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
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Chapter 5, Problem 5.1P

(a)

Interpretation Introduction

Interpretation:

For the given reaction, the standard Gibbs energy has to be calculated by calculating the standard reaction enthalpy and entropy.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

ΔrG0=ΔrH0-TΔrS0where, ΔrG0-standardreactionGibbsenergy, ΔrH0-standardreactionenthalpy, T-temperature, ΔrS0-standardreactionentropy.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given reaction,

HCl(g)+NH3(g)NH4Cl(s)

Calculate the value of ΔrH0 for the given reaction,

HCl(g)+NH3(g)NH4Cl(s)ΔrH0=productsvΔfH0-reactantsvΔfH0=[(-314.43)-((-92.31)+(294.1))]kJmol-1=-516.22kJmol-1

Calculate the value of ΔrS0 for the given reaction,

HCl(g)+NH3(g)NH4Cl(s)ΔrS0=productsvSm0-reactantsvSm0=[94.6-(186.91+238.97)]JK-1mol-1=-331.28JK-1mol-1

Calculate the reaction Gibbs energy for the given reaction,

ΔrG0=ΔrH0-TΔrS0=-516.22kJmol-1-(298K×-331.28JK-1mol-1)=-516.22kJmol-1+98.72kJmol-1=-417.5kJmol-1

Therefore, the standard Gibbs energy of the given reaction is -417.5kJmol-1.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, the standard Gibbs energy has to be calculated by calculating the standard reaction enthalpy and entropy.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

ΔrG0=ΔrH0-TΔrS0where, ΔrG0-standardreactionGibbsenergy, ΔrH0-standardreactionenthalpy, T-temperature, ΔrS0-standardreactionentropy.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given reaction,

2Al2O3(s)+3Si(s)3SiO2(s)+4Al(s)

Calculate the value of ΔrH0 for the given reaction,

2Al2O3(s)+3Si(s)3SiO2(s)+4Al(s)ΔrH0=productsvΔfH0-reactantsvΔfH0=[((3×-910.94)+(4×0))-((2×-1675.7)+(3×0))]kJmol-1=[-2732.82+3351.4]kJmol-1=618.58kJmol-1

Calculate the value of ΔrS0 for the given reaction,

2Al2O3(s)+3Si(s)3SiO2(s)+4Al(s)ΔrS0=productsvSm0-reactantsvSm0=[((3×41.84)+(4×28.33))-((2×50.92)+(3×18.83))]JK-1mol-1=[238.84-158.33]JK-1mol-1=80.51JK-1mol-1

Calculate the reaction Gibbs energy for the given reaction,

ΔrG0=ΔrH0-TΔrS0=-618.58kJmol-1-(298K×80.51JK-1mol-1)=-618.58kJmol-1-23.99kJmol-1=-642.57kJmol-1

Therefore, the standard Gibbs energy of the given reaction is -642.57kJmol-1.

(c)

Interpretation Introduction

Interpretation:

For the given reaction, the standard Gibbs energy has to be calculated by calculating the standard reaction enthalpy and entropy.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

ΔrG0=ΔrH0-TΔrS0where, ΔrG0-standardreactionGibbsenergy, ΔrH0-standardreactionenthalpy, T-temperature, ΔrS0-standardreactionentropy.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given reaction,

Fe(s)+H2S(g)FeS(s)+H2(g)

Calculate the value of ΔrH0 for the given reaction,

Fe(s)+H2S(g)FeS(s)+H2(g)ΔrH0=productsvΔfH0-reactantsvΔfH0=[(-100.0+0)-((0)+(-20.63))]kJmol-1=-70.37kJmol-1

Calculate the value of ΔrS0 for the given reaction,

Fe(s)+H2S(g)FeS(s)+H2(g)ΔrS0=productsvSm0-reactantsvSm0=[(60.29+130.684)-(27.28+205.79)]JK-1mol-1=-42.1JK-1mol-1

Calculate the reaction Gibbs energy for the given reaction,

ΔrG0=ΔrH0-TΔrS0=-70.37kJmol-1-(298K×-42.1JK-1mol-1)=-70.37kJmol-1+12.55kJmol-1=-57.82kJmol-1

Therefore, the standard Gibbs energy of the given reaction is -57.82kJmol-1.

(d)

Interpretation Introduction

Interpretation:

For the given reaction, the standard Gibbs energy has to be calculated by calculating the standard reaction enthalpy and entropy.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

ΔrG0=ΔrH0-TΔrS0where, ΔrG0-standardreactionGibbsenergy, ΔrH0-standardreactionenthalpy, T-temperature, ΔrS0-standardreactionentropy.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given reaction,

FeS2(s)+2H2(g)Fe(s)+2H2S(g)

Calculate the value of ΔrH0 for the given reaction,

FeS2(s)+2H2(g)Fe(s)+2H2S(g)ΔrH0=productsvΔfH0-reactantsvΔfH0=[(0+2×-20.63)-(-178.2+(2×0))]kJmol-1=136.94kJmol-1

Calculate the value of ΔrS0 for the given reaction,

FeS2(s)+2H2(g)Fe(s)+2H2S(g)ΔrS0=productsvSm0-reactantsvSm0=[(27.28+2×205.79)-(52.93+2×130.684)]JK-1mol-1=314.298JK-1mol-1

Calculate the reaction Gibbs energy for the given reaction,

ΔrG0=ΔrH0-TΔrS0=136.94kJmol-1-(298K×314.298JK-1mol-1)=136.94kJmol-1-93.66kJmol-1=43.28kJmol-1

Therefore, the standard Gibbs energy of the given reaction is 43.28kJmol-1.

(e)

Interpretation Introduction

Interpretation:

For the given reaction, the standard Gibbs energy has to be calculated by calculating the standard reaction enthalpy and entropy.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

ΔrG0=ΔrH0-TΔrS0where, ΔrG0-standardreactionGibbsenergy, ΔrH0-standardreactionenthalpy, T-temperature, ΔrS0-standardreactionentropy.

(e)

Expert Solution
Check Mark

Explanation of Solution

Given reaction,

2H2O2(l)+H2S(g)H2SO4(l)+2H2(g)

Calculate the value of ΔrH0 for the given reaction,

2H2O2(l)+H2S(g)H2SO4(l)+2H2(g)ΔrH0=productsvΔfH0-reactantsvΔfH0=[(-813.99+(2×0))-((2×-187.78)+-20.63)]kJmol-1=-417.8kJmol-1

Calculate the value of ΔrS0 for the given reaction,

2H2O2(l)+H2S(g)H2SO4(l)+2H2(g)ΔrS0=productsvSm0-reactantsvSm0=[(20.1+(2×130.684))-((2×109.6)+37.99)]JK-1mol-1=24.28JK-1mol-1

Calculate the reaction Gibbs energy for the given reaction,

ΔrG0=ΔrH0-TΔrS0=-417.8kJmol-1-(298K×24.28JK-1mol-1)=-417.8kJmol-1-7.24kJmol-1=-425.04kJmol-1

Therefore, the standard Gibbs energy of the given reaction is -425.04kJmol-1.

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Chapter 5 Solutions

Elements Of Physical Chemistry

Ch. 5 - Prob. 5D.1STCh. 5 - Prob. 5D.2STCh. 5 - Prob. 5D.3STCh. 5 - Prob. 5D.4STCh. 5 - Prob. 5D.5STCh. 5 - Prob. 5E.1STCh. 5 - Prob. 5E.2STCh. 5 - Prob. 5F.1STCh. 5 - Prob. 5F.2STCh. 5 - Prob. 5F.3STCh. 5 - Prob. 5F.4STCh. 5 - Prob. 5F.5STCh. 5 - Prob. 5G.1STCh. 5 - Prob. 5G.2STCh. 5 - Prob. 5G.3STCh. 5 - Prob. 5H.1STCh. 5 - Prob. 5H.2STCh. 5 - Prob. 5H.3STCh. 5 - Prob. 5I.1STCh. 5 - Prob. 5I.2STCh. 5 - Prob. 5I.3STCh. 5 - Prob. 5I.4STCh. 5 - Prob. 5I.5STCh. 5 - Prob. 5I.6STCh. 5 - Prob. 5J.1STCh. 5 - Prob. 5J.2STCh. 5 - Prob. 5J.3STCh. 5 - Prob. 5J.4STCh. 5 - Prob. 5J.5STCh. 5 - Prob. 5A.1ECh. 5 - Prob. 5A.2ECh. 5 - Prob. 5A.3ECh. 5 - Prob. 5A.4ECh. 5 - Prob. 5A.5ECh. 5 - Prob. 5A.6ECh. 5 - Prob. 5A.7ECh. 5 - Prob. 5A.8ECh. 5 - Prob. 5A.9ECh. 5 - Prob. 5A.10ECh. 5 - Prob. 5A.11ECh. 5 - Prob. 5A.12ECh. 5 - Prob. 5A.13ECh. 5 - Prob. 5B.1ECh. 5 - Prob. 5B.2ECh. 5 - Prob. 5B.3ECh. 5 - Prob. 5B.4ECh. 5 - Prob. 5B.5ECh. 5 - Prob. 5C.1ECh. 5 - Prob. 5C.2ECh. 5 - Prob. 5C.3ECh. 5 - Prob. 5C.4ECh. 5 - Prob. 5D.1ECh. 5 - Prob. 5D.2ECh. 5 - Prob. 5D.3ECh. 5 - Prob. 5D.4ECh. 5 - Prob. 5E.1ECh. 5 - Prob. 5E.2ECh. 5 - Prob. 5F.4ECh. 5 - Prob. 5G.1ECh. 5 - Prob. 5G.2ECh. 5 - Prob. 5G.3ECh. 5 - Prob. 5H.1ECh. 5 - Prob. 5H.2ECh. 5 - Prob. 5H.3ECh. 5 - Prob. 5H.4ECh. 5 - Prob. 5I.1ECh. 5 - Prob. 5I.2ECh. 5 - Prob. 5I.3ECh. 5 - Prob. 5I.4ECh. 5 - Prob. 5J.1ECh. 5 - Prob. 5J.2ECh. 5 - Prob. 5J.3ECh. 5 - Prob. 5J.4ECh. 5 - Prob. 5.1DQCh. 5 - Prob. 5.2DQCh. 5 - Prob. 5.3DQCh. 5 - Prob. 5.4DQCh. 5 - Prob. 5.5DQCh. 5 - Prob. 5.6DQCh. 5 - Prob. 5.7DQCh. 5 - Prob. 5.8DQCh. 5 - Prob. 5.9DQCh. 5 - Prob. 5.10DQCh. 5 - Prob. 5.11DQCh. 5 - Prob. 5.12DQCh. 5 - Prob. 5.13DQCh. 5 - Prob. 5.14DQCh. 5 - Prob. 5.15DQCh. 5 - Prob. 5.16DQCh. 5 - Prob. 5.17DQCh. 5 - Prob. 5.1PCh. 5 - Prob. 5.2PCh. 5 - Prob. 5.3PCh. 5 - Prob. 5.4PCh. 5 - Prob. 5.5PCh. 5 - Prob. 5.6PCh. 5 - Prob. 5.7PCh. 5 - Prob. 5.8PCh. 5 - Prob. 5.9PCh. 5 - Prob. 5.10PCh. 5 - Prob. 5.11PCh. 5 - Prob. 5.12PCh. 5 - Prob. 5.13PCh. 5 - Prob. 5.14PCh. 5 - Prob. 5.15PCh. 5 - Prob. 5.16PCh. 5 - Prob. 5.17PCh. 5 - Prob. 5.18PCh. 5 - Prob. 5.19PCh. 5 - Prob. 5.20PCh. 5 - Prob. 5.21PCh. 5 - Prob. 5.22PCh. 5 - Prob. 5.23PCh. 5 - Prob. 5.24PCh. 5 - Prob. 5.25PCh. 5 - Prob. 5.27PCh. 5 - Prob. 5.28PCh. 5 - Prob. 5.29PCh. 5 - Prob. 5.30PCh. 5 - Prob. 5.31PCh. 5 - Prob. 5.32PCh. 5 - Prob. 5.33PCh. 5 - Prob. 5.34PCh. 5 - Prob. 5.35PCh. 5 - Prob. 5.36PCh. 5 - Prob. 5.37PCh. 5 - Prob. 5.38PCh. 5 - Prob. 5.39PCh. 5 - Prob. 5.40PCh. 5 - Prob. 5.41PCh. 5 - Prob. 5.42PCh. 5 - Prob. 5.43PCh. 5 - Prob. 5.44PCh. 5 - Prob. 5.45PCh. 5 - Prob. 5.46PCh. 5 - Prob. 5.47PCh. 5 - Prob. 5.48PCh. 5 - Prob. 5.49PCh. 5 - Prob. 5.50PCh. 5 - Prob. 5.51PCh. 5 - Prob. 5.52PCh. 5 - Prob. 5.53PCh. 5 - Prob. 5.54PCh. 5 - Prob. 5.55PCh. 5 - Prob. 5.56PCh. 5 - Prob. 5.58PCh. 5 - Prob. 5.59PCh. 5 - Prob. 5.60PCh. 5 - Prob. 5.61PCh. 5 - Prob. 5.62PCh. 5 - Prob. 5.63PCh. 5 - Prob. 5.64PCh. 5 - Prob. 5.66PCh. 5 - Prob. 5.67PCh. 5 - Prob. 5.68PCh. 5 - Prob. 5.69PCh. 5 - Prob. 5.70PCh. 5 - Prob. 5.71PCh. 5 - Prob. 5.72PCh. 5 - Prob. 5.73PCh. 5 - Prob. 5.74PCh. 5 - Prob. 5.76PCh. 5 - Prob. 5.77PCh. 5 - Prob. 5.78PCh. 5 - Prob. 5.79PCh. 5 - Prob. 5.1PRCh. 5 - Prob. 5.2PRCh. 5 - Prob. 5.3PRCh. 5 - Prob. 5.4PR
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