(a) Interpretation: Energy possessed by a mole of photon should be calculated and expressed with the units of kilojoules and kilocalories. Concept introduction: A photon can be introduced as a piece of energy which has no mass. The relationship between the energy of a photon and its wavelength can be expressed with the following Planck-Einstein equation represented as follows: E = h f

Question

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Answer 1

Question

Chapter 19, Problem 23P

Interpretation Introduction

(a)

Interpretation:

Energy possessed by a mole of photon should be calculated and expressed with the units of kilojoules and kilocalories.

Concept introduction:

A photon can be introduced as a piece of energy which has no mass. The relationship between the energy of a photon and its wavelength can be expressed with the following Planck-Einstein equation represented as follows:

E=hf

Expert Solution

Answer to Problem 23P

119.71 kJ mol−1 / 28.61 kcal mol−1

Explanation of Solution

A photon can be introduced as a piece of energy which has no mass. The speed of a photon is equal to the speed of the light. The relationship between the energy of a photon and its wavelength can be expressed with the Planck-Einstein equation represented as follows:

E=hf

Here,

E = energy of a photon

h = Planck constant

f = frequency of a photon

Since, the frequency of the photon is not given, the equation of

c=fλ should be used, where,

c = speed of the light

f = frequency of the light

λ = wavelength of the light

Using

c=fλ, f=c/λ

Substituting f=c/λ for f in E=hf

E=hc/λ

= (6.6260 × 10−34) m2kg s−1× (3× 108) ms−1(1000×10 −9) m

Since, (1 nm=1×10−9m)

=19.878×10−20J=1.9878×10−19J

Energy of a photon in kilojoules,=1.9878×10−19×10−3kJ

Since,

(1J=1×10−3kJ)

=1.9878×10−22kJ

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, energy of mole of photon in kiloJoules,

=1.978×10-22kJ×6.022×1023mol-1=119.7053 kJ mol-1=119.71 kJ mol-1

Energy of a photon in kilojoules =1.9878×10−19kJ

Energy of a photon in kilocalories,

=1.9878×10-19×0.000239 kcal(1J=0.000239 kcal)=4.7508×10-23kcal

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, Energy of moles of photon in kilocalories,

=4.7908×10 -23kcal×6.022×1023=28.6091 kcal mol -1=28.61 kcal mol-1

Einstein is also a unit of energy.

Einstein =1kJ mol-1

Thus, energy of mole of photon in kilo Joules = 119.71 kJ mol−1

Energy of moles of photon in kilocalories = 28.61 kcal mol−1

Interpretation Introduction

(b)

Interpretation:

The maximum increase in the redox potential induced by 1000 nm photon needs to be determined.

Concept introduction:

One electron volt is the energy need to move an electron between one-volt potential difference.

One electron volt =1.602×10−19J

Expert Solution

Answer to Problem 23P

1.24 V

Explanation of Solution

The maximum increase in a redox potential can be calculated as follows:

Maximum increase in redox potential

=Energy possess by a photonEnergy need to move an electron

Putting the values,

=1 .9878×10 -19 J1 .602×10 -19J=1.24 V

Interpretation Introduction

(c)

Interpretation:

Number of photons need to overcome the Gibbs free energy should be calculated.

Concept introduction:

In a simple definition, Gibbs free energy of a reaction is the energy associated with that particular chemical reaction.

Phosphorylation is the conversion of ATP from ADP.

Expert Solution

Answer to Problem 23P

0.42 photons.

Explanation of Solution

Phosphorylation is the conversion of ATP from ADP.

ADP+PO43−→Energy from electronsATP

ATP- Adenosine triphosphate

ADP-Adenosine diphosphate

Therefore,

Energy needed for the phosphorylation reaction = Energy needed to convert a mole of ADP to ATP = 50 kJ mol−1.

Number of ADP in one mole of ADP = Avogadro’s number of ADP = 6.022×1023.

Energy needed to convert one ADP to ATP,

= 50 kJ6.022×1023.

Maximum energy that one 1000 nm photon possess

=1.978×10−22kJ.

Minimum number of 1000 nm photons needed for the phosphorylation reaction,

=Total energy to convert one ADP to ATPEnergy of one 1000 nm photon

Putting the values,

= 50kJ mo l -1 6.022×1 0 23 1.978×10 -22kJ=0.4176 photons=0.42 photons

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Answer 2

Question

Chapter 19, Problem 23P

Interpretation Introduction

(a)

Interpretation:

Energy possessed by a mole of photon should be calculated and expressed with the units of kilojoules and kilocalories.

