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

Which of the following statements regarding Gibb' energy change is correct? (a) If \(\Delta G\) is negative \((<0)\), the process is \(n_{0}\). spontaneous. (b) If \(\Delta G\) is positive \((>0)\), the process is spontancous. (c) If \(\Delta G\) is negative \((<0)\), the process is spontaneous. (d) If \(\Delta G\) is positive \((>0)\), the process is in equilibrium.

Short Answer

Expert verified
The correct statement regarding Gibbs' energy change is (c) If \(\Delta G\) is negative \(\(<0)\), the process is spontaneous.

Step by step solution

01

Understanding Gibbs Free Energy

Gibbs Free Energy, represented by \(\Delta G\), is a thermodynamic property used to predict the direction of chemical reactions. A negative \(\Delta G\) indicates a spontaneous process, while a positive \(\Delta G\) indicates a non-spontaneous process. Equilibrium is characterized by \(\Delta G = 0\).
02

Analyzing Statement (a)

Statement (a) says that if \(\Delta G < 0\), the process is non-spontaneous. This is incorrect because a negative \(\Delta G\) indicates the process is actually spontaneous.
03

Analyzing Statement (b)

Statement (b) claims that if \(\Delta G > 0\), the process is spontaneous. This is the opposite of the correct relationship; a positive \(\Delta G\) means the process is not spontaneous.
04

Analyzing Statement (c)

Statement (c) correctly states that if \(\Delta G < 0\), the process is spontaneous. This is in line with the definition of Gibbs Free Energy.
05

Analyzing Statement (d)

Statement (d) suggests that if \(\Delta G > 0\), the process is in equilibrium. This is incorrect since a non-zero \(\Delta G\) indicates a non-equilibrium state. Equilibrium occurs at \(\Delta G = 0\).

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.

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In essence, it describes how thermal energy is converted to and from other forms of energy, and how it affects matter. The key concept in thermodynamics related to our discussion is the Second Law, which states that in any closed system, the entropy of the system will either remain the same or increase over time. Entropy can be thought of as the measure of disorder or randomness in a system.

In the context of Gibbs Free Energy, which we denote as \(\Delta G\), thermodynamics helps us predict the direction of chemical reactions and the spontaneity of processes. The value of \(\Delta G\) is derived from the enthalpy (heat content) of the system and its entropy, at a particular temperature. Gibbs Free Energy ties directly into the Second Law of Thermodynamics by considering the energy and entropy changes of a system to determine process spontaneity.

To make this concept more digestible, consider a simple analogy: If you have a room full of air molecules at one end and you remove the barrier, the molecules will spontaneously spread out to fill the entire room, increasing the entropy. Similarly, a chemical reaction that leads to an increase in disorder (entropy) and releases energy (enthalpy) will likely be spontaneous.
Spontaneous Process
A spontaneous process in thermodynamics is one that occurs without needing to be driven by an external force or energy. This doesn't necessarily mean that the process happens quickly; rather, it means that the process is energetically favorable and can proceed on its own given enough time.

In terms of Gibbs Free Energy, a spontaneous process is characterized by a negative \(\Delta G\), meaning the system loses free energy. It’s a sign that the system is moving towards a state of equilibrium or a lower energy state.

Think of a ball rolling down a hill; once it starts rolling, it will continue to do so without additional pushing, because this is the energetically favorable direction. This parallels how chemical reactions move in the direction where the free energy decreases, which is also toward stability. Therefore, statement (c) from the original exercise correctly identifies a negative \(\Delta G\) as indicative of a spontaneous reaction, while the misconception in statement (a) can lead to confusion. Simplifying complex concepts like this is key to improving student understanding.
Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction proceeds in both the forward and reverse directions at the same rate. At this point, the concentrations of reactants and products remain constant over time, though they are not necessarily equal. This balanced state is a dynamic one—not static, as commonly mistaken—since reactions continue to occur but with no net change in concentration.

At equilibrium, the Gibbs Free Energy change, \(\Delta G\), is zero. This signifies that there is no further net release or absorption of free energy; the system has reached its maximum amount of entropy without an input of external energy. This is an important distinction: a positive or negative \(\Delta G\) means the system has not yet reached equilibrium and that there is an inherent 'drive' towards equilibrium. Therefore, statement (d) in the exercise is incorrect.

Understanding equilibrium is crucial for grasping many chemical processes, including biochemical interactions in the body, reactions in industrial settings, and balancing environmental systems. To reinforce this learning, an exercise improvement could be to have students predict the direction of a reaction when given \(\Delta G\) values and explain in their own words how equilibrium relates to the spontaneity of a process.

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

At \(373 \mathrm{~K}\), steam and water are in equilibrium and \(\Delta H=40.98 \mathrm{~kJ} \mathrm{~mol}^{-1}\). What will be \(\Delta S\) for conversion of water into steam? \(\mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow \mathrm{H}_{2} \mathrm{O}_{(g)}\) (a) \(109.8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(31^{-1} \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(21.93 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (d) \(326 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

When the system does not exchange heat with the surroundings, the process is (a) isothermal (b) adiabatic (c) thermal (d) isochoric.

For a reaction, \(\mathrm{CaCO}_{3(\mathrm{~s})} \rightarrow \mathrm{CaO}_{(0)}+\mathrm{CO}_{2(g)}\) \(\Delta_{j} H^{\circ}(\mathrm{CaO})=-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta_{f} H^{\circ}\left(\mathrm{CO}_{2}\right)=-393.5 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}\) and \(\Delta_{f} H^{\circ}\left(\mathrm{CaCO}_{3}\right)=-1206.9 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}\) Which of the following is a correct statement? (a) A large amount of heat is evolved during the decomposition of \(\mathrm{CaCO}_{3}\). (b) Decomposition of \(\mathrm{CaCO}_{3}\) is an endothermic process and heat is provided for decomposition. (c) The amount of heat evolved cannot be calculated from the data provided. (d) \(\Delta_{r} H^{\circ}=\Sigma \Delta_{f} H^{\circ}\) (reactants) \(-\Sigma \Delta_{f} H^{\circ}\) (products)

A system absorbs \(50 \mathrm{~kJ}\) heat and does \(20 \mathrm{~kJ}\) of work. What is the net change in the internal energy of the system? (a) Increase by \(30 \mathrm{~kJ}\) (b) Decrease by \(30 \mathrm{~kJ}\) (c) Increase by \(70 \mathrm{~kJ}\) (d) Decrease by \(70 \mathrm{~kJ}\)

What is the entropy change when 1 mole oxygen gas expands isothermally and reversibly from an initial volume of \(10 \mathrm{~L}\) to \(100 \mathrm{~L}\) at \(300 \mathrm{~K} ?\) (a) \(19.14 \mathrm{JK}^{-1}\) (b) \(109.12] \mathrm{K}^{-1}\) (c) \(29.12 \mathrm{JK}^{-1}\) (d) \(10 \mathrm{~K}^{-1}\)

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