Chemistry
Chemistry
9th Edition
ISBN: 9781133611097
Author: Steven S. Zumdahl
Publisher: Cengage Learning
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Chapter 17, Problem 53E

From data in Appendix 4, calculate ∆H°, ∆S°, and ∆G° for each of the following reactions at 25°C.

a. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

b. 6 CO 2 ( g ) + 6 H 2 O ( l ) C 6 H 12 O 6 ( s ) Glucose + 6 O 2 ( g )

c. P4O10(s) + 6H2O(l) → 4H3PO4(s)

d. HCl(g) + NH3(g) → NH4Cl(s)

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The values of ΔS,ΔH and ΔG is to be calculated at 25°C in each case.

Concept introduction: The expression for ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

The expression for ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

The expression for ΔG is,

ΔG=npΔG(product)nfΔG(reactant)

Explanation of Solution

Explanation

The stated reaction is,

CH4(g)+2O2(g)CO2(g)+2H2O(g)

Refer to Appendix 4 .

The value of ΔS(J/Kmol) for the given reactant and product is,

Molecules ΔS(J/Kmol)
O2(g) 205
CH4(g) 186
CO2(g) 214
H2O(g) 189

The formula of ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

Where,

  • ΔS is the standard entropy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔS(product) is the standard entropy of product at a pressure of 1atm .
  • ΔS(reactant) is the standard entropy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔS=npΔS(product)nfΔS(reactant)=[(214)+2(189){2(205)+(186)}]J/K=4J/K_

The value of ΔH(kJ/mol) for the given reactant and product is,

Molecules ΔH(kJ/mol)
O2(g) 0
CH4(g) 75
CO2(g) 393.5
H2O(g) 242

The formula of ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

Where,

  • ΔH is the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔH(product) is the standard enthalpy of product at a pressure of 1atm .
  • ΔH(reactant) is the standard enthalpy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔH=npΔH(product)nfΔH(reactant)=[(393.5)+2(242){(75)+2(0)}]kJ=802.5kJ_

The value of ΔG(kJ/mol) for the given reactant and product is,

Molecules ΔG(kJ/mol)
O2(g) 0
CH4(g) 51
CO2(g) 394
H2O(g) 229

The formula of ΔG is,

ΔG=npΔG(product)nfΔG(reactant)

Where,

  • ΔG is the standard Gibb’s free energy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔG(product) is the standard Gibb’s free energy of product at a pressure of 1atm .
  • ΔG(reactant) is the standard Gibb’s free energy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔG=npΔG(product)nfΔG(reactant)=[(394)+2(229){(51)+2(0)}]kJ=801kJ_

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The values of ΔS,ΔH and ΔG is to be calculated at 25°C in each case.

Concept introduction: The expression for ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

The expression for ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

The expression for ΔG is,

ΔG=npΔG(product)nfΔG(reactant)

Explanation of Solution

Explanation

The stated reaction is,

6CO2(g)+6H2O(g)C6H12O6(s)+6O2(g)

Refer to Appendix 4 .

The value of ΔS(J/Kmol) for the given reactant and product is,

Molecules ΔS(J/Kmol)
O2(g) 205
C6H12O6(s) 212
CO2(g) 214
H2O(l) 70

The formula of ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

Where,

  • ΔS is the standard entropy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔS(product) is the standard entropy of product at a pressure of 1atm .
  • ΔS(reactant) is the standard entropy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔS=npΔS(product)nfΔS(reactant)=[(212)+6(205){6(214)+6(70)}]J/K=262J/K_

The value of ΔH(kJ/mol) for the given reactant and product is,

Molecules ΔH(kJ/mol)
O2(g) 0
C6H12O6(s) 1275
CO2(g) 393.5
H2O(l) 286

The formula of ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

Where,

  • ΔH is the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔH(product) is the standard enthalpy of product at a pressure of 1atm .
  • ΔH(reactant) is the standard enthalpy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔH=npΔH(product)nfΔH(reactant)=[(1275)+6(0){6(393.5)+6(286)}]kJ=2802kJ_

The value of ΔG(kJ/mol) for the given reactant and product is,

Molecules ΔG(kJ/mol)
O2(g) 0
C6H12O6(s) 911
CO2(g) 394
H2O(l) 237

The formula of ΔG is,

ΔG=npΔG(product)nfΔG(reactant)

Where,

  • ΔG is the standard Gibb’s free energy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔG(product) is the standard Gibb’s free energy of product at a pressure of 1atm .
  • ΔG(reactant) is the standard Gibb’s free energy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔG=npΔG(product)nfΔG(reactant)=[(911)+6(0){6(394)+6(237)}]kJ=2875kJ_

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The values of ΔS,ΔH and ΔG is to be calculated at 25°C in each case.

