Appendix A: Calculating the Free Energy of ATP Hydrolysis
Here are the details for how to calculate the average free energy available from ATP hydrolysis under typical in
vitro steady-state ATP turnover conditions. Assumptions: [Mg2+] = 0.005 M, [ATP]o = 0.005 M, 0–0.1 fraction
of hydrolysis products ADP and Pi, T = 298.15 K (25 C), pH 7.0, G = –35,590 J mol-1 (Shikama 1971), 0.2
M ionic strength. It is especially noted that the G (the standard free energy change) varies as a function of
Mg2+ concentration, pH, and ionic strength (Shikama 1971). For energy unit conversion, 1 kcal mol-1 equals
4,187 J mol-1. R is the gas constant at 8.314 J K-1 mol-1.
Eq. A1
ATP + H2 O ⇌ ADP + Pi
Eq. A2
∆G𝑟𝑥𝑛 = ∆G° + 𝑅𝑇 ln 𝑄
Eq. A3
[ADP][Pi ]
𝑄 =
[ATP][H2 O]
Since the concentration of H2O (e.g. ~55.2 M in a buffer consisting of 0.02 M HEPES, pH 7.0 and 0.2 M ionic
strength) is vastly in excess and is significantly greater than the initial ATP concentration, we can assume its
concentration remains constant for Eq. A3. Grxn can be expressed as a function of the fraction of ATP
hydrolyzed (f):
Eq. A4
[𝐴𝑇𝑃] = [𝐴𝑇𝑃]0 − √[𝐴𝐷𝑃][𝑃𝑖 ]
Eq. A5
√[ADP][Pi ]
𝑓 =
[ATP]o
Eq. A6
2
𝑓 [ATP]o
𝑄(𝑓) =
(1 − 𝑓)[H2 O]
Eq. A7
𝑓 2 [ATP]o
∆G𝑟𝑥𝑛 = ∆G° + 𝑅𝑇 ln
(1 − 𝑓)[H2 O]
In order to determine the Grxn during steady-state ATP turnover, we integrate Eq. A7 with the fraction of ATP
hydrolyzed ( = 0.1):
Eq. A8
𝛽
2
1
𝑓
[ATP]
o
𝑎𝑣𝑔
∆G𝑟𝑥𝑛 = lim ∫ ( ) (∆G° + 𝑅𝑇 ln
) 𝑑𝑓
𝛼→0 𝛼
(1 − 𝑓)[H2 O]
𝛽
Using the fundamental theorem of calculus and L’Hôpital’s rule, we simplify Eq. A8 to:
Eq. A9
2 [ATP]
𝛽
ln(1 − 𝛽)
o
𝑎𝑣𝑔
∆G𝑟𝑥𝑛 = ∆G° + 𝑅𝑇 [ln (
)+
− 1]
(1 − 𝛽)[H2 O]
𝛽
𝑎𝑣𝑔
Substituting values for the constants G, R, T, and  in Eq. A9, ∆G𝑟𝑥𝑛 = −74,911 J mol−1 . Converting the
units of average free energy into pN nm per molecule:
Eq. A10
−1
12
−1
9
−1
−74,911
N
m
mol
∗
10
pN
N
∗
10
nm
m
𝑎𝑣𝑔
∆G𝑟𝑥𝑛 =
= −124.4 pN nm
6.022 x 1023 molecules mol−1