Question

In the citric acid cycle, malate is dehydrogenated to oxaloacetate in a highly endergonic reaction with a ΔG’o of +30 kJ mol-1: L‐malate + NAD+ ⇌ oxaloacetate + NADH + H+ Calculate the equilibrium constant K’eq of this reaction. What is the implication of this result?

Answer 1

Answer :  The value of $K_{eq}$ of this reaction is, $5.51\times 10^{-6}$

At equilibrium, [L-malate] > [oxaloacetate]

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy  = +30 kJ/mol = +30000 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$  = equilibrium constant  = ?

The given reaction is:

$\text{L-malate}+NAD^+\rightleftharpoons \text{oxaloacetate}+NADH+H^+$

$\Delta G^o=-RT\times \ln K_{eq}$

$+30000J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=5.51\times 10^{-6}$

Therefore, the value of $K_{eq}$ of this reaction is, $5.51\times 10^{-6}$

As, the value of $K_{eq}$ < 1 that means the reaction mixture contains reactants.

At equilibrium, [L-malate] > [oxaloacetate]