# Solutions to Haynie Ch. 1

September 3, 2017 | Author: Dharin Detama | Category: Electron, Chemistry, Physical Sciences, Science, Nature

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a solution for biological thermodynamic...

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temperature decreases, leaves die. They also change of color, reflecting an underlying change in the distribution of pigment molecules. In late autumn tree leaves are brown, when no pigment molecules are being produced. 5. Energy of a 470 nm photon = E470 = (6.63 × 10–34 J s) × (2.998 × 108 m s–1) / (470 × 10–9 m) = 4.2 × 10–19 J. E700 = E470 × 470 / 700 = 2.8 × 10–19 J. 3.5 × 10–19 J photon–1 × (6.02 × 1023 photons mol–1) = 2.1 × 105 J mol–1. 1 cal = 4.184 J. 2.1 × 105 J mol–1 / (7 × 103 cal mol–1 × 4.184 J cal–1) = 7.2. That is, a single visible wavelength photon corresponds to an energy about seven times greater than the energy released upon hydrolysis of one molecule ATP to ADP under standard state conditions. 6. 1000 µE s–1 = 103 × 10–6 E s–1 × 6.02 × 1023 photons E–1 × 36 × 102 s hr–1 = 217 × 1022 photons hr–1. 0.4 efficiency × 217 × 1022 photons hr–1 = 86.7 × 1022 effective photons hr–1. 86.7 × 1016 photons hr–1 / (8 photons / molecule of CO2 fixed) = 10.8 × 1022 molecules of CO2 fixed hr–1. 10.8 × 1022 molecules of CO2 hr–1 × (1 mol CH2O produced / 1 mol CO2 consumed) × 30 g CH2O mol–1 / (6.02 × 1023 mol–1) = 1.8 × 10–1 mol of CH2O hr–1 = 5.4 g of CH2O produced. 7. At least 2870 kJ mol–1 are needed to synthesize glucose. The energy of a 700 nm photon is E = hc/λ = 6.63 × 10–34 J s × 2.998 × 108 m s–1 / (700 × 10–9 m) = 0.0284 × 10–17 J One mole of such photons has an energy of 0.02839 × 10–17 J × 6.02 × 1023 = 171 kJ. The energy required to fix one mole of CO2 is 2870 kJ / 6 = 478 kJ. The number of 700 nm photons required is 478 kJ mol–1 / 171 kJ mol–1 = 2.80. Because 3–4 times that many photons are required, the efficiency of the process is 1/4–1/3, or 25–33%. In comparison with industrial chemical processes, the level of efficiency of biotic glucose synthesis is very good. 8. Be creative!