Question

19. (3 points) The standard reducing potential for'NADH is -0.32 V, and the standard réducin potential for oxygen is +0.82 V (Faradays constant 96500 JV-1mol-1, R 8.314 JmolK1 a. Using these values calculate the free energy released upon the transfer of electrons from the electron donor to the final electron acceptor. b. We discussed in class that one mole of NADH can produce 2.5 moles of ATP, and assuming that ?Go for the synthesis of ATP is + 30.5 kJ/mol, and using the free energ obtained in part a, calculate the efficiency of ATP synthesis by ATP synthas

Answer 1

Ans. #a.

Step 1: The net reaction can be written as the sum of following two half-reactions: Steps for Writing the coupled REDOX Reaction: L Write the oxidation-half reaction. Since the standard redox potential (E) Oxidation half: NADH H NAD" + 2H+ + 2e Eor = 0.320 V → Reduction half 1/2 022H 2e-> Net Reaction: NADH +1/202-H. > 0.820 V are given for reduction (gain of electrons) in standard redox-potential table, ΔΕ0-1.140V the oxidation-half reaction is reverse of standard reduction # when a reaction is reversed, the sign of its redox potential is reversed, too. The superscript E represents "reverse of standard redox potential NAD +HO Step 2: The standard Gibb's free energy change for the reaction is given by- -Equation 1 Write the reduction-half reaction as such. No need of changing the sign of Eo for it. mL Add the E0 values of two half-reactions WITHOUT changing their sign (because it has already been done in step #1). Where, n number of electrons transferred s constant= 96.5 kJ mol-1 V-1 Putting the values in equation 1- 2 x 96.500 kJ mol v x 1.140 v Hence, ΔG' - 220.02 kJ mol

#b. Total energy conserved in form of ATP =

dG^0’ of ATP synthesis x Moles of ATP formed

= (30.5 kJ / mol) x 2.5 mol

= 76.5 kJ

Now,

% efficiency of ATP synthesis =

(Energy conserved in ATP / Energy from Oxidation of 1 mol NADH) x100

            = (76.5 kJ / 220.02 kJ) x 100

            = 37.87 %