In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:

Glu + ATP → G6P + ADP

ΔG^°₂₉₈ = -17kJ

In this process, ATP becomes ADP summarized by the following equation:

ATP → ADP ΔG^°₂₉₈ = -30kJ

Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:

Glu → G6PΔG^°₂₉₈ = ?

Gibbs free energy, also known as Gibbs function, Gibbs energy, or free enthalpy, is a term used to measure the topmost quantum of work done in a thermodynamic system when temperature and pressure remain constant.

Use the formula for the free energy change

\(\Delta G = {G_f}({\rm{ products }}) - {G_f}({\rm{ reactants }})\)

For the first reaction,\(\Delta G\)is:

\(\Delta {G_1} = {G_f}(G6P) + {G_f}(ADP) - {G_f}(Glu) - {G_f}(ATP)\)

We can rearrange it like this:

\(\Delta {G_1} = {G_f}(ADP) - {G_f}(ATP) + {G_f}(G6P) - {G_f}(Glu)\)

For the second reaction:

\(\Delta {G_2} = {G_f}(ADP) - {G_f}(ATP)\)

And for the third:

\(\Delta {G_3} = {G_f}(G6P) - {G_f}(Glu)\)

In the problem, we know the values for\(\Delta {G_1}\)and\(\Delta {G_2}\)and we need to calculate the\(\Delta {G_3}\).

If we look closely at\(\Delta {G_1}\), we can see that inside the expression we have the expressions for both\(\Delta {G_2}\)and\(\Delta {G_3}\). Therefore, we can write:

\(\begin{array}{l}\Delta {G_1} = \Delta {G_2} + \Delta {G_3}\\\Delta {G_3} = \Delta {G_1} - \Delta {G_2}\\\Delta G = ( - 17 - ( - 30))kJ\\\Delta G = 13kJ\end{array}\)

The standard free energy change for the reaction in question is\(13\;{\rm{kJ}}\).

Since it is positive, the reaction is not spontaneous and that is why ATP is needed.

Its releasing of a phosphate group introduces a new spontaneous reaction and thus "pushes" the overall free energy change below zero.