   # The formation of diamond from graphite is a process of considerable importance. (a) Using data in Appendix L, calculate Δ r S °, Δ r H °, and Δ r G ° for this process at 25 °C. (b) The calculations will suggest that this process is not possible at any temperature. However, the synthesis of diamonds by this reaction is a commercial process. How can this contradiction be rationalized? ( Note: In the industrial synthesis, high pressure and high temperatures are used.) ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18, Problem 80SCQ
Textbook Problem
1 views

## The formation of diamond from graphite is a process of considerable importance. (a) Using data in Appendix L, calculate ΔrS°, ΔrH°, and ΔrG° for this process at 25 °C. (b) The calculations will suggest that this process is not possible at any temperature. However, the synthesis of diamonds by this reaction is a commercial process. How can this contradiction be rationalized? (Note: In the industrial synthesis, high pressure and high temperatures are used.)

(a)

Interpretation Introduction

Interpretation:

The value of ΔrH°, ΔrS° and ΔrGo for the formation of diamond at given temperature at given temperature.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It is related to entropy and entropy by the following expression,

ΔrGorHo-TΔrSo

Here, ΔrH° is the change in enthalpy and ΔrS° is the change in entropy.

### Explanation of Solution

The value of ΔrH°, ΔrS° and ΔrGo for formation of diamond is calculated below.

Given:

Refer to Appendix L for the values of standard entropies and enthalpies.

The standard entropy value for C(graphite) is 5.6 J/Kmol.

The standard entropy value for C(diamond) is 2.377 J/Kmol

The standard enthalpy value for C(graphite) is 0 kJ/mol.

The standard enthalpy value for C(diamond) is 1.8 kJ/mol.

The given reaction is,

C(graphite)C(diamond)

ΔrH°=fH°(products)fH°(reactants)[(1 mol C(diamond)/mol-rxn)ΔfH°[C(diamond)]-(1 mol C(graphite)/mol-rxn)ΔfH°[C(graphite)]]

Substitute the values,

ΔrH°[(1 mol C(diamond)/mol-rxn)(1.8 kJ/mol)-(1 mol C(graphite)/mol-rxn)(0 kJ/mol)]= 1

(b)

Interpretation Introduction

Interpretation:

The contradiction in production of diamond from graphite with the calculations should be rationalized.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It is related to entropy and entropy by the following expression,

ΔrGorHo-TΔrSo

Here, ΔrH° is the change in enthalpy and ΔrS° is the change in entropy.

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
Cells a. are self-contained, living units. b. serve the bodys needs but have few needs of their own. c. remain ...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

Define the following terms: a. chromosome b. chromatin

Human Heredity: Principles and Issues (MindTap Course List)

What energy sources on Earth cannot be thought of as stored sunlight?

Horizons: Exploring the Universe (MindTap Course List)

A rope with mass mr is attached to a block with mass mb as in Figure P5.42. The block rests on a frictionless, ...

Physics for Scientists and Engineers, Technology Update (No access codes included)

The density of 18.0%HCl is 1.09g/mL. Calculate its molarity.

Introductory Chemistry: An Active Learning Approach

What is the difference between sand spits and bay mouth bars?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin 