The hydroxyl radical ( OH ) plays an important role in atmospheric chemistry. It is highly reactive and has a tendency to combine with an H atom from other compounds, causing them to break up. Thus OH is sometimes called a "detergent" radical because it helps to clean up the atmosphere. (a) Draw the Lewis structure for the radical. (b) Refer to Table 8.6 and explain why the radical has a high affinity for H atoms. (c) Estimate the enthalpy change for the following reaction: OH ( g ) + CH 4 ( g ) → CH 3 ( g ) + H 2 O ( g ) (d) The radical is generated when sunlight hits water vapor. Calculate the maximum wavelength (in nm) required to break an O—H bond in H 2 O .

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 8, Problem 142AP

The hydroxyl radical $( OH )$ plays an important role in atmospheric chemistry. It is highly reactive and has a tendency to combine with an H atom from other compounds, causing them to break up. Thus $OH$ is sometimes called a "detergent" radical because it helps to clean up the atmosphere. (a) Draw the Lewis structure for the radical. (b) Refer to Table 8.6 and explain why the radical has a high affinity for H atoms. (c) Estimate the enthalpy change for the following reaction:

$OH ( g ) + CH 4 ( g ) → CH 3 ( g ) + H 2 O ( g )$

(d) The radical is generated when sunlight hits water vapor. Calculate the maximum wavelength (in nm) required to break an $O—H$ bond in $H 2 O$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The Lewis structure of hydroxyl radical is to be drawn; the reason for radical’s high affinity for H atoms, enthalpy change for the given reaction, and the maximum wavelength required to break given bond in given compound are to be calculated.

Concept introduction:

The standard enthalpy of a reaction is the enthalpy change that occurs under standard conditions.

The standard enthalpy of a reaction is determined using the equation as given below:

$ΔH°rxn=∑ nΔHf°(products)−∑ mΔHf°(reactants)$

Here, the stoichiometric coefficients are represented by m for reactants and n for products, and enthalpy of formation at standard conditions is represented by $ΔHf°$ .

The standard enthalpy of formation is the amount of heat change that arises as a result of one mole of compound formation at the standard states from its integral elements.

The value of enthalpy of formation of an element is zero at its most stable state.

Bond Enthalpy is the energy required to break a bond of 1 mole of a substance. It is given in $KJ/mol$ .

The Lewis dot symbols contain dots, which give information about valence electrons.

According to Planck’s energy–frequency law, energy is given by:

$E=hcλ$

Here, h is Planck’s constant, $c$ is speed of light, and $λ$ is wavelength of radiation.

$h = 6.626×10−34Jsc = 3.0×108 m/s$

Answer to Problem 142AP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 1

(b) The reason for radical $OH$ to have a strong affinity toward hydrogen is the bond enthalpy value of oxygen–hydrogen bond, which is very high.

(c) $−46 KJ/mol$

(d) $260 nm$

Explanation of Solution

a) Lewis structure for radical (OH)

The skeletal structure for $OH$ radical is as follows:

$O−H$

The number of valence electrons is as follows:

$n=(O atom)+(H atoms)=6+1=7$

Now, only two electrons are used in bond formation. So, the number of remaining electrons is 5.

The oxygen atom has only one electron used in bond formation, so it will have 5 remaining electrons.

So, the Lewis structure for $OH$ radical is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 2

b) The radical has high affinity for H atom, refer table 8.6.

From table 8.6, the bond enthalpy of oxygen–hydrogen bond: $BE(O−H)=460 kJ/mol$

Since the value of oxygen–hydrogen bond is very high, the radical $OH$ has very strong affinity toward hydrogen to complete its octet.

Hence, the reason for radical $OH$ to have a strong affinity toward hydrogen is the high bond enthalpy value of oxygen–hydrogen bond.

c) The enthalpy change for the following given reaction:

$OH(g)+CH4(g)→CH3(g)+H2O(g)$

The enthalpy of reaction can be calculated as follows:

$ΔH°rxn=∑ BE(reactants)−∑ BE(products)$

From table 8.6, the enthalpy of formation values are as follows:

$BE[ C−H ]=414 KJ/molBE[ O−H ]=460 KJ/mol$

Now, the standard enthalpy of the given reaction is as follows:

$ΔH°rxn= { 4BE[ C−H ] + BE[ O−H ] }−{ 3BE[ C−H ]+ 2BE[ O−H ] } = { BE[ C−H ] }−{ BE[ O−H ] }$

Substitute $414 KJ/mol$ for $BE[ C−H ]$ and $460 KJ/mol$ for $BE[ O−H ]$ in the above equation

$ΔH°rxn=[ 414 KJ/mol ]−[ 460 KJ/mol ]=−46 KJ/mol$

Hence, the enthalpy of the given reaction from the bond enthalpy values is $−46 KJ/mol$ .

d) The maximum wavelength (in nm) required to break an $O−H$ bond in water.

