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* 10^(-6)`

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

Explanation :

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

`\Delta G^o=-RT* \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* \ln K_(eq)`

`+30000J/mol=-(8.314J/K.mol)* (298K)* \ln K_(eq)`

`K_(eq)=5.51* 10^(-6)`

Therefore, the value of
`K_(eq)` of this reaction is,
`5.51* 10^(-6)`

As, the value of
`K_(eq)` < 1 that means the reaction mixture contains reactants.

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