Consider the following metabolic reaction:
3-phosphoglycerate → 2-phosphoglycerate δg°’ = +4.40 kj/mol
what is the δg for this reaction when the concentration of 2-phosphoglycerate is 0.290 mm and the concentration of 3-phosphoglycerate is 2.90 mm at 37°c?

ΔG = -1.53 kJ/mol

Explanation:

The given reaction is:

3-Phosphoglycerate → 2-Phosphoglycerate

The standard Gibbs free energy, ΔG°=+4.40 kJ

[2-Phosphoglycerate] = 0.290 mM

[3-Phosphoglycerate] = 2.90 mM

Temperature T = 37 C = 310 K

The standard Gibbs free energy, ΔG° is related to the free energy change ΔG at a given temperature by the following equation:

In this reaction:

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :