Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Chapter 3, Problem 3.56E

Determine the entropy of formation, Δ f S , of the following compounds. Assume 25 ° C . (a) H 2 O ( l ) (b) H 2 O ( g ) (c) Fe 2 ( SO 4 ) 3 (d) Al 2 O 3

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The entropy of formation of H2O(l) is to be determined.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

Answer to Problem 3.56E

The entropy of formation of H2O(l) is 163.34J/K.

Explanation of Solution

The formation reaction of H2O(l) is shown below.

H2(g)+12O2(g)H2O(l)

The entropy of formation of H2O(l) is calculated by the formula as shown below.

ΔfS=ΔS°(H2O(l))(ΔS°(H2(g))+12ΔS°(O2(g))) …(1)

Where,

ΔS°(H2O(l)) is the standard entropy of H2O at 298K.

ΔS°(H2(g)) is the standard entropy of H2 at 298K.

ΔS°(O2(g)) is the standard entropy of O2 at 298K.

The standard entropy of H2O(l) at 298K is 69.91J/molK.

The standard entropy of H2 at 298K is 130.68J/molK.

The standard entropy of O2 at 298K is 205.14J/molK.

Substitute the standard entropy of H2O, H2, O2 in equation (1).

ΔfS=69.91J/molK(130.68J/molK+12×205.14J/molK)=69.91J/K130.68J/K102.57J/K=163.34J/K

Conclusion

The entropy of formation of H2O(l) is 163.34J/K.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The entropy of formation of H2O(g) is to be determined.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

Answer to Problem 3.56E

The entropy of formation of H2O(g) is 44.42J/K.

Explanation of Solution

The formation reaction of H2O(g) is shown below,

H2(g)+12O2(g)H2O(g)

The entropy of formation of H2O(g) is calculated by the formula as shown below.

ΔfS=ΔS°(H2O(g))(ΔS°(H2(g))+12ΔS°(O2(g))) …(2)

Where,

ΔS°(H2O(g)) is the standard entropy of H2O at 298K.

ΔS°(H2(g)) is the standard entropy of H2 at 298K.

ΔS°(O2(g)) is the standard entropy of O2 at 298K.

The standard entropy of H2O(g) at 298K is 188.83J/molK.

The standard entropy of H2 at 298K is 130.68J/molK.

The standard entropy of O2 at 298K is 205.14J/molK.

Substitute the entropy enthalpy of H2O, H2, O2 in equation (2).

ΔfS=188.83J/molK(130.68J/molK+12×205.14J/molK)=188.83J/K130.68J/K102.57J/K=44.42J/K

Conclusion

The entropy of formation of H2O(g) is 44.42J/K.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The entropy of formation of Fe2(SO4)3 is to be determined.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

Answer to Problem 3.56E

The entropy of formation of Fe2(SO4)3 is 550.25J/K.

Explanation of Solution

The formation reaction of Fe2(SO4)3 is shown below.

Fe2O3(s)+3SO3(g)Fe2(SO4)3(s)

The entropy of formation of Fe2(SO4)3 is calculated by the formula as shown below.

ΔfS=ΔS°(Fe2(SO4)3(s))(ΔS°(Fe2O3(s))+3ΔS°(SO3(g))) … (3)

Where,

ΔS°(Fe2(SO4)3(s)) is the standard entropy of Fe2(SO4)3(s) at 298K.

ΔS°(Fe2O3(s)) is the standard entropy of Fe2O3(s) at 298K.

ΔS°(SO3(g)) is the standard entropy of SO3(g) at 298K.

The standard entropy of Fe2(SO4)3(s) at 298K is 307.46J/molK.

The standard entropy of Fe2O3(s) at 298K is 87.4J/molK.

The standard entropy of SO3(g) at 298K is 256.77J/molK.

Substitute the standard entropy of Fe2(SO4)3(s), Fe2O3(s), SO3(g) in equation (3).

ΔfS=307.46J/molK(87.4J/molK+3×256.77J/molK)=307.46J/K87.4J/K770.31J/K=550.25J/K

Conclusion

The entropy of formation of Fe2(SO4)3 is 550.25J/K.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

The entropy of formation of Al2O3 is to be determined.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

Answer to Problem 3.56E

The entropy of formation of Al2O3 is 313.39J/K.

Explanation of Solution

The formation reaction of Al2O3 is shown below.

2Al(s)+32O2(g)Al2O3(s)

The entropy of formation of Al2O3 is calculated by the formula as shown below.

ΔfS=ΔS°(Al2O3(s))(2ΔS°(Al(s))+32ΔS°(O2(g))) … (4)

Where,

ΔS°(Al2O3(s)) is the standard entropy of Al2O3(s) at 298K.

ΔS°(Al(s)) is the standard entropy of Al(s) at 298K.

ΔS°(O2(g)) is the standard entropy of O2(g) at 298K.

The standard entropy of Al2O3(s) at 298K is 50.92J/molK.

The standard entropy of Al(s) at 298K is 28.30J/molK.

The standard entropy of O2(g) at 298K is 205.14J/molK.

Substitute the standard entropy of Al2O3(s), Al(s), O2(g) in equation (4).

ΔfS=50.92J/molK(2×28.30J/molK+32×205.14J/molK)=50.92J/K56.6J/K307.71J/K=313.39J/K

Conclusion

The entropy of formation of Al2O3 is 313.39J/K.

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Chapter 3 Solutions

Physical Chemistry

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