Concept introduction:

A photon can be introduced as a piece of energy which has no mass. The relationship between the energy of a photon and its wavelength can be expressed with the following Planck-Einstein equation represented as follows:

E=hf

Expert Solution

Answer to Problem 23P

119.71 kJ mol−1 / 28.61 kcal mol−1

Explanation of Solution

A photon can be introduced as a piece of energy which has no mass. The speed of a photon is equal to the speed of the light. The relationship between the energy of a photon and its wavelength can be expressed with the Planck-Einstein equation represented as follows:

E=hf

Here,

E = energy of a photon

h = Planck constant

f = frequency of a photon

Since, the frequency of the photon is not given, the equation of

c=fλ should be used, where,

c = speed of the light

f = frequency of the light

λ = wavelength of the light

Using

c=fλ, f=c/λ

Substituting f=c/λ for f in E=hf

E=hc/λ

= (6.6260 × 10−34) m2kg s−1× (3× 108) ms−1(1000×10 −9) m

Since, (1 nm=1×10−9m)

=19.878×10−20J=1.9878×10−19J

Energy of a photon in kilojoules,=1.9878×10−19×10−3kJ

Since,

(1J=1×10−3kJ)

=1.9878×10−22kJ

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, energy of mole of photon in kiloJoules,

=1.978×10-22kJ×6.022×1023mol-1=119.7053 kJ mol-1=119.71 kJ mol-1

Energy of a photon in kilojoules =1.9878×10−19kJ

Energy of a photon in kilocalories,

=1.9878×10-19×0.000239 kcal(1J=0.000239 kcal)=4.7508×10-23kcal

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, Energy of moles of photon in kilocalories,

=4.7908×10 -23kcal×6.022×1023=28.6091 kcal mol -1=28.61 kcal mol-1

Einstein is also a unit of energy.

Einstein =1kJ mol-1

Thus, energy of mole of photon in kilo Joules = 119.71 kJ mol−1

Energy of moles of photon in kilocalories = 28.61 kcal mol−1

Interpretation Introduction

(b)

Interpretation:

The maximum increase in the redox potential induced by 1000 nm photon needs to be determined.

Concept introduction:

One electron volt is the energy need to move an electron between one-volt potential difference.

One electron volt =1.602×10−19J

Expert Solution

Answer to Problem 23P

1.24 V

Explanation of Solution

The maximum increase in a redox potential can be calculated as follows:

Maximum increase in redox potential

=Energy possess by a photonEnergy need to move an electron

Putting the values,

=1 .9878×10 -19 J1 .602×10 -19J=1.24 V

Interpretation Introduction

(c)

Interpretation:

Number of photons need to overcome the Gibbs free energy should be calculated.

Concept introduction:

In a simple definition, Gibbs free energy of a reaction is the energy associated with that particular chemical reaction.

Phosphorylation is the conversion of ATP from ADP.

Expert Solution

Answer to Problem 23P

0.42 photons.

Explanation of Solution

Phosphorylation is the conversion of ATP from ADP.

ADP+PO43−→Energy from electronsATP

ATP- Adenosine triphosphate

ADP-Adenosine diphosphate

Therefore,

Energy needed for the phosphorylation reaction = Energy needed to convert a mole of ADP to ATP = 50 kJ mol−1.

Number of ADP in one mole of ADP = Avogadro’s number of ADP = 6.022×1023.

Energy needed to convert one ADP to ATP,

= 50 kJ6.022×1023.

Maximum energy that one 1000 nm photon possess

=1.978×10−22kJ.

Minimum number of 1000 nm photons needed for the phosphorylation reaction,

=Total energy to convert one ADP to ATPEnergy of one 1000 nm photon

Putting the values,

= 50kJ mo l -1 6.022×1 0 23 1.978×10 -22kJ=0.4176 photons=0.42 photons

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Answer 3

Question

Chapter 19, Problem 23P

Interpretation Introduction

(a)

Interpretation:

Energy possessed by a mole of photon should be calculated and expressed with the units of kilojoules and kilocalories.

