2-phosphoglycerate(2pg) is converted to phosphoenolpyruvate (pep) by the enzyme enolase. the standard free energy change(deltago’) for this reaction is +1.7 kj/mol. if the cellular concentrations are 2pg = 0.5 mm and pep = 0.1 mm, what is the free energy change at 37 oc for the reaction 2pg ↔ pep? (a) 5.8 kj/mol(b) -5.8 kj/mol(c) +2.4 kj/mol(d) -2.4 kj/mol(e) -4146.4 kj/mol(f) +4146.4 kj/mol

The correct option is: (D) -2.4 kJ/mol

Explanation:

Chemical reaction involved: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K    (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol)     (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient:

To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:

Therefore, the Gibb's free energy change at 37° C (310.15 K): ΔG = (-2.45 kJ/mol)

the answer is either b or c