   Chapter 17, Problem 53E

Chapter
Section
Textbook Problem

# 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)

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

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/K⋅mol) 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

(b)

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)

(c)

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)

(d)

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)

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