Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 17, Problem 17.79QP

(a)

Interpretation Introduction

Interpretation: The values of standard Gibbs free energy (ΔGrxnο) for the given reactions are to be calculated. The balanced chemical equation of the overall reaction obtained by coupling the given two reactions is to be stated. Also, whether the coupled reaction is spontaneous under standard condition is to be explained.

Concept introduction: Gibbs free energy is a thermodynamic quantity that is used to calculate the maximum work of reversible reaction performed by a system. It is equal to the difference between the enthalpy and the product of entropy at absolute temperature.

To determine: The values of ΔGrxnο for the given reactions.

(a)

Expert Solution
Check Mark

Answer to Problem 17.79QP

Solution

The value of ΔGrxnο for the decomposition reaction of methane is 50.8kJ_ .

The value of ΔGrxnο for oxidation reaction of carbon is 394.4kJ_ .

Explanation of Solution

Explanation

Given

The decomposition reaction of methane is,

CH4(g)C(s)+2H2(g)

The reaction consumes one mole of methane to produce one mole of carbon and two moles of hydrogen.

The ΔGfο value of CH4(g) is 50.8kJ/mol .

The ΔGfο value of C(s) is 0.0kJ/mol .

The ΔGfο value of H2(g) is 0.0kJ/mol .

The change in standard free energy of the reaction, ΔGrxnο , using the standard free energy of formation is calculated from the given formula,

ΔGrxnο=nproductsΔGf.productsοnreactantsΔGf.reactantsο

Where,

  • n is the number of moles of for products and reactants.
  • ΔGfο is the standard free energy of formation of reactants and products.

Substitute the values of nproducts , ΔGf.productsο , nreactants and ΔGf.reactantsο in the above formula to calculate the ΔGrxnο for the reaction.

ΔGrxnο=[(nC×ΔGCο+nH2×ΔGH2ο)][nCH4×ΔGCH4ο]=[(1×(0.0kJ/mol)+2×0.0kJ/mol)][1×(50.8kJ/mol)]=[0.0kJ+0.0kJ](50.8kJ)=0.0kJ+50.8kJ

Simplify the above equation,

ΔGrxnο=50.8kJ_

Hence, the value of ΔGrxnο for the reaction is 50.8kJ_ .

The oxidation reaction of carbon is,

C(s)+O2(g)CO2(g)

The reaction consumes one mole of carbon and one mole of oxygen gas to produce one mole of carbon dioxide.

The ΔGfο value of CO2(g) is 394.4kJ/mol .

The ΔGfο value of C(s) is 0.0kJ/mol .

The ΔGfο value of O2(g) is 0.0kJ/mol .

The change in standard free energy of the reaction, ΔGrxnο , using the standard free energy of formation is calculated from the given formula,

ΔGrxnο=nproductsΔGf.productsοnreactantsΔGf.reactantsο

Where,

  • n is the number of moles of for products and reactants.
  • ΔGfο is the standard free energy of formation of reactants and products.

Substitute the values of nproducts , ΔGf.productsο , nreactants and ΔGf.reactantsο in the above formula to calculate the ΔGrxnο for the reaction.

ΔGrxnο=[nCO2×ΔGCO2ο][(nC×ΔGCο+nO2×ΔGO2ο)]=[1×(394.4kJ/mol)][(1×(0.0kJ/mol)+1×0.0kJ/mol)]=394.4kJ0.0kJ=394.4kJ_

Hence, the value of ΔGrxnο for the reaction is 394.4kJ_ .

(b)

Interpretation Introduction

To determine: The balanced chemical equation of the overall reaction obtained by coupling the given two reactions and if the coupled reaction is spontaneous under standard condition.

(b)

Expert Solution
Check Mark

Answer to Problem 17.79QP

Solution

The overall balanced chemical equation is,

CH4(g)+O2(g)CO2(g)+2H2(g)

The value of ΔGoverallο is 343.6kJ_ . The value of ΔGoverallο<0 , the reaction is spontaneous under standard condition.

Explanation of Solution

Explanation

The decomposition reaction of methane is,

CH4(g)C(s)+2H2(g) (1)

The oxidation reaction of carbon is,

C(s)+O2(g)CO2(g) (2)

The two reactions are combined to give overall reaction as,

CH4(g)+O2(g)CO2(g)+2H2(g)

Hence, the overall reaction is,

CH4(g)+O2(g)CO2(g)+2H2(g)

The ΔGrxnο values of the coupled reactions are additive. Hence, their values are added to give ΔGrxnο of the overall reaction. The change in free energy of the overall reaction is considered to be ΔGoverallο .

The change in free energy of the decomposition reaction of methane is considered to ΔG1ο and the change in free energy of the oxidation reaction of carbon is considered to be ΔG2ο . The value of ΔG1ο is 50.8kJ and the value of ΔG2ο is 394.4kJ as calculated in part (a) and (b) respectively. The overall change in free energy is calculated as,

ΔGoverallο=ΔG1ο+ΔG2ο

Substitute the values of ΔG1ο and ΔG2ο in the above expression to calculate the value of ΔGoverallο .

ΔGoverallο=50.8kJ+(394.4kJ)=50.8kJ394.4kJ=343.6kJ_

When ΔG<0 then the reaction is spontaneous and when ΔG>0 the reaction is non spontaneous. If ΔG=0 then the reaction is at equilibrium.

Hence, the value of ΔGoverallο is 343.6kJ_ . The value of ΔGoverallο<0 , the reaction is spontaneous under standard condition.

Conclusion

  1. a. The value of ΔGrxnο for the decomposition reaction of methane is 50.8kJ_ and the value of ΔGrxnο for oxidation reaction of carbon is 394.4kJ_ .
  2. b. The value of ΔGoverallο is 343.6kJ_ which is less than zero shows the reaction is spontaneous under standard condition

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Chapter 17 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 17 - Prob. 17.3VPCh. 17 - Prob. 17.4VPCh. 17 - Prob. 17.5VPCh. 17 - Prob. 17.6VPCh. 17 - Prob. 17.7VPCh. 17 - Prob. 17.8VPCh. 17 - Prob. 17.9QPCh. 17 - Prob. 17.10QPCh. 17 - Prob. 17.11QPCh. 17 - Prob. 17.12QPCh. 17 - Prob. 17.13QPCh. 17 - Prob. 17.14QPCh. 17 - Prob. 17.15QPCh. 17 - Prob. 17.16QPCh. 17 - Prob. 17.17QPCh. 17 - Prob. 17.18QPCh. 17 - Prob. 17.19QPCh. 17 - Prob. 17.20QPCh. 17 - Prob. 17.21QPCh. 17 - Prob. 17.22QPCh. 17 - Prob. 17.23QPCh. 17 - Prob. 17.24QPCh. 17 - Prob. 17.25QPCh. 17 - Prob. 17.26QPCh. 17 - Prob. 17.27QPCh. 17 - Prob. 17.28QPCh. 17 - Prob. 17.29QPCh. 17 - Prob. 17.30QPCh. 17 - Prob. 17.31QPCh. 17 - Prob. 17.32QPCh. 17 - Prob. 17.33QPCh. 17 - Prob. 17.34QPCh. 17 - Prob. 17.35QPCh. 17 - Prob. 17.36QPCh. 17 - Prob. 17.37QPCh. 17 - Prob. 17.38QPCh. 17 - Prob. 17.39QPCh. 17 - Prob. 17.40QPCh. 17 - Prob. 17.41QPCh. 17 - Prob. 17.42QPCh. 17 - Prob. 17.43QPCh. 17 - Prob. 17.44QPCh. 17 - Prob. 17.45QPCh. 17 - Prob. 17.46QPCh. 17 - Prob. 17.47QPCh. 17 - Prob. 17.48QPCh. 17 - Prob. 17.49QPCh. 17 - Prob. 17.50QPCh. 17 - Prob. 17.51QPCh. 17 - Prob. 17.52QPCh. 17 - Prob. 17.53QPCh. 17 - Prob. 17.54QPCh. 17 - Prob. 17.55QPCh. 17 - Prob. 17.56QPCh. 17 - Prob. 17.57QPCh. 17 - Prob. 17.58QPCh. 17 - Prob. 17.59QPCh. 17 - Prob. 17.60QPCh. 17 - Prob. 17.61QPCh. 17 - Prob. 17.62QPCh. 17 - Prob. 17.63QPCh. 17 - Prob. 17.64QPCh. 17 - Prob. 17.65QPCh. 17 - Prob. 17.66QPCh. 17 - Prob. 17.67QPCh. 17 - Prob. 17.68QPCh. 17 - Prob. 17.69QPCh. 17 - Prob. 17.70QPCh. 17 - Prob. 17.71QPCh. 17 - Prob. 17.72QPCh. 17 - Prob. 17.73QPCh. 17 - Prob. 17.74QPCh. 17 - Prob. 17.75QPCh. 17 - Prob. 17.76QPCh. 17 - Prob. 17.77QPCh. 17 - Prob. 17.78QPCh. 17 - Prob. 17.79QPCh. 17 - Prob. 17.80QPCh. 17 - Prob. 17.81QPCh. 17 - Prob. 17.82QPCh. 17 - Prob. 17.83QPCh. 17 - Prob. 17.84QPCh. 17 - Prob. 17.85QPCh. 17 - Prob. 17.86QPCh. 17 - Prob. 17.87QPCh. 17 - Prob. 17.88APCh. 17 - Prob. 17.89APCh. 17 - Prob. 17.90APCh. 17 - Prob. 17.91APCh. 17 - Prob. 17.92APCh. 17 - Prob. 17.93APCh. 17 - Prob. 17.94APCh. 17 - Prob. 17.95APCh. 17 - Prob. 17.96APCh. 17 - Prob. 17.97APCh. 17 - Prob. 17.98APCh. 17 - Prob. 17.99APCh. 17 - Prob. 17.100APCh. 17 - Prob. 17.101APCh. 17 - Prob. 17.102APCh. 17 - Prob. 17.103APCh. 17 - Prob. 17.104APCh. 17 - Prob. 17.105AP
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