Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated. Concept introduction: The standard enthalpy of the reaction is calculated by the formula, Δ H ∘ reaction = Δ H ∘ formation ( product ) − Δ H ∘ formation ( reactant ) The change in standard Gipp’s free energy of the reaction is calculated as, Δ G ∘ = Δ H ∘ − T Δ S ∘ To determine: Standard enthalpy of the reaction and change in standard entropy of the given reaction.

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Answer 1

Question

Chapter 20, Problem 87CWP

(a)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: Standard enthalpy of the reaction and change in standard entropy of the given reaction.

(a)

Expert Solution

Answer to Problem 87CWP

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol_ and 134 J/K⋅mol_ , respectively.

Explanation of Solution

Hydrogen gas is produced by reacting graphite with water.

C(s)+H2O(g)→CO(g)+H2(g)

The standard enthalpy of formation of H2O(g) is −242 kJ/mol .

The standard enthalpy of formation of CO(g) is −110.5 kJ/mol .

The standard enthalpy of formation of C(s) and H2(g) is zero since the standard enthalpy of formation of free elements in their standard state is zero.

The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

Therefore, the above equation becomes,

ΔH∘reaction=ΔH∘formation(CO(g))−ΔH∘formation(H2O(g))

Substitute the value of ΔH∘formation(CO(g)) and ΔH∘formation(H2O(g)) in the above equation.

ΔH∘reaction=(−110.5 kJ/mol)−(−242 kJ/mol)=−110.5 kJ/mol+242 kJ/mol=+131.5 kJ/mol_

Therefore, the standard enthalpy of the reaction is +131.5 kJ/mol .

The standard entropy of H2O(g) is 189 J/K⋅mol .

The standard entropy of CO(g) is 198 J/K⋅mol .

The standard entropy of C(s) is 6 J/K⋅mol .

The standard entropy of H2(g) is 131 J/K⋅mol .

The standard entropy change of the reaction is calculated by the formula,

ΔS∘reaction=S∘(product)−S∘(reactant)

Therefore, the above equation becomes,

ΔS∘reaction=[S∘(CO(g))+S∘(H2(g))]−[S∘(H2O(g))+S∘(C(s))]

Substitute the value of S∘(CO(g)) , S∘(H2(g)) , S∘(H2O(g)) and S∘(C(s)) in the above equation.

ΔS∘reaction=[198 J/K⋅mol+131 J/K⋅mol]−[189 J/K⋅mol+6 J/K⋅mol]=329 J/K⋅mol−195 J/K⋅mol=134 J/K⋅mol_

Therefore, the standard entropy change of the reaction is 134 J/K⋅mol_ .

(b)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: The temperature at which the change in standard Gipp’s energy of the given reaction is zero.

(b)

Expert Solution

Answer to Problem 87CWP

The temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

Explanation of Solution

Given

The change in standard Gipp’s energy of the given reaction is zero.

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol and 134 J/K⋅mol , respectively.

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

Substitute the value of ΔG∘ , ΔH∘ and ΔS∘ in the above equation.

0=+131.5×103 J/mol−T×134 J/K⋅molT×134 J/K⋅mol=131.5×103 J/molT=131.5×103 J/mol134 J/K⋅mol=981.3 K_

Therefore, the temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

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Answer 2

Question

Chapter 20, Problem 87CWP

(a)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: Standard enthalpy of the reaction and change in standard entropy of the given reaction.

(a)

Expert Solution

Answer to Problem 87CWP

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol_ and 134 J/K⋅mol_ , respectively.

Explanation of Solution

Hydrogen gas is produced by reacting graphite with water.

C(s)+H2O(g)→CO(g)+H2(g)

The standard enthalpy of formation of H2O(g) is −242 kJ/mol .

The standard enthalpy of formation of CO(g) is −110.5 kJ/mol .

The standard enthalpy of formation of C(s) and H2(g) is zero since the standard enthalpy of formation of free elements in their standard state is zero.

The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

Therefore, the above equation becomes,

ΔH∘reaction=ΔH∘formation(CO(g))−ΔH∘formation(H2O(g))

Substitute the value of ΔH∘formation(CO(g)) and ΔH∘formation(H2O(g)) in the above equation.

