For the coupled reaction given below, the equilibrium constants 

(Keq)(K_{eq})(Keq​) for equation $[1]$ and equation $[2]$ are 270 and 890, respectively.

Glucose 6-phosphate+H2O→Glucose+Pi[1]\text{Glucose 6-phosphate} + H_2O \rightarrow \text{Glucose} + P_i \quad [1]Glucose 6-phosphate+H2​O→Glucose+Pi​[1]ATP+Glucose→ADP+Glucose 6-phosphate[2]\text{ATP} + \text{Glucose} \rightarrow \text{ADP} + \text{Glucose 6-phosphate} \quad [2]

ATP+Glucose→ADP+Glucose 6-phosphate[2]

The standard free energy of hydrolysis of ATP at 25°C is:

Correct option is C

The given question requires calculating the standard free energy of hydrolysis of ATP at 25°C using the equilibrium constants provided.

Given Data:

Let's calculate the standard free energy of hydrolysis of ATP.

The calculated standard free energy of hydrolysis of ATP at 25°C is approximately -30.7 kJ/mol.

Correct Answer:

Option 3: (-30 to -32) kJ/mol

Solution:

Using the Gibbs free energy equation:

​ΔG∘=−RTlnK_eq​

Substituting the values:

​ΔG∘=−(8.314×298)×ln⁡(270×890)

ΔG∘≈−30.7 kJ/mol​

Thus, the correct answer is option 3.​