From table 8.6, the bond enthalpy for oxygen–hydrogen bond is as follows:

$BE(O−H)=460 kJ/mol$

This can be written as the energy of oxygen–hydrogen bond as

$E(O−H)=460×103 J/mol×16.022×1023 bonds/mol=460×1036.022×1023 J/bonds=76.39×10−20 J/bonds=7.64×10−19 J/bonds$

According to Planck’s energy–frequency law, energy is given as follows:

$E=hcλ$

This can be written as follows:

$λ=hcE$

Substitute $c$ for $3×108m/s$ , $6.626×10−34Js$ for $h$ , and $7.64×10−19 J$ for $E$ in the above equation

$λ=(6.626×10−34Js) ×3×108 m/s(7.64×10−19 J)=19.9×10−267.64×10−19 m=2.6×10−7 m=260 nm$

Hence, the maximum wavelength required to break the oxygen–hydrogen bond is $2.6×10−7 nm$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

20,0 Complete the electron pushing mechanism to y drawing the necomery unicaciones and carved on for Step 1: Add curved arms for the tint step, traiment with NalilĻ. The Nation 458 Step 2: Added for the second step, inalment with), how the "counterion bar Step 3: Daw the products of the last simplom organic and one incoganic spacient, including all nonbonding

please provide the structure for this problem, thank you!

Draw the Fischer projection from the skeletal structure shown below. HO OH OH OH OH H Q Drawing Atoms, Bonds and Rings Charges I ☐ T HO H H OH HO I CH2OH H OH Drag H OH -CH2OH CHO -COOH Undo Reset Remove Done

Answer 2

Textbook Question

Chapter 8, Problem 142AP

The hydroxyl radical $( OH )$ plays an important role in atmospheric chemistry. It is highly reactive and has a tendency to combine with an H atom from other compounds, causing them to break up. Thus $OH$ is sometimes called a "detergent" radical because it helps to clean up the atmosphere. (a) Draw the Lewis structure for the radical. (b) Refer to Table 8.6 and explain why the radical has a high affinity for H atoms. (c) Estimate the enthalpy change for the following reaction:

$OH ( g ) + CH 4 ( g ) → CH 3 ( g ) + H 2 O ( g )$

(d) The radical is generated when sunlight hits water vapor. Calculate the maximum wavelength (in nm) required to break an $O—H$ bond in $H 2 O$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The Lewis structure of hydroxyl radical is to be drawn; the reason for radical’s high affinity for H atoms, enthalpy change for the given reaction, and the maximum wavelength required to break given bond in given compound are to be calculated.

Concept introduction:

The standard enthalpy of a reaction is the enthalpy change that occurs under standard conditions.

The standard enthalpy of a reaction is determined using the equation as given below:

$ΔH°rxn=∑ nΔHf°(products)−∑ mΔHf°(reactants)$

Here, the stoichiometric coefficients are represented by m for reactants and n for products, and enthalpy of formation at standard conditions is represented by $ΔHf°$ .

The standard enthalpy of formation is the amount of heat change that arises as a result of one mole of compound formation at the standard states from its integral elements.

The value of enthalpy of formation of an element is zero at its most stable state.

Bond Enthalpy is the energy required to break a bond of 1 mole of a substance. It is given in $KJ/mol$ .

The Lewis dot symbols contain dots, which give information about valence electrons.

According to Planck’s energy–frequency law, energy is given by:

$E=hcλ$

Here, h is Planck’s constant, $c$ is speed of light, and $λ$ is wavelength of radiation.

$h = 6.626×10−34Jsc = 3.0×108 m/s$

Answer to Problem 142AP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 1

(b) The reason for radical $OH$ to have a strong affinity toward hydrogen is the bond enthalpy value of oxygen–hydrogen bond, which is very high.

(c) $−46 KJ/mol$

(d) $260 nm$

Explanation of Solution

a) Lewis structure for radical (OH)

The skeletal structure for $OH$ radical is as follows:

$O−H$

The number of valence electrons is as follows:

$n=(O atom)+(H atoms)=6+1=7$

Now, only two electrons are used in bond formation. So, the number of remaining electrons is 5.

The oxygen atom has only one electron used in bond formation, so it will have 5 remaining electrons.

So, the Lewis structure for $OH$ radical is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 2

b) The radical has high affinity for H atom, refer table 8.6.

From table 8.6, the bond enthalpy of oxygen–hydrogen bond: $BE(O−H)=460 kJ/mol$

Since the value of oxygen–hydrogen bond is very high, the radical $OH$ has very strong affinity toward hydrogen to complete its octet.

Hence, the reason for radical $OH$ to have a strong affinity toward hydrogen is the high bond enthalpy value of oxygen–hydrogen bond.

c) The enthalpy change for the following given reaction:

$OH(g)+CH4(g)→CH3(g)+H2O(g)$

The enthalpy of reaction can be calculated as follows:

$ΔH°rxn=∑ BE(reactants)−∑ BE(products)$

From table 8.6, the enthalpy of formation values are as follows:

$BE[ C−H ]=414 KJ/molBE[ O−H ]=460 KJ/mol$

Now, the standard enthalpy of the given reaction is as follows:

$ΔH°rxn= { 4BE[ C−H ] + BE[ O−H ] }−{ 3BE[ C−H ]+ 2BE[ O−H ] } = { BE[ C−H ] }−{ BE[ O−H ] }$

Substitute $414 KJ/mol$ for $BE[ C−H ]$ and $460 KJ/mol$ for $BE[ O−H ]$ in the above equation

$ΔH°rxn=[ 414 KJ/mol ]−[ 460 KJ/mol ]=−46 KJ/mol$

Hence, the enthalpy of the given reaction from the bond enthalpy values is $−46 KJ/mol$ .

d) The maximum wavelength (in nm) required to break an $O−H$ bond in water.

From table 8.6, the bond enthalpy for oxygen–hydrogen bond is as follows:

$BE(O−H)=460 kJ/mol$

This can be written as the energy of oxygen–hydrogen bond as

$E(O−H)=460×103 J/mol×16.022×1023 bonds/mol=460×1036.022×1023 J/bonds=76.39×10−20 J/bonds=7.64×10−19 J/bonds$

According to Planck’s energy–frequency law, energy is given as follows:

$E=hcλ$

This can be written as follows:

$λ=hcE$

Substitute $c$ for $3×108m/s$ , $6.626×10−34Js$ for $h$ , and $7.64×10−19 J$ for $E$ in the above equation

$λ=(6.626×10−34Js) ×3×108 m/s(7.64×10−19 J)=19.9×10−267.64×10−19 m=2.6×10−7 m=260 nm$

Hence, the maximum wavelength required to break the oxygen–hydrogen bond is $2.6×10−7 nm$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 8, Problem 142AP

The hydroxyl radical $( OH )$ plays an important role in atmospheric chemistry. It is highly reactive and has a tendency to combine with an H atom from other compounds, causing them to break up. Thus $OH$ is sometimes called a "detergent" radical because it helps to clean up the atmosphere. (a) Draw the Lewis structure for the radical. (b) Refer to Table 8.6 and explain why the radical has a high affinity for H atoms. (c) Estimate the enthalpy change for the following reaction:

$OH ( g ) + CH 4 ( g ) → CH 3 ( g ) + H 2 O ( g )$

(d) The radical is generated when sunlight hits water vapor. Calculate the maximum wavelength (in nm) required to break an $O—H$ bond in $H 2 O$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The Lewis structure of hydroxyl radical is to be drawn; the reason for radical’s high affinity for H atoms, enthalpy change for the given reaction, and the maximum wavelength required to break given bond in given compound are to be calculated.

Concept introduction:

The standard enthalpy of a reaction is the enthalpy change that occurs under standard conditions.

The standard enthalpy of a reaction is determined using the equation as given below:

$ΔH°rxn=∑ nΔHf°(products)−∑ mΔHf°(reactants)$

Here, the stoichiometric coefficients are represented by m for reactants and n for products, and enthalpy of formation at standard conditions is represented by $ΔHf°$ .

The standard enthalpy of formation is the amount of heat change that arises as a result of one mole of compound formation at the standard states from its integral elements.

The value of enthalpy of formation of an element is zero at its most stable state.

Bond Enthalpy is the energy required to break a bond of 1 mole of a substance. It is given in $KJ/mol$ .

The Lewis dot symbols contain dots, which give information about valence electrons.

According to Planck’s energy–frequency law, energy is given by:

$E=hcλ$

Here, h is Planck’s constant, $c$ is speed of light, and $λ$ is wavelength of radiation.