ΔH∘reaction=(−110.5 kJ/mol)−(−242 kJ/mol)=−110.5 kJ/mol+242 kJ/mol=+131.5 kJ/mol_

Therefore, the standard enthalpy of the reaction is +131.5 kJ/mol .

The standard entropy of H2O(g) is 189 J/K⋅mol .

The standard entropy of CO(g) is 198 J/K⋅mol .

The standard entropy of C(s) is 6 J/K⋅mol .

The standard entropy of H2(g) is 131 J/K⋅mol .

The standard entropy change of the reaction is calculated by the formula,

ΔS∘reaction=S∘(product)−S∘(reactant)

Therefore, the above equation becomes,

ΔS∘reaction=[S∘(CO(g))+S∘(H2(g))]−[S∘(H2O(g))+S∘(C(s))]

Substitute the value of S∘(CO(g)) , S∘(H2(g)) , S∘(H2O(g)) and S∘(C(s)) in the above equation.

ΔS∘reaction=[198 J/K⋅mol+131 J/K⋅mol]−[189 J/K⋅mol+6 J/K⋅mol]=329 J/K⋅mol−195 J/K⋅mol=134 J/K⋅mol_

Therefore, the standard entropy change of the reaction is 134 J/K⋅mol_ .

(b)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: The temperature at which the change in standard Gipp’s energy of the given reaction is zero.

(b)

Expert Solution

Answer to Problem 87CWP

The temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

Explanation of Solution

Given

The change in standard Gipp’s energy of the given reaction is zero.

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol and 134 J/K⋅mol , respectively.

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

Substitute the value of ΔG∘ , ΔH∘ and ΔS∘ in the above equation.

0=+131.5×103 J/mol−T×134 J/K⋅molT×134 J/K⋅mol=131.5×103 J/molT=131.5×103 J/mol134 J/K⋅mol=981.3 K_

Therefore, the temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

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Answer 3

Question

Chapter 20, Problem 87CWP

(a)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: Standard enthalpy of the reaction and change in standard entropy of the given reaction.

(a)

Expert Solution

Answer to Problem 87CWP

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol_ and 134 J/K⋅mol_ , respectively.

Explanation of Solution

Hydrogen gas is produced by reacting graphite with water.

C(s)+H2O(g)→CO(g)+H2(g)

The standard enthalpy of formation of H2O(g) is −242 kJ/mol .

The standard enthalpy of formation of CO(g) is −110.5 kJ/mol .

The standard enthalpy of formation of C(s) and H2(g) is zero since the standard enthalpy of formation of free elements in their standard state is zero.

The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

Therefore, the above equation becomes,

ΔH∘reaction=ΔH∘formation(CO(g))−ΔH∘formation(H2O(g))

Substitute the value of ΔH∘formation(CO(g)) and ΔH∘formation(H2O(g)) in the above equation.

ΔH∘reaction=(−110.5 kJ/mol)−(−242 kJ/mol)=−110.5 kJ/mol+242 kJ/mol=+131.5 kJ/mol_

Therefore, the standard enthalpy of the reaction is +131.5 kJ/mol .

The standard entropy of H2O(g) is 189 J/K⋅mol .

The standard entropy of CO(g) is 198 J/K⋅mol .

The standard entropy of C(s) is 6 J/K⋅mol .

The standard entropy of H2(g) is 131 J/K⋅mol .

The standard entropy change of the reaction is calculated by the formula,

ΔS∘reaction=S∘(product)−S∘(reactant)

Therefore, the above equation becomes,

ΔS∘reaction=[S∘(CO(g))+S∘(H2(g))]−[S∘(H2O(g))+S∘(C(s))]

Substitute the value of S∘(CO(g)) , S∘(H2(g)) , S∘(H2O(g)) and S∘(C(s)) in the above equation.

ΔS∘reaction=[198 J/K⋅mol+131 J/K⋅mol]−[189 J/K⋅mol+6 J/K⋅mol]=329 J/K⋅mol−195 J/K⋅mol=134 J/K⋅mol_

Therefore, the standard entropy change of the reaction is 134 J/K⋅mol_ .

(b)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: The temperature at which the change in standard Gipp’s energy of the given reaction is zero.