Concept introduction:

A photon can be introduced as a piece of energy which has no mass. The relationship between the energy of a photon and its wavelength can be expressed with the following Planck-Einstein equation represented as follows:

E=hf

Expert Solution

Answer to Problem 23P

119.71 kJ mol−1 / 28.61 kcal mol−1

Explanation of Solution

A photon can be introduced as a piece of energy which has no mass. The speed of a photon is equal to the speed of the light. The relationship between the energy of a photon and its wavelength can be expressed with the Planck-Einstein equation represented as follows:

E=hf

Here,

E = energy of a photon

h = Planck constant

f = frequency of a photon

Since, the frequency of the photon is not given, the equation of

c=fλ should be used, where,

c = speed of the light

f = frequency of the light

λ = wavelength of the light

Using

c=fλ, f=c/λ

Substituting f=c/λ for f in E=hf

E=hc/λ

= (6.6260 × 10−34) m2kg s−1× (3× 108) ms−1(1000×10 −9) m

Since, (1 nm=1×10−9m)

=19.878×10−20J=1.9878×10−19J

Energy of a photon in kilojoules,=1.9878×10−19×10−3kJ

Since,

(1J=1×10−3kJ)

=1.9878×10−22kJ

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, energy of mole of photon in kiloJoules,

=1.978×10-22kJ×6.022×1023mol-1=119.7053 kJ mol-1=119.71 kJ mol-1

Energy of a photon in kilojoules =1.9878×10−19kJ

Energy of a photon in kilocalories,

=1.9878×10-19×0.000239 kcal(1J=0.000239 kcal)=4.7508×10-23kcal

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, Energy of moles of photon in kilocalories,

=4.7908×10 -23kcal×6.022×1023=28.6091 kcal mol -1=28.61 kcal mol-1

Einstein is also a unit of energy.

Einstein =1kJ mol-1

Thus, energy of mole of photon in kilo Joules = 119.71 kJ mol−1

Energy of moles of photon in kilocalories = 28.61 kcal mol−1

Interpretation Introduction

(b)

Interpretation:

The maximum increase in the redox potential induced by 1000 nm photon needs to be determined.

Concept introduction:

One electron volt is the energy need to move an electron between one-volt potential difference.

One electron volt =1.602×10−19J

Expert Solution

Answer to Problem 23P

1.24 V

Explanation of Solution

The maximum increase in a redox potential can be calculated as follows:

Maximum increase in redox potential

=Energy possess by a photonEnergy need to move an electron

Putting the values,

=1 .9878×10 -19 J1 .602×10 -19J=1.24 V

Interpretation Introduction

(c)

Interpretation:

Number of photons need to overcome the Gibbs free energy should be calculated.

Concept introduction:

In a simple definition, Gibbs free energy of a reaction is the energy associated with that particular chemical reaction.

Phosphorylation is the conversion of ATP from ADP.

Expert Solution

Answer to Problem 23P

0.42 photons.

Explanation of Solution

Phosphorylation is the conversion of ATP from ADP.

ADP+PO43−→Energy from electronsATP

ATP- Adenosine triphosphate

ADP-Adenosine diphosphate

Therefore,

Energy needed for the phosphorylation reaction = Energy needed to convert a mole of ADP to ATP = 50 kJ mol−1.

Number of ADP in one mole of ADP = Avogadro’s number of ADP = 6.022×1023.

Energy needed to convert one ADP to ATP,

= 50 kJ6.022×1023.

Maximum energy that one 1000 nm photon possess

=1.978×10−22kJ.

Minimum number of 1000 nm photons needed for the phosphorylation reaction,

=Total energy to convert one ADP to ATPEnergy of one 1000 nm photon

Putting the values,

= 50kJ mo l -1 6.022×1 0 23 1.978×10 -22kJ=0.4176 photons=0.42 photons

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Answer 4

Question

Chapter 19, Problem 23P

Interpretation Introduction

(a)

Interpretation:

Energy possessed by a mole of photon should be calculated and expressed with the units of kilojoules and kilocalories.

