1 Suppose for simplicity that the formation of a glucose molecule from carbon dioxide gas and liquid water were to take place without requiring any potential energy barrier to be surmounted. The chemical equation for the conversion is:

    6CO2 + 6H2O + Light → C6H12O6 + 6O2

What is the minimum number of photons that could provide sufficient energy, considering light with a wavelength of 485 nm (the most intense of the visible range)? Suggest why the actual number of photons required significantly exceeds this minimum.

    Avogadro constant NA = 6.02 x 1023 mol-1

    Planck constant h = 6.63 x 10-34 J s

    speed of light c = 3.00 x 108 m s-1.

The relevant enthalpies of formation under standard conditions are:

    CO2 -394 kJ mol-1

    H2O -286 kJ mol-1

    glucose -1273 kJ mol-1.

2 Calculate the number of photons arriving in a chlorophyll molecule per second, given that the light-absorbing conjugated structure of the pyrrole ring unit has dimensions of around 0.70 x 0.85 nm2. See Slide 9 and use the formula.

ñ = Iσλ/hc

3 Describe qualitatively the differences between electronic energy transfer and electron transfer.


1 We can write equations for the formation of water, carbon dioxide and glucose from their elements in natural form at standard temperature and pressure, denoting enthalpy changes by ΔH:

    ① C (solid) + O2 (gas) → CO2 (gas); ΔH = -394 kJ mol-1

    ② H2 (gas) + ½O2 (gas) → H2O (water); ΔH = -286 kJ mol-1

    ③ 6C (solid) + 6H2 (gas) + 3O2 (gas) → C6H12O6 (solid); ΔH = -1273 kJ mol-1

Take ③ - 6 x ① - 6 x ② to give

    6CO2 (gas) +6H2O (water) → C6H12O6 (solid) + 6O2 (gas).

Similarly for the enthalpies the net energy required is:

    (-1273) – 6 x (-394 -286) = 2807 kJ mol-1

(Note: the enthalpy of formation of gaseous oxygen is zero because the molecule is in its naturally occurring state).

    2807 kJ mol-1 ÷ 6.02 x 1023mol-1= 4.66 x 10-18 J

    Photon energy at 485 nm = hc/l

         = 4.10 x 10-19 J


      4.66 x 10-18 J ÷ 4.10 x 10-19 J = 11.36

Hence the minimum number of photons is 12. Actually, most of the absorption of sunlight occurs at longer wavelengths, and as we shall see some losses are necessarily built into the photosynthetic process, so the true number is closer to 48.

2 Let us consider that the conjugated structure of the pyrrole ring represents the part of the molecule that absorbs the light; it has dimensions of around 0.70 nm x 0.85 nm, say an area of 0.60 nm2. Since this can be positioned at all angles with respect to the input radiation, we have to average over three dimensions, using a cos-squared function to account for the random orientations; this gives a value of 1/3 so that the mean effective area presented by each molecule is 0.20 nm2 .

So with say I = 1 kW m-2, and considering light with a wavelength of 485 nm, the most intense of the visible range, we have

Problem Equation 2

In fact most of this light falls outside the absorption range; the effective number is at most one tenth of this amount (see the absorption profile), so not more than 50 per second. And then accounting for non-equatorial regions, daytime average, cloud cover, tree canopy screening, there will be another at least one or two orders of magnitude reduction. So the number of effective photons per second is about one.

3 The main difference between energy transfer and electron transfer is that in the case of energy transfer an excited state relocates from one chromophore to another, giving rise to a net energy change, but not a net change in the charge distribution. In the case of electron transfer, an electron relocates from one chromophore to another giving rise to net change in charge distribution between the chromophores.

Full answers would also require molecular orbital diagrams similar to those on Slides 15 and 17.