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Lecture13, 2-8-08 Equilibrium constant

Lecture13, 2-8-08 Equilibrium constant - Calculate G for...

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Calculate Δ G for photosynthesis of glucose, C 12 O 6 H 12 •Find the Δ G f 0 values of reactants and products Δ G 0 photosyn = {[-911 + 0] - [6(-395) + 6(-237)]} kJ Δ G 0 photosyn =+2881 kJ per mole of glucose 6 CO 2 (g) + 6 H 2 O(l) C 6 H 12 O 6 (s) + 6 O 2 (g) -237 kJ/mol -395 kJ/mol 0 -911 kJ/mol Δ G f 0 H 2 O(l) CO 2 (g) O 2 (g) C 6 H 12 O 6 (s)

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equilibrium position 6CO 2 + 6H 2 O C 6 H 12 O 6 + 6O 2 G reactants G products G 0 =2881 kJ/mol Standard Conditions Interpretation of large positive value of G 0 photosyn Reactants Products
Plants know how to make glucose Δ G 0 photosyn = +2881 kJ/mol What’s missing in this cartoon? CO 2 (g) H 2 O (l) C 6 H 12 O 6 Where’s the O 2 ? Plant’s don’t use O 2. Why do they make it? O 2 Energy

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6 CO 2 (g) + 6 H 2 O(l) C 6 H 12 O 6 (s) + 6 O 2 (g) Let’s look at a half reaction for CO 2 reduction 6 CO 2 (g)+24 e - +24 H + C 6 H 12 O 6 (s)+6 H 2 O( l ) Where does a plant find e - and H + ? By making oxygen! 12 H 2 O 24 H + + 6 O 2 + 24 e - It all adds up to the photosynthesis reaction 6 CO 2 (g) + 6 H 2 O(l) C 6 H 12 O 6 (s) + 6 O 2 (g) Why do plants make oxygen? (We are mighty glad they do!)
Shedding some light on Photosynthesis The experiment: Mix CO 2 and water Question 1 : What happens? Answer 1 : A little bit of H 2 CO 3 is formed and a little bit of acid dissociation occurs • H 2 O + CO 2 H 2 CO 3 H + + HCO 3 - Next, expose the solution to sunlight Question 2 : What happens? Answer 2 : Maybe heats the solution a little, no glucose formed There’s more to this than just providing energy. Why do plants have green leaves?

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Lecture13, 2-8-08 Equilibrium constant - Calculate G for...

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