Concept introduction:

A photon can be introduced as a piece of energy which has no mass. The relationship between the energy of a photon and its wavelength can be expressed with the following Planck-Einstein equation represented as follows:

E=hf

Expert Solution

Answer to Problem 23P

119.71 kJ mol−1 / 28.61 kcal mol−1

Explanation of Solution

A photon can be introduced as a piece of energy which has no mass. The speed of a photon is equal to the speed of the light. The relationship between the energy of a photon and its wavelength can be expressed with the Planck-Einstein equation represented as follows:

E=hf

Here,

E = energy of a photon

h = Planck constant

f = frequency of a photon

Since, the frequency of the photon is not given, the equation of

c=fλ should be used, where,

c = speed of the light

f = frequency of the light

λ = wavelength of the light

Using

c=fλ, f=c/λ

Substituting f=c/λ for f in E=hf

E=hc/λ

= (6.6260 × 10−34) m2kg s−1× (3× 108) ms−1(1000×10 −9) m

Since, (1 nm=1×10−9m)

=19.878×10−20J=1.9878×10−19J

Energy of a photon in kilojoules,=1.9878×10−19×10−3kJ

Since,

(1J=1×10−3kJ)

=1.9878×10−22kJ

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, energy of mole of photon in kiloJoules,

=1.978×10-22kJ×6.022×1023mol-1=119.7053 kJ mol-1=119.71 kJ mol-1

Energy of a photon in kilojoules =1.9878×10−19kJ

Energy of a photon in kilocalories,

=1.9878×10-19×0.000239 kcal(1J=0.000239 kcal)=4.7508×10-23kcal

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, Energy of moles of photon in kilocalories,

=4.7908×10 -23kcal×6.022×1023=28.6091 kcal mol -1=28.61 kcal mol-1

Einstein is also a unit of energy.

Einstein =1kJ mol-1

Thus, energy of mole of photon in kilo Joules = 119.71 kJ mol−1

Energy of moles of photon in kilocalories = 28.61 kcal mol−1

Interpretation Introduction

(b)

Interpretation:

The maximum increase in the redox potential induced by 1000 nm photon needs to be determined.

Concept introduction:

One electron volt is the energy need to move an electron between one-volt potential difference.

One electron volt =1.602×10−19J

Expert Solution

Answer to Problem 23P

1.24 V

Explanation of Solution

The maximum increase in a redox potential can be calculated as follows:

Maximum increase in redox potential

=Energy possess by a photonEnergy need to move an electron

Putting the values,

=1 .9878×10 -19 J1 .602×10 -19J=1.24 V

Interpretation Introduction

(c)

Interpretation:

Number of photons need to overcome the Gibbs free energy should be calculated.

Concept introduction:

In a simple definition, Gibbs free energy of a reaction is the energy associated with that particular chemical reaction.

Phosphorylation is the conversion of ATP from ADP.

Expert Solution

Answer to Problem 23P

0.42 photons.

Explanation of Solution

Phosphorylation is the conversion of ATP from ADP.

ADP+PO43−→Energy from electronsATP

ATP- Adenosine triphosphate

ADP-Adenosine diphosphate

Therefore,

Energy needed for the phosphorylation reaction = Energy needed to convert a mole of ADP to ATP = 50 kJ mol−1.

Number of ADP in one mole of ADP = Avogadro’s number of ADP = 6.022×1023.

Energy needed to convert one ADP to ATP,

= 50 kJ6.022×1023.

Maximum energy that one 1000 nm photon possess

=1.978×10−22kJ.

Minimum number of 1000 nm photons needed for the phosphorylation reaction,

=Total energy to convert one ADP to ATPEnergy of one 1000 nm photon

Putting the values,

= 50kJ mo l -1 6.022×1 0 23 1.978×10 -22kJ=0.4176 photons=0.42 photons

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Answer 5

Chapter 19, Problem 23P

Interpretation Introduction

(a)

Interpretation:

Energy possessed by a mole of photon should be calculated and expressed with the units of kilojoules and kilocalories.

