/*! 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 55 For a certain chemical reaction,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For a certain chemical reaction, \(\Delta H^{\circ}=-40.0 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-150.0 \mathrm{~J} / \mathrm{K} .(\mathbf{a})\) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Short Answer

Expert verified
(a) Decrease in system disorder. (b) Increase in surroundings' disorder. (c) \( \Delta G^{\circ} = 4.7 \, \mathrm{kJ} \). (d) Reaction is not spontaneous.

Step by step solution

01

Determine Change in System's Disorder

To evaluate the change in the disorder of the system, we examine the sign of \( \Delta S^{\circ} \). Here, \( \Delta S^{\circ} = -150.0 \, \text{J/K} \), meaning the system's disorder decreases.
02

Assess Surroundings’ Disorder

The change in the surroundings’ disorder is determined by the sign of \( \Delta H^{\circ} \). Here, \( \Delta H^{\circ} = -40.0 \, \text{kJ} \), which implies that the heat is released to the surroundings, thus increasing the surroundings' disorder.
03

Calculate \( \Delta G^{\circ} \) for the Reaction

Use the equation \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). Convert \( \Delta S^{\circ} \) to kJ by dividing by 1000: \( \Delta S^{\circ} = -0.150 \, \text{kJ/K} \). Substituting the values: \[ \Delta G^{\circ} = -40.0 \, \text{kJ} - 298 \, \text{K} \times (-0.150 \, \text{kJ/K}) = -40.0 \, \text{kJ} + 44.7 \, \text{kJ} =4.7 \, \text{kJ} \].
04

Determine the Spontaneity of the Reaction

A reaction is spontaneous if \( \Delta G^{\circ} \) is negative. Here, \( \Delta G^{\circ} = 4.7 \, \text{kJ} \), which is positive, indicating the reaction is not spontaneous at 298 K under standard conditions.

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.

Standard Entropy Change
Entropy is a fundamental concept in chemistry, helping to describe the degree of disorder or randomness in a system. The standard entropy change, denoted as \( \Delta S^{\circ} \), tells us how the disorder of a system changes during a reaction. In the given exercise, we have \( \Delta S^{\circ} = -150.0 \, \text{J/K} \). The negative sign indicates a decrease in entropy, meaning the reaction results in a more ordered system. When molecules in a reaction become more structured or align in an ordered fashion, \( \Delta S^{\circ} \) becomes negative.

However, it is also important to consider the entropy change of the surroundings. Despite the system's decrease in entropy, the overall process, especially exothermic reactions which release heat (as observed with a negative \( \Delta H^{\circ} \)), can increase the disorder, or entropy, in the surroundings. Thus, even if entropy decreases within the system, the total change involving both system and surroundings can reveal much about how energy dispersion affects the universe as a whole.
Standard Enthalpy Change
Enthalpy change, \( \Delta H^{\circ} \), reflects the heat exchange between a system and its surroundings at constant pressure. In the problem, \( \Delta H^{\circ} = -40.0 \, \text{kJ} \), suggesting an exothermic reaction. An exothermic process releases heat, thus increasing the entropy of the surroundings, even as the system itself may become more ordered from the perspective of standard entropy change.

The release of energy to the surroundings during an exothermic reaction has consequences beyond simple heat flow. It introduces energy into the surroundings, increasing molecular motion and effectively enhancing the randomness of the particles in the surroundings. This is why an exothermic reaction, like the one described, can run counter to the system's decrease in entropy, further playing into the great balancing act that thermodynamics describes.
Spontaneity of Reactions
Spontaneity in chemical reactions is determined by the sign of the Gibbs Free Energy change, \( \Delta G^{\circ} \). Calculated via \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \), it considers both the enthalpy and entropy components of a reaction. In our problem, substituting the given values, \( \Delta G^{\circ} = 4.7 \, \text{kJ} \), is positive.

A positive \( \Delta G^{\circ} \) indicates non-spontaneity under standard conditions. Essentially, it means the reaction as described is unlikely to proceed on its own at 298 K. The interplay between energy release and storage (enthalpy and entropy) informs us whether the universe 'wants' the reaction to occur without external influence. Spontaneity is not just about energy favorability but also how energy spreads or becomes more chaotic on a whole, reaffirming the delicate balance in thermodynamics.

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

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>1\) at \(25^{\circ} \mathrm{C} .(\mathbf{c})\) In each case, indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{FeO}(s)\) (ii) \(\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}(g)\) (iii) \(\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\) (iv) \(\mathrm{CO}_{2}(g)+\mathrm{CaO}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)\)

Isomers are moleculesthat have the samechemical formula but different arrangements of atoms, as shown here for two isomers of pentane, \(\mathrm{C}_{5} \mathrm{H}_{12} .\) (a) Do you expect a significant difference in the enthalpy of combustion of the two isomers? Explain. (b) Which isomer do you expect to have the higher standard molar entropy? Explain. \([\) Section 19.4\(]\)

A standard air conditioner involves a refrigerant that is typically now a fluorinated hydrocarbon, such as \(\mathrm{CH}_{2} \mathrm{~F}_{2}\). An air- conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram). During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (a) What is the sign of \(q\) for the expansion? (b) What is the sign of \(q\) for the compression? (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why? (d) Imagine that a sample of liquid refrigerant undergoes expansion followed by compression, so that it is back to its original state. Would you expect that to be a reversible process? (e) Suppose that a house and its exterior are both initially at \(31^{\circ} \mathrm{C}\). Some time after the air conditioner is turned on, the house is cooled to \(24^{\circ} \mathrm{C}\). Is this process spontaneous or nonspontaneous?

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(\mathrm{CO}(g)+\mathrm{H}_{2}(g) \longrightarrow C(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{O}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{HCl}(g)+\mathrm{NH}_{3}(g)\) (d) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g)\)

Does the entropy of the system increase, decrease, or stay the same when (a) a solid melts, (b) a gas liquefies, (c) a solid sublimes?

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.