If the E°' for the NAD/NADH half cell is -0.32V and for the fumarate/succinate half cell is 0.031V, calculate the ΔG°' for the reaction NADH + H+ + fumarate → NAD+ + succinate given F = 96.5 kJ/Vmol.
-12.55 kJ/mol | ||
-67.74 kJ/mol | ||
67.74 kJ/mol | ||
-55.78 kJ/mol | ||
55.78 kJ/mol |
E=-0.32+0.031 V
=0.289V
∆Go
’ = –nF ∆Eo
∆Go
’ = –2(96.5 kJ V–1 mol –1)(0.289V) = – 55.77kJ /mol
Therefore answer is -55.78kj/mol
If the E°' for the NAD/NADH half cell is -0.32V and for the fumarate/succinate half cell...
What would ΔG∘′ be for an enzyme that oxidizes succinate with NAD+ instead of FAD? Oxidant Reductant n E∘′(V) NAD++H++2e− ⇌ NADH 2 −0.32 Fumarate+2H++2e− ⇌ Succinate 2 0.03 Use the following equation and F = 96.485 kJ/mol. ΔG°’ = ‒nFΔE°’ Express your answer to two significant figures and include the appropriate units ΔG′= value w/ units please
E Volts) 6 co2/glucose 0.43] -0.5 I-0.42] -0.4 NAD /NADH 0.32] -03 -0.2 -0.1 Fumarate/succinate 0.03] 0.0 [0.10] +0.1 2 H /H2 I- 0.27] FADIFADH2 0.18] S°/H2S CoQ/CoQH2 +0.2 Cyt C (Fe )Cyt C (Fe+) [0.25] +0.3- +0.4 +0.5 +0.6 No,INO2 [0.421 Fe3+/Fe2+ 4 0,H20 (0.82108 10.77) +07
1. Consider the following 2 half reactions Fumarate +2H + 2e- --> succinate E= 0.031 Pyruvate +2H + 2e- --> lactate E= -0.185 a. In each reaction identify the oxidized form and the reduced form. b. Write the overall reaction c. Identify what is getting oxidized and what is getting reduced. d. Calculate deltaE e. Calculate deltaE under the following conditions; fumarate= 0.18mM, succinate= 0.32 mM, pyruvate= 0.25mM, lactate= 0.28nM at 37 degrees.
E'° of the NAD+/NADH half reaction is –0.32 V. The E'° of the Acetic acid/acetaldehyde half reaction is –0.58 V. What is the E'° of the spontaneous reaction?
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