Concept introduction:

A photon can be introduced as a piece of energy which has no mass. The relationship between the energy of a photon and its wavelength can be expressed with the following Planck-Einstein equation represented as follows:

E=hf

Expert Solution

Answer to Problem 23P

119.71 kJ mol−1 / 28.61 kcal mol−1

Explanation of Solution

A photon can be introduced as a piece of energy which has no mass. The speed of a photon is equal to the speed of the light. The relationship between the energy of a photon and its wavelength can be expressed with the Planck-Einstein equation represented as follows:

E=hf

Here,

E = energy of a photon

h = Planck constant

f = frequency of a photon

Since, the frequency of the photon is not given, the equation of

c=fλ should be used, where,

c = speed of the light

f = frequency of the light

λ = wavelength of the light

Using

c=fλ, f=c/λ

Substituting f=c/λ for f in E=hf

E=hc/λ

= (6.6260 × 10−34) m2kg s−1× (3× 108) ms−1(1000×10 −9) m

Since, (1 nm=1×10−9m)

=19.878×10−20J=1.9878×10−19J

Energy of a photon in kilojoules,=1.9878×10−19×10−3kJ

Since,

(1J=1×10−3kJ)

=1.9878×10−22kJ

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, energy of mole of photon in kiloJoules,

=1.978×10-22kJ×6.022×1023mol-1=119.7053 kJ mol-1=119.71 kJ mol-1

Energy of a photon in kilojoules =1.9878×10−19kJ

Energy of a photon in kilocalories,

=1.9878×10-19×0.000239 kcal(1J=0.000239 kcal)=4.7508×10-23kcal

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, Energy of moles of photon in kilocalories,

=4.7908×10 -23kcal×6.022×1023=28.6091 kcal mol -1=28.61 kcal mol-1

Einstein is also a unit of energy.

Einstein =1kJ mol-1

Thus, energy of mole of photon in kilo Joules = 119.71 kJ mol−1

Energy of moles of photon in kilocalories = 28.61 kcal mol−1

Interpretation Introduction

(b)

Interpretation:

The maximum increase in the redox potential induced by 1000 nm photon needs to be determined.

Concept introduction:

One electron volt is the energy need to move an electron between one-volt potential difference.

One electron volt =1.602×10−19J

Expert Solution

Answer to Problem 23P

1.24 V

Explanation of Solution

The maximum increase in a redox potential can be calculated as follows:

Maximum increase in redox potential

=Energy possess by a photonEnergy need to move an electron

Putting the values,

=1 .9878×10 -19 J1 .602×10 -19J=1.24 V

Interpretation Introduction

(c)

Interpretation:

Number of photons need to overcome the Gibbs free energy should be calculated.

Concept introduction:

In a simple definition, Gibbs free energy of a reaction is the energy associated with that particular chemical reaction.

Phosphorylation is the conversion of ATP from ADP.

Expert Solution

Answer to Problem 23P

0.42 photons.

Explanation of Solution

Phosphorylation is the conversion of ATP from ADP.

ADP+PO43−→Energy from electronsATP

ATP- Adenosine triphosphate

ADP-Adenosine diphosphate

Therefore,

Energy needed for the phosphorylation reaction = Energy needed to convert a mole of ADP to ATP = 50 kJ mol−1.

Number of ADP in one mole of ADP = Avogadro’s number of ADP = 6.022×1023.

Energy needed to convert one ADP to ATP,

= 50 kJ6.022×1023.

Maximum energy that one 1000 nm photon possess

=1.978×10−22kJ.

Minimum number of 1000 nm photons needed for the phosphorylation reaction,

=Total energy to convert one ADP to ATPEnergy of one 1000 nm photon

Putting the values,

= 50kJ mo l -1 6.022×1 0 23 1.978×10 -22kJ=0.4176 photons=0.42 photons

Answer 6

Chapter 19, Problem 23P

Interpretation Introduction

(a)

Interpretation:

Energy possessed by a mole of photon should be calculated and expressed with the units of kilojoules and kilocalories.

Concept introduction:

A photon can be introduced as a piece of energy which has no mass. The relationship between the energy of a photon and its wavelength can be expressed with the following Planck-Einstein equation represented as follows:

E=hf

Expert Solution

Answer to Problem 23P

119.71 kJ mol−1 / 28.61 kcal mol−1

Explanation of Solution

A photon can be introduced as a piece of energy which has no mass. The speed of a photon is equal to the speed of the light. The relationship between the energy of a photon and its wavelength can be expressed with the Planck-Einstein equation represented as follows:

E=hf

Here,

E = energy of a photon

h = Planck constant

f = frequency of a photon

Since, the frequency of the photon is not given, the equation of

c=fλ should be used, where,

c = speed of the light

f = frequency of the light

λ = wavelength of the light

Using

c=fλ, f=c/λ

Substituting f=c/λ for f in E=hf

E=hc/λ

= (6.6260 × 10−34) m2kg s−1× (3× 108) ms−1(1000×10 −9) m

Since, (1 nm=1×10−9m)

