Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198727873
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
bartleby

Videos

Question
Book Icon
Chapter 2, Problem 2.10PR

(a)

Interpretation Introduction

Interpretation:

An expression for the difference between the change in enthalpy and the change in internal energy for a gas phase process has to be determined assuming all the species behave as perfect gas.

Concept introduction:

Kirchhoff’s law:

This law states that the variation of change of enthalpy of a reaction with temperature at constant pressure is equal to the change in specific heat capacity at constant temperature of the system.

Internal energy:

Internal energy of a system is the total energy contained in the system.  It keeps an account for the loss and gain of energy of the system due to changes in internal state.  It is dependent on temperature and pressure.

It is denoted as U.

From 1st law of thermodynamics,

  ΔU=W+Q

Where,

ΔU= Change in the internal energy of system.

W= Energy transferred as the form of work to system

Q= Energy transferred as the form of heat to system

Enthalpy:

Enthalpy is a property of a thermodynamic system that is equal to the sum of the internal energy of the system and the product of pressure and volume.  For a closed system where transfer of matter between system and surroundings is prohibited, for the processes that occur at constant pressure, the heat absorbed or released equals to the change in enthalpy.

It is denoted as H. The molar enthalpy is defined as the enthalpy per mole. It is denoted as Hm.

From thermodynamics,

  H=U+PV

Where,

H= enthalpy of system

U= internal energy of system

P= pressure of system

V= volume of system

Specific heat capacity at constant pressure:

Specific heat capacity at constant pressure can be defined as the amount of energy required to increase the temperature of a substance by 1οC.

It is denoted as Cp. The molar specific heat capacity is defined as the specific heat capacity per mole and it is denoted by Cp,m.

  Cp=(HT)p

Heat capacity at constant volume:

Specific heat capacity at constant volume is defined as the amount of heat required to increase the temperature by 1οC keeping the system at constant volume.

It is denoted as Cv

  Cv=(UT)v

U= internal energy of system

T= temperature

(a)

Expert Solution
Check Mark

Explanation of Solution

Given that for perfect gas Cp=52R.

Applying the Kirchhoff’s law,

d(ΔH)P=ΔCPdT

d(ΔU)v=ΔCvdT

Thus by integration of both,

  ΔH=ΔCpTΔU=ΔCvT

Thus the difference is,

  ΔH-ΔU=(ΔCp-ΔCv)TΔH-ΔU=(52-32)RT[Cp-Cv=R]ΔH-ΔU=RT

The difference between the change in enthalpy and the change in internal energy for a gas phase process is RT.

(b)

Interpretation Introduction

Interpretation:

For ionization ΔHion-ΔUion=52RT that has to be established.

Concept introduction:

Ionization enthalpy:

Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion

Ionization energy:

Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound valence electron of an isolated neutral gas molecule.

Kirchhoff’s law:

This law states that the variation of change of enthalpy of a reaction with temperature at constant pressure is equal to the change in specific heat capacity at constant temperature of the system.

(b)

Expert Solution
Check Mark

Explanation of Solution

According to Kirchhoff’s law,

d(ΔH)P=ΔCPdT

Applying integration,

H(0)H(T)(ΔH)P=ΔCP0TdTΔH(T)-ΔH(0)=ΔCPTΔHion-ΔUion=52RT[ΔCP=52RΔUion=ionisationenergy]

Thus it has been established that, ΔHion-ΔUion=52RT.

(c)

Interpretation Introduction

Interpretation:

Difference between the standard enthalpy of ionization of Ca(g) to Ca2+(g) and the accompanying change in internal energy at 25οC has to be calculated.

Concept introduction:

Ionization enthalpy:

Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion

(c)

Expert Solution
Check Mark

Answer to Problem 2.10PR

The difference between the standard enthalpy of ionization of Ca(g) to Ca2+(g) and the accompanying change in internal energy at 25οC is 6193.93Jmol1.

Explanation of Solution

From part (b) it can be concluded that,

ΔHion-ΔUion=52RT

According to the question, difference between the standard enthalpy of ionization of Ca(g) to Ca2+(g) and the accompanying change in internal energy at 25οC has to be calculated.

Given that,

  T=(25+273)KT=298K

Hence,

  ΔHion-ΔUion=52RTΔHion-ΔUion=52×8.314Jmol1K1×298KΔHion-ΔUion=6193.93Jmol1

Hence the difference between the standard enthalpy of ionization of Ca(g) to Ca2+(g) and the accompanying change in internal energy at 25οC is 6193.93Jmol1.

(d)

Interpretation Introduction

Interpretation:

For electron gain ΔHeg-ΔUeg=52RT has to be established.

Concept introduction:

Electron gain enthalpy:

Electron gain enthalpy can be defined as the amount of energy released when an electron is added to an isolated gaseous atom for electron gain enthalpy the sign convention is negative.

Electron affinity:

Electron affinity can be defined as the change in energy of a neutral atom in the gaseous phase when an electron is added to the atom to form the negative ion.

Kirchhoff’s law:

This law states that the variation of change of enthalpy of a reaction with temperature at constant pressure is equal to the change in specific heat capacity at constant temperature of the system.

(d)

Expert Solution
Check Mark

Explanation of Solution

According to Kirchhoff’s law,

d(ΔH)P=ΔCPdT

Applying integration,

  H(0)H(T)(ΔH)P=ΔCP0TdTΔH(T)-ΔH(0)=ΔCPTΔHeg-ΔUeg=52RT[ΔCP=52R due to electron gainΔUeg=electronaffinity]

Thus it has been established that, ΔHeg-ΔUeg=52RT.

(e)

Interpretation Introduction

Interpretation:

Difference between the standard electron gain enthalpy of Br(g) and the corresponding change in the internal energy at 25οC has to be calculated.

Concept introduction:

Electron gain enthalpy:

Electron gain enthalpy can be defined as the amount of energy released when an electron is added to an isolated gaseous atom for electron gain enthalpy the sign convention is negative.

(e)

Expert Solution
Check Mark

Answer to Problem 2.10PR

The difference between the standard electron gain enthalpy of Br(g) and the corresponding change in the internal energy at 25οC is 6193.93Jmol-1.

Explanation of Solution

From the part (d) it can be concluded that,

  ΔHeg-ΔUeg=52RT

Now according to the question the difference between the standard electron gain enthalpy of Br(g) and the corresponding change in the internal energy at 25οC has to be calculated.

Given that,

  T=(25+273)KT=298K

Hence,

  ΔHeg-ΔUeg=52RTΔHeg-ΔUeg=52×8.314Jmol-1K-1×298KΔHeg-ΔUeg=6193.93Jmol-1

Hence the difference between the standard electron gain enthalpy of Br(g) and the corresponding change in the internal energy at 25οC is 6193.93Jmol-1.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 2 Solutions

Elements Of Physical Chemistry

Ch. 2 - Prob. 2D.2STCh. 2 - Prob. 2E.1STCh. 2 - Prob. 2E.2STCh. 2 - Prob. 2E.3STCh. 2 - Prob. 2F.1STCh. 2 - Prob. 2F.2STCh. 2 - Prob. 2F.3STCh. 2 - Prob. 2F.4STCh. 2 - Prob. 2F.5STCh. 2 - Prob. 2F.6STCh. 2 - Prob. 2A.2ECh. 2 - Prob. 2A.3ECh. 2 - Prob. 2A.4ECh. 2 - Prob. 2A.5ECh. 2 - Prob. 2A.6ECh. 2 - Prob. 2A.7ECh. 2 - Prob. 2A.8ECh. 2 - Prob. 2B.1ECh. 2 - Prob. 2B.2ECh. 2 - Prob. 2B.3ECh. 2 - Prob. 2B.4ECh. 2 - Prob. 2B.5ECh. 2 - Prob. 2C.1ECh. 2 - Prob. 2C.2ECh. 2 - Prob. 2D.1ECh. 2 - Prob. 2D.2ECh. 2 - Prob. 2D.3ECh. 2 - Prob. 2D.4ECh. 2 - Prob. 2D.5ECh. 2 - Prob. 2D.6ECh. 2 - Prob. 2E.1ECh. 2 - Prob. 2E.2ECh. 2 - Prob. 2E.3ECh. 2 - Prob. 2E.4ECh. 2 - Prob. 2E.5ECh. 2 - Prob. 2E.6ECh. 2 - Prob. 2E.7ECh. 2 - Prob. 2E.8ECh. 2 - Prob. 2E.9ECh. 2 - Prob. 2F.1ECh. 2 - Prob. 2F.2ECh. 2 - Prob. 2F.3ECh. 2 - Prob. 2F.4ECh. 2 - Prob. 2F.5ECh. 2 - Prob. 2F.6ECh. 2 - Prob. 2F.7ECh. 2 - Prob. 2F.8ECh. 2 - Prob. 2F.9ECh. 2 - Prob. 2F.10ECh. 2 - Prob. 2.1DQCh. 2 - Prob. 2.2DQCh. 2 - Prob. 2.3DQCh. 2 - Prob. 2.4DQCh. 2 - Prob. 2.5DQCh. 2 - Prob. 2.6DQCh. 2 - Prob. 2.7DQCh. 2 - Prob. 2.8DQCh. 2 - Prob. 2.9DQCh. 2 - Prob. 2.10DQCh. 2 - Prob. 2.11DQCh. 2 - Prob. 2.12DQCh. 2 - Prob. 2.13DQCh. 2 - Prob. 2.14DQCh. 2 - Prob. 2.15DQCh. 2 - Prob. 2.16DQCh. 2 - Prob. 2.1PCh. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - Prob. 2.4PCh. 2 - Prob. 2.5PCh. 2 - Prob. 2.6PCh. 2 - Prob. 2.7PCh. 2 - Prob. 2.8PCh. 2 - Prob. 2.9PCh. 2 - Prob. 2.10PCh. 2 - Prob. 2.12PCh. 2 - Prob. 2.13PCh. 2 - Prob. 2.14PCh. 2 - Prob. 2.15PCh. 2 - Prob. 2.16PCh. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Prob. 2.19PCh. 2 - Prob. 2.20PCh. 2 - Prob. 2.21PCh. 2 - Prob. 2.22PCh. 2 - Prob. 2.23PCh. 2 - Prob. 2.25PCh. 2 - Prob. 2.1PRCh. 2 - Prob. 2.2PRCh. 2 - Prob. 2.3PRCh. 2 - Prob. 2.4PRCh. 2 - Prob. 2.5PRCh. 2 - Prob. 2.6PRCh. 2 - Prob. 2.8PRCh. 2 - Prob. 2.9PRCh. 2 - Prob. 2.10PR
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
The Laws of Thermodynamics, Entropy, and Gibbs Free Energy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=8N1BxHgsoOw;License: Standard YouTube License, CC-BY