/*! 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 44 The enthalpy of vaporization of ... [FREE SOLUTION] | 91影视

91影视

The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boiling point \(\left(78^{\circ} \mathrm{C}\right) .\) Determine \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{surr}},\) and \(\Delta S_{\mathrm{univ}}\) when 1.00 mole of ethanol is vaporized at \(78^{\circ} \mathrm{C}\) and 1.00 atm.

Short Answer

Expert verified
In conclusion, when 1 mole of ethanol is vaporized at 78掳C and 1 atm, the change in entropy of the system (螖S_sys) is \(110.16 J/mol\cdot K\), the change in entropy of the surroundings (螖S_surr) is \(-110.16 J/mol\cdot K\), and the change in entropy of the universe (螖S_univ) is \(0 J/mol\cdot K\).

Step by step solution

01

Convert enthalpy of vaporization to J/mol

We are given that the enthalpy of vaporization of ethanol is 38.7 kJ/mol. To convert this value to J/mol, we will multiply by 1000: 38.7 kJ/mol * 1000 J/kJ = 38,700 J/mol Now, the enthalpy of vaporization of ethanol is 38,700 J/mol.
02

Calculate 螖S_sys

To calculate the change in entropy of the system (螖S_sys), we will use the formula: 螖S_sys = 螖Hvap/T First, we need to convert the boiling point temperature from Celsius to Kelvin by adding 273.15: T = 78掳C + 273.15 = 351.15 K Now, we can calculate 螖S_sys: 螖S_sys= (38,700 J/mol) / (351.15 K)= 110.16 J/mol路K The change in entropy of the system is 110.16 J/mol路K.
03

Calculate 螖S_surr

To calculate the change in entropy of the surroundings (螖S_surr), we will use the following formula: 螖S_surr = -螖Hvap/T Using the previously calculated values for 螖Hvap (38,700 J/mol) and T(351.15 K): 螖S_surr = -(38,700 J/mol) / (351.15 K) = -110.16 J/mol路K The change in entropy of the surroundings is -110.16 J/mol路K.
04

Calculate 螖S_univ

To calculate the change in entropy of the universe (螖S_univ), we will add 螖S_sys and 螖S_surr: 螖S_univ = 螖S_sys + 螖S_surr = 110.16 J/mol路K + (-110.16 J/mol路K) = 0 J/mol路K The change in entropy of the universe is 0 J/mol路K. In conclusion, the change in entropy of the system (螖S_sys) is 110.16 J/mol路K, the change in entropy of the surroundings (螖S_surr) is -110.16 J/mol路K, and the change in entropy of the universe (螖S_univ) is 0 J/mol路K when 1 mole of ethanol is vaporized at 78掳C and 1 atm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Entropy Change
Entropy is a fascinating concept in thermodynamics. It helps determine the feasibility of a process by gauging disorder or randomness. When calculating entropy change, we're usually looking at changes during physical transformations. For instance, when ethanol vaporizes, its molecules move from a liquid state (with restricted movement) to a gaseous state (with more freedom to move). This increases system entropy because gaseous molecules are more disordered than liquid molecules.

The change in entropy (\(\Delta S_{\mathrm{sys}}\) is calculated using the enthalpy of vaporization (\(\Delta H_{\text{vap}}\)) and the temperature in Kelvin (\(T\)). The formula is:
  • \(\Delta S_{\text{sys}} = \frac{\Delta H_{\text{vap}}}{T}\)
This formula indicates that the entropy change is directly proportional to the heat transferred and inversely proportional to the temperature. For ethanol, vaporizing at its boiling point shows us how thermal energy changes the degree of randomness within the system.
System and Surroundings
In thermodynamics, the universe is often divided into the system and its surroundings. The 鈥渟ystem鈥 is the part we are focused on, while the "surroundings" include everything else. When dealing with problems like calculating entropy during a phase change, it's crucial to distinguish between these two categories.

The system here refers to the ethanol undergoing vaporization. Its entropy change (\(\Delta S_{\text{sys}}\)) gives a positive value as the molecules gain freedom and disorder. On the other hand, the surroundings experience an opposite, but equal change in entropy (\(\Delta S_{\text{surr}}\)). This is because as the ethanol absorbs heat from its surroundings, the available energy in the surroundings decreases resulting in a negative entropy change. The formula used is:
  • \(\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{vap}}}{T}\)
The universe's net entropy change, calculated as \(\Delta S_{\text{univ}}\), is the combination of both, thus in a reversible process at equilibrium it results in zero: \(\Delta S_{\text{univ}} = 0\). This highlights how system and surroundings are intricately related.
Temperature Conversion
Temperature conversion is a key step in solving many thermodynamic problems. Different temperature scales can be used in scientific investigations. In chemistry and physics, temperature in Kelvin is often used because it starts from absolute zero.

To convert temperatures from Celsius to Kelvin, simply add 273.15. This conversion is essential because thermodynamic formulas usually require temperatures to be in Kelvin for accuracy and standardization. In our ethanol example, the boiling point is 78掳C, which converts to:
  • \(78^{\circ} \text{C} + 273.15 = 351.15 \text{K}\)
Always ensure your temperatures are correctly converted to avoid mistakes when calculating entropy or other thermodynamic properties.
Spontaneous Processes
Spontaneity in processes is determined by the overall change in the universe's entropy. A process is spontaneous when it can occur without any outside intervention. For many processes at constant pressure and temperature, we assess spontaneity by calculating the change in the universe's entropy \(\Delta S_{\text{univ}}\).

In our ethanol vaporization example, \(\Delta S_{\text{univ}}\) equals zero (0 J/mol路K), demonstrating it is a reversible process at the boiling point. While \(\Delta S_{\text{sys}}\) and \(\Delta S_{\text{surr}}\) are equal and opposite, they balance out, indicating equilibrium rather than true spontaneity. However, if \(\Delta S_{\text{univ}}\) were positive, it would point to a spontaneous direction, for such a process energy from the surroundings would freely assist in vaporization.
  • Positive \(\Delta S_{\text{univ}}\) \(\rightarrow\) spontaneous
  • Zero \(\Delta S_{\text{univ}}\) \(\rightarrow\) equilibrium
  • Negative \(\Delta S_{\text{univ}}\) \(\rightarrow\) non-spontaneous
The critical takeaway is that positive increase in entropy in the universe indicates that energy naturally disperses, emphasizing the preferred nature of spontaneous actions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When the environment is contaminated by a toxic or potentially toxic substance (for example, from a chemical spill or the use of insecticides), the substance tends to disperse. How is this consistent with the second law of thermodynamics? In terms of the second law, which requires the least work: cleaning the environment after it has been contaminated or trying to prevent the contamination before it occurs? Explain.

The synthesis of glucose directly from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) and the synthesis of proteins directly from amino acids are both nonspontaneous processes under standard conditions. Yet it is necessary for these to occur for life to exist. In light of the second law of thermodynamics, how can life exist?

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.

Consider the reaction: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ for which \(\Delta H\) is \(-233 \mathrm{kJ}\) and \(\Delta S\) is \(-424 \mathrm{J} / \mathrm{K}\) . a. Calculate the free energy change for the reaction \((\Delta G)\) at 393 \(\mathrm{K} .\) b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

For a liquid, which would you expect to be larger, \(\Delta S_{\text { fusion }}\) or \(\Delta S_{\text { evaporation }} ?\) Why?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.