Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 20, Problem 95CWP

(a)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔHreaction=ΔHformation(product)ΔHformation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG=ΔHTΔS

To determine: Standard enthalpy of the reaction and change in standard entropy of the given reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 95CWP

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5kJ/mol_ and 134J/Kmol_ , respectively.

Explanation of Solution

Hydrogen gas is produced by reacting graphite with water.

C(s)+H2O(g)CO(g)+H2(g)

The standard enthalpy of formation of H2O(g) is 242kJ/mol .

The standard enthalpy of formation of CO(g) is 110.5kJ/mol .

The standard enthalpy of formation of C(s) and H2(g) is zero since the standard enthalpy of formation of free elements in their standard state is zero.

The standard enthalpy of the reaction is calculated by the formula,

ΔHreaction=ΔHformation(product)ΔHformation(reactant)

Therefore, the above equation becomes,

ΔHreaction=ΔHformation(CO(g))ΔHformation(H2O(g))

Substitute the value of ΔHformation(CO(g)) and ΔHformation(H2O(g)) in the above equation.

ΔHreaction=(110.5kJ/mol)(242kJ/mol)=110.5kJ/mol+242kJ/mol=+131.5kJ/mol_

Therefore, the standard enthalpy of the reaction is +131.5kJ/mol .

The standard entropy of H2O(g) is 189J/Kmol .

The standard entropy of CO(g) is 198J/Kmol .

The standard entropy of C(s) is 6J/Kmol .

The standard entropy of H2(g) is 131J/Kmol .

The standard entropy change of the reaction is calculated by the formula,

ΔSreaction=S(product)S(reactant)

Therefore, the above equation becomes,

ΔSreaction=[S(CO(g))+S(H2(g))][S(H2O(g))+S(C(s))]

Substitute the value of S(CO(g)) , S(H2(g)) , S(H2O(g)) and S(C(s)) in the above equation.

ΔSreaction=[198J/Kmol+131J/Kmol][189J/Kmol+6J/Kmol]=329J/Kmol195J/Kmol=134J/Kmol_

Therefore, the standard entropy change of the reaction is 134J/Kmol_ .

(b)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔHreaction=ΔHformation(product)ΔHformation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG=ΔHTΔS

To determine: The temperature at which the change in standard Gipp’s energy of the given reaction is zero.

(b)

Expert Solution
Check Mark

Answer to Problem 95CWP

The temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3K_

Explanation of Solution

Given

The change in standard Gipp’s energy of the given reaction is zero.

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5kJ/mol and 134J/Kmol , respectively.

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG=ΔHTΔS

Substitute the value of ΔG , ΔH and ΔS in the above equation.

0=+131.5×103J/molT×134J/KmolT×134J/Kmol=131.5×103J/molT=131.5×103J/mol134J/Kmol=981.3K_

Therefore, the temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3K_

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Chapter 20 Solutions

Chemistry

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