(b)

Expert Solution

Answer to Problem 87CWP

The temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

Explanation of Solution

Given

The change in standard Gipp’s energy of the given reaction is zero.

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol and 134 J/K⋅mol , respectively.

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

Substitute the value of ΔG∘ , ΔH∘ and ΔS∘ in the above equation.

0=+131.5×103 J/mol−T×134 J/K⋅molT×134 J/K⋅mol=131.5×103 J/molT=131.5×103 J/mol134 J/K⋅mol=981.3 K_

Therefore, the temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

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Answer 4

Question

Chapter 20, Problem 87CWP

(a)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: Standard enthalpy of the reaction and change in standard entropy of the given reaction.

(a)

Expert Solution

Answer to Problem 87CWP

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol_ and 134 J/K⋅mol_ , respectively.

Explanation of Solution

Hydrogen gas is produced by reacting graphite with water.

C(s)+H2O(g)→CO(g)+H2(g)

The standard enthalpy of formation of H2O(g) is −242 kJ/mol .

The standard enthalpy of formation of CO(g) is −110.5 kJ/mol .

The standard enthalpy of formation of C(s) and H2(g) is zero since the standard enthalpy of formation of free elements in their standard state is zero.

The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

Therefore, the above equation becomes,

ΔH∘reaction=ΔH∘formation(CO(g))−ΔH∘formation(H2O(g))

Substitute the value of ΔH∘formation(CO(g)) and ΔH∘formation(H2O(g)) in the above equation.

ΔH∘reaction=(−110.5 kJ/mol)−(−242 kJ/mol)=−110.5 kJ/mol+242 kJ/mol=+131.5 kJ/mol_

Therefore, the standard enthalpy of the reaction is +131.5 kJ/mol .

The standard entropy of H2O(g) is 189 J/K⋅mol .

The standard entropy of CO(g) is 198 J/K⋅mol .

The standard entropy of C(s) is 6 J/K⋅mol .

The standard entropy of H2(g) is 131 J/K⋅mol .

The standard entropy change of the reaction is calculated by the formula,

ΔS∘reaction=S∘(product)−S∘(reactant)

Therefore, the above equation becomes,

ΔS∘reaction=[S∘(CO(g))+S∘(H2(g))]−[S∘(H2O(g))+S∘(C(s))]

Substitute the value of S∘(CO(g)) , S∘(H2(g)) , S∘(H2O(g)) and S∘(C(s)) in the above equation.

ΔS∘reaction=[198 J/K⋅mol+131 J/K⋅mol]−[189 J/K⋅mol+6 J/K⋅mol]=329 J/K⋅mol−195 J/K⋅mol=134 J/K⋅mol_

Therefore, the standard entropy change of the reaction is 134 J/K⋅mol_ .

(b)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: The temperature at which the change in standard Gipp’s energy of the given reaction is zero.

(b)

Expert Solution

Answer to Problem 87CWP

The temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

Explanation of Solution

Given

The change in standard Gipp’s energy of the given reaction is zero.

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol and 134 J/K⋅mol , respectively.

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

Substitute the value of ΔG∘ , ΔH∘ and ΔS∘ in the above equation.

0=+131.5×103 J/mol−T×134 J/K⋅molT×134 J/K⋅mol=131.5×103 J/molT=131.5×103 J/mol134 J/K⋅mol=981.3 K_

Therefore, the temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

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Answer 5

Chapter 20, Problem 87CWP

(a)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: Standard enthalpy of the reaction and change in standard entropy of the given reaction.

(a)

Expert Solution

Answer to Problem 87CWP

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol_ and 134 J/K⋅mol_ , respectively.

Explanation of Solution

Hydrogen gas is produced by reacting graphite with water.

C(s)+H2O(g)→CO(g)+H2(g)

The standard enthalpy of formation of H2O(g) is −242 kJ/mol .

The standard enthalpy of formation of CO(g) is −110.5 kJ/mol .

The standard enthalpy of formation of C(s) and H2(g) is zero since the standard enthalpy of formation of free elements in their standard state is zero.

The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

Therefore, the above equation becomes,

ΔH∘reaction=ΔH∘formation(CO(g))−ΔH∘formation(H2O(g))

Substitute the value of ΔH∘formation(CO(g)) and ΔH∘formation(H2O(g)) in the above equation.

