Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 36

The enthalpy of vaporization of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) is \(31.4\) \(\mathrm{kJ} / \mathrm{mol}\) at its boiling point \(\left(61.7^{\circ} \mathrm{C}\right) .\) Determine \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{sur}}\), and \(\Delta S_{\text {univ }}\) when \(1.00 \mathrm{~mol}\) chloroform is vaporized at \(61.7^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\)

Short Answer

Expert verified
The change in entropy for the system (ΔS_sys) when 1.00 mol chloroform vaporizes at 61.7°C and 1.00 atm is approximately 93.9 J / (mol K). The change in entropy for the surroundings (ΔS_sur) is approximately -93.9 J / (mol K). The change in entropy for the universe (ΔS_univ) is 0 J / (mol K), which indicates that the process is at equilibrium.

Step by step solution

01

Calculate ΔS_sys (Change in Entropy of the System)

: To calculate the change in entropy for the system (ΔS_sys), we can use the formula: ΔS_sys = q_sys / T Where q_sys is the heat absorbed by the system at constant pressure, and T is the temperature in Kelvin. We have the enthalpy of vaporization (ΔH_vap) given as 31.4 kJ/mol. The heat absorbed (q_sys) is equal to the enthalpy of vaporization multiplied by the number of moles: q_sys = ΔH_vap × n Converting the boiling point from Celsius to Kelvin: T = 61.7°C + 273.15 = 334.85 K Now, plug in the given values and calculate ΔS_sys: ΔS_sys = (31.4 kJ/mol × 1.00 mol) / 334.85 K
02

Convert ΔS_sys units

: To obtain the result in J / (mol K), we need to convert kJ to J: ΔS_sys = (31.4 × 10^3 J/mol) / 334.85 K ΔS_sys ≈ 93.9 J / (mol K)
03

Calculate ΔS_sur (Change in Entropy of the Surroundings)

: To calculate the change in entropy for the surroundings (ΔS_sur), we can use the formula: ΔS_sur = -q_sur / T Here, q_sur (heat gained by the surroundings) is equal to -q_sys (heat absorbed by the system) because the energy is conserved. Therefore, the formula becomes: ΔS_sur = -(-31.4 kJ/mol × 1.00 mol) / 334.85 K
04

Convert ΔS_sur units

: As before, we convert kJ to J for ΔS_sur: ΔS_sur = (31.4 × 10^3 J/mol) / 334.85 K ΔS_sur ≈ -93.9 J / (mol K)
05

Calculate ΔS_univ (Change in Entropy of the Universe)

: To calculate the change in entropy for the universe (ΔS_univ), we can sum the changes in entropy for the system and surroundings: ΔS_univ = ΔS_sys + ΔS_sur ΔS_univ = 93.9 J / (mol K) - 93.9 J / (mol K) ΔS_univ = 0 J / (mol K) The change in entropy for the universe is 0 J / (mol K) when 1.00 mol chloroform is vaporized at 61.7°C and 1.00 atm, which indicates that the process is at equilibrium.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 ammonia \(\left(\mathrm{NH}_{3}\right.\) ), the enthalpy of fusion is \(5.65 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of fusion is \(28.9 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). a. Will \(\mathrm{NH}_{3}(s)\) spontaneously melt at \(200 . \mathrm{K}\) ? b. What is the approximate melting point of ammonia?

Consider the reaction $$2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\). b. Would the reaction be more spontaneous at high or low temperatures?

Is \(\Delta S_{\text {surr }}\) favorable or unfavorable for exothermic reactions? Endothermic reactions? Explain.

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C}\), the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

When most biologic enzymes are heated, they lose their catalytic activity. The change Original enzyme \(\longrightarrow\) new form that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation
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.

Sign-up for free