present in the Earth's early atmosphere. These simple molecules then assembled into more and more complex molecules, such as DNA and RNA. Is this process consistent with the second law of thermodynamics? Explain your answer.

List the following substances in order of increasing molar entropy at \(298 \mathrm{~K}: \mathrm{H}_{2} \mathrm{O}(\mathrm{l}), \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{s}), \mathrm{C}\) (s, diamond). Explain your reasoning.

Three isomeric alkenes have the formula \(\mathrm{C}_{4} \mathrm{H}_{8}\) (see the following table). (a) Draw Lewis structures of these compounds. (b) Calculate \(\Delta G^{\circ}, \Delta H^{\circ}\), and \(\Delta S^{\circ}\) for the three reactions that interconvert each pair of compounds. (c) Which isomer is the most stable? (d) Rank the isomers in order of decreasing \(S_{m}{ }^{\circ}\). \(\begin{array}{lrc} \text { Compound } & \Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\left(\mathrm{kJ} \cdot \mathrm{mol}^{-1}\right) & \Delta \mathrm{G}_{\mathrm{f}}^{\circ}\left(\mathrm{kJ} \cdot \mathrm{mol}^{-1}\right) \\ \hline \text { 2-methylpropene } & -16.90 & +58.07 \\ \text { cis-2-butene } & -6.99 & +65.86 \\ \text { trans-2-butene } & -11.17 & +62.97 \\ \hline \end{array}\)

Determine whether titanium dioxide can be reduced by carbon at \(1000 . \mathrm{K}\) in each of the following reactions: (a) \(\mathrm{TiO}_{2}\) (s) \(+2 \mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s})+2 \mathrm{CO}(\mathrm{g})\) (b) \(\mathrm{TiO}_{2}\) (s) \(+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s})+\mathrm{CO}_{2}\) (g) given that, at \(1000 . \mathrm{K}, \Delta G_{f}{ }^{\circ}\left(\mathrm{CO}\right.\), g) \(=-200 . \mathrm{kJ} \cdot \mathrm{mol}^{-1}\), \(\Delta G_{f}^{\circ}\left(\mathrm{CO}_{2}, \mathrm{~g}\right)=-396 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} ;\) and \(\Delta G_{f}{ }^{\circ}\left(\mathrm{TiO}_{2}, \mathrm{~s}\right)=\) \(-762 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\).

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