To determine the Δ r G ° for the reaction of silver with HX where X is a halogen. Concept introduction: The Gibbs free energy or the free energy change is a thermodynamic quantity represented by Δ r G ∘ . It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is: Δ r G ° = ∑ n Δ f G ° ( products ) − ∑ n Δ f G ° ( reactants ) It is related to entropy and entropy by the following expression, Δ r G ∘ = Δ r H ∘ − T Δ r S ∘ Here, Δ r H ∘ is the change in enthalpy and Δ r S ∘ is the change in entropy.

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Answer 1

Question

Chapter 21, Problem 92GQ

Interpretation Introduction

Interpretation:

To determine the ΔrG° for the reaction of silver with HX where X is a halogen.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrG∘. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)

It is related to entropy and entropy by the following expression,

ΔrG∘=ΔrH∘−TΔrS∘

Here, ΔrH∘ is the change in enthalpy and ΔrS∘ is the change in entropy.

Expert Solution & Answer

Answer to Problem 92GQ

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

Explanation of Solution

The ΔrG∘ for the reaction of silver with hydrogen halide is calculated below.

Given:

Refer to Appendix L for the values of standard free energies.

For X=F,

The given reaction is,

Ag(s)+HF(g)→AgF(s)+12H2(g)

The ΔfG° for AgF(s) is −193.8 kJ/mol.

The ΔfG° for HF(g) is −273.2 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgF(s)/mol-rxn)ΔfG°[AgF(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HF(g)/mol-rxn)ΔfG°[HF(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgF(s)/mol-rxn)(−193.8 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HF(g)/mol-rxn)(−273.2 kJ/mol)] ] =+79.4 kJ/mol-rxn

For X=Cl,

The given reaction is,

Ag(s)+HCl(g)→AgCl(s)+12H2(g)

The ΔfG° for AgCl(s) is −109.76 kJ/mol.

The ΔfG° for HCl(g) is −95.09 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgCl(s)/mol-rxn)ΔfG°[AgCl(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HCl(g)/mol-rxn)ΔfG°[HCl(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgCl(s)/mol-rxn)(−109.76 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HCl(g)/mol-rxn)(−95.09 kJ/mol)] ] =−14.67 kJ/mol-rxn

For X=Br,

The given reaction is,

Ag(s)+HBr(g)→AgBr(s)+12H2(g)

The ΔfG° for AgBr(s) is −96.90 kJ/mol.

The ΔfG° for HBr(g) is −53.45 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgBr(s)/mol-rxn)ΔfG°[AgBr(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HBr(g)/mol-rxn)ΔfG°[HBr(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgBr(s)/mol-rxn)(−96.90 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HBr(g)/mol-rxn)(−53.45 kJ/mol)] ] =−43.45 kJ/mol-rxn

For X=I,

The given reaction is,

Ag(s)+HI(g)→AgI(s)+12H2(g)

The ΔfG° for AgI(s) is −66.19 kJ/mol.

The ΔfG° for HI(g) is −1.56 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgI(s)/mol-rxn)ΔfG°[AgI(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HI(g)/mol-rxn)ΔfG°[HI(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgI(s)/mol-rxn)(−66.19 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HI(g)/mol-rxn)(−1.56 kJ/mol)] ] =−64.63 kJ/mol-rxn

The free energy change for the reaction of silver with hydrogen fluoride is positive and the reaction is not spontaneous. The rest three reactions are spontaneous.

Conclusion

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

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. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. < cleavage Bond A • CH3 + 26. t cleavage 2°C• +3°C• Bond C Cleavage CH3 ZC '2°C. 26. E Strongest 3°C. 2C. Gund Largest BDE weakest bond In that molecule a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest C bond Produces A Weakest Bond Most Strongest Bond Stable radical Strongest Gund produces least stable radicals b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 人 8°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. methyl radical •CH3 formed in bund A Cleavage

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Answer 2

Question

Chapter 21, Problem 92GQ

Interpretation Introduction

Interpretation:

To determine the ΔrG° for the reaction of silver with HX where X is a halogen.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrG∘. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)

It is related to entropy and entropy by the following expression,

ΔrG∘=ΔrH∘−TΔrS∘

Here, ΔrH∘ is the change in enthalpy and ΔrS∘ is the change in entropy.

Expert Solution & Answer

Answer to Problem 92GQ

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

Explanation of Solution

The ΔrG∘ for the reaction of silver with hydrogen halide is calculated below.

