How do you calculate the free energy change for the acetyl phospahate hydrolysis in a solution of 2 mM acetate, 2 mM phosphate and 3 nM acetyl phosphate. The standard state free energy of hydolysis of acetyl phosphate is /\G = -42.3 kJ/mol.

To calculate the free energy change (\(\Delta G\)) for the hydrolysis of acetyl phosphate in the given solution, you can use the equation:

\(\Delta G = \Delta G^\circ + RT \cdot \ln(Q)\)

Where:
- \(\Delta G\) is the free energy change for the reaction
- \(\Delta G^\circ\) is the standard state free energy change
- \(R\) is the gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K))
- \(T\) is the temperature in Kelvin
- \(Q\) is the reaction quotient

First, convert the standard state free energy change to the same units as the concentration. Since the concentrations are given in millimoles (mM) and nanomoles (nM), convert \(\Delta G^\circ\) to kJ/mol:

\(\Delta G^\circ = -42.3 \, \text{kJ/mol}\)

Next, substitute the given values into the equation. The temperature \(T\) needs to be in Kelvin, so you'll need to convert it if it's given in a different unit:

\(\Delta G = -42.3 \, \text{kJ/mol} + (0.008314 \, \text{kJ/(mol·K)}) \cdot \ln(Q)\)

Now, calculate the reaction quotient \(Q\) by dividing the product of the concentrations of the products by the product of the concentrations of the reactants:

\(Q = \frac{{[\text{acetate}] \cdot [\text{phosphate}]}}{{[\text{acetyl phosphate}]}}\)

Substitute the given concentrations into the equation:

\(Q = \frac{{2 \, \text{mM} \cdot 2 \, \text{mM}}}{{3 \, \text{nM}}}\)

To perform this calculation, convert the given concentrations to the same unit (either millimoles or nanomoles):

\(Q = \frac{{0.002 \, \text{mol/L} \cdot 0.002 \, \text{mol/L}}}{{3 \times 10^{-9} \, \text{mol/L}}}\)

Now, substitute the calculated value of \(Q\) into the earlier equation to find the \(\Delta G\) value. Make sure the temperature is also in Kelvin:

\(\Delta G = -42.3 \, \text{kJ/mol} + (0.008314 \, \text{kJ/(mol·K)}) \cdot \ln(0.001333 \, \text{mol/L})\)

Simplify the equation and evaluate the natural logarithm:

\(\Delta G = -42.3 \, \text{kJ/mol} + 0.008314 \, \text{kJ/(mol·K)} \cdot \ln(0.001333)\)

Finally, calculate \(\Delta G\) using a calculator:

\(\Delta G = -42.3 \, \text{kJ/mol} - 14.581\, \text{kJ/mol}\)

\(\Delta G = -56.881\, \text{kJ/mol}\)

So, the free energy change (\(\Delta G\)) for the acetyl phosphate hydrolysis in the given solution is approximately -56.881 kJ/mol.