Metabolic Pathway Equilibrium Simulator

Set each enzyme's ΔG°′ and ΔG^‡ and the starting concentrations, then watch the intermediates move toward equilibrium in a closed flask at 25 °C.

Reaction mechanism

Steps

Temperature °C

Example

Energies in kJ/mol, intermediate concentrations in mM, inhibitor [I] and K_i in µM, rate constants in s⁻¹.

Concentrations over time

Hold axes

t =

Free-energy profile

Values

How the simulator calculates

Every calculation uses the reaction temperature T set at the top: 25 °C (298.15 K). At this temperature RT = 2.479 kJ/mol and k_BT/h = 6.21 × 10¹² s⁻¹. The default, 25 °C, is the standard temperature for ΔG°′.

Each step S ⇌ P is a reversible first-order reaction with net rate v = k_f[S] − k_r[P].

The equilibrium constant comes from the standard free energy change: K′ = exp(−ΔG°′/RT).

Rate constants come from each barrier through the Eyring equation: k_f = (k_BT/h) exp(−ΔG^‡/RT). Every additional 5.71 kJ/mol of barrier (RT ln 10) makes a step 10 times slower. The reverse rate constant is k_r = k_f/K′, the same as a reverse barrier of ΔG^‡ − ΔG°′, so ΔG^‡ can never be set below ΔG°′.

ΔG°′ and ΔG^‡ are treated as independent of temperature. At 25 °C this is exact; at other temperatures it is the usual approximation when ΔH°′ is not known. Changing T changes RT and k_BT/h, and with them K′, the rate constants, and every ΔG.

At equilibrium every step satisfies [P]/[S] = K′ while the total amount of material stays constant, so [A]_eq = total ÷ (1 + K′₁ + K′₁K′₂ + …), and each later intermediate follows from the K′ values. Activation energies change how fast the flask gets there, never where it ends up.

The actual free energy change of a step is ΔG = ΔG°′ + RT ln Q, where Q = [P]/[S]. It is zero for every step at equilibrium. In the actual profile, each intermediate sits at G°′ + RT ln[concentration], so the drop between neighbors is exactly that step's ΔG.

A reversible inhibitor binds the free enzyme with dissociation constant K_i. With the enzyme far from saturation, only the free fraction is active, so both rate constants are divided by 1 + [I]/K_i. That is the same as raising both barriers by RT ln(1 + [I]/K_i), and K′ does not change. An irreversible inhibitor sets both rate constants of its step to zero, so the intermediates on each side settle into separate equilibria.

The curves are exact solutions of the rate equations, computed from the eigenvalues and eigenvectors of the rate matrix, not step-by-step approximations.