All question types · Kinetics
Michaelis–Menten rate
UKChO Round 2 · IChO · unreviewed
Steady state on ES, with [E] = [E]₀ − [ES], gives v = k₂[E]₀[S]/(K_M + [S]) = V_max[S]/(K_M + [S]).
Worked example
Find the initial rate of an enzyme-catalysed reaction at [S] = 9.7 × 10⁻⁴ mol dm⁻³, given K_M = 0.0017 mol dm⁻³ and V_max = 1.2 × 10⁻⁴ mol dm⁻³ s⁻¹.
1. Multiply V_max × [S] → V_max[S]
= 0.1164 mol² m⁻⁶ s⁻¹
2. Add K_M + [S]
= 0.002670 mol dm⁻³
3. Divide V_max[S] ÷ (K_M + [S]) → v
= 4.360 × 10⁻⁵ mol dm⁻³ s⁻¹
Answer: v = 4.36 × 10⁻⁵ mol dm⁻³ s⁻¹
Method
Steady state on ES, with [E] = [E]₀ − [ES], gives v = k₂[E]₀[S]/(K_M + [S]) = V_max[S]/(K_M + [S]). At [S] = K_M the rate is exactly half of V_max.
Common mistakes
- That ignores saturation: v = V_max[S]/(K_M + [S]), not V_max[S]/K_M.
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