Concept introduction: The expression for ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

The expression for ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

The expression for ΔG is,

ΔG=npΔG(product)nfΔG(reactant)

Explanation of Solution

Explanation

The stated reaction is,

P4O10(s)+6H2O(g)4H3PO4(s)

Refer to Appendix 4 .

The value of ΔS(J/Kmol) for the given reactant and product is,

Molecules ΔS(J/Kmol)
H3PO4(s) 110
P4O10(s) 229
H2(l) 70

The formula of ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

Where,

  • ΔS is the standard entropy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔS(product) is the standard entropy of product at a pressure of 1atm .
  • ΔS(reactant) is the standard entropy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔS=npΔS(product)nfΔS(reactant)=[4(110){(229)+6(70)}]J/K=209J/K_

The value of ΔH(kJ/mol) for the given reactant and product is,

Molecules ΔH(kJ/mol)
H3PO4(s) 1279
P4O10(s) 2984
H2(l) 286

The formula of ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

Where,

  • ΔH is the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔH(product) is the standard enthalpy of product at a pressure of 1atm .
  • ΔH(reactant) is the standard enthalpy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔH=npΔH(product)nfΔH(reactant)=[4(1279){(2984)+6(286)}]kJ=416kJ_

The value of ΔG(kJ/mol) for the given reactant and product is,

Molecules ΔG(kJ/mol)
H3PO4(s) 1119
P4O10(s) 2698
H2(l) 237

The formula of ΔG is,

ΔG=npΔG(product)nfΔG(reactant)

Where,

  • ΔG is the standard Gibb’s free energy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔG(product) is the standard Gibb’s free energy of product at a pressure of 1atm .
  • ΔG(reactant) is the standard Gibb’s free energy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔG=npΔG(product)nfΔG(reactant)=[4(1119){(2698)+6(237)}]kJ=356kJ_

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The values of ΔS,ΔH and ΔG is to be calculated at 25°C in each case.

Concept introduction: The expression for ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

The expression for ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

The expression for ΔG is,

ΔG=npΔG(product)nfΔG(reactant)

Explanation of Solution

Explanation

The stated reaction is,

P4O10(s)+6H2O(g)4H3PO4(s)

Refer to Appendix 4 .

The value of ΔS(J/Kmol) for the given reactant and product is,

Molecules ΔS(J/Kmol)
NH4Cl 96
HCl(g) 187
NH3 (g) 193

The formula of ΔS is,

ΔS=npΔS(product)nfΔS(reactant)

Where,

  • ΔS is the standard entropy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔS(product) is the standard entropy of product at a pressure of 1atm .
  • ΔS(reactant) is the standard entropy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔS=npΔS(product)nfΔS(reactant)=[(96){(193)+(187)}]J/K=284J/K_

The value of ΔH(kJ/mol) for the given reactant and product is,

Molecules ΔH(kJ/mol)
NH4Cl 314
HCl(g) 92
NH3 (g) 46

The formula of ΔH is,

ΔH=npΔH(product)nfΔH(reactant)

Where,

  • ΔH is the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔH(product) is the standard enthalpy of product at a pressure of 1atm .
  • ΔH(reactant) is the standard enthalpy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔH=npΔH(product)nfΔH(reactant)=[(314){(92)+(46)}]kJ=176kJ_

The value of ΔG(kJ/mol) for the given reactant and product is,

Molecules ΔG(kJ/mol)
NH4Cl 203
HCl(g) 95
NH3 (g) 17

The formula of ΔG is,

ΔG=npΔG(product)nfΔG(reactant)

Where,

  • ΔG is the standard Gibb’s free energy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔG(product) is the standard Gibb’s free energy of product at a pressure of 1atm .
  • ΔG(reactant) is the standard Gibb’s free energy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔG=npΔG(product)nfΔG(reactant)=[(203){(95)+(17)}]kJ=91kJ_

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Chapter 17 Solutions

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