/*! 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 36 Diethyl ether (known simply as e... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
Entropy change is approximately 115.89 J/K.

Step by step solution

01

Recall the Formula for Entropy Change

The entropy change \( \Delta S \) when a substance vaporizes can be calculated using the formula: \( \Delta S = \frac{\Delta H_{\text{vap}}}{T} \), where \( \Delta H_{\text{vap}} \) is the heat of vaporization and \( T \) is the temperature in Kelvin.
02

Convert Temperature to Kelvin

Convert the boiling point of diethyl ether from Celsius to Kelvin using the formula: \( T_{K} = T_{C} + 273.15 \). For diethyl ether, \( T = 35.6 + 273.15 = 308.75 \text{ K} \).
03

Calculate Entropy Change for 1 mol of Diethyl Ether

Use the formula from Step 1 to find the entropy change for 1 mol of diethyl ether: \( \Delta S = \frac{26.7 \text{ kJ/mol}}{308.75 \text{ K}} \). Convert 26.7 kJ to J: \( 26.7 \times 10^3 \text{ J/mol} \). Hence, \( \Delta S = \frac{26.7 \times 10^3}{308.75} \approx 86.49 \text{ J/(mol K)} \).
04

Calculate Entropy Change for 1.34 mol of Diethyl Ether

Multiply the entropy change per mol by 1.34 to find the total entropy change: \( \Delta S_{\text{total}} = 86.49 \text{ J/(mol K)} \times 1.34 \text{ mol} = 115.89 \text{ J/K} \).

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 91Ó°ÊÓ!

Key Concepts

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

Diethyl Ether
Diethyl ether, often referred to simply as ether, is a colorless, highly volatile liquid commonly used in laboratories. With the chemical formula \((\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{O}\), it is known for its ability to dissolve a wide range of substances, making it an excellent solvent. Ether has a history of use as an anesthetic, given its effectiveness in inducing unconsciousness for medical procedures.

Despite its widespread use, diethyl ether is highly flammable and must be handled with caution. In addition to its flammability, it can form explosive peroxides upon prolonged exposure to air. Therefore, proper storage and handling procedures are vital. This versatility and the importance of safety are why ether remains a staple in chemistry labs worldwide.
Heat of Vaporization
The heat of vaporization is an essential property reflecting the energy required to convert a substance from a liquid to a gas at its boiling point. For diethyl ether, this value is \(26.7 \text{ kJ/mol}\).

When a liquid changes to a vapor, molecules move from a structured // state to one of higher freedom, increasing the substance's energy. This energy is absorbed without a temperature change, contributing to entropy. To calculate the entropy change during vaporization, we use the heat of vaporization to imply energy input into the system.
  • Important to note: Heat of vaporization is temperature-dependent and typically decreases with rising temperature.
  • This concept helps us understand the balancing of energy exchanges in chemical processes, as seen in phase transitions of substances like diethyl ether.
Boiling Point
The boiling point of a substance is the temperature at which its liquid form turns into gas. Diethyl ether's boiling point is precisely \(35.6^{\circ} \text{C}\).

At this specific temperature, the vapor pressure of diethyl ether equals the external pressure, allowing for bubbles to form within the liquid and not just at the surface. This transition represents a physical change in state.
  • The boiling point is crucial in the industrial and medicinal use of ether due to rapid vaporization at relatively low temperatures.
  • Changes in atmospheric pressure can alter the boiling point; this is why boiling points are usually stated for standard atmospheric pressure.
Understanding the boiling point is essential in processes that involve heating substances to their evaporation thresholds.
Kelvin Temperature Conversion
Converting temperatures from Celsius to Kelvin is a routine yet vital step in chemical calculations and thermodynamic equations. The formula for conversion is straightforward: \(T_{K} = T_{C} + 273.15\).

Using diethyl ether as an example, its boiling point in Celsius is \(35.6^{\circ} \text{C}\). By adding \(273.15\), we convert this to Kelvin, resulting in \(308.75 \text{ K}\). This conversion is crucial because the Kelvin scale, unlike the Celsius scale, starts at absolute zero, thereby providing a more consistent framework for scientific calculations.
  • Kelvin is the standard unit of temperature in scientific calculations because it avoids negative values, ensuring precision in mathematical equations.
  • Temperature conversions also assist in direct comparisons of thermal properties across different scientific disciplines.
Proper application of this conversion is key in exercises involving thermodynamic equations, like calculating entropy changes.

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

For each of the following reactions, decide whether there is an increase or a decrease in entropy. Why do you think so? (No calculations are needed.) a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\) c. \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) d. \(\mathrm{Li}_{3} \mathrm{~N}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{LiOH}(a q)+\mathrm{NH}_{3}(g)\)

Silver carbonate, \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\), is a light yellow compound that decomposes when heated to give silver oxide and carbon dioxide: $$ \mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g) $$ A researcher measured the partial pressure of carbon dioxide over a sample of silver carbonate at \(220^{\circ} \mathrm{C}\) and found that it was \(1.37\) atm. Calculate the partial pressure of carbon dioxide at \(25^{\circ} \mathrm{C}\). The standard enthalpies of formation of silver carbonate and silver oxide at \(25^{\circ} \mathrm{C}\) are \(-505.9 \mathrm{~kJ} / \mathrm{mol}\) and \(-31.05 \mathrm{~kJ} / \mathrm{mol}\), respectively. Make any reasonable assumptions in your calculations. State the assumptions that you make, and note why you think they are reasonable.

Consider a sample of water at \(25^{\circ} \mathrm{C}\) in a beaker in a room at \(50{ }^{\circ} \mathrm{C} .\) a. What change do you expect to observe in the water sample? Would this be a spontaneous process or not? b. What are the enthalpy and entropy changes for this change in the water sample? (Just indicate the sign of the changes.) Explain your answers. c. Does the entropy of the water increase or decrease during the change? How do you know? d. Is there a change in free energy for the water sample? If so, indicate the sign of the free-energy change and explain how you arrived at your answer. Consider the same sample of water, but starting at \(75^{\circ} \mathrm{C}\) in a room at \(50^{\circ} \mathrm{C}\). e. What change would you observe in the water sample? Would this change be a spontaneous process or not? f. What are the enthalpy and entropy changes for the water sample? (Just indicate the sign of the changes.) Explain your answers. g. Does the entropy of the water increase or decrease during the change? How do you know? h. Is there a change in free energy for the water sample? If so, indicate the sign of the free-energy change and explain how you arrived at your answer. Finally, consider the same sample of water, starting at \(50^{\circ} \mathrm{C}\) in a room at \(50^{\circ} \mathrm{C}\). i. What would you observe in the water sample? Is this a spontaneous process? j. What are the enthalpy and entropy changes for the water sample? (Just indicate the sign of the changes.) Be sure to justify your answer. k. Did the entropy of the water increase or decrease during the change? How do you know? I. Can you determine the exact free-energy change of the water in this case? If you can make this determination, what is the significance of this value?

Give the expression for the thermodynamic equilibrium constant for each of the following reactions. a. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) b. \(\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\) c. \(2 \mathrm{Li}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{Li}^{+}(a q)+2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2}(g)\)

Explain how the free energy changes as a spontaneous reaction occurs. Show by means of a diagram how \(G\) changes with the extent of reaction.

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.