Respuesta :
Answer : The value of [tex]K_{eq}[/tex] of this reaction is, [tex]5.51\times 10^{-6}[/tex]
At equilibrium, [L-malate] > [oxaloacetate]
Explanation :
The relation between the equilibrium constant and standard Gibbs free energy is:
[tex]\Delta G^o=-RT\times \ln K_{eq}[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = +30 kJ/mol = +30000 J/mol
R = gas constant = 8.314 J/K.mol
T = temperature = [tex]25^oC=273+25=298K[/tex]
[tex]K_{eq}[/tex] = equilibrium constant = ?
The given reaction is:
[tex]\text{L-malate}+NAD^+\rightleftharpoons \text{oxaloacetate}+NADH+H^+[/tex]
[tex]\Delta G^o=-RT\times \ln K_{eq}[/tex]
[tex]+30000J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}[/tex]
[tex]K_{eq}=5.51\times 10^{-6}[/tex]
Therefore, the value of [tex]K_{eq}[/tex] of this reaction is, [tex]5.51\times 10^{-6}[/tex]
As, the value of [tex]K_{eq}[/tex] < 1 that means the reaction mixture contains reactants.
At equilibrium, [L-malate] > [oxaloacetate]