=19.878×10−20J=1.9878×10−19J

Energy of a photon in kilojoules,=1.9878×10−19×10−3kJ

Since,

(1J=1×10−3kJ)

=1.9878×10−22kJ

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, energy of mole of photon in kiloJoules,

=1.978×10-22kJ×6.022×1023mol-1=119.7053 kJ mol-1=119.71 kJ mol-1

Energy of a photon in kilojoules =1.9878×10−19kJ

Energy of a photon in kilocalories,

=1.9878×10-19×0.000239 kcal(1J=0.000239 kcal)=4.7508×10-23kcal

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, Energy of moles of photon in kilocalories,

=4.7908×10 -23kcal×6.022×1023=28.6091 kcal mol -1=28.61 kcal mol-1

Einstein is also a unit of energy.

Einstein =1kJ mol-1

Thus, energy of mole of photon in kilo Joules = 119.71 kJ mol−1

Energy of moles of photon in kilocalories = 28.61 kcal mol−1

Answer 7

Chapter 19, Problem 23P

Interpretation Introduction

(a)

Interpretation:

Energy possessed by a mole of photon should be calculated and expressed with the units of kilojoules and kilocalories.

Concept introduction:

A photon can be introduced as a piece of energy which has no mass. The relationship between the energy of a photon and its wavelength can be expressed with the following Planck-Einstein equation represented as follows:

E=hf

Answer 8

Expert Solution

Answer to Problem 23P

119.71 kJ mol−1 / 28.61 kcal mol−1

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

A photon can be introduced as a piece of energy which has no mass. The speed of a photon is equal to the speed of the light. The relationship between the energy of a photon and its wavelength can be expressed with the Planck-Einstein equation represented as follows:

E=hf

Here,

E = energy of a photon

h = Planck constant

f = frequency of a photon

Since, the frequency of the photon is not given, the equation of

c=fλ should be used, where,

c = speed of the light

f = frequency of the light

λ = wavelength of the light

Using

c=fλ, f=c/λ

Substituting f=c/λ for f in E=hf

E=hc/λ

= (6.6260 × 10−34) m2kg s−1× (3× 108) ms−1(1000×10 −9) m

Since, (1 nm=1×10−9m)

=19.878×10−20J=1.9878×10−19J

Energy of a photon in kilojoules,=1.9878×10−19×10−3kJ

Since,

(1J=1×10−3kJ)

=1.9878×10−22kJ

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, energy of mole of photon in kiloJoules,

=1.978×10-22kJ×6.022×1023mol-1=119.7053 kJ mol-1=119.71 kJ mol-1

Energy of a photon in kilojoules =1.9878×10−19kJ

Energy of a photon in kilocalories,

=1.9878×10-19×0.000239 kcal(1J=0.000239 kcal)=4.7508×10-23kcal

Number of photons in a mole = Avogadro number of photons =6.022×1023

Therefore, Energy of moles of photon in kilocalories,

=4.7908×10 -23kcal×6.022×1023=28.6091 kcal mol -1=28.61 kcal mol-1

Einstein is also a unit of energy.

Einstein =1kJ mol-1

Thus, energy of mole of photon in kilo Joules = 119.71 kJ mol−1

Energy of moles of photon in kilocalories = 28.61 kcal mol−1

Answer 13

Interpretation Introduction

(b)

Interpretation:

The maximum increase in the redox potential induced by 1000 nm photon needs to be determined.

Concept introduction:

One electron volt is the energy need to move an electron between one-volt potential difference.

One electron volt =1.602×10−19J

Expert Solution

Answer to Problem 23P

1.24 V

Explanation of Solution

The maximum increase in a redox potential can be calculated as follows:

Maximum increase in redox potential

=Energy possess by a photonEnergy need to move an electron

Putting the values,

=1 .9878×10 -19 J1 .602×10 -19J=1.24 V

Answer 14

Interpretation Introduction

(b)

Interpretation:

The maximum increase in the redox potential induced by 1000 nm photon needs to be determined.

Concept introduction:

One electron volt is the energy need to move an electron between one-volt potential difference.

One electron volt =1.602×10−19J

Answer 15

Expert Solution

Answer to Problem 23P

1.24 V

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The maximum increase in a redox potential can be calculated as follows:

Maximum increase in redox potential

=Energy possess by a photonEnergy need to move an electron

Putting the values,

=1 .9878×10 -19 J1 .602×10 -19J=1.24 V

Answer 20

Interpretation Introduction

(c)

Interpretation:

Number of photons need to overcome the Gibbs free energy should be calculated.

Concept introduction:

In a simple definition, Gibbs free energy of a reaction is the energy associated with that particular chemical reaction.

Phosphorylation is the conversion of ATP from ADP.

Expert Solution

Answer to Problem 23P

0.42 photons.

Explanation of Solution

Phosphorylation is the conversion of ATP from ADP.

ADP+PO43−→Energy from electronsATP

ATP- Adenosine triphosphate

ADP-Adenosine diphosphate

Therefore,

Energy needed for the phosphorylation reaction = Energy needed to convert a mole of ADP to ATP = 50 kJ mol−1.

Number of ADP in one mole of ADP = Avogadro’s number of ADP = 6.022×1023.

Energy needed to convert one ADP to ATP,

= 50 kJ6.022×1023.

Maximum energy that one 1000 nm photon possess

=1.978×10−22kJ.

Minimum number of 1000 nm photons needed for the phosphorylation reaction,

=Total energy to convert one ADP to ATPEnergy of one 1000 nm photon

Putting the values,

= 50kJ mo l -1 6.022×1 0 23 1.978×10 -22kJ=0.4176 photons=0.42 photons

Answer 21

Interpretation Introduction

(c)

Interpretation:

Number of photons need to overcome the Gibbs free energy should be calculated.

Concept introduction:

In a simple definition, Gibbs free energy of a reaction is the energy associated with that particular chemical reaction.

Phosphorylation is the conversion of ATP from ADP.

Answer 22

Expert Solution

Answer to Problem 23P

0.42 photons.

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Phosphorylation is the conversion of ATP from ADP.

ADP+PO43−→Energy from electronsATP

ATP- Adenosine triphosphate

ADP-Adenosine diphosphate

Therefore,

Energy needed for the phosphorylation reaction = Energy needed to convert a mole of ADP to ATP = 50 kJ mol−1.

Number of ADP in one mole of ADP = Avogadro’s number of ADP = 6.022×1023.

Energy needed to convert one ADP to ATP,

= 50 kJ6.022×1023.

Maximum energy that one 1000 nm photon possess

=1.978×10−22kJ.

Minimum number of 1000 nm photons needed for the phosphorylation reaction,

=Total energy to convert one ADP to ATPEnergy of one 1000 nm photon

Putting the values,

= 50kJ mo l -1 6.022×1 0 23 1.978×10 -22kJ=0.4176 photons=0.42 photons

Answer 27

Ch. 19 - Prob. 1P Ch. 19 - Prob. 2P Ch. 19 - Prob. 3P Ch. 19 - Prob. 4P Ch. 19 - Prob. 5P Ch. 19 - Prob. 6P Ch. 19 - Prob. 7P Ch. 19 - Prob. 8P Ch. 19 - Prob. 9P Ch. 19 - Prob. 10P

Answer to Problem 23P

Explanation of Solution

(b) Interpretation: The maximum increase in the redox potential induced by 1000 nm photon needs to be determined. Concept introduction: One electron volt is the energy need to move an electron between one-volt potential difference. One electron volt =1.602×10−19J

Answer to Problem 23P

Explanation of Solution

(c) Interpretation: Number of photons need to overcome the Gibbs free energy should be calculated. Concept introduction: In a simple definition, Gibbs free energy of a reaction is the energy associated with that particular chemical reaction. Phosphorylation is the conversion of ATP from ADP.

Answer to Problem 23P

Explanation of Solution

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Chapter 19 Solutions

(b)

Interpretation:

The maximum increase in the redox potential induced by 1000 nm photon needs to be determined.

Concept introduction:

One electron volt is the energy need to move an electron between one-volt potential difference.

One electron volt =1.602×10−19J

(c)

Interpretation:

Number of photons need to overcome the Gibbs free energy should be calculated.

Concept introduction:

In a simple definition, Gibbs free energy of a reaction is the energy associated with that particular chemical reaction.

Phosphorylation is the conversion of ATP from ADP.