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snow_lady [41]
3 years ago
8

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
Chemistry
1 answer:
pogonyaev3 years ago
4 0

Answer:

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

Explanation:

<u>Chemical reaction involved</u>: 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: Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2

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

\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}

\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)

\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)

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

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