Thermodynamics of photosynthesis cellular respiration PDF

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Thermodynamics of Photosynthesis & Cellular Respiration C6H12O6 + 6O2 ⇌ 6H2O + 6CO2 ΔH o (kJ/mol)

ΔS o J/(mol*K)

ΔGo (kJ/mol)

C6H12O6

-1271

209.2

-63.6

O2

0

0

0

H2O

-285.8

70

-21.1

CO2

-393.5

213.4

-64

Cellular Respiration — a spontaneous process whereby glucose is broken down into carbon dioxide and water releasing energy to be used as work o o Δ Gosys = 6ΔGoH 2O + 6ΔGCO2 − Δ GC6H 12O2 = 6 (− 2 1.1 kJ /mol) + 6 (− 64 kJ /mol) − (− 63.6 kJ /mol)

ΔGosys =− 447 kJ /mol o Δ Sosys = 6ΔS oH 2O + 6ΔSCO2 − Δ SoC6H 12O2 = 6(70 J/(mol * K)) + 6(213.4 J/(mol * K)) − (209.2 J/(mol * K))

ΔS osys = 1491.2 J/(mol * K ) Intuitively it is (relatively) clear how this process satisfies the second law: ΔS univ > 0 for a spontaneous process, since the entropy of the system will increase. This is (mostly) because 7 molecules are being reacted to produce 12 molecules, resulting in a larger number of microstates available to the system due to the reaction: Photosynthesis — energy harvested from incoming light ( E = hν ) is used to drive the nonspontaneous synthesis of glucose from water and carbon dioxide o o o Δ Gosys = ΔGC6H 12O2 − (6ΔGH 2O + 6 ΔGCO2 ) =− 63.6 kJ /mol − (6(− 21.1 kJ /mol) + 6(− 64 kJ /mol))

ΔGosys = 447 kJ /mol o o o = (209.2 J/(mol * K )) − (6(70 J/(mol * K)) + 6(213.4 J/(mol * K))) Δ Sosys = ΔS C6H 12O2 − (6ΔS H 2O + 6ΔS CO2 )

ΔS osys =− 1491.2 J/(mol * K) But how does photosynthesis satisfy the second law? The entropy of the system is decreasing here, (mostly) because 12 small molecules are combining to produce 7 larger molecules. It turns out that the light harvesting process is highly inefficient and generates a lot of heat, increasing the entropy of the surroundings allowing this process to be spontaneous....