Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
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Chapter 2, Problem 2.23P

(a)

Interpretation Introduction

Interpretation:

Standard enthalpies of the given reaction whether increases (or) decreases with increasing temperature should be decided.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system Ηsys) can be calculated by the following equation.

ΔHrxn = ΔH°produdcts- ΔH°reactants

Where,

ΔH°reactants is the standard entropy of the reactants

ΔH°produdcts is the standard entropy of the products

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

  ΔGo=ΔHo-TΔSowhere,ΔGo-standardfreeenergy changeΔHo-standardenthalpychangeΔSo-standardentropy changeand T-temperature.

(a)

Expert Solution
Check Mark

Explanation of Solution

According to second law of thermodynamic changes in enthalpy

  ΔH=ΔU+ΔnRT[1]

Where

  (Δn)sum of stoichiometric coefficient of productssum of stoichiometric coefficient ofReactants

Given enthalpy reaction is,

  2H2(g)+O2(g)2H2O(g)

Plugging in (Δn) in equation (1),

  Δn=(2)(2+1)=1

Therefore, the ΔH=ΔU+ΔnRT as temperature increase and (ΔH) value is decreases

(b)

Interpretation Introduction

Interpretation:

Standard enthalpies of the given reaction whether increases (or) decreases with increasing temperature should be decided.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system Ηsys) can be calculated by the following equation.

ΔHrxn = ΔH°produdcts- ΔH°reactants

Where,

ΔH°reactants is the standard entropy of the reactants

ΔH°produdcts is the standard entropy of the products

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

  ΔGo=ΔHo-TΔSowhere,ΔGo-standardfreeenergy changeΔHo-standardenthalpychangeΔSo-standardentropy changeand T-temperature.

(b)

Expert Solution
Check Mark

Explanation of Solution

Second law of thermodynamic changes in enthalpy

  ΔH=ΔU+ΔnRT[1]

Where

  (Δn)sum of stoichiometric coefficient of productssum of stoichiometric coefficient ofReactants

Given enthalpy reaction is,

  CH4(g)+2O2(g)CO2(g)+2H2O(g)

Plugging in (Δn) in equation (1),

  Δn=(2+1)(2+1)=0

Hence, the internal energy (ΔU) depended only on temperature; it is increase as temperature increase and same for (ΔH).

As temperature increase and (ΔH) value is Increases

(c)

Interpretation Introduction

Interpretation:

Standard enthalpies of the given reaction whether increases (or) decreases with increasing temperature should be decided.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system Ηsys) can be calculated by the following equation.

ΔHrxn = ΔH°produdcts- ΔH°reactants

Where,

ΔH°reactants is the standard entropy of the reactants

ΔH°produdcts is the standard entropy of the products

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

ΔGo=ΔHo-TΔSowhere,ΔGo-standardfreeenergy changeΔHo-standardenthalpychangeΔSo-standardentropy changeand T-temperature.

(c)

Expert Solution
Check Mark

Explanation of Solution

Second law of thermodynamic changes in enthalpy

  ΔH=ΔU+ΔnRT[1]

Where

  (Δn)sum of stoichiometric coefficient of productssum of stoichiometric coefficient ofReactants

Given enthalpy reaction is,

  N2(g)+3H2(g)2NH3(g)

Put in (Δn) in equation (1),

  Δn=(2)(3+1)=2

Therefore, the ΔH=ΔU+ΔnRT as temperature increase and (ΔH) value is decreases

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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
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