Question

Standard Reduction Potential of Some Biologically Important Half-Reactions, at pH 7.0 and 25°C (298 K) Half-reaction E' (V) 2H+ + 2e — H, (at standard conditions, pH 0) 0.000 Crotonyl-CoA + 2H+ + 2e butyryl-CoA -0.015 Oxaloacetate?- + 2H+ + 2e malate? -0.166 Pyruvate + 2H+ + 2e lactate -0.185 Acetaldehyde + 2H+ + 2e-ethanol -0.197 FAD + 2H+ + 2e FADH -0.219 Glutathione + 2H+ + 2e → 2 reduced glutathione -0.23 S + 2H+ + 2e ►H,S -0.243 Lipoic acid + 2H+ + 2e dihydrolipoic acid -0.29 NAD+ + H+ + 2e NADH -0.320 NADP+ + H+ + 2e NADPH -0.324 Acetoacetate + 2H+ + 2e → B-hydroxybutyrate -0.346 a-ketoglutarate + CO2 + 2H+ + 2e" - isocitrate -0.38 2H+ + 2e H, (at pH 7) -0.414 Ferredoxin (Fe3+) + ferredoxin (Fe2+) -0.432 Source: Data mostly from Loach, PA. (1976) In Handbook of Biochemistry and Molecular Biology, 3rd edn (Fasman, G.D. ed.), Physical and Chemical Data, Vol. I, pp. 122-130, CRC Press, Boca Raton, FL This is the value for free FAD; FAD bound to a specific flavoprotein (for example succinate dehydrogenase) has a dit- ferent that depends on its protein environments.

For the reaction: L-Malate + NAD⁺ ← → Oxaloacetate + NADH

ΔE' = -0.154

n=2

T=298 K

using the equation

ln (Keq) = (nFΔE'৹)/(RT)

Find Keq?

I know this sounds just plug and chug but am confused about the signs of the numbers when plugged in. Please only answer if you're of what you're doing!

Answer 1

$\Delta G = -nF\Delta E$ ................ (a)

Also, $\Delta G = -RTlnKeq$ ..............(b)

Equating (a) and (b),

$-RTlnKeq = -nF\Delta E$

or

$lnKeq = \frac{nF\Delta E}{RT}$

or

$lnKeq = \frac{2*96500*(-0.154)}{8.314*298}$

or

or

Keq = $e^{-11.966}$

or

Keq = 6.35 x 10^-6

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