ΔH∘reaction=(−110.5 kJ/mol)−(−242 kJ/mol)=−110.5 kJ/mol+242 kJ/mol=+131.5 kJ/mol_

Therefore, the standard enthalpy of the reaction is +131.5 kJ/mol .

The standard entropy of H2O(g) is 189 J/K⋅mol .

The standard entropy of CO(g) is 198 J/K⋅mol .

The standard entropy of C(s) is 6 J/K⋅mol .

The standard entropy of H2(g) is 131 J/K⋅mol .

The standard entropy change of the reaction is calculated by the formula,

ΔS∘reaction=S∘(product)−S∘(reactant)

Therefore, the above equation becomes,

ΔS∘reaction=[S∘(CO(g))+S∘(H2(g))]−[S∘(H2O(g))+S∘(C(s))]

Substitute the value of S∘(CO(g)) , S∘(H2(g)) , S∘(H2O(g)) and S∘(C(s)) in the above equation.

ΔS∘reaction=[198 J/K⋅mol+131 J/K⋅mol]−[189 J/K⋅mol+6 J/K⋅mol]=329 J/K⋅mol−195 J/K⋅mol=134 J/K⋅mol_

Therefore, the standard entropy change of the reaction is 134 J/K⋅mol_ .

(b)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: The temperature at which the change in standard Gipp’s energy of the given reaction is zero.

(b)

Expert Solution

Answer to Problem 87CWP

The temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

Explanation of Solution

Given

The change in standard Gipp’s energy of the given reaction is zero.

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol and 134 J/K⋅mol , respectively.

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

Substitute the value of ΔG∘ , ΔH∘ and ΔS∘ in the above equation.

0=+131.5×103 J/mol−T×134 J/K⋅molT×134 J/K⋅mol=131.5×103 J/molT=131.5×103 J/mol134 J/K⋅mol=981.3 K_

Therefore, the temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

Answer 6

Chapter 20, Problem 87CWP

(a)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: Standard enthalpy of the reaction and change in standard entropy of the given reaction.

(a)

Expert Solution

Answer to Problem 87CWP

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol_ and 134 J/K⋅mol_ , respectively.

Explanation of Solution

Hydrogen gas is produced by reacting graphite with water.

C(s)+H2O(g)→CO(g)+H2(g)

The standard enthalpy of formation of H2O(g) is −242 kJ/mol .

The standard enthalpy of formation of CO(g) is −110.5 kJ/mol .

The standard enthalpy of formation of C(s) and H2(g) is zero since the standard enthalpy of formation of free elements in their standard state is zero.

The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

Therefore, the above equation becomes,

ΔH∘reaction=ΔH∘formation(CO(g))−ΔH∘formation(H2O(g))

Substitute the value of ΔH∘formation(CO(g)) and ΔH∘formation(H2O(g)) in the above equation.

ΔH∘reaction=(−110.5 kJ/mol)−(−242 kJ/mol)=−110.5 kJ/mol+242 kJ/mol=+131.5 kJ/mol_

Therefore, the standard enthalpy of the reaction is +131.5 kJ/mol .

The standard entropy of H2O(g) is 189 J/K⋅mol .

The standard entropy of CO(g) is 198 J/K⋅mol .

The standard entropy of C(s) is 6 J/K⋅mol .

The standard entropy of H2(g) is 131 J/K⋅mol .

The standard entropy change of the reaction is calculated by the formula,

ΔS∘reaction=S∘(product)−S∘(reactant)

Therefore, the above equation becomes,

ΔS∘reaction=[S∘(CO(g))+S∘(H2(g))]−[S∘(H2O(g))+S∘(C(s))]

Substitute the value of S∘(CO(g)) , S∘(H2(g)) , S∘(H2O(g)) and S∘(C(s)) in the above equation.

ΔS∘reaction=[198 J/K⋅mol+131 J/K⋅mol]−[189 J/K⋅mol+6 J/K⋅mol]=329 J/K⋅mol−195 J/K⋅mol=134 J/K⋅mol_

Therefore, the standard entropy change of the reaction is 134 J/K⋅mol_ .

