Chemistry Labs

Problem 2

The Hill reaction dissects photosynthesis. (a) Write the overall equation of plant photosynthesis, reducing COX2\ce{CO2} to {CHX2O}\{\ce{CH2O}\}. (b) Hill found that isolated chloroplasts do not evolve OX2\ce{O2} in light even with COX2\ce{CO2}, but do so upon adding potassium ferrioxalate KX3[Fe(CX2OX4)X3]\ce{K3[Fe(C2O4)3]} (with excess oxalate) without COX2\ce{CO2}. Give the oxidant and reducing agent in natural photosynthesis and in the Hill reaction. (c) Hill measured evolved OX2\ce{O2} with haemoglobin (Hb binds OX2\ce{O2} 1:1, initial [Hb]=0.6×10−4[\text{Hb}] = 0.6 \times 10^{-4} mol dm−3^{-3}); at [FeXIII]=2.0×10−4[\ce{Fe^{III}}] = 2.0 \times 10^{-4} mol dm−3^{-3} the HbO2_2 fraction saturates at about 75 %. Estimate the Fe/OX2\ce{O2} ratio, write the Hill reaction equation, and calculate its ΔG\Delta G at T=298T = 298 K, p(OX2)=1p(\ce{O2}) = 1 mmHg, pH = 8, standard concentrations of other species (E∘E^{\circ}: OX2+4 HX++4 eX−→2 HX2O\ce{O2 + 4H+ + 4e- -> 2H2O} +1.23 V; [Fe(CX2OX4)X3]X3−+eX−→[Fe(CX2OX4)X3]X4−\ce{[Fe(C2O4)3]^{3-} + e- -> [Fe(C2O4)3]^{4-}} +0.05 V). Is it spontaneous? (d) Isolated chloroplasts were irradiated 2 h with 672 nm light of 0.503 mJ s−1^{-1}, producing 47.6 mm3^3 OX2\ce{O2} (10 °C, 740 mmHg). Calculate the quantum requirement (photons per electron transferred). (e) Conclusions: are water oxidation and COX2\ce{CO2} reduction spatially separated? Is OX2\ce{O2} produced from COX2\ce{CO2}? Does water oxidation require light? Do most chlorophylls participate directly? Does each photon transfer one electron?
Step 4 of 5: Quantum requirement
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Eabs=0.503×10−3×7200=3.62 J;nphot=3.62hcNA/λ=3.621.78×105=2.03×10−5 mol;ne=4 n(OX2)=4×2.00×10−6;Φreq=2.03×10−58.00×10−6=2.5E_{\text{abs}} = 0.503 \times 10^{-3} \times 7200 = 3.62\ \text{J};\quad n_{\text{phot}} = \dfrac{3.62}{hc N_A/\lambda} = \dfrac{3.62}{1.78 \times 10^{5}} = 2.03 \times 10^{-5}\ \text{mol};\quad n_{e} = 4\,n(\ce{O2}) = 4 \times 2.00 \times 10^{-6};\quad \Phi_{req} = \dfrac{2.03 \times 10^{-5}}{8.00 \times 10^{-6}} = 2.5
Analysis

One mole of 672 nm photons carries NAhc/λ=1.78×105N_A hc/\lambda = 1.78 \times 10^{5} J; the absorbed 3.62 J is 2.03×10−52.03 \times 10^{-5} mol photons. The 47.6 mm3^3 of OX2\ce{O2} is 2.00×10−62.00 \times 10^{-6} mol, i.e. 8.00×10−68.00 \times 10^{-6} mol of electrons, so 2.52.5 photons are required per electron transferred.