/*! 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 79 Diamond and graphite are two for... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 79

Diamond and graphite are two forms of elemental carbon. Under the appropriate conditions they will be in equilibrium with each other: \(C_{\text {diamond }} \rightleftarrows C_{\text {graphite }}\) If graphite is subjected to very high pressure and temperature, it will convert into the diamond form. (a) Is the above equilibrium reaction exothermic or endothermic? Explain how you know. (b) Which form, graphite or diamond, has the higher density? (Hint: Think about what increasing the pressure of a gas does to its density. It works the same for the solid and liquid phases as well.)

Short Answer

Expert verified
(a) The equilibrium reaction between diamond and graphite is endothermic. This can be deduced from the fact that at high temperature and pressure, graphite converts to diamond, indicating that heat is absorbed in the process. (b) Diamond has a higher density than graphite. When pressure is increased, the density of substances generally increases, and since graphite converts to diamond under high pressure, it implies that diamond is denser than graphite.

Step by step solution

01

(a) Determine if the reaction is exothermic or endothermic

To determine if the reaction is exothermic or endothermic, we need to analyze the conditions under which the phase transition occurs. The problem states that at very high pressure and temperature, graphite turns into diamond. Recall that an exothermic reaction is one where heat is released and an endothermic reaction is where heat is absorbed. According to Le Ch芒telier's principle, a system in equilibrium will shift in such a way as to counteract a change in conditions applied to the system. In this case, the equilibrium undergoes an increase in both pressure and temperature. If the reaction were exothermic, then increasing the temperature would shift the equilibrium towards the reactant, graphite, as the system tries to absorb the extra heat. However, we are given that graphite converts to diamond under these conditions, which implies that the reaction is endothermic. Thus, at high temperature and pressure, the equilibrium shifts towards the product, diamond, as the system absorbs the extra heat.
02

(b) Determine which form has higher density

To determine which form of carbon has the higher density, we can use the hint provided. We are told to think about what increasing the pressure of a gas does to its density and apply the same principle to solid and liquid phases. Recall that when the pressure of a gas increases, its volume generally decreases, leading to an increase in density. We are given that at high pressure, graphite converts to diamond. This means that increasing the pressure of graphite makes it denser, resulting in the diamond form. Therefore, diamond has a higher density than graphite.

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 91影视!

Key Concepts

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

Exothermic and Endothermic Reactions
Understanding whether a reaction is exothermic or endothermic is crucial when studying chemical equilibrium. To put it simply, an exothermic reaction releases heat to its surroundings, often causing the temperature of the environment to rise. An everyday example would be a hand warmer packet. On the flip side, an endothermic reaction absorbs heat, which can result in a cooling effect, similar to what happens when certain salts dissolve in water.

Applying this knowledge to the conversion of graphite to diamond, we know high pressure and temperature are required. For the reaction to proceed under these conditions, it suggests that the transformation of graphite into diamond consumes heat and is, therefore, endothermic. The consumption of heat at high temperatures helps shift the reaction towards diamond, explaining why diamond forms under such extreme conditions.
Le Ch芒telier's Principle
Imagine sitting comfortably on a seesaw in balance, and suddenly someone adds a weight to one end. Naturally, you'd try to regain your balance. This is, in essence, what Le Ch芒telier's principle implies for chemical reactions in equilibrium. When a change is introduced to a system at equilibrium, the system will adjust to counteract that change and reestablish a new balance.

When we apply this concept to the diamond and graphite equilibrium, increasing temperature or pressure is akin to adding weight to the seesaw. Le Ch芒telier's principle leads us to predict that the reaction will shift in a direction to absorb the additional heat (in the case of temperature) or to reduce volume (in the case of pressure). Therefore, the formation of diamond from graphite under high pressure and temperature can be considered a direct consequence of Le Ch芒telier's principle at work.
Density of Carbon Forms
Density, defined as mass per unit volume, is a fundamental property used to characterize substances. Carbon, being an element, can exist in different structures or allotropes, notably diamond and graphite, which have different densities because of how the carbon atoms are arranged in the lattice.

In the case of diamond and graphite, diamond's structure is highly compact and three-dimensional, leading to its greater density. This is consistent with the notion that increasing pressure on graphite, a less dense allotrope, transforms it into the more densely packed form of diamond. Thus, understanding the relationship between pressure and density helps us ascertain why diamond is the denser form of carbon.
Phase Transition

Carbon Allotropes and Phase Transition

A phase transition refers to the change of a substance from one state of matter to another, such as solid to liquid, or, in the case of carbon, from one solid form (graphite) to another solid form (diamond). These transitions often involve changes in energy, structure, and other physical properties.

Transitioning from graphite to diamond is not the typical melting or freezing process we might be more familiar with; instead, it is a transition between two solid phases. It is indicative of the incredible versatility and diversity of carbon's bonding capabilities. The phase transition between graphite and diamond is a result of rearranging carbon atoms into a different crystalline structure, a process that is entirely dependent on environmental conditions such as pressure and temperature.

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

Consider the gas-phase reaction \(3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{O}_{3}(g) .\) Suppose \(K_{\mathrm{eq}}\) for this reaction is \(\sim 1\) (it is not, but assume it is for this problem). Suppose you want pure ozone \(\left(\mathrm{O}_{3}\right)\) that is uncontaminated with oxygen \(\left(\mathrm{O}_{2}\right)\). (a) Why can't you simply remove the oxygen from the reaction vessel once the reaction has come to equilibrium to obtain pure ozone? (b) In fact, \(K_{\text {eq }}\) for this reaction at room temperature is \(2.5 \times 10^{-29}\). Knowing this, how important would you say Le Ch芒telier's principle is for this reaction when it comes to influencing the amount of ozone present at equilibrium? Explain.

The equilibrium concentrations for the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) at \(2000^{\circ} \mathrm{C}\) are \(\left[\mathrm{N}_{2}\right]=\) \(0.25 \mathrm{M} ;\left[\mathrm{O}_{2}\right]=1.2 \mathrm{M} ;[\mathrm{NO}]=0.011 \mathrm{M} .\) What is the value of \(K_{\mathrm{eq}}\) for this reaction?

What is "dynamic" about the equilibrium that is established when a sparingly soluble salt is added to water?

What does a value of \(K_{\text {eq }}\) less than \(10^{-3}\) imply? Prove that your answer is correct by using the general expression \(K_{\text {eq }}=\) [Products \(] /[\) Reactants \(]\).

The solubility of iron(II) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{2}\), is \(1.43 \times 10^{-3} \mathrm{~g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) (a) Write a balanced equation for the solubility equilibrium. (b) Write the expression for \(K_{\mathrm{sp}}\) and calculate its value. (c) \(\mathrm{A} 60.0 \mathrm{~mL}\) sample of \(3.00 \times 10^{-3} \mathrm{M} \mathrm{FeSO}_{4}\) solution is added to \(60.0 \mathrm{~mL}\) of \(4.00 \times 10^{-6} \mathrm{M} \mathrm{NaOH}\) solution. Does a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{2}\) form? (Hint: Use the concentration values for \(\mathrm{Fe}^{2+}\) and \(\mathrm{OH}^{-}\) and plug them into the \(K_{\mathrm{sp}}\) expression. If the value you get is larger than \(K_{\mathrm{sp}}\) ' precipitation will occur.)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

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.