Q16.4-62E
Question
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 → G6P ΔG°298 = ?
Step-by-Step Solution
VerifiedThe change of free energy is \(\Delta G = 13kJ\)
ATP is needed to make this reaction spontaneous.
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.