1. Using Table 10-2 (Standard Reduction Potential Values) and the reaction below

Question

1. Using Table 10-2 (Standard Reduction Potential Values) and the reaction below calculate the standard free energy change (AG") associated with the transfer of 2 electrons between NADH and O2-(4 pts) Reaction: %02 + NADH H,O + NAD, 2. Write the equation for the reaction catalyzed by the enzyme lactate dehydrogenase resulting in lactate production. Then, calculate its equilibrium constant (K'e) at 25'C, given the standard reduction potentials (E) in Table 10.2. (Egn 2pts; Cale 4 pts; Show all work).

Explanation / Answer

a) Here the two half reactions are-

1/2 O2 + 2H+ + 2e => H2O......E'o = +0.816V (acceptor)

NAD+ + 2H+ + 2e => NADH + H+ ........E'o = -0.32V (donor)

Therefore, deltaE'o = E'oacceptor - E'odonor = 0.816 - (-0.32) = 1.14V

Now free energy change deltaG'o =- nFdeltaE'o

Here n is the no. of electrons involved (in our case it is 2); F is Faraday's constant which is 23 kcal/mol*V.

Therefore, deltaG'o = -2 * 23 * 1.14 = -52 kcal/mol.

(b) The reaction which is mentioned in question is conversion of pyruvate to lactate. The reaction is-

pyruvate + NADH + H+ => lactate + NAD+

Now the half reactions are-

pyruvate + 2H+ + 2e => lactate........E'o = -0.19V (acceptor)

NAD+ + 2H+ + 2e => NADH + H+ ........E'o = -0.32V (donor).

Now we calculate deltaE'o as above

So, deltaE'o = E'oacceptor - E'odonor = -0.19 - (-0.32) = 0.13V

Now, deltaG'o =-nFdeltaE'o = -2 * 23 * 0.13 = -5.98 kcal/mol. or -5980cal/mol

Again, we know that deltaG'o =-RTlnKeq or deltaG'o =- 2.303RT log Keq.

Here R is gas contant which is equal to 1.98 cal/K-mol, and T is 25C or 298K.

After putting all the values we get-

-5980cal/mol = -2.303 *1.98cal/K-mol * 298 K * log Keq => log Keq = 4.4 => Keq = 2.5 x 10^4

Thanks!

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