glucose 6‑phosphate+H2O⟶glucose+Pi glucose 6‑phosphate+H2O⟶glucose+Pi K′eq1=270 K′eq1=270 ATP+glucose⟶ADP+glucose 6‑phosphate ATP+glucose⟶ADP+glucose 6‑phosphate K′eq2=890 K′eq2=890 Using this information for equilibrium constants determined at 25∘C, 25∘C, calculate the standard free energy of hydrolysis of ATP. standard free energy:

-30.7 kj/mol

Explanation:

The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula: ∆Go ' = - RTln K'eq

where,

R = - 8.315 J / mo

T = 298 K

For reaction,

1. K′eq1=270,

∆Go ' = - RTln K'eq

= - 8.315 x 298 x ln 270

= - 8.315 x 298 x 5.59

= - 13,851.293 J / mo

= - 13.85 kj/mol

2. K′eq2=890

∆Go ' = - RTln K'eq

= - 8.315 x 298 x ln 890

= - 8.315 x 298 x 6.79

= - 16.82 kj/mol

therefore, total standard free energy

= - 13.85 + (-16.82)

= - 30.7 kj/mol

Thus, - 30.7 kj/mol is the correct answer.