Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For each of the following processes, indicate whether the signs of \(\Delta S\) and \(\Delta H\) are expected to be positive, negative, or about zero. (a) A solid sublimes. (b) The temperature of a sample of \(\mathrm{Co}(s)\) is lowered from \(60^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\). (c) Ethyl alcohol evaporates from a beaker. (d) A diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\).

Short Answer

Expert verified
(a) ∆S: positive, ∆H: positive; (b) ∆S: negative, ∆H: negative; (c) ∆S: positive, ∆H: positive; (d) ∆S: positive, ∆H: positive; (e) ∆S: positive, ∆H: negative.

Step by step solution

01

(a) A solid sublimes

The entropy change (∆S) should be positive due to the increase in randomness as the solid turns into gas. The enthalpy change (∆H) should also be positive since heat is absorbed to break the bonds in the solid.
02

(b) The temperature of a sample of Co(s) is lowered from 60°C to 25°C

The entropy change (∆S) should be negative due to the decrease in randomness as the solid cools down. The enthalpy change (∆H) should be negative as well since heat is removed from the solid during cooling.
03

(c) Ethyl alcohol evaporates from a beaker

The entropy change (∆S) should be positive due to the increase in randomness as the liquid turns into gas. The enthalpy change (∆H) should be positive since heat is absorbed to break the intermolecular forces in the liquid.
04

(d) A diatomic molecule dissociates into atoms

The entropy change (∆S) should be positive due to the increase in randomness as the molecule dissociates into individual atoms. The enthalpy change (∆H) should be positive since energy is required to break the molecular bonds.
05

(e) A piece of charcoal is combusted to form CO₂(g) and H₂O(g)

The entropy change (∆S) should be positive due to the increase in randomness with the formation of gaseous products. The enthalpy change (∆H) should be negative since the combustion process releases energy (exothermic reaction).

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 Vaia!

Key Concepts

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

Entropy
Entropy is a measure of disorder or randomness in a system. In thermodynamics, it helps predict the direction of spontaneous processes. When a system's entropy increases, it means there's more molecular disorder.

Consider the example of solid sublimation: when a solid changes directly into a gas, it moves from an ordered state to a more disordered state. This means the entropy change (\(\Delta S\) ) is positive, as gas particles have more freedom to move around compared to solids.

Whenever substances change phase from solid to liquid or liquid to gas, entropy also increases due to greater molecular movement. During chemical reactions like dissociation or combustion involving gaseous products, \(\Delta S\) is often positive too, indicative of the increase in disorder.
Enthalpy
Enthalpy is a concept in thermodynamics that represents the total heat content of a system. It's often associated with heat changes during phase transitions and chemical reactions.

For endothermic processes, where heat is absorbed, \(\Delta H\) is positive. This occurs when phase changes require breaking of bonds, such as in sublimation and evaporation. Energy input is necessary to overcome the attractive forces between molecules, thus absorbing heat.

On the contrary, in exothermic processes like combustion, \(\Delta H\) is negative since energy is released as heat. The formation of products requires less energy than is released, making it energetically favorable. Understanding whether \(\Delta H\) is positive or negative helps in predicting whether a reaction absorbs or releases energy.
Phase Changes
Phase changes refer to the transformation of a substance from one state of matter to another, such as solid to liquid, liquid to gas, or vice versa. Each phase change involves energy exchange and changes in entropy.

During phase changes like melting, vaporization, and sublimation, energy is absorbed to break bonds, leading to an increase in disorder. Therefore, processes like evaporation and sublimation are associated with positive entropy (\(\Delta S\) ) and enthalpy (\(\Delta H\) ) changes.

Conversely, when substances move from higher to lower energy states, such as freezing or condensation, the process releases energy while decreasing randomness, resulting in negative \(\Delta S\) and \(\Delta H\) . Phase transitions are crucial to understanding energy flow in matter.
Dissociation Reactions
Dissociation reactions occur when molecules break apart into smaller entities such as ions or atoms. This process is significant, especially in chemistry, for understanding reaction pathways and energy dynamics.

During dissociation, entropy (\(\Delta S\) ) increases as a single entity separates into multiple parts, leading to increased randomness or disorder in the system. This is why dissociating diatomic molecules into individual atoms results in a positive \(\Delta S\) .

The enthalpy (\(\Delta H\) ) during such reactions is positive, because breaking chemical bonds requires an input of energy. Recognizing how dissociation impacts \(\Delta H\) and \(\Delta S\) aids in predicting the spontaneity and energy requirements of chemical reactions.

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

The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8{ }^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=29.6 \mathrm{~kJ} /\) mol. (a) When \(\mathrm{Br}_{2}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(1.00 \mathrm{~mol}\) of \(\mathrm{Br}_{2}(l)\) is vaporized at \(58.8{ }^{\circ} \mathrm{C}\).

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a liquid hydrocarbon at room temperature that is the primary constituent of gasoline. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the controlled oxidation of methane: $$ \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. (b) How is \(\Delta G^{\circ}\) for the reaction expected to vary with increasing temperature? (c) Calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). Under standard conditions, is the reaction spontaneous at this temperature? (d) Is there a temperature at which the reaction would be at equilibrium under standard conditions and that is low enough so that the compounds involved are likely to be stable?

In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) or \(\mathrm{C}_{2} \mathrm{H}_{6}(g),(\mathbf{b}) \mathrm{CO}_{2}(g)\) or \(\mathrm{CO}(g) ?\) Explain.

The relationship between the temperature of a reaction, its standard enthalpy change, and the equilibrium constant at that temperature can be expressed as the following linear equation: $$ \ln K=\frac{-\Delta H^{\circ}}{R T}+\text { constant } $$ (a) Explain how this equation can be used to determine \(\Delta H^{\circ}\) experimentally from the equilibrium constants at several different temperatures. (b) Derive the preceding equation using relationships given in this chapter. To what is the constant equal?

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.

Sign-up for free