Chapter 21, Problem 16P

Interpretation Introduction

(a)

To Calculate:

The value of $\Delta G$ for the plastoquinone by 870nm light.

Introduction:

Oxygen is produced during the photosynthesis and this oxygen evolution rate is measured. Method of calibration is used having the hydrogen peroxide and the précised reading of the oxygen concentration is ensured by using the catalyst. The oxygen that is above the air saturated level is measured.

Explanation of Solution

The light quantum energy is determined using the given formula,

  $E=N\frac{hc}{\lambda }$

Where,

H=Planck’s constant

c=light speed

N=avogardo’s number

  $\lambda$ =wave length

For one mole of photon 870nm we have,

  $E=(6.02\times 10^{23})\frac{(6.626\times 10^{-34}Jsec)(3\times 10^{2}m/sec)}{870nm}\times 10^{9}nm/m$

  $=1375\text{47J=137}\text{.5kJ}$

As 2 moles of quanta is considered,

  $137.5\times 2=27\text{5}\text{.0kJ}$

The free change of energy for the reaction is calculated by determining the change of the standard potential,

  $\Delta E_0^{\text{'}}=acceptor\Delta E_0^{\text{'}}-donor\Delta E_0^{\text{'}}$

  $\begin{array}{l}=0.07V-0.5V\\ =-0.43V\end{array}$

  $\Delta G^{0\text{'}}$ is calculated using,

  $\begin{array}{l}\Delta G^{0\text{'}}=-nF\Delta E_0^{\text{'}}\\ n=2\\ F=96,485kj/Vmol\end{array}$

  $\begin{array}{l}=-2(96.485kj/V.mol)(-0.43V)\\ =-83.0k\text{J}/mol\end{array}$

Now G is calculated as,

  $-275.\text{0kJ+83}\text{.0kJ}\Rightarrow -19\text{2kJ}$

Interpretation Introduction

(b)

To predict:

For the plastoquinone by 700 nm light $\Delta G$ is calculated.

Introduction:

Oxygen is produced during the photosynthesis and this oxygen evolution rate is measured. Method of calibration is used having the hydrogen peroxide and the précised reading of the oxygen concentration is ensured by using the catalyst. The oxygen that is above the air saturated level is measured.

Explanation of Solution

The light quantum energy is determined using the given formula,

  $E=N\frac{hc}{\lambda }$

Where,

H=Planck’s constant

c=light speed

N=Avogadro’s number

  $\lambda$ =wave length

For one mole of photon 700nm we have,

  $E=(6.02\times {10}^{23})\frac{\text{(6}{\text{.626×10}}^{\text{-34}}{\text{ Jsec)(3×10}}^{\text{2}}\text{ m/sec)}}{\text{700 nm}}{\text{×10}}^{\text{9}}\text{ nm/m}$

  $=170951\text{ J=171}\text{.0 kJ}$

As 2 moles of quanta is considered,

  $171\times 2=342.\text{0 kJ}$

The free change of energy for the reaction is calculated by determining the change of the standard potential,

  $\Delta E_0^{\text{'}}=\text{acceptor}\Delta E_0^{\text{'}}-\text{donor}\Delta E_0^{\text{'}}$

  $\begin{array}{l}=0.07V-0.5V\\ =-0.43V\end{array}$

  $\Delta G^{0\text{'}}$ is calculated using,

  $\begin{array}{l}\Delta G^{0\text{'}}=-nF\Delta E_0^{\text{'}}\\ n=2\\ F=96,485\text{ kJ/Vmol}\end{array}$

  $\begin{array}{l}=-2(\text{96}\text{.485kJ/V}\text{.mol)(-0}\text{.43V)}\\ \text{=--83}\text{.0kJ/mol}\end{array}$

Now G is calculated as,

  $-342.0\text{kJ+83}\text{.0kJ=-256kJ}$

Interpretation Introduction

(c)

To predict:

For the plastoquinone by 680 nm light $\Delta G$ is calculated.

Introduction:

Oxygen is produced during the photosynthesis and this oxygen evolution rate is measured. Method of calibration is used having the hydrogen peroxide and the précised reading of the oxygen concentration is ensured by using the catalyst. The oxygen that is above the air saturated level is measured.

Explanation of Solution

The light quantum energy is determined using the given formula,

  $E=N\frac{hc}{\lambda }$

Where,

h=Planck’s constant

c=light speed

N=Avogadro’s number

  $\lambda$ =wave length

For one mole of photon 680nm we have,

  $E=(6.02\times {10}^{23})\frac{(6.626\times 10^{-34}Jsec)(3\times 10^{2}m/sec)}{680nm}\times {10}^9\text{nm/m}$

  $=175979J\Rightarrow 176.0kJ$

As 2 moles of quanta is considered,

  $176*2=352.0kJ$

The free change of energy for the reaction is calculated by determining the change of the standard potential,

  $\Delta E_0^{\text{'}}=acceptor\Delta E_0^{\text{'}}-donor\Delta E_0^{\text{'}}$

  $\begin{array}{l}=0.07V-0.5V\\ =-0.43V\end{array}$

  $\Delta G^{0\text{'}}$ is calculated using,

  $\begin{array}{l}\Delta G^{0\text{'}}=-nF\Delta E_0^{\text{'}}\\ n=2\\ F=96,4\text{85kJ/Vmol}\end{array}$

  $\begin{array}{l}=-2(96.485k\text{J}/V.mol)(-0.43V)\\ =--83.0k\text{J}/mol\end{array}$

Now G is calculated as,

  $-3\text{52}\text{.2kJ+83}\text{.0kJ=-269kJ}$