Given:

Refer to Appendix L for the values of standard free energies.

For X=F,

The given reaction is,

Ag(s)+HF(g)→AgF(s)+12H2(g)

The ΔfG° for AgF(s) is −193.8 kJ/mol.

The ΔfG° for HF(g) is −273.2 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgF(s)/mol-rxn)ΔfG°[AgF(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HF(g)/mol-rxn)ΔfG°[HF(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgF(s)/mol-rxn)(−193.8 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HF(g)/mol-rxn)(−273.2 kJ/mol)] ] =+79.4 kJ/mol-rxn

For X=Cl,

The given reaction is,

Ag(s)+HCl(g)→AgCl(s)+12H2(g)

The ΔfG° for AgCl(s) is −109.76 kJ/mol.

The ΔfG° for HCl(g) is −95.09 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgCl(s)/mol-rxn)ΔfG°[AgCl(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HCl(g)/mol-rxn)ΔfG°[HCl(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgCl(s)/mol-rxn)(−109.76 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HCl(g)/mol-rxn)(−95.09 kJ/mol)] ] =−14.67 kJ/mol-rxn

For X=Br,

The given reaction is,

Ag(s)+HBr(g)→AgBr(s)+12H2(g)

The ΔfG° for AgBr(s) is −96.90 kJ/mol.

The ΔfG° for HBr(g) is −53.45 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgBr(s)/mol-rxn)ΔfG°[AgBr(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HBr(g)/mol-rxn)ΔfG°[HBr(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgBr(s)/mol-rxn)(−96.90 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HBr(g)/mol-rxn)(−53.45 kJ/mol)] ] =−43.45 kJ/mol-rxn

For X=I,

The given reaction is,

Ag(s)+HI(g)→AgI(s)+12H2(g)

The ΔfG° for AgI(s) is −66.19 kJ/mol.

The ΔfG° for HI(g) is −1.56 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgI(s)/mol-rxn)ΔfG°[AgI(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HI(g)/mol-rxn)ΔfG°[HI(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgI(s)/mol-rxn)(−66.19 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HI(g)/mol-rxn)(−1.56 kJ/mol)] ] =−64.63 kJ/mol-rxn

The free energy change for the reaction of silver with hydrogen fluoride is positive and the reaction is not spontaneous. The rest three reactions are spontaneous.

Conclusion

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

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Answer 3

Question

Chapter 21, Problem 92GQ

Interpretation Introduction

Interpretation:

To determine the ΔrG° for the reaction of silver with HX where X is a halogen.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrG∘. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)

It is related to entropy and entropy by the following expression,

ΔrG∘=ΔrH∘−TΔrS∘

Here, ΔrH∘ is the change in enthalpy and ΔrS∘ is the change in entropy.

Expert Solution & Answer

Answer to Problem 92GQ

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

Explanation of Solution

The ΔrG∘ for the reaction of silver with hydrogen halide is calculated below.

Given:

Refer to Appendix L for the values of standard free energies.

For X=F,

The given reaction is,

Ag(s)+HF(g)→AgF(s)+12H2(g)

The ΔfG° for AgF(s) is −193.8 kJ/mol.

The ΔfG° for HF(g) is −273.2 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgF(s)/mol-rxn)ΔfG°[AgF(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HF(g)/mol-rxn)ΔfG°[HF(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgF(s)/mol-rxn)(−193.8 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HF(g)/mol-rxn)(−273.2 kJ/mol)] ] =+79.4 kJ/mol-rxn

For X=Cl,

The given reaction is,

Ag(s)+HCl(g)→AgCl(s)+12H2(g)

The ΔfG° for AgCl(s) is −109.76 kJ/mol.

The ΔfG° for HCl(g) is −95.09 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgCl(s)/mol-rxn)ΔfG°[AgCl(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HCl(g)/mol-rxn)ΔfG°[HCl(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgCl(s)/mol-rxn)(−109.76 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HCl(g)/mol-rxn)(−95.09 kJ/mol)] ] =−14.67 kJ/mol-rxn

For X=Br,

The given reaction is,

Ag(s)+HBr(g)→AgBr(s)+12H2(g)

The ΔfG° for AgBr(s) is −96.90 kJ/mol.

The ΔfG° for HBr(g) is −53.45 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgBr(s)/mol-rxn)ΔfG°[AgBr(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HBr(g)/mol-rxn)ΔfG°[HBr(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgBr(s)/mol-rxn)(−96.90 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HBr(g)/mol-rxn)(−53.45 kJ/mol)] ] =−43.45 kJ/mol-rxn

For X=I,

The given reaction is,

Ag(s)+HI(g)→AgI(s)+12H2(g)

The ΔfG° for AgI(s) is −66.19 kJ/mol.

The ΔfG° for HI(g) is −1.56 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgI(s)/mol-rxn)ΔfG°[AgI(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HI(g)/mol-rxn)ΔfG°[HI(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgI(s)/mol-rxn)(−66.19 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HI(g)/mol-rxn)(−1.56 kJ/mol)] ] =−64.63 kJ/mol-rxn

The free energy change for the reaction of silver with hydrogen fluoride is positive and the reaction is not spontaneous. The rest three reactions are spontaneous.

Conclusion

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

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Answer 4

Question

Chapter 21, Problem 92GQ

Interpretation Introduction

Interpretation:

To determine the ΔrG° for the reaction of silver with HX where X is a halogen.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrG∘. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)

It is related to entropy and entropy by the following expression,

ΔrG∘=ΔrH∘−TΔrS∘

Here, ΔrH∘ is the change in enthalpy and ΔrS∘ is the change in entropy.

Expert Solution & Answer

Answer to Problem 92GQ

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

Explanation of Solution

The ΔrG∘ for the reaction of silver with hydrogen halide is calculated below.

Given:

Refer to Appendix L for the values of standard free energies.

For X=F,

The given reaction is,

Ag(s)+HF(g)→AgF(s)+12H2(g)

The ΔfG° for AgF(s) is −193.8 kJ/mol.

The ΔfG° for HF(g) is −273.2 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgF(s)/mol-rxn)ΔfG°[AgF(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HF(g)/mol-rxn)ΔfG°[HF(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgF(s)/mol-rxn)(−193.8 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HF(g)/mol-rxn)(−273.2 kJ/mol)] ] =+79.4 kJ/mol-rxn

For X=Cl,

The given reaction is,

Ag(s)+HCl(g)→AgCl(s)+12H2(g)

The ΔfG° for AgCl(s) is −109.76 kJ/mol.

The ΔfG° for HCl(g) is −95.09 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgCl(s)/mol-rxn)ΔfG°[AgCl(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HCl(g)/mol-rxn)ΔfG°[HCl(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgCl(s)/mol-rxn)(−109.76 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HCl(g)/mol-rxn)(−95.09 kJ/mol)] ] =−14.67 kJ/mol-rxn

For X=Br,

The given reaction is,

Ag(s)+HBr(g)→AgBr(s)+12H2(g)

The ΔfG° for AgBr(s) is −96.90 kJ/mol.

The ΔfG° for HBr(g) is −53.45 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgBr(s)/mol-rxn)ΔfG°[AgBr(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HBr(g)/mol-rxn)ΔfG°[HBr(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgBr(s)/mol-rxn)(−96.90 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HBr(g)/mol-rxn)(−53.45 kJ/mol)] ] =−43.45 kJ/mol-rxn

For X=I,

The given reaction is,

Ag(s)+HI(g)→AgI(s)+12H2(g)

The ΔfG° for AgI(s) is −66.19 kJ/mol.

The ΔfG° for HI(g) is −1.56 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgI(s)/mol-rxn)ΔfG°[AgI(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HI(g)/mol-rxn)ΔfG°[HI(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgI(s)/mol-rxn)(−66.19 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HI(g)/mol-rxn)(−1.56 kJ/mol)] ] =−64.63 kJ/mol-rxn

The free energy change for the reaction of silver with hydrogen fluoride is positive and the reaction is not spontaneous. The rest three reactions are spontaneous.

Conclusion

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

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Answer 5

Chapter 21, Problem 92GQ

Interpretation Introduction

Interpretation:

To determine the ΔrG° for the reaction of silver with HX where X is a halogen.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrG∘. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)

It is related to entropy and entropy by the following expression,

ΔrG∘=ΔrH∘−TΔrS∘

Here, ΔrH∘ is the change in enthalpy and ΔrS∘ is the change in entropy.

Expert Solution & Answer

Answer to Problem 92GQ

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

Explanation of Solution

The ΔrG∘ for the reaction of silver with hydrogen halide is calculated below.

Given:

Refer to Appendix L for the values of standard free energies.

For X=F,

The given reaction is,

Ag(s)+HF(g)→AgF(s)+12H2(g)

The ΔfG° for AgF(s) is −193.8 kJ/mol.

The ΔfG° for HF(g) is −273.2 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgF(s)/mol-rxn)ΔfG°[AgF(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HF(g)/mol-rxn)ΔfG°[HF(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgF(s)/mol-rxn)(−193.8 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HF(g)/mol-rxn)(−273.2 kJ/mol)] ] =+79.4 kJ/mol-rxn

For X=Cl,

The given reaction is,

Ag(s)+HCl(g)→AgCl(s)+12H2(g)

The ΔfG° for AgCl(s) is −109.76 kJ/mol.

The ΔfG° for HCl(g) is −95.09 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgCl(s)/mol-rxn)ΔfG°[AgCl(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HCl(g)/mol-rxn)ΔfG°[HCl(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgCl(s)/mol-rxn)(−109.76 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HCl(g)/mol-rxn)(−95.09 kJ/mol)] ] =−14.67 kJ/mol-rxn

For X=Br,

The given reaction is,

Ag(s)+HBr(g)→AgBr(s)+12H2(g)

The ΔfG° for AgBr(s) is −96.90 kJ/mol.

The ΔfG° for HBr(g) is −53.45 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgBr(s)/mol-rxn)ΔfG°[AgBr(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HBr(g)/mol-rxn)ΔfG°[HBr(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgBr(s)/mol-rxn)(−96.90 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HBr(g)/mol-rxn)(−53.45 kJ/mol)] ] =−43.45 kJ/mol-rxn

For X=I,

The given reaction is,

Ag(s)+HI(g)→AgI(s)+12H2(g)

The ΔfG° for AgI(s) is −66.19 kJ/mol.

The ΔfG° for HI(g) is −1.56 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgI(s)/mol-rxn)ΔfG°[AgI(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HI(g)/mol-rxn)ΔfG°[HI(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgI(s)/mol-rxn)(−66.19 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HI(g)/mol-rxn)(−1.56 kJ/mol)] ] =−64.63 kJ/mol-rxn

The free energy change for the reaction of silver with hydrogen fluoride is positive and the reaction is not spontaneous. The rest three reactions are spontaneous.

Conclusion

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

Answer 6

Chapter 21, Problem 92GQ

Interpretation Introduction

Interpretation:

To determine the ΔrG° for the reaction of silver with HX where X is a halogen.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrG∘. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)

It is related to entropy and entropy by the following expression,

ΔrG∘=ΔrH∘−TΔrS∘

Here, ΔrH∘ is the change in enthalpy and ΔrS∘ is the change in entropy.

Expert Solution & Answer

Answer to Problem 92GQ

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

Explanation of Solution

The ΔrG∘ for the reaction of silver with hydrogen halide is calculated below.

Given:

Refer to Appendix L for the values of standard free energies.

For X=F,

The given reaction is,

Ag(s)+HF(g)→AgF(s)+12H2(g)

The ΔfG° for AgF(s) is −193.8 kJ/mol.

The ΔfG° for HF(g) is −273.2 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgF(s)/mol-rxn)ΔfG°[AgF(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HF(g)/mol-rxn)ΔfG°[HF(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgF(s)/mol-rxn)(−193.8 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HF(g)/mol-rxn)(−273.2 kJ/mol)] ] =+79.4 kJ/mol-rxn

For X=Cl,

The given reaction is,

Ag(s)+HCl(g)→AgCl(s)+12H2(g)

The ΔfG° for AgCl(s) is −109.76 kJ/mol.

The ΔfG° for HCl(g) is −95.09 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgCl(s)/mol-rxn)ΔfG°[AgCl(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HCl(g)/mol-rxn)ΔfG°[HCl(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgCl(s)/mol-rxn)(−109.76 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HCl(g)/mol-rxn)(−95.09 kJ/mol)] ] =−14.67 kJ/mol-rxn

For X=Br,

The given reaction is,

Ag(s)+HBr(g)→AgBr(s)+12H2(g)

The ΔfG° for AgBr(s) is −96.90 kJ/mol.

The ΔfG° for HBr(g) is −53.45 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgBr(s)/mol-rxn)ΔfG°[AgBr(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HBr(g)/mol-rxn)ΔfG°[HBr(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgBr(s)/mol-rxn)(−96.90 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HBr(g)/mol-rxn)(−53.45 kJ/mol)] ] =−43.45 kJ/mol-rxn

For X=I,

The given reaction is,

Ag(s)+HI(g)→AgI(s)+12H2(g)

The ΔfG° for AgI(s) is −66.19 kJ/mol.

The ΔfG° for HI(g) is −1.56 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgI(s)/mol-rxn)ΔfG°[AgI(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HI(g)/mol-rxn)ΔfG°[HI(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgI(s)/mol-rxn)(−66.19 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HI(g)/mol-rxn)(−1.56 kJ/mol)] ] =−64.63 kJ/mol-rxn

The free energy change for the reaction of silver with hydrogen fluoride is positive and the reaction is not spontaneous. The rest three reactions are spontaneous.

Conclusion

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

Answer 7

Chapter 21, Problem 92GQ

Interpretation Introduction

Interpretation:

To determine the ΔrG° for the reaction of silver with HX where X is a halogen.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrG∘. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)

It is related to entropy and entropy by the following expression,

ΔrG∘=ΔrH∘−TΔrS∘

Here, ΔrH∘ is the change in enthalpy and ΔrS∘ is the change in entropy.

Answer 8

Expert Solution & Answer

Answer to Problem 92GQ

The ΔrG∘ for the reaction of silver with hydrogen fluoride is +79.4 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen chloride is −14.67 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen bromide is −43.45 kJ/mol-rxn.

The ΔrG∘ for the reaction of silver with hydrogen iodide is −64.63 kJ/mol-rxn.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The ΔrG∘ for the reaction of silver with hydrogen halide is calculated below.

Given:

Refer to Appendix L for the values of standard free energies.

For X=F,

The given reaction is,

Ag(s)+HF(g)→AgF(s)+12H2(g)

The ΔfG° for AgF(s) is −193.8 kJ/mol.

The ΔfG° for HF(g) is −273.2 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgF(s)/mol-rxn)ΔfG°[AgF(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HF(g)/mol-rxn)ΔfG°[HF(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgF(s)/mol-rxn)(−193.8 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HF(g)/mol-rxn)(−273.2 kJ/mol)] ] =+79.4 kJ/mol-rxn

For X=Cl,

The given reaction is,

Ag(s)+HCl(g)→AgCl(s)+12H2(g)

The ΔfG° for AgCl(s) is −109.76 kJ/mol.

The ΔfG° for HCl(g) is −95.09 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgCl(s)/mol-rxn)ΔfG°[AgCl(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HCl(g)/mol-rxn)ΔfG°[HCl(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgCl(s)/mol-rxn)(−109.76 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HCl(g)/mol-rxn)(−95.09 kJ/mol)] ] =−14.67 kJ/mol-rxn

For X=Br,

The given reaction is,

Ag(s)+HBr(g)→AgBr(s)+12H2(g)

The ΔfG° for AgBr(s) is −96.90 kJ/mol.

The ΔfG° for HBr(g) is −53.45 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgBr(s)/mol-rxn)ΔfG°[AgBr(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HBr(g)/mol-rxn)ΔfG°[HBr(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgBr(s)/mol-rxn)(−96.90 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HBr(g)/mol-rxn)(−53.45 kJ/mol)] ] =−43.45 kJ/mol-rxn

For X=I,

The given reaction is,

Ag(s)+HI(g)→AgI(s)+12H2(g)

The ΔfG° for AgI(s) is −66.19 kJ/mol.

The ΔfG° for HI(g) is −1.56 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for Ag(s) is 0 kJ/mol.

The expression for free energy change is,

ΔrG°=∑nΔfG°(products)−∑nΔfG°(reactants)=[[(1 mol AgI(s)/mol-rxn)ΔfG°[AgI(s)]+(0.5 mol H2(g)/mol-rxn)ΔfG°[H2(g)]]−[(1 mol Ag(s)/mol-rxn)ΔfG°[Ag(s)]+(1 mol HI(g)/mol-rxn)ΔfG°[HI(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol AgI(s)/mol-rxn)(−66.19 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol)]−[(1 mol Ag(s)/mol-rxn)(0 kJ/mol)+(1 mol HI(g)/mol-rxn)(−1.56 kJ/mol)] ] =−64.63 kJ/mol-rxn

The free energy change for the reaction of silver with hydrogen fluoride is positive and the reaction is not spontaneous. The rest three reactions are spontaneous.