/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Ethanethiol \(\left(\mathrm{C}_{... [FREE SOLUTION] | 91影视

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Ethanethiol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ; \text { also called ethyl mercaptan) is }\right.\) commonly added to natural gas to provide the 鈥渞otten egg鈥 smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is 27.5 \(\mathrm{kJ} / \mathrm{mol}\) . What is the entropy of vaporization for this substance?

Short Answer

Expert verified
The entropy of vaporization for ethanethiol is approximately 89.2 J/mol K.

Step by step solution

01

Recall the entropy of vaporization formula

We will use the equation for the entropy of vaporization, which is given by: \[\Delta S_{vap} = \frac{\Delta H_{vap}}{T}\] Where: - \(\Delta S_{vap}\) is the entropy of vaporization - \(\Delta H_{vap}\) is the heat of vaporization (given in the problem) - \(T\) is the temperature in Kelvin at which the vaporization occurs (boiling point)
02

Convert given temperature to Kelvin

We are given the boiling point of ethanethiol as \(35^{\circ} \mathrm{C}\). To convert it to Kelvin, we will add 273.15 to the Celsius temperature: \[T = 35 + 273.15 = 308.15 \, \mathrm{K}\]
03

Calculate the entropy of vaporization

Now, we can substitute the given heat of vaporization and the calculated temperature in Kelvin into the formula to find the entropy of vaporization: \[\Delta S_{vap}=\frac{27.5 \, \mathrm{kJ/mol}}{308.15 \, \mathrm{K}}\] First, we need to convert the heat of vaporization to J/mol. Since 1 kJ = 1000 J: \[\Delta H_{vap} = 27.5 \, \mathrm{kJ/mol} \times \frac{1000 \, \mathrm{J}}{1 \, \mathrm{kJ}} = 27500 \, \mathrm{J/mol}\] Now substitute the value of \(\Delta H_{vap}\) in J/mol and \(T\) in the entropy of vaporization formula: \[\Delta S_{vap} = \frac{27500 \, \mathrm{J/mol}}{308.15 \, \mathrm{K}} = 89.2 \, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\] The entropy of vaporization for ethanethiol is approximately 89.2 J/mol K.

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Key Concepts

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

Heat of Vaporization
The heat of vaporization is a key thermodynamic property that defines the amount of energy required to transform a given quantity of a substance from a liquid to a gas at its boiling point. It is typically measured in kilojoules per mole (kJ/mol). For ethanethiol, this process requires 27.5 kJ/mol, meaning that energy must be supplied to overcome intermolecular forces that hold the liquid molecules together.
Here are some important points to consider about heat of vaporization:
  • It is always a positive value because heat is absorbed during vaporization.
  • Higher heat of vaporization implies stronger intermolecular forces.
  • Understanding this concept helps in determining energy needs for processes like distillation and evaporation.
By knowing the heat of vaporization, we can better understand and calculate related properties such as the entropy of vaporization, which provides insight into the disorder increase during the phase transition.
Boiling Point
The boiling point of a substance is the temperature at which its vapor pressure equals the external pressure, causing the liquid to turn into a vapor. For ethanethiol, this occurs at 35掳C. At this point:
  • The molecules have enough energy to overcome atmospheric pressure.
  • Vapor bubbles form within the liquid.
  • The temperature is constant until all the liquid has vaporized.
The boiling point can vary with atmospheric conditions. At higher altitudes, the boiling point is lower because of decreased external pressure.
Knowing the boiling point is crucial in calculations related to thermodynamics as it defines the conditions under which the transformations occur, such as in the calculation of entropy of vaporization. It also facilitates understanding of phenomena and applications like cooking and industrial reactions.
Temperature Conversion
Temperature plays a critical role in chemical processes, and sometimes it needs to be converted from one scale to another for calculations. The original exercise involves converting 35掳C, the boiling point of ethanethiol, to Kelvin. In everyday scientific calculations, the Kelvin scale is preferred:
  • Unlike Celsius, Kelvin starts at absolute zero, where theoretically molecular motion ceases.
  • This makes it the preferred unit for thermodynamic equations.
Converting is simple: add 273.15 to the Celsius temperature. Thus, 35掳C converts to:\[ T = 35 + 273.15 = 308.15 \, \text{K} \]This conversion is critical for accurately using formulas that measure changes related to heat and energy, as these often need temperatures in Kelvin for precision. Understanding temperature conversion ensures correct inputs in formulas, avoiding potential errors in calculations and experiments.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships and conversions between heat and other forms of energy. Key to understanding properties like the entropy and heat of vaporization, thermodynamics helps us grasp how substances change states and the energy dynamics involved. The original problem highlights several core principles:
Key thermodynamic concepts include:
  • Conservation of energy, where energy is not created or destroyed but converted.
  • Entropy, representing the disorder or randomness in a system, which increases during vaporization.
  • The heat of vaporization, illustrating the energy input required for phase changes.
Through thermodynamic processes, chemists and engineers can predict how different variables affect the state and energy of materials. This understanding supports applications from chemical manufacturing to environmental science, allowing for innovation and efficient resource use. By grasping thermodynamics, students comprehend the energetic and molecular transformations underlying natural and industrial phenomena.

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Most popular questions from this chapter

Consider a weak acid, HX. If a \(0.10-M\) solution of \(\mathrm{HX}\) has a pH of 5.83 at \(25^{\circ} \mathrm{C},\) what is \(\Delta G^{\circ}\) for the acid's dissociation reaction at \(25^{\circ} \mathrm{C} ?\)

For the sublimation of iodine at \(25^{\circ} \mathrm{C}\) $$\mathrm{I}_{2}(s) \rightarrow \mathrm{I}_{2}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) are, respectively, 62 \(\mathrm{kJ}\) and 19 \(\mathrm{kJ}\) . Estimate the temperature at which iodine sublimes. Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

The equilibrium constant for a certain reaction increases by a factor of 6.67 when the temperature is increased from 300.0 \(\mathrm{K}\) to 350.0 \(\mathrm{K}\) . Calculate the standard change in enthalpy \(\left(\Delta H^{\circ}\right)\) for this reaction (assuming \(\Delta H^{\circ}\) is temperature-independent).

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2} .\) Consider the following reactions and approximate standard free energy changes: $$\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} \quad \Delta G^{\circ}=-70 \mathrm{kJ}$$ $$\mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} \quad \Delta G^{\circ}=-80 \mathrm{kJ} $$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

Acrylonitrile is the starting material used in the manufacture of acrylic fibers (U.S. annual production capacity is more than two million pounds). Three industrial processes for the production of acrylonitrile are given below. Using data from Appendix 4, calculate \(\Delta S^{\circ}, \Delta H^{p},\) and \(\Delta G^{\circ}\) for each process. For part a, assume that \(T=25^{\circ} \mathrm{C} ;\) for part \(\mathrm{b}, T=70 .^{\circ} \mathrm{C} ;\) and for part \(\mathrm{c}, T=700 .^{\circ} \mathrm{C}\) Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

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