Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 18, Problem 98QP
Interpretation Introduction

Interpretation:

The equilibrium pressure constant and temperature for the given reaction is to be calculated.

Concept introduction:

All the energy available with the system, utilized in doing useful work, is called Gibbs free energy.

Entropy is the direct measurement of randomness or disorder. It is an extensive property and a state function.

The enthalpy of the system defined as the sum of the internal energy and the product of the pressure and the volume. It is a state function and an extensive property.

The standard Gibbs free energy change for the given reaction at temperature is calculated using the following expression:

ΔGo=ΔHoT×ΔSo

The standard enthalpy change of the reaction, ΔHrxno, is calculated using the following expression:

ΔHrxno= nΔHfo(product) mΔHfo(reactant)

The standard entropy change for this reaction is calculated using the following expression:

ΔSrxno= nSo(product) mSo(reactant)

The equilibrium pressure constant is calculated using expression as follows:

ΔGo=RTlnKp

Expert Solution & Answer
Check Mark

Answer to Problem 98QP

Solution: The equilibrium pressure constant for the given reaction is 36.

At a temperature higher than 981K, the reaction favors the formation of H2O(g)

and CO(g). The value Kp remains same.

Explanation of Solution

Given information: The reaction is as follows:

H2O(g)+CO(g)CO2(g)+H2(g).

Temperature, T=300oC.

The equation for the reaction of H2O(g) with CO(g) is as follow:

H2O(g)+CO(g)CO2(g)+H2(g).

The standard Gibbs free energy change for the given reaction at temperature 300oC is calculated using the following expression:

ΔGo=ΔHoT×ΔSo.

Here, ΔHo is the standard enthalpy change, T is the temperature, ΔSo is the standard entropy change and ΔGo is the standard Gibbs free energy change.

The standard enthalpy change of the system, ΔHo, is calculated using the following expression:

ΔHo=ΔHrxno.

The standard enthalpy change of the reaction, ΔHrxno, is calculated using the following expression:

ΔHrxno= nΔHfo(product) mΔHfo(reactant).

Here, ΔHrxno is the standard enthalpy change for the reaction, ΔHfo is the standard enthalpy change for the formation of the substance, n is the stoichiometric coefficient of products, and m is the stoichiometric coefficient of reactants.

The enthalpy change for the reaction is as follows:

ΔHrxno=(ΔHfo[ CO2(g) ]+ΔHfo[ H2(g) ])(ΔHfo[ H2O(g) ]+ΔHfo[ CO(g) ]).

From appendix 2, the standard enthalpy changes of formation for the substances are as follows:

ΔHfo[ CO(g) ]=110.5kJ/mol,

ΔHfo[ H2(g) ]=0kJ/mol,

ΔHfo[ CO2(g) ]=393.5kJ/mol,

ΔHfo[ H2O(g) ]=241.8kJ/mol.

Substitute the standard enthalpy change of the formation value of the substance in the above expression.

ΔHrxno=[ (393.5kJ/mol)+(0) ][ (241.8kJ/mol)+(110.5kJ/mol) ]=41.2kJ/molΔHo=41.2kJ/mol.

Therefore, the standard enthalpy changes for the given reaction are 41.2kJ/mol.

The entropy change of the system, ΔSo, is calculated using the following expression:

ΔSo=ΔSrxno

The standard entropy change for this reaction is calculated using the following expression:

ΔSrxno= nSo(product) mSo(reactant).

Here, ΔSrxno is the standard entropy change for the reaction, ΔSo is the standard entropy change of the substance, n is the stoichiometric coefficient of products, and m is the stoichiometric coefficient of reactants.

The entropy change for the reaction is as follows:

ΔSrxno=(So[ CO2(g) ]+So[ H2(g) ])(So[ H2O(g) ]+So[ CO(g) ]).

From appendix2, the standard entropy values of the substances are as follows:

So[ CO(g) ]=197.9J/Kmol;

So[ H2(g) ]=131.0J/Kmol;

So[ H2O(g) ]=188.7kJ/mol;

So[ CO2(g) ]=213.79J/Kmol.

Substitute the standard entropy values of the substances in the above expression.

ΔSrxno=[ (213.79J/Kmol)+(131.0J/Kmol) ][ (188.7J/Kmol)+(197.9J/Kmol) ].=42.0J/KmolΔSo=42.0J/Kmol

Therefore, the standard entropy change for the given reaction is 42.0J/mol.

The standard Gibbs free energy change for the given reaction at temperature 300oC is calculated using the following expression:

ΔGo=ΔHoT×ΔSo.

Here, ΔHo is the standard enthalpy changes, T is the temperature, ΔSo is the standard entropy change and ΔGo is the standard Gibbs free energy change.

Substitutes the value of ΔHo, T and ΔSo in the above expression.

ΔGo=(41.2×103J/mol)(573K)×(42.0J/mol)=1.71×104J/mol.

Therefore, standard Gibbs free energy change for the given reaction is 1.71×104J/mol.

The equilibrium pressure constant is calculated using expression as follows:

ΔGo=RTlnKp

Here, R

is the constant, T

is the temperature, Kp is the equilibrium pressure constant and ΔGo is the Gibbs free energy change.

Substitute the value of R, T, and ΔGo in the above expression.

lnKp=(1.71×104J/mol(8.314J/Kmol)(573K))lnKp=3.59Kp=e3.59Kp=36

Therefore, the equilibrium pressure constant for the given reaction is 36.

Entropy change for this reaction is negative. The Gibbs free energy is positive at higher temperature.

The temperature is calculated using the following expression:

ΔGo=ΔHoT×ΔSo.

Here, ΔHo is the standard enthalpy changes, T is the temperature, ΔSo is the standard entropy change and ΔGo is the standard Gibbs free energy change.

Substitute the value of ΔHo, ΔGo and ΔSo in the above expression.

0=ΔHoT×ΔSoT=ΔHoΔSo=41.2×103J/mol40.×103J/mol=981K.

Therefore, ΔGo is zero is 981K. Above this temperature ΔGo is positive and Kp

will be smaller than one. The reaction moves in the backward direction. The temperature higher than 981K the reaction favors the formation of H2O(g) and CO(g).

On adding a catalyst to the reaction, the value of Kp does not increase. The catalyst increases both, the rate of forward and the rate of backward reaction. The value Kp remains the same.

Conclusion

The equilibrium pressure constant for the given reaction is 36. Temperature higher than 981K the reaction favours the formation of H2O(g) and CO(g). The value of Kp remains the same.

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Chapter 18 Solutions

Chemistry

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