Problem 2

The Hill reaction dissects photosynthesis. (a) Write the overall equation of plant photosynthesis, reducing

\ce{CO2}

\{\ce{CH2O}\}

. (b) Hill found that isolated chloroplasts do not evolve

\ce{O2}

in light even with

\ce{CO2}

, but do so upon adding potassium ferrioxalate

\ce{K3[Fe(C2O4)3]}

(with excess oxalate) without

\ce{CO2}

. Give the oxidant and reducing agent in natural photosynthesis and in the Hill reaction. (c) Hill measured evolved

\ce{O2}

with haemoglobin (Hb binds

\ce{O2}

1:1, initial

[\text{Hb}] = 0.6 \times 10^{-4}

mol dm

^{-3}

); at

[\ce{Fe^{III}}] = 2.0 \times 10^{-4}

mol dm

^{-3}

the HbO

_2

fraction saturates at about 75 %. Estimate the Fe/

\ce{O2}

ratio, write the Hill reaction equation, and calculate its

\Delta G

T = 298

p(\ce{O2}) = 1

mmHg, pH = 8, standard concentrations of other species (

E^{\circ}

\ce{O2 + 4H+ + 4e- -> 2H2O}

+1.23 V;

\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}

, producing 47.6 mm

^3

\ce{O2}

(10 °C, 740 mmHg). Calculate the quantum requirement (photons per electron transferred). (e) Conclusions: are water oxidation and

\ce{CO2}

reduction spatially separated? Is

\ce{O2}

produced from

\ce{CO2}

? Does water oxidation require light? Do most chlorophylls participate directly? Does each photon transfer one electron?

Step 3 of 5: Gibbs energy and spontaneity

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E^{\circ} = 0.05 - 1.23 = -1.18\ \text{V};\quad \Delta G = -nF E^{\circ} + RT\ln\!ig(p_{\ce{O2}}[\ce{H+}]^{4}\big) \approx 454 + RT\ln(10^{-3} \times 10^{-32}) \approx +257\ \text{kJ mol}^{-1} > 0

Analysis

The cell emf is $-1.18$ V, giving $\Delta G^{\circ} = -4 \times 96485 \times (-1.18) = +454$ kJ mol $^{-1}$ ; the reaction quotient $Q = p_{\ce{O2}}[\ce{H+}]^4 = 10^{-3} \times 10^{-32} = 10^{-35}$ lowers it to $\approx +257$ kJ mol $^{-1}$ — still strongly endergonic, so the reaction is not spontaneous and light is required.