Photosynthesis is believed to be an efficient way of light energy conversion. Let’s check this state — Organic Chemistry Chemistry Question
Efficiency of photosynthesis
Photosynthesis is believed to be an efficient way of light energy conversion. Let’s check this statement from various points of view. Consider the overall chemical equation of photosynthesis performed by green plants in the form:
H2O + CO2 → CH2O + O2
where CH2O denotes the formed carbohydrates. Though glucose is not the main organic product of photosynthesis, it is quite common to consider CH2O as 1/6(glucose). Using the information presented below, answer the following questions.
Necessary information:
Average (over 24 h) solar energy absorbed by Moscow region in summer time – 150 W⋅m–2;
Moscow area – 1070 km2, percentage of green plants area – 18 %;
MSU campus area – 1.7 km2, percentage of green plants area – 54 %;
green plants utilize ~10 % of the available solar energy (average wavelength is 680 nm)
Calculate the standard enthalpy and standard Gibbs energy of the above reaction at 298 K. Assuming that the reaction is driven by light energy only, determine the minimum number of photons necessary to produce one molecule of oxygen.
Model Answer
2.1 H2O + CO2 → CH2O + O2.
The process is reverse to combustion of 1/6(glucose), hence:
∆H°298 = –1/6 ∆H°c,298(C6H12O6) = 467.5 kJ mol–1
Standard entropy change in the reaction:
∆S°298 = 1/6 S°298(C6H12O6) + S°298(O2) – S°298(H2O) – S°298(CO2) = –43.7 J K–1 mol–1
Standard Gibbs energy change:
∆G°298 = ∆H°298 – 298⋅∆S°298 = 467.5 – 298×(–43.7⋅10–3) = 480.5 kJ mol–1
Energy of 1 mol of photons with wavelength of 680 nm:
E = (hc/λ)NA = 6.63⋅10–34 × 3.00⋅108 × 6.02⋅1023 / (680⋅10–9) = 176 kJ mol–1
The minimum number of photons necessary to supply more energy than E = 480.5 kJ mol–1 is 3.
Standard Gibbs energy corresponds to standard partial pressures of all gases (1 bar). In atmosphere, the average partial pressure of oxygen is 0.21 bar and that of carbon dioxide –3⋅10–4 bar. Calculate the Gibbs energy of the above reaction under these conditions (temperature 298 K).
Model Answer
2.2 Energy of 10 mol of photons absorbed by green plants is 176⋅10 = 1760 kJ. Of this amount 480.5 kJ is converted to Gibbs energy. The efficiency of the solar energy conversion by green plants can be estimated as 480.5 / 1760 ⋅ 100 % = 27 %.
Actually, liberation of one oxygen molecule by green plants requires not less than 10 photons. What percent of the absorbed solar energy is stored in the form of Gibbs energy? This value can be considered as the efficiency of the solar energy conversion.
Model Answer
2.3 Total solar energy absorbed:
a) Moscow area: E = 1070 ⋅ 106 m2 ⋅ 150 J s–1⋅m–2 ⋅ (10 × 86400) s = 1.4 ⋅ 1017 J.
b) MSU campus: E = 1.7 ⋅ 106 m2 ⋅ 150 J s–1⋅m–2 ⋅ (5 × 3600) s = 4.6 ⋅ 1012 J.
Number of photons N = (E / Em) NA:
a) Moscow area: N = 4.8 ⋅ 1035.
b) MSU campus: N = 1.6 ⋅ 1031.
Solar energy utilized by green plants and converted to chemical energy:
a) Moscow area: Eutil = 1.4 ⋅ 1017× (18% / 100%) ×(10% / 100%) ×(27% / 100%) = 6.8 ⋅ 1014 J
b) MSU campus: Eutil = 4.6 ⋅ 1012 × (54% / 100%) × (10% / 100%) ×⋅ (27% / 100%) = 6.7 ⋅ 1010 J
Quantity of photosynthesis products n(CH2O) = Eutil / ∆G°r 298
a) Moscow area: n(CH2O) = n(O2) = 1.4 ⋅ 109 mol
m(CH2O) = n⋅M = 1.4 ⋅ 109 mol ⋅ 0.03 kg / mol = 4.2 ⋅ 107 kg
V(O2) = n⋅Vm = 1.4 ⋅ 109 mol ⋅ 0.0244 m3 / mol = 3.4 ⋅ 107 m3
b) MSU campus: n(CH2O) = n(O2) = 1.4 ⋅ 105 mol
m(CH2O) = n⋅M = 1.4 ⋅ 105 mol × 0.03 kg / mol = 4200 kg
V(O2) = n⋅Vm = 1.4 ⋅ 105 mol × 0.0244 m3 / mol = 3400 m3
How many photons will be absorbed and how much biomass (in kg) and oxygen (in m3 at 25 oC and 1 atm) will be formed:
a) in Moscow during 10 days of IChO;
b) in the MSU campus during the practical examination (5 hours)?
Model Answer
2.4 Percent of solar energy converted to chemical energy:
a) Moscow area: (18% / 100%) × (10% / 100%) × (27% / 100%) = 0.005 = 0.5 %
b) MSU campus: (54% / 100%) ⋅ (10% / 100%) ⋅ (27% / 100%) = 0.015 = 1.5 %
What percent of the solar energy absorbed by the total area will be converted to chemical energy:
a) in Moscow;
b) in MSU?
This is another measure of photosynthesis efficiency.