Answer 7

Chapter 20, Problem 87CWP

(a)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: Standard enthalpy of the reaction and change in standard entropy of the given reaction.

Answer 8

(a)

Expert Solution

Answer to Problem 87CWP

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol_ and 134 J/K⋅mol_ , respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Hydrogen gas is produced by reacting graphite with water.

C(s)+H2O(g)→CO(g)+H2(g)

The standard enthalpy of formation of H2O(g) is −242 kJ/mol .

The standard enthalpy of formation of CO(g) is −110.5 kJ/mol .

The standard enthalpy of formation of C(s) and H2(g) is zero since the standard enthalpy of formation of free elements in their standard state is zero.

The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

Therefore, the above equation becomes,

ΔH∘reaction=ΔH∘formation(CO(g))−ΔH∘formation(H2O(g))

Substitute the value of ΔH∘formation(CO(g)) and ΔH∘formation(H2O(g)) in the above equation.

ΔH∘reaction=(−110.5 kJ/mol)−(−242 kJ/mol)=−110.5 kJ/mol+242 kJ/mol=+131.5 kJ/mol_

Therefore, the standard enthalpy of the reaction is +131.5 kJ/mol .

The standard entropy of H2O(g) is 189 J/K⋅mol .

The standard entropy of CO(g) is 198 J/K⋅mol .

The standard entropy of C(s) is 6 J/K⋅mol .

The standard entropy of H2(g) is 131 J/K⋅mol .

The standard entropy change of the reaction is calculated by the formula,

ΔS∘reaction=S∘(product)−S∘(reactant)

Therefore, the above equation becomes,

ΔS∘reaction=[S∘(CO(g))+S∘(H2(g))]−[S∘(H2O(g))+S∘(C(s))]

Substitute the value of S∘(CO(g)) , S∘(H2(g)) , S∘(H2O(g)) and S∘(C(s)) in the above equation.

ΔS∘reaction=[198 J/K⋅mol+131 J/K⋅mol]−[189 J/K⋅mol+6 J/K⋅mol]=329 J/K⋅mol−195 J/K⋅mol=134 J/K⋅mol_

Therefore, the standard entropy change of the reaction is 134 J/K⋅mol_ .

Answer 13

(b)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: The temperature at which the change in standard Gipp’s energy of the given reaction is zero.

(b)

Expert Solution

Answer to Problem 87CWP

The temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

Explanation of Solution

Given

The change in standard Gipp’s energy of the given reaction is zero.

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol and 134 J/K⋅mol , respectively.

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

Substitute the value of ΔG∘ , ΔH∘ and ΔS∘ in the above equation.

0=+131.5×103 J/mol−T×134 J/K⋅molT×134 J/K⋅mol=131.5×103 J/molT=131.5×103 J/mol134 J/K⋅mol=981.3 K_

Therefore, the temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

Answer 14

(b)

Interpretation Introduction

Interpretation: Standard enthalpy of the reaction and change in standard entropy of the given reaction are to be determined and the temperature at which the change in standard Gibb’s energy is zero, is to be calculated.

Concept introduction: The standard enthalpy of the reaction is calculated by the formula,

ΔH∘reaction=ΔH∘formation(product)−ΔH∘formation(reactant)

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

To determine: The temperature at which the change in standard Gipp’s energy of the given reaction is zero.

Answer 15

(b)

Expert Solution

Answer to Problem 87CWP

The temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given

The change in standard Gipp’s energy of the given reaction is zero.

Standard enthalpy of the reaction and change in standard entropy of the given reaction is +131.5 kJ/mol and 134 J/K⋅mol , respectively.

The change in standard Gipp’s free energy of the reaction is calculated as,

ΔG∘=ΔH∘−TΔS∘

Substitute the value of ΔG∘ , ΔH∘ and ΔS∘ in the above equation.

0=+131.5×103 J/mol−T×134 J/K⋅molT×134 J/K⋅mol=131.5×103 J/molT=131.5×103 J/mol134 J/K⋅mol=981.3 K_

Therefore, the temperature at which the change in standard Gipp’s energy of the given reaction is zero is 981.3 K_

Answer 20

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Answer to Problem 87CWP

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Answer to Problem 87CWP

Explanation of Solution

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