Problem #2023-0001: Assume the following intracellular conditions: [ATP] = 10 mM, [ADP] = 0.8 mM, [Pi] = 3.3 mM, pH = 7.0, 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]
Given:
ΔG°′ = −30.5 kJ/mol; R = 8.314 J/mol·K; T = 37°C; [ADP] = 0.8 mM; [Pi] = 3.3 mM; [ATP] = 10 mM (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 (0.8 mM * 1 M/1000 mM = 0.0008 M)
Change [Pi] from mM to M (3.3 mM * 1 M/1000 mM = 0.0033 M)
Change [ATP] from mM to M (10 mM * 1 M/1000 mM = 0.01 M)
Step 3: Substitution & Solve
First, solve Q = [ADP][Pi]/[ATP]
Q = 0.0008 * 0.0033 / 0.01 = 0.000264
(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 (0.000264)
ΔG = −30,500 J/mol + (8.314 J/mol·K * 310.15 K) * (-8.23956)
ΔG = −30,500 J/mol + (−21,246 J/mol)
ΔG = −51,746 J/mol = −51.7 kJ/mol