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Chapter 21, Problem 114GQ
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### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Metals generally react with hydrogen halides to give the metal halide and hydrogen. Determine whether this is true for silver by calculating ΔrG° for the reaction with each of the hydrogen halides.Ag(s) + HX(g) → AgX(s) + ½ H2(g)The required free energies of formation are:

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=ΔrHTΔrS

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

Explanation

The Î”rGâˆ˜ 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,

â€‚Â Ag(s)+HF(g)â†’AgF(s)+12H2(g)

The Î”fGÂ° for AgF(s) is âˆ’193.8Â kJ/mol.

The Î”fGÂ° for HF(g) is âˆ’273.2Â kJ/mol.

The Î”fGÂ° for H2(g) is 0Â kJ/mol.

The Î”fGÂ° for Ag(s) is 0Â kJ/mol.

The expression for free energy change is,

Î”rGÂ°=âˆ‘nÎ”fGÂ°(products)âˆ’âˆ‘nÎ”fGÂ°(reactants)=[[(1Â molÂ AgF(s)/mol-rxn)Î”fGÂ°[AgF(s)]+(0.5Â molÂ H2(g)/mol-rxn)Î”fGÂ°[H2(g)]]âˆ’[(1Â molÂ Ag(s)/mol-rxn)Î”fGÂ°[Ag(s)]+(1Â molÂ HF(g)/mol-rxn)Î”fGÂ°[HF(g)]]Â ]Â

Substitute the values,

Î”rGÂ°=[[(1Â molÂ AgF(s)/mol-rxn)(âˆ’193.8Â kJ/mol)+(0.5Â molÂ H2(g)/mol-rxn)(0Â kJ/mol)]âˆ’[(1Â molÂ Ag(s)/mol-rxn)(0Â kJ/mol)+(1Â molÂ HF(g)/mol-rxn)(âˆ’273.2Â kJ/mol)]Â ]Â =+79.4Â kJ/mol-rxn

For X=Cl,

The given reaction is,

â€‚Â Ag(s)+HCl(g)â†’AgCl(s)+12H2(g)

The Î”fGÂ° for AgCl(s) is âˆ’109.76Â kJ/mol.

The Î”fGÂ° for HCl(g) is âˆ’95.09Â kJ/mol.

The Î”fGÂ° for H2(g) is 0Â kJ/mol.

The Î”fGÂ° for Ag(s) is 0Â kJ/mol.

The expression for free energy change is,

Î”rGÂ°=âˆ‘nÎ”fGÂ°(products)âˆ’âˆ‘nÎ”fGÂ°(reactants)=[[(1Â molÂ AgCl(s)/mol-rxn)Î”fGÂ°[AgCl(s)]+(0.5Â molÂ H2(g)/mol-rxn)Î”fGÂ°[H2(g)]]âˆ’[(1Â molÂ Ag(s)/mol-rxn)Î”fGÂ°[Ag(s)]+(1Â molÂ HCl(g)/mol-rxn)Î”fGÂ°[HCl(g)]]Â ]Â

Substitute the values,

Î”rGÂ°=[[(1Â molÂ AgCl(s)/mol-rxn)(âˆ’109.76Â kJ/mol)+(0.5Â molÂ H2(g)/mol-rxn)(0Â kJ/mol)]âˆ’[(1Â molÂ Ag(s)/mol-rxn)(0Â kJ/mol)+(1Â molÂ HCl(g)/mol-rxn)(âˆ’95.09Â kJ/mol)]Â ]Â =âˆ’14.67Â kJ/mol-rxn

For X=Br,

The given reaction is,

â€‚Â Ag(s)+HBr(g)â†’AgBr(s)+12H2(g)

The Î”fGÂ° for AgBr(s) is âˆ’96

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