$h = 6.626×10−34Jsc = 3.0×108 m/s$

Answer to Problem 142AP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 1

(b) The reason for radical $OH$ to have a strong affinity toward hydrogen is the bond enthalpy value of oxygen–hydrogen bond, which is very high.

(c) $−46 KJ/mol$

(d) $260 nm$

Explanation of Solution

a) Lewis structure for radical (OH)

The skeletal structure for $OH$ radical is as follows:

$O−H$

The number of valence electrons is as follows:

$n=(O atom)+(H atoms)=6+1=7$

Now, only two electrons are used in bond formation. So, the number of remaining electrons is 5.

The oxygen atom has only one electron used in bond formation, so it will have 5 remaining electrons.

So, the Lewis structure for $OH$ radical is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 2

b) The radical has high affinity for H atom, refer table 8.6.

From table 8.6, the bond enthalpy of oxygen–hydrogen bond: $BE(O−H)=460 kJ/mol$

Since the value of oxygen–hydrogen bond is very high, the radical $OH$ has very strong affinity toward hydrogen to complete its octet.

Hence, the reason for radical $OH$ to have a strong affinity toward hydrogen is the high bond enthalpy value of oxygen–hydrogen bond.

c) The enthalpy change for the following given reaction:

$OH(g)+CH4(g)→CH3(g)+H2O(g)$

The enthalpy of reaction can be calculated as follows:

$ΔH°rxn=∑ BE(reactants)−∑ BE(products)$

From table 8.6, the enthalpy of formation values are as follows:

$BE[ C−H ]=414 KJ/molBE[ O−H ]=460 KJ/mol$

Now, the standard enthalpy of the given reaction is as follows:

$ΔH°rxn= { 4BE[ C−H ] + BE[ O−H ] }−{ 3BE[ C−H ]+ 2BE[ O−H ] } = { BE[ C−H ] }−{ BE[ O−H ] }$

Substitute $414 KJ/mol$ for $BE[ C−H ]$ and $460 KJ/mol$ for $BE[ O−H ]$ in the above equation

$ΔH°rxn=[ 414 KJ/mol ]−[ 460 KJ/mol ]=−46 KJ/mol$

Hence, the enthalpy of the given reaction from the bond enthalpy values is $−46 KJ/mol$ .

d) The maximum wavelength (in nm) required to break an $O−H$ bond in water.

From table 8.6, the bond enthalpy for oxygen–hydrogen bond is as follows:

$BE(O−H)=460 kJ/mol$

This can be written as the energy of oxygen–hydrogen bond as

$E(O−H)=460×103 J/mol×16.022×1023 bonds/mol=460×1036.022×1023 J/bonds=76.39×10−20 J/bonds=7.64×10−19 J/bonds$

According to Planck’s energy–frequency law, energy is given as follows:

$E=hcλ$

This can be written as follows:

$λ=hcE$

Substitute $c$ for $3×108m/s$ , $6.626×10−34Js$ for $h$ , and $7.64×10−19 J$ for $E$ in the above equation

$λ=(6.626×10−34Js) ×3×108 m/s(7.64×10−19 J)=19.9×10−267.64×10−19 m=2.6×10−7 m=260 nm$

Hence, the maximum wavelength required to break the oxygen–hydrogen bond is $2.6×10−7 nm$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 8, Problem 142AP

The hydroxyl radical $( OH )$ plays an important role in atmospheric chemistry. It is highly reactive and has a tendency to combine with an H atom from other compounds, causing them to break up. Thus $OH$ is sometimes called a "detergent" radical because it helps to clean up the atmosphere. (a) Draw the Lewis structure for the radical. (b) Refer to Table 8.6 and explain why the radical has a high affinity for H atoms. (c) Estimate the enthalpy change for the following reaction:

$OH ( g ) + CH 4 ( g ) → CH 3 ( g ) + H 2 O ( g )$

(d) The radical is generated when sunlight hits water vapor. Calculate the maximum wavelength (in nm) required to break an $O—H$ bond in $H 2 O$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The Lewis structure of hydroxyl radical is to be drawn; the reason for radical’s high affinity for H atoms, enthalpy change for the given reaction, and the maximum wavelength required to break given bond in given compound are to be calculated.

Concept introduction:

The standard enthalpy of a reaction is the enthalpy change that occurs under standard conditions.

The standard enthalpy of a reaction is determined using the equation as given below:

$ΔH°rxn=∑ nΔHf°(products)−∑ mΔHf°(reactants)$

Here, the stoichiometric coefficients are represented by m for reactants and n for products, and enthalpy of formation at standard conditions is represented by $ΔHf°$ .

The standard enthalpy of formation is the amount of heat change that arises as a result of one mole of compound formation at the standard states from its integral elements.

The value of enthalpy of formation of an element is zero at its most stable state.

Bond Enthalpy is the energy required to break a bond of 1 mole of a substance. It is given in $KJ/mol$ .

The Lewis dot symbols contain dots, which give information about valence electrons.

According to Planck’s energy–frequency law, energy is given by:

$E=hcλ$

Here, h is Planck’s constant, $c$ is speed of light, and $λ$ is wavelength of radiation.

$h = 6.626×10−34Jsc = 3.0×108 m/s$

Answer to Problem 142AP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 1

(b) The reason for radical $OH$ to have a strong affinity toward hydrogen is the bond enthalpy value of oxygen–hydrogen bond, which is very high.

(c) $−46 KJ/mol$

(d) $260 nm$

Explanation of Solution

a) Lewis structure for radical (OH)

The skeletal structure for $OH$ radical is as follows:

$O−H$

The number of valence electrons is as follows:

$n=(O atom)+(H atoms)=6+1=7$

Now, only two electrons are used in bond formation. So, the number of remaining electrons is 5.

The oxygen atom has only one electron used in bond formation, so it will have 5 remaining electrons.

So, the Lewis structure for $OH$ radical is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 2

b) The radical has high affinity for H atom, refer table 8.6.

From table 8.6, the bond enthalpy of oxygen–hydrogen bond: $BE(O−H)=460 kJ/mol$

Since the value of oxygen–hydrogen bond is very high, the radical $OH$ has very strong affinity toward hydrogen to complete its octet.

Hence, the reason for radical $OH$ to have a strong affinity toward hydrogen is the high bond enthalpy value of oxygen–hydrogen bond.

c) The enthalpy change for the following given reaction:

$OH(g)+CH4(g)→CH3(g)+H2O(g)$

The enthalpy of reaction can be calculated as follows:

$ΔH°rxn=∑ BE(reactants)−∑ BE(products)$

From table 8.6, the enthalpy of formation values are as follows:

$BE[ C−H ]=414 KJ/molBE[ O−H ]=460 KJ/mol$

Now, the standard enthalpy of the given reaction is as follows:

$ΔH°rxn= { 4BE[ C−H ] + BE[ O−H ] }−{ 3BE[ C−H ]+ 2BE[ O−H ] } = { BE[ C−H ] }−{ BE[ O−H ] }$

Substitute $414 KJ/mol$ for $BE[ C−H ]$ and $460 KJ/mol$ for $BE[ O−H ]$ in the above equation

$ΔH°rxn=[ 414 KJ/mol ]−[ 460 KJ/mol ]=−46 KJ/mol$

Hence, the enthalpy of the given reaction from the bond enthalpy values is $−46 KJ/mol$ .

d) The maximum wavelength (in nm) required to break an $O−H$ bond in water.

From table 8.6, the bond enthalpy for oxygen–hydrogen bond is as follows:

$BE(O−H)=460 kJ/mol$

This can be written as the energy of oxygen–hydrogen bond as

$E(O−H)=460×103 J/mol×16.022×1023 bonds/mol=460×1036.022×1023 J/bonds=76.39×10−20 J/bonds=7.64×10−19 J/bonds$

According to Planck’s energy–frequency law, energy is given as follows:

$E=hcλ$

This can be written as follows:

$λ=hcE$

Substitute $c$ for $3×108m/s$ , $6.626×10−34Js$ for $h$ , and $7.64×10−19 J$ for $E$ in the above equation

$λ=(6.626×10−34Js) ×3×108 m/s(7.64×10−19 J)=19.9×10−267.64×10−19 m=2.6×10−7 m=260 nm$

Hence, the maximum wavelength required to break the oxygen–hydrogen bond is $2.6×10−7 nm$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The Lewis structure of hydroxyl radical is to be drawn; the reason for radical’s high affinity for H atoms, enthalpy change for the given reaction, and the maximum wavelength required to break given bond in given compound are to be calculated.

Concept introduction:

The standard enthalpy of a reaction is the enthalpy change that occurs under standard conditions.

The standard enthalpy of a reaction is determined using the equation as given below:

$ΔH°rxn=∑ nΔHf°(products)−∑ mΔHf°(reactants)$

Here, the stoichiometric coefficients are represented by m for reactants and n for products, and enthalpy of formation at standard conditions is represented by $ΔHf°$ .

The standard enthalpy of formation is the amount of heat change that arises as a result of one mole of compound formation at the standard states from its integral elements.

The value of enthalpy of formation of an element is zero at its most stable state.

Bond Enthalpy is the energy required to break a bond of 1 mole of a substance. It is given in $KJ/mol$ .

The Lewis dot symbols contain dots, which give information about valence electrons.

According to Planck’s energy–frequency law, energy is given by:

$E=hcλ$

Here, h is Planck’s constant, $c$ is speed of light, and $λ$ is wavelength of radiation.

$h = 6.626×10−34Jsc = 3.0×108 m/s$

Answer to Problem 142AP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 1

(b) The reason for radical $OH$ to have a strong affinity toward hydrogen is the bond enthalpy value of oxygen–hydrogen bond, which is very high.

(c) $−46 KJ/mol$

(d) $260 nm$

Explanation of Solution

a) Lewis structure for radical (OH)

The skeletal structure for $OH$ radical is as follows:

$O−H$

The number of valence electrons is as follows:

$n=(O atom)+(H atoms)=6+1=7$

Now, only two electrons are used in bond formation. So, the number of remaining electrons is 5.

The oxygen atom has only one electron used in bond formation, so it will have 5 remaining electrons.

So, the Lewis structure for $OH$ radical is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 2

b) The radical has high affinity for H atom, refer table 8.6.

From table 8.6, the bond enthalpy of oxygen–hydrogen bond: $BE(O−H)=460 kJ/mol$

Since the value of oxygen–hydrogen bond is very high, the radical $OH$ has very strong affinity toward hydrogen to complete its octet.

Hence, the reason for radical $OH$ to have a strong affinity toward hydrogen is the high bond enthalpy value of oxygen–hydrogen bond.

c) The enthalpy change for the following given reaction:

$OH(g)+CH4(g)→CH3(g)+H2O(g)$

The enthalpy of reaction can be calculated as follows:

$ΔH°rxn=∑ BE(reactants)−∑ BE(products)$

From table 8.6, the enthalpy of formation values are as follows:

$BE[ C−H ]=414 KJ/molBE[ O−H ]=460 KJ/mol$

Now, the standard enthalpy of the given reaction is as follows:

$ΔH°rxn= { 4BE[ C−H ] + BE[ O−H ] }−{ 3BE[ C−H ]+ 2BE[ O−H ] } = { BE[ C−H ] }−{ BE[ O−H ] }$

Substitute $414 KJ/mol$ for $BE[ C−H ]$ and $460 KJ/mol$ for $BE[ O−H ]$ in the above equation

$ΔH°rxn=[ 414 KJ/mol ]−[ 460 KJ/mol ]=−46 KJ/mol$

Hence, the enthalpy of the given reaction from the bond enthalpy values is $−46 KJ/mol$ .

d) The maximum wavelength (in nm) required to break an $O−H$ bond in water.

From table 8.6, the bond enthalpy for oxygen–hydrogen bond is as follows:

$BE(O−H)=460 kJ/mol$

This can be written as the energy of oxygen–hydrogen bond as

$E(O−H)=460×103 J/mol×16.022×1023 bonds/mol=460×1036.022×1023 J/bonds=76.39×10−20 J/bonds=7.64×10−19 J/bonds$

According to Planck’s energy–frequency law, energy is given as follows:

$E=hcλ$

This can be written as follows:

$λ=hcE$

Substitute $c$ for $3×108m/s$ , $6.626×10−34Js$ for $h$ , and $7.64×10−19 J$ for $E$ in the above equation

$λ=(6.626×10−34Js) ×3×108 m/s(7.64×10−19 J)=19.9×10−267.64×10−19 m=2.6×10−7 m=260 nm$

Hence, the maximum wavelength required to break the oxygen–hydrogen bond is $2.6×10−7 nm$ .

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The Lewis structure of hydroxyl radical is to be drawn; the reason for radical’s high affinity for H atoms, enthalpy change for the given reaction, and the maximum wavelength required to break given bond in given compound are to be calculated.

Concept introduction:

The standard enthalpy of a reaction is the enthalpy change that occurs under standard conditions.

The standard enthalpy of a reaction is determined using the equation as given below:

$ΔH°rxn=∑ nΔHf°(products)−∑ mΔHf°(reactants)$

Here, the stoichiometric coefficients are represented by m for reactants and n for products, and enthalpy of formation at standard conditions is represented by $ΔHf°$ .

The standard enthalpy of formation is the amount of heat change that arises as a result of one mole of compound formation at the standard states from its integral elements.

The value of enthalpy of formation of an element is zero at its most stable state.

Bond Enthalpy is the energy required to break a bond of 1 mole of a substance. It is given in $KJ/mol$ .

The Lewis dot symbols contain dots, which give information about valence electrons.

According to Planck’s energy–frequency law, energy is given by:

$E=hcλ$

Here, h is Planck’s constant, $c$ is speed of light, and $λ$ is wavelength of radiation.

$h = 6.626×10−34Jsc = 3.0×108 m/s$

Answer to Problem 142AP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 1

(b) The reason for radical $OH$ to have a strong affinity toward hydrogen is the bond enthalpy value of oxygen–hydrogen bond, which is very high.

(c) $−46 KJ/mol$

(d) $260 nm$

Explanation of Solution

a) Lewis structure for radical (OH)

The skeletal structure for $OH$ radical is as follows:

$O−H$

The number of valence electrons is as follows:

$n=(O atom)+(H atoms)=6+1=7$

Now, only two electrons are used in bond formation. So, the number of remaining electrons is 5.

The oxygen atom has only one electron used in bond formation, so it will have 5 remaining electrons.

So, the Lewis structure for $OH$ radical is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 2

b) The radical has high affinity for H atom, refer table 8.6.

From table 8.6, the bond enthalpy of oxygen–hydrogen bond: $BE(O−H)=460 kJ/mol$

Since the value of oxygen–hydrogen bond is very high, the radical $OH$ has very strong affinity toward hydrogen to complete its octet.

Hence, the reason for radical $OH$ to have a strong affinity toward hydrogen is the high bond enthalpy value of oxygen–hydrogen bond.

c) The enthalpy change for the following given reaction:

$OH(g)+CH4(g)→CH3(g)+H2O(g)$

The enthalpy of reaction can be calculated as follows:

$ΔH°rxn=∑ BE(reactants)−∑ BE(products)$

From table 8.6, the enthalpy of formation values are as follows:

$BE[ C−H ]=414 KJ/molBE[ O−H ]=460 KJ/mol$

Now, the standard enthalpy of the given reaction is as follows:

$ΔH°rxn= { 4BE[ C−H ] + BE[ O−H ] }−{ 3BE[ C−H ]+ 2BE[ O−H ] } = { BE[ C−H ] }−{ BE[ O−H ] }$

Substitute $414 KJ/mol$ for $BE[ C−H ]$ and $460 KJ/mol$ for $BE[ O−H ]$ in the above equation

$ΔH°rxn=[ 414 KJ/mol ]−[ 460 KJ/mol ]=−46 KJ/mol$

Hence, the enthalpy of the given reaction from the bond enthalpy values is $−46 KJ/mol$ .

d) The maximum wavelength (in nm) required to break an $O−H$ bond in water.

From table 8.6, the bond enthalpy for oxygen–hydrogen bond is as follows:

$BE(O−H)=460 kJ/mol$

This can be written as the energy of oxygen–hydrogen bond as

$E(O−H)=460×103 J/mol×16.022×1023 bonds/mol=460×1036.022×1023 J/bonds=76.39×10−20 J/bonds=7.64×10−19 J/bonds$

According to Planck’s energy–frequency law, energy is given as follows:

$E=hcλ$

This can be written as follows:

$λ=hcE$

Substitute $c$ for $3×108m/s$ , $6.626×10−34Js$ for $h$ , and $7.64×10−19 J$ for $E$ in the above equation

$λ=(6.626×10−34Js) ×3×108 m/s(7.64×10−19 J)=19.9×10−267.64×10−19 m=2.6×10−7 m=260 nm$

Hence, the maximum wavelength required to break the oxygen–hydrogen bond is $2.6×10−7 nm$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

The Lewis structure of hydroxyl radical is to be drawn; the reason for radical’s high affinity for H atoms, enthalpy change for the given reaction, and the maximum wavelength required to break given bond in given compound are to be calculated.

Concept introduction:

The standard enthalpy of a reaction is the enthalpy change that occurs under standard conditions.

The standard enthalpy of a reaction is determined using the equation as given below:

$ΔH°rxn=∑ nΔHf°(products)−∑ mΔHf°(reactants)$

Here, the stoichiometric coefficients are represented by m for reactants and n for products, and enthalpy of formation at standard conditions is represented by $ΔHf°$ .

The standard enthalpy of formation is the amount of heat change that arises as a result of one mole of compound formation at the standard states from its integral elements.

The value of enthalpy of formation of an element is zero at its most stable state.

Bond Enthalpy is the energy required to break a bond of 1 mole of a substance. It is given in $KJ/mol$ .

The Lewis dot symbols contain dots, which give information about valence electrons.

According to Planck’s energy–frequency law, energy is given by:

$E=hcλ$

Here, h is Planck’s constant, $c$ is speed of light, and $λ$ is wavelength of radiation.

$h = 6.626×10−34Jsc = 3.0×108 m/s$

Answer 12

Answer to Problem 142AP

Solution:

(a)

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 1

(b) The reason for radical $OH$ to have a strong affinity toward hydrogen is the bond enthalpy value of oxygen–hydrogen bond, which is very high.

(c) $−46 KJ/mol$

(d) $260 nm$

Answer 13

Explanation of Solution

a) Lewis structure for radical (OH)

The skeletal structure for $OH$ radical is as follows:

$O−H$

The number of valence electrons is as follows:

$n=(O atom)+(H atoms)=6+1=7$

Now, only two electrons are used in bond formation. So, the number of remaining electrons is 5.

The oxygen atom has only one electron used in bond formation, so it will have 5 remaining electrons.

So, the Lewis structure for $OH$ radical is as follows:

CHEMISTRY >CUSTOM<, Chapter 8, Problem 142AP , additional homework tip 2

b) The radical has high affinity for H atom, refer table 8.6.

From table 8.6, the bond enthalpy of oxygen–hydrogen bond: $BE(O−H)=460 kJ/mol$

Since the value of oxygen–hydrogen bond is very high, the radical $OH$ has very strong affinity toward hydrogen to complete its octet.

Hence, the reason for radical $OH$ to have a strong affinity toward hydrogen is the high bond enthalpy value of oxygen–hydrogen bond.

c) The enthalpy change for the following given reaction:

$OH(g)+CH4(g)→CH3(g)+H2O(g)$

The enthalpy of reaction can be calculated as follows:

$ΔH°rxn=∑ BE(reactants)−∑ BE(products)$

From table 8.6, the enthalpy of formation values are as follows:

$BE[ C−H ]=414 KJ/molBE[ O−H ]=460 KJ/mol$

Now, the standard enthalpy of the given reaction is as follows:

$ΔH°rxn= { 4BE[ C−H ] + BE[ O−H ] }−{ 3BE[ C−H ]+ 2BE[ O−H ] } = { BE[ C−H ] }−{ BE[ O−H ] }$

Substitute $414 KJ/mol$ for $BE[ C−H ]$ and $460 KJ/mol$ for $BE[ O−H ]$ in the above equation

$ΔH°rxn=[ 414 KJ/mol ]−[ 460 KJ/mol ]=−46 KJ/mol$

Hence, the enthalpy of the given reaction from the bond enthalpy values is $−46 KJ/mol$ .

d) The maximum wavelength (in nm) required to break an $O−H$ bond in water.

From table 8.6, the bond enthalpy for oxygen–hydrogen bond is as follows:

$BE(O−H)=460 kJ/mol$

This can be written as the energy of oxygen–hydrogen bond as

$E(O−H)=460×103 J/mol×16.022×1023 bonds/mol=460×1036.022×1023 J/bonds=76.39×10−20 J/bonds=7.64×10−19 J/bonds$

According to Planck’s energy–frequency law, energy is given as follows:

$E=hcλ$

This can be written as follows:

$λ=hcE$

Substitute $c$ for $3×108m/s$ , $6.626×10−34Js$ for $h$ , and $7.64×10−19 J$ for $E$ in the above equation

$λ=(6.626×10−34Js) ×3×108 m/s(7.64×10−19 J)=19.9×10−267.64×10−19 m=2.6×10−7 m=260 nm$

Hence, the maximum wavelength required to break the oxygen–hydrogen bond is $2.6×10−7 nm$ .

Answer 14

Ch. 8.1 - Practice ProblemATTEMPT Write Lewis dot symbols...Ch. 8.1 - Practice Problem BUILD Indicate the charge on...Ch. 8.1 - Practice ProblemCONCEPTUALIZE For each of the...Ch. 8.1 - 8.1.1 Using only a periodic table, determine the...Ch. 8.1 - 8.1.2 Using only a periodic table, determine the...Ch. 8.1 - To which group does element X belong if its Lewis...Ch. 8.1 - Prob. 4CP Ch. 8.2 - Prob. 1PPA Ch. 8.2 - Practice ProblemBUILD Arrange the compounds NaF,...Ch. 8.2 - Practice ProblemCONCEPTUALIZE Common ions of four...

Concept explainers

Videos

Answer to Problem 142AP

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions