Problem #2023-0002: Assume the following intracellular conditions: [ATP] = 15 mM, [ADP] = 0.5 mM, [Pi] = 7.9 mM, pH = 7.4, and temperature = 37°C. Given the ΔG°′ for ATP hydrolysis is −30.5 kJ/mol, what is the overall ΔG (kJ/mol) for ATP hydrolysis under these cellular conditions?
Step 1: Background Info Needed
Type of Problem:
ΔG calculation (chemical thermodynamics)
Reaction:
ATP + H2O → ADP + Pi
Equation(s):
ΔG = ΔG°′ + RT ln Q*
Q* = ([ADP][Pi]/[ATP]) * ([H+] @ pH 7.4/[H+] @ pH 7.0)
Given:
ΔG°′ = −30.5 kJ/mol; R = 8.314 J/mol·K; T = 37°C; [ADP] = 0.5 mM; [Pi] = 7.9 mM; [ATP] = 15 mM; pH = 7.4 (remember, since [H2O] is extraordinarily high (55 M), we can assume it remains unchanged by the reaction; therefore, [H2O] is not included in the expression)
Step 2: Unit Conversion
Change ΔG°′ from kJ/mol to J/mol (−30.5 kJ/mol * 1000 J/kJ = −30,500 J/mol)
Change T from Celsius to Kelvin (37°C + 273.15 = 310.15 K)
Change [ADP] from mM to M (0.5 mM * 1 M/1000 mM = 0.0005 M)
Change [Pi] from mM to M (7.9 mM * 1 M/1000 mM = 0.0079 M)
Change [ATP] from mM to M (15 mM * 1 M/1000 mM = 0.015 M)
[H+] @ pH 7.4 = 10-7.4 = 3.981 x 10-8 M
[H+] @ pH 7.0 = 10-7.0 = 1 x 10-7 M
Step 3: Substitution & Solve
First, solve Q* = ([ADP][Pi]/[ATP]) * ([H+] @ pH 7.4/[H+] @ pH 7.0)
Q* = (0.0005*0.0079/0.015) * (3.981 x 10-8/1 x 10-7) = 0.000104835
(note: when solving Q*, as long as the unit for concentration is M for each reactant and product, then the activity coefficient of each reactant and product (1 M-1; not shown) will render Q* unitless)
Second, solve ΔG = ΔG°′ + RT ln Q*
ΔG = −30,500 J/mol + (8.314 J/mol·K * 310.15 K) ln (0.000104835)
ΔG = −30,500 J/mol + (8.314 J/mol·K * 310.15 K) * (-9.16312)
ΔG = −30,500 J/mol + (−23,628 J/mol)
ΔG = −54,128 J/mol = −54.1 kJ/mol