SME
Biochemistry · Problem type
Coupled reactions and K′
Use this for “Fructose 6-phosphate + Pᵢ → fructose 1,6-bisphosphate has ΔG°′ = +16.3 kJ/mol. Coupled to ATP hydrolysis (−30.5 kJ/mol), find the total ΔG°′ and K′ at 25 °C.”
Diagram
ΔG°′₁: A + Pᵢ + ATP → B + ATP, ++13.8 kJ/mol
ΔG°′₂: B + ATP → B + ADP + Pᵢ, −30.5 kJ/mol
ΔG°′ = ΔG°′₁ + ΔG°′₂
ΔG°′ = +13.8 + (−30.5) = −16.7 kJ/mol
Down the ladder: ΔG°′ < 0, so it runs forward on its own (exergonic).
This is an example. Type your own values to start a new problem.
ΔG°′ (kJ/mol)
=
ΔG°′₁ (kJ/mol)
+
(
ΔG°′₂ (kJ/mol)
)
T
Temperature
example · K
K′
Equilibrium constant
calculated
Formulas
ΔG°′ = ΔG°′₁ + ΔG°′₂
−16.7 = 13.8 + (−30.5)
K′
=
e
−ΔG°′ × 1,000 ÷ (8.314 ×
T
)
843
=
e
−(−16.7) × 1,000 ÷ (8.314 × 298.15)
Assumptions
•
Two reactions that share an intermediate (here Pᵢ) add: their equations and their ΔG°′ values sum.
•
K′ = e−ΔG°′ ÷ RT with ΔG°′ in J/mol and R = 8.314 J/(mol·K); the prime means pH 7.
•
A + Pᵢ → B is the uphill reaction; ATP → ADP + Pᵢ, −30.5 kJ/mol, is the usual downhill partner.
Step-by-step
WE KNOW
ΔG°′₁ = +13.8 kJ/mol, ΔG°′₂ = −30.5 kJ/mol, T = 298.15 K
FIND
total standard free energy (ΔG°′), equilibrium constant (K′)
Step 1 · Find total standard free energy (ΔG°′)
Use ΔG°′ = ΔG°′₁ + ΔG°′₂
Adding the two equations adds their free energies; Pᵢ made in one is used in the other.
ΔG°′ = 13.8 + (−30.5)
ΔG°′ = −16.7 kJ/mol
Step 2 · Find equilibrium constant (K′)
Use K′ = e−ΔG°′ × 1,000 ÷ (8.314 × T)
Put ΔG°′ into J/mol, divide −ΔG°′ by RT, and raise e to it.
K′
=
e
−(−16.7) × 1,000 ÷ (8.314 × 298.15)
K′
=
e
−(−16.7) × 1,000 ÷ 2,478.8191
K′
=
e
16.7 × 1,000 ÷ 2,478.8191
K′
=
e
16,700 ÷ 2,478.8191
K′
=
e
6.737
K′ = 843 → K′ > 1: the coupled reaction runs forward; ATP pays for the uphill step.
Check
Put the numbers back in. Both sides should match.
−16.7 = 13.8 + (−30.5) ✓
843
=
e
−(−16.7) × 1,000 ÷ (8.314 × 298.15)
✓
Topic check
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