Study Guide with Student Solutions Manual and Problems Book for Garrett/Grisham's Biochemistry, 6th
Study Guide with Student Solutions Manual and Problems Book for Garrett/Grisham's Biochemistry, 6th
6th Edition
ISBN: 9798214167343
Author: GARRETT
Publisher: Cengage Learning US
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Chapter 21, Problem 16P
Interpretation Introduction

(a)

To Calculate:

The value of ΔG 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.

Expert Solution
Check Mark

Explanation of Solution

The light quantum energy is determined using the given formula,

  E=Nhcλ

Where,

H=Planck’s constant

c=light speed

N=avogardo’s number

  λ =wave length

For one mole of photon 870nm we have,

  E=(6.02×1023)(6.626×10-34Jsec)(3×102m/sec)870nm×109nm/m

  =137547J=137.5kJ

As 2 moles of quanta is considered,

  137.5×2=275.0kJ

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

  ΔE0'=acceptorΔE0'donorΔE0'

  =0.07V0.5V=0.43V

  ΔG0' is calculated using,

  ΔG0'=nFΔE0'n=2F=96,485kj/Vmol

  =2(96.485kj/V.mol)(-0.43V)=-83.0kJ/mol

Now G is calculated as,

  275.0kJ+83.0kJ192kJ

Interpretation Introduction

(b)

To predict:

For the plastoquinone by 700 nm light ΔG 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.

Expert Solution
Check Mark

Explanation of Solution

The light quantum energy is determined using the given formula,

  E=Nhcλ

Where,

H=Planck’s constant

c=light speed

N=Avogadro’s number

  λ =wave length

For one mole of photon 700nm we have,

  E=(6.02×1023)(6.626×10-34 Jsec)(3×102 m/sec)700 nm×109 nm/m

  =170951 J=171.0 kJ

As 2 moles of quanta is considered,

  171×2=342.0 kJ

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

  ΔE0'=acceptorΔE0'donorΔE0'

  =0.07V0.5V=0.43V

  ΔG0' is calculated using,

  ΔG0'=nFΔE0'n=2F=96,485 kJ/Vmol

  =2(96.485kJ/V.mol)(-0.43V)=--83.0kJ/mol

Now G is calculated as,

  342.0kJ+83.0kJ=-256kJ

Interpretation Introduction

(c)

To predict:

For the plastoquinone by 680 nm light ΔG 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.

Expert Solution
Check Mark

Explanation of Solution

The light quantum energy is determined using the given formula,

  E=Nhcλ

Where,

h=Planck’s constant

c=light speed

N=Avogadro’s number

  λ =wave length

For one mole of photon 680nm we have,

  E=(6.02×1023)(6.626×10-34Jsec)(3×102m/sec)680nm×109nm/m

  =175979J176.0kJ

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,

  ΔE0'=acceptorΔE0'donorΔE0'

  =0.07V0.5V=0.43V

  ΔG0' is calculated using,

  ΔG0'=nFΔE0'n=2F=96,485kJ/Vmol

  =2(96.485kJ/V.mol)(-0.43V)=--83.0kJ/mol

Now G is calculated as,

  352.2kJ+83.0kJ=-269kJ

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