Library / Solution chemistry, equilibria and kinetics / SCH-42
Solution chemistry, equilibria and kinetics
Time of maximum intermediate concentration
Time at which the intermediate B peaks in the irreversible scheme A → B → C starting from pure A.
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- SCH-42
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- Solution chemistry, equilibria and kinetics
- Result
- tmax [s]
The intermediate B peaks when its formation rate k₁[A] equals its removal rate k₂[B]. Setting dB/dt = 0 in the consecutive-reaction solution gives tmax = ln(k₂/k₁)/(k₂ − k₁), and for equal rate constants tmax = 1/k₁. The faster the second step compared with the first, the earlier and lower the peak. It applies to irreversible first-order steps starting from pure A.
Time