Calculate the equilibrium constant at 298.15 K for the reaction.

Question

Oxaloacetate2−(aq)+NADH(aq)+H+(aq)malate2−(aq)+NAD+(aq)

The standard potential has been calculated at .164 V

Calculate the equilibrium constant at 298.15 K for the reaction.

Answer 1

OA^2-(aq) + NADH(aq) + H⁺(aq) ==> malate^2-(aq) + NAD⁺(aq)

Given standard reduction potential for the reaction = 0.164 V

Asked for Keq

Use the equation -nFE = -RT ln K

n = moles of electrons transferred = 2

F = Faraday constant = 96485 C/mole e-

E = cell potential = 0.164 V

R = constant = 8.314 J/Kmol

T = temperature in K = 298K

K = equilibrium constant = ?

-(2)(96485)(0.164) = -(8.314)(298) ln K

-31647 = -2478 ln K

ln K = 12.77

K = 351,512

Keq = 3.52x10⁵