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Question 4 (1 point) Given the following reaction, what is the AGC for the spontaneous reaction in kJ/mol? Faradays constant

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Answer #1

NADH ------------> NAD^+ + H^+ + 2e^-            E0 = 0.320v

NO3^- + 2H^+ +2e^- ------> NO2^- + H2O        E0 = 0.421v

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NADH + NO3^- + H^+ ------------> NAD^+ + NO2^- + H2O   E0cell = 0.741v

n = 2

\DeltaG0    = -nE0cell*F

              = -2*0.741*96500

             = -143013J/mole

            = -143KJ/mole >>>answer

8 mol of Acetyl-CoA

24 mol of NADH

8 mol of FADH2

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