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 to Problem 23P

$\text{119}{\text{.71 kJ mol}}^{-1}$ / $28.61{\text{ kcal mol}}^{-1}$

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 $$

$c=f\lambda$ should be used, where,

$c$ = speed of the light

$f$ = frequency of the light

$\lambda$ = wavelength of the light

Using $$

$c=f\lambda$, $f=c/\lambda$

Substituting $f=c/\lambda$ for f in $E=hf$$$

$E=hc/\lambda$

= $\frac{(6.6260\times {10}^{-34})m^{2}kg{s}^{-1}\times \left(3\times {10}^8\right)m{s}^{-1}}{\begin{array}{l}(1000\times {10}^{-9})m\\ \end{array}}$

Since, $\left(1nm=1\times {10}^{-9}m\right)$

$\begin{array}{l}=19.878\times {10}^{-20}J\\ =1.9878\times {10}^{-19}J\\ \\ \end{array}$

Energy of a photon in kilojoules,$=1.9878\times {10}^{-19}\times {10}^{-3}kJ$

Since,

$\left(1J=1\times {10}^{-3}kJ\right)$

$=1.9878\times {10}^{-22}kJ$

$$

Number of photons in a mole = Avogadro number of photons $=6.022\times {10}^{23}$

Therefore, energy of mole of photon in kiloJoules,

$\begin{array}{l}\text{=1}{\text{.978×10}}^{\text{-22}}\text{kJ×6}{\text{.022×10}}^{\text{23}}{\text{mol}}^{\text{-1}}\\ \text{=119}{\text{.7053 kJ mol}}^{\text{-1}}\\ \text{=119}{\text{.71 kJ mol}}^{\text{-1}}\end{array}$

Energy of a photon in kilojoules $=1.9878\times {10}^{-19}kJ$

Energy of a photon in kilocalories,

$\begin{array}{l}\text{=1}{\text{.9878×10}}^{\text{-19}}\text{×0}\text{.000239 kcal}\\ \left(\text{1J=0}\text{.000239 kcal}\right)\\ \text{=4}{\text{.7508×10}}^{\text{-23}}\text{kcal}\end{array}$

Number of photons in a mole = Avogadro number of photons $=6.022\times {10}^{23}$

Therefore, Energy of moles of photon in kilocalories,

$\begin{array}{l}\text{=4}{\text{.7908×10}}^{{}^{\text{-23}}}\text{kcal×6}{\text{.022×10}}^{\text{23}}\\ \text{=28}{\text{.6091 kcal mol}}^{{}^{\text{-1}}}\\ \text{=28}{\text{.61 kcal mol}}^{\text{-1}}\end{array}$

Einstein is also a unit of energy.

Einstein $=1{\text{kJ mol}}^{{}^{-1}}$

Thus, energy of mole of photon in kilo Joules = $119.71{\text{ kJ mol}}^{-1}$

Energy of moles of photon in kilocalories = $28.61{\text{ kcal mol}}^{-1}$

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 $=1.602\times {10}^{-19}\text{J}$

Answer to Problem 23P

$1.24\text{ V}$

Explanation of Solution

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

$\text{Maximum increase in redox potential}$

$\text{=}\frac{\text{Energy possess by a photon}}{\text{Energy need to move an electron}}$

Putting the values,

$\begin{array}{l}\text{=}\frac{\text{1}{\text{.9878×10}}^{{}^{\text{-19}}}\text{J}}{\text{1}{\text{.602×10}}^{\text{-19}}\text{J}}\\ \text{=1}\text{.24 V}\end{array}$

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 to Problem 23P

$\text{0}\text{.42 photons}\text{.}$

Explanation of Solution

Phosphorylation is the conversion of ATP from ADP.

$\text{ADP}+{\text{PO}}_4{}^{3-}\stackrel{\text{Energy from electrons}}{\to }\text{ATP}$

ATP- Adenosine triphosphate

ADP-Adenosine diphosphate

Therefore,

Energy needed for the phosphorylation reaction = Energy needed to convert a mole of ADP to ATP = $50{\text{ kJ mol}}^{-1}$.

Number of ADP in one mole of ADP = Avogadro’s number of ADP = $6.022\times {10}^{23}$.

Energy needed to convert one ADP to ATP,

= $\frac{50\text{ kJ}}{6.022\times {10}^{23}}$.

Maximum energy that one $1000nm$ photon possess

$=1.978\times {10}^{{}^{-22}}kJ$.

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

$=\frac{\text{Total energy to convert one ADP to ATP}}{\text{Energy of one }1000nm\text{ photon}}$

Putting the values,

$\begin{array}{l}=\frac{\frac{50kJmo{l}^{-1}}{6.022\times 10^{{}^{23}}}}{1.978\times 10^{-22}kJ}\\ =0.4176photons\\ =0.42photons\end{array}$