Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN: 9781305079373
Author: William L. Masterton, Cecile N. Hurley
Publisher: Cengage Learning
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Chapter 16, Problem 20QAP

Use Table 16.1 to calculate ΔS° for each of the following reactions.

(a) 5NO ( g ) + 3 MnO 4 ( a q ) + 4 H + ( a q ) 5 NO 3 ( a q ) + 3 Mn 2 + ( a q ) + 2 H 2 O ( l ) (b) 6Fe 2 + ( a q ) + CrO 4 2 ( a q ) + 8 H + ( a q ) Cr ( s ) + 6 Fe 3 + ( a q ) + 4 H 2 O ( l )

(c) C 2 H 2 ( g ) + 5 2 O 2 ( g ) 2 CO 2 ( g ) + H 2 O ( g )

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The entropy change ( ΔS°) for the following reaction should be calculated.

5NO(g)+3MnO42(aq)+4H+(aq)5NO3(aq) + 3Mn2+(aq)+2H2O(l)

Concept introduction:

Entropy is defined as the ratio of thermal energy to the temperature which is unavailable for work done. It is also defined as the measure of disorder of molecule of a system. It is an extensive property and state function.

Entropy is related with the number of microstates for a system and microstate is defined as the number of ways for the system to be arranged.

The standard entropy change at room temperature is equal to the difference between the standard entropy of reactant and standard entropy of product.

Answer to Problem 20QAP

ΔS°=976.1 J/Kmol

Explanation of Solution

Given process is: 5NO(g)+3MnO42(aq)+4H+(aq)5NO3(aq) + 3Mn2+(aq)+2H2O(l)

The mathematical expression for the standard entropy value at room temperature is:

ΔS°=ΔS°298=nS°(products)pS°(reactants)

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The value of standard entropy for NO(g) is 210.7 J/Kmol

The value of standard entropy for MnO42(aq) is 191.2 J/Kmol

The value of standard entropy for H+(aq) is 0.0 J/Kmol

The value of standard entropy for NO3(aq) is 146.4 J/Kmol

The value of standard entropy for Mn2+(aq) is -73.6 J/Kmol

The value of standard entropy for H2O(l) is 69.9 J/Kmol

Put the values, we get:

ΔS°=(5×S°(NO3(aq))+3×S°(Mn2+(aq))+2×S°(H2O(l)))(5×S°(NO(g))+3×S°(MnO42(aq))+4×S°(H+(aq)))

ΔS°=[(5×146.4+3×(73.6)+2×(69.9)(5×210.7+3×191.2+4×0.0)] J/Kmol

ΔS°=[6511627.1] J/Kmol

ΔS°=976.1 J/Kmol

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The entropy change ( ΔS°) for the following reaction should be calculated.

6Fe2+(aq)+CrO42(aq)+8H+(aq)Cr (s) + 6Fe3+(aq)+4H2O(l)

Concept introduction:

Entropy is defined as the ratio of thermal energy to the temperature which is unavailable for work done. It is also defined as the measure of disorder of molecule of a system. It is an extensive property and state function.

Entropy is related with the number of microstates for a system and microstate is defined as the number of ways for the system to be arranged.

The standard entropy change at room temperature is equal to the difference between the standard entropy of reactant and standard entropy of product.

Answer to Problem 20QAP

ΔS°=819 J/Kmol

Explanation of Solution

Given process is: 6Fe2+(aq)+CrO42(aq)+8H+(aq)Cr (s) + 6Fe3+(aq)+4H2O(l)

The mathematical expression for the standard entropy value at room temperature is:

ΔS°=ΔS°298=nS°(products)pS°(reactants)

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The value of standard entropy for Fe2+(aq) is 137.2 J/Kmol

The value of standard entropy for CrO42(aq) is 50.2 J/Kmol

The value of standard entropy for H+(aq) is 0.0 J/Kmol

The value of standard entropy for Cr (s) is

Fe3+(g)+23.8 J/Kmol

The value of standard entropy for is 315.9 J/Kmol

The value of standard entropy for H2O(l) is 69.9 J/Kmol

Put the values, we get:

6Fe2+(aq)+CrO42(aq)+8H+(aq)Cr (s) + 6Fe3+(aq)+4H2O(l)

ΔS°=(1×S°(Cr(s))+6×S°(Fe3+(aq))+4×S°(H2O(l)))(6×S°(Fe2+(aq))+1×S°(CrO42(aq))+8×S°(H+(aq)))

ΔS°=[(1×23.8+6×(315.9)+4×69.9)(6×(137.2)+1×50.2+8×0.0)] J/Kmol

ΔS°=[1592+773] J/Kmol

ΔS°=819 J/Kmol

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The entropy change ( ΔS°) for the following reaction should be calculated.

C2H2(g)+52O2(g)2CO2(g)+H2O(g)

Concept introduction:

Entropy is defined as the ratio of thermal energy to the temperature which is unavailable for work done. It is also defined as the measure of disorder of molecule of a system. It is an extensive property and state function.

Entropy is related with the number of microstates for a system and microstate is defined as the number of ways for the system to be arranged.

The standard entropy change at room temperature is equal to the difference between the standard entropy of reactant and standard entropy of product.

Answer to Problem 20QAP

ΔS°=97.4 J/Kmol

Explanation of Solution

Given process is: C2H2(g)+52O2(g)2CO2(g)+H2O(g)

The mathematical expression for the standard entropy value at room temperature is:

ΔS°=ΔS°298=nS°(products)pS°(reactants)

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The value of standard entropy for C2H2(g) is 200.8 J/Kmol

The value of standard entropy for O2(g) is 205.0 J/Kmol

The value of standard entropy for CO2(g) is 213.6 J/Kmol

The value of standard entropy for H2O(g) is 188.7 J/Kmol

Put the values, we get:

ΔS°=(2×S°(CO2(g))+1×S°(H2O(g)))(1×S°(C2H2(g))+52×S°(O2(g)))

ΔS°=[(2×213.6+1×188.7)(1×200.8+52×205.0)] J/Kmol

ΔS°=[615.9713.3] J/Kmol

ΔS°=97.4 J/Kmol

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Chapter 16 Solutions

Chemistry: Principles and Reactions

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