General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 18, Problem 18.27QP

(a)

Interpretation Introduction

Interpretation: From the given data of ΔGfo, ΔGoandKp have to be calculated.

Concept Introduction:

Gibbs free energy: The amount of energy that is required to perform the maximum amount of reversible work by a system at constant temperature and pressure.

Symbol = G, Unit = Joule (J)/mol.

ΔGo-Standard Gibbs free energy change.

Gibbs free energy change of formation is the change of Gibbs free energy that is needed for the formation of 1 mole of the substance in its standard state from its constituent elements.

ΔGfo-Standard Gibbs free energy change of formation.

Relation between ΔGoandΔGfo:

ΔGo=Gfo(products)-Gfo(reactants)

where,Gfo(products)isthetotalsumofGibbsfreeenergychangeofproducts.Gfo(reactants)isthetotalsumofGibbsfreeenergychangeofreactantsΔGfoisthestandardchangeinthegibbsfreeenergyofformation.

mis the number of moles of the products.nis the number of moles of the reactants.

Relation between ΔGoandKp:

ΔGo=-RTlnKp

Where,

R=Universalgasconstant.=8.314JK1mol-1T=Temperature.Kp=Equilibriumconstant.

(a)

Expert Solution
Check Mark

Answer to Problem 18.27QP

ΔGo=39000J/mol

Kp=1.4578×1007

Explanation of Solution

The given reaction is:

PCl5(g)PCl3(g)+Cl2(g)

Given:

ΔGfo for Cl2(g) is 0.

ΔGfo for PCl3(g) is 286kJ/mol

ΔGfo for PCl5(g) is 325kJ/mol

Substituting these values in the formula shown below:

ΔGo=Gfo(products)-Gfo(reactants)=[(1×-286kJ/mol)+(1×0kJ/mol)](1×-325kJ/mol)=-286kJ/mol+325kJ/mol=39kJ/mol

Therefore, ΔGo=39kJ/mol

Converting ΔGo from kJ/mol to J/mol:

1kJ=1000J39kJ/mol=1000J1kJ×39kJ/mol=39000J/mol

Therefore, ΔGo=39000J/mol

Given: T=Temperature=25oC

Converting the temperature from oC into Kelvin:

T=25oC+273=298K

It is known that ΔGo=-RTlnKp

This expression is rearranged as follows to find Kp:

Kp=exp(ΔGoRT)=exp(39000J/mol8.314JK1mol-1×298K)=1.4578×1007

Therefore, Kp=1.4578×1007

Conclusion

From the given data of ΔGfo, ΔGoandKp have been calculated.

(b)

Interpretation Introduction

Interpretation: From the given data of partial pressures, ΔG has to be calculated.

Concept Introduction:

Partial pressure: The hypothetical pressure occupied by a gas as alone from the entire volume of the original mixture of gases at the same temperature is said to be the partial pressure of that gas.

Gibbs free energy: The amount of energy that is required to perform the maximum amount of reversible work by a system at constant temperature and pressure.

Symbol = G, Unit = Joule (J)/mol.

ΔG- Gibbs free energy change.

Relation between ΔGoandKp:

ΔG=-RTlnK

Where,

R=Universalgasconstant.=8.314JK1mol-1T=Temperature.K=Equilibriumconstant.

The equilibrium constant for gaseous reactions can be either as Kp or as Kc. Kp is defined as equilibrium constant of a reaction in terms of partial pressures whereas Kc is defined as equilibrium constant of a reaction in terms of concentration.

(b)

Expert Solution
Check Mark

Answer to Problem 18.27QP

ΔG=92268.25Jmol-1

Explanation of Solution

The given reaction is:

PCl5(g)PCl3(g)+Cl2(g)

Given data:

PPCl5=0.0029atmPPCl3=0.27atmPCl2=0.40atm

Finding Kp:

Kp=PPCl3×PCl2PPCl5=0.27atm×0.40atm0.0029atm=37.2414

Therefore, Kp=37.2414

Given: T=Temperature=25oC

Converting the temperature from oC into Kelvin:

T=25oC+273=298K

It is known that ΔG=-RTlnK

The equilibrium constant in the expression of ΔG is considered to be the equilibrium constant in terms of partial pressure which means K is considered as Kp.

Finding ΔG by substituting Kp in the expression of ΔG:

ΔG=-RTlnK=8.314JK1mol-1×298K×37.2414=92268.25Jmol-1

Therefore, ΔG=92268.25Jmol-1

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Chapter 18 Solutions

General Chemistry

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