Question

Answer the following questions (shown in the attached image)

1) What does this change in reducing power represent in kilojoules per mole of electrons?

2) You have incorrectly calculated the energy of 840 nm light. What is the energy of 840 nm light in kilojoules per einstein (i.e., mole) of photons?

3) What fraction of the absorbed light energy is trapped in the form of reducing power? Enter as a decimal. fraction:

When the photosystem of green sulfur bacteria absorbs light at 840 nm, the standard reduction potential of P(840) changes from about 0.1 V to about –0.9 V. What does this change in reducing power represent in kilojoules per mole of electrons? kJ AG' ® electrons = mol You have incorrectly calculated the energy of 840 nm light. What is the energy of 840 nm light in kilojoules per einstein (i.e., mole) of photons? Ephotons = kJ einstein What fraction of the absorbed light energy is trapped in the form of reducing power? Enter as a decimal. fraction:

Answer 1

1.

We use the following equation to calculate ΔG from potential V

ΔG=-n*F*V (unit:J/mol)

where n=number of electrons involved

F=Faraday's constant=96485.33 C mol−1

Here we are assuming that 1 electron transfer is taking place, hence n=1

In the first case, V=0.1 v

ΔG=-n*F*V

ΔG1=-1*96485.33*0.1 J/mol

ΔG1=-9648.533 J/mol

In the second case, V=-0.9V

ΔG2=-1*96485.33*-0.9 J/mol

ΔG2=86836.797 J/mol

Change in ΔG due to change in V = ΔG2 - ΔG1 = 86836.797 - (-9648.533) J/mol = 96485.33 J/mol

Change in ΔG due to change in reducing power =  96.48533 KJ/mol

2.

The energy of a photon is given by

E = hc/λ

where h=Planck's constant , c=speed of light in SI units , λ=wavelength

Since we want energy per mole, we multiply with Avagadro number (NA = 6.022*1023)

The energy of a mole of photon is given by

E = (NA*hc)/λ

Substituting values,

h=6.626× 10-34 m2 kg / s

c=3×108 m / s

λ=840×10-9 m

NA = 6.022*1023

you get

E=142.5 KJ/Einstein

3.

Fraction of light energy trapped= 96.48533 /  142.5 = 0.677