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The value of ?G�\' for the conversion of 3-phosphoglycerate to 2-phosphoglycerate (2PG) is 4.4 kJ/mol. If the concentration of 3-phosphoglycerate at equilibrium is 2.35 mM, what is the concentration of 2-phosphoglycerate? Assume a temperature of 25.0 �C. The constant R = 8.3145 J/(mol�K)

User Dumamilk
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Answer:

0.3983 mM is the concentration of 2-phosphoglycerate.

Step-by-step 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 =
K_c=(x)/(2.35 mM)


\Delta G^o=-RT\ln K_(c)


4400 J/mol=-8.3145 J/mol K* 298.15 K* \ln [(x)/(2.35 mM)]


\ln [(x)/(2.35 mM)]=(4400 J/mol)/(-8.3145 J/mol K* 298.15 K)


\ln [(x)/(2.35 mM)]=-1.7750


(x)/(2.35 mM)=0.1695

x = 0.3983 mM

0.3983 mM is the concentration of 2-phosphoglycerate.

User Sam Hosseini
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