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Chapter 18: Problem 76

Acetone, \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\), boils at \(56^{\circ} \mathrm{C}\). The heat of vaporization of acetone at this temperature is \(29.1 \mathrm{~kJ} / \mathrm{mol}\). What is the entropy change when \(1 \mathrm{~mol}\) of liquid acetone vaporizes at \(56^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The entropy change is 88.4 J/mol路K.

Step by step solution

01

Understanding the Concept

The entropy change \( \Delta S \) for the vaporization process can be determined using the formula \( \Delta S = \frac{\Delta H}{T} \), where \( \Delta H \) is the heat of vaporization and \( T \) is the temperature in Kelvin.
02

Converting Temperature

Convert the given boiling temperature from Celsius to Kelvin. Use the conversion formula: \(T(K) = T(^{\circ}C) + 273.15\). Substitute the boiling point of acetone: \( T(K) = 56 + 273.15 = 329.15\). Thus, \( T = 329.15 \) K.
03

Calculating Entropy Change

Use the formula \( \Delta S = \frac{\Delta H}{T} \) to calculate the entropy change.Substitute the values: \( \Delta H = 29.1\, \mathrm{kJ/mol} \) and \( T = 329.15\, \text{K} \). \[\Delta S = \frac{29.1 \, \mathrm{kJ/mol}}{329.15 \, \mathrm{K}} = 0.0884 \, \mathrm{kJ/mol\cdot K}\] Convert \( \Delta S \) to \( \mathrm{J/mol\cdot K} \) by multiplying by 1000: \[ \Delta S = 88.4 \, \mathrm{J/mol\cdot K} \]

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Key Concepts

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

Entropy Change
Entropy change is a fundamental concept in thermodynamics. It measures the amount of disorder or randomness in a system. When a liquid vaporizes, its molecules move from a more ordered state (liquid) to a less ordered state (vapor). This process increases the entropy of the system.To calculate the entropy change \( \Delta S \), we use:\[ \Delta S = \frac{\Delta H}{T} \]where:
  • \( \Delta H \): Heat of vaporization
  • \( T \): Temperature in Kelvin
This formula shows that the entropy change depends on the heat absorbed and the temperature at which the change occurs.For acetone vaporization, the calculated entropy change is \( 88.4 \, \mathrm{J/mol \cdot K} \).
Vaporization Process
The vaporization process is when a liquid turns into a vapor. This transition is crucial in understanding how energy and matter interact. It occurs at a specific temperature called the boiling point. During vaporization, energy, known as the heat of vaporization, is absorbed by the liquid to break intermolecular forces and change its state. Key Points of Vaporization:
  • Molecule movement increases
  • System entropy increases
  • Heat is absorbed
For acetone, this process begins at its boiling point of \( 56^{\circ} \mathrm{C} \). Understanding this process is critical for applications in chemical and industrial processes.
Heat of Vaporization
The heat of vaporization is the energy required to convert a liquid into a gas at its boiling point, without changing the temperature. It is a crucial parameter in characterizing substances and their interactions.For acetone, the heat of vaporization is \( 29.1 \, \mathrm{kJ/mol} \).Factors affecting the heat of vaporization:
  • The strength of intermolecular forces: Stronger forces require more energy.
  • Molecular weight: Larger molecules generally need more energy.
These factors determine how much energy is needed for a substance to vaporize. It's essential in designing cooling and heating processes.
Temperature Conversion
Temperature conversion is an important skill in scientific calculations because thermodynamic equations often use Kelvin.The conversion from Celsius to Kelvin is straightforward:\[ T(K) = T(^{\circ}C) + 273.15 \]This formula shifts the Celsius scale, which is more intuitive for daily use, to the Kelvin scale, used in scientific contexts because it starts from absolute zero.For acetone, the boiling point in Celsius (\(56^{\circ} \mathrm{C}\)) becomes \(329.15 \, \mathrm{K}\).Understanding this conversion ensures accuracy and consistency in thermodynamic calculations involving temperature.

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Most popular questions from this chapter

Diethyl ether (known simply as ether), \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\), is a solvent and anesthetic. The heat of vaporization of diethyl ether at its boiling point \(\left(35.6^{\circ} \mathrm{C}\right)\) is \(26.7 \mathrm{~kJ} / \mathrm{mol}\). What is the entropy change when \(1.34 \mathrm{~mol}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) vaporizes at its boiling point?

a. Why are some reactions exothermic and others endothermic? b. Discuss the driving force in a spontaneous reaction that is highly exothermic and in one that is endothermic.

A reaction has a \(\Delta G=-10.0 \mathrm{~kJ}\) and a \(\Delta H=-20.0 \mathrm{~kJ}\). If \(\Delta S=-1.82 \times 10^{3} \mathrm{~J} / \mathrm{K}\), what was the temperature at which the reaction occurred? a. \(6.04 \mathrm{~K}\) b. \(6.04 \times 10^{-3} \mathrm{~K}\) c. \(5.49 \times 10^{-3} \mathrm{~K}\) d. \(5.49 \mathrm{~K}\) e. \(8.55 \mathrm{~K}\)

\(K_{s p}\) for silver chloride at \(25.0^{\circ} \mathrm{C}\) is \(1.782 \times 10^{-10}\). At \(35.0^{\circ} \mathrm{C}, K_{s p}^{\circ}\) is \(4.159 \times 10^{-10} .\) What are \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction?

On the basis of \(\Delta G^{\circ}\) for each of the following reactions, decide whether the reaction is spontaneous or nonspontaneous as written. Or, if you expect an equilibrium mixture with significant amounts of both reactants and products, say SO. a. \(\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \longrightarrow 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta G^{\circ}=-91 \mathrm{~kJ}\) b. \(2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) ; \Delta G^{\circ}=-211 \mathrm{~kJ}\) c. \(\mathrm{HCOOH}(l) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) ; \Delta G^{\circ}=119 \mathrm{~kJ}\) d. \(\mathrm{I}_{2}(s)+\mathrm{Br}_{2}(l) \longrightarrow 2 \operatorname{IBr}(g) ; \Delta G^{\circ}=7.5 \mathrm{~kJ}\) e. \(\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) ; \Delta G^{\circ}=92 \mathrm{~kJ}\)
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