Answer:
0.3983 mM is the concentration of 2-phosphoglycerate.
Explanation:
3-phosphoglycerate ⇄ 2-phosphoglycerate , ΔG° = 4.4 kJ/mol
Concentration of 3-phosphoglycerate at equilibrium = 2.35 mM
Concentration of 2-phosphoglycerate at equilibrium = x
Equilibrium constant of the reaction at 25°C = [tex]K_c=\frac{x}{2.35 mM}[/tex]
[tex]\Delta G^o=-RT\ln K_{c}[/tex]
[tex]4400 J/mol=-8.3145 J/mol K\times 298.15 K\times \ln [\frac{x}{2.35 mM}][/tex]
[tex]\ln [\frac{x}{2.35 mM}]=\frac{4400 J/mol}{-8.3145 J/mol K\times 298.15 K}[/tex]
[tex]\ln [\frac{x}{2.35 mM}]=-1.7750[/tex]
[tex]\frac{x}{2.35 mM}=0.1695[/tex]
x = 0.3983 mM
0.3983 mM is the concentration of 2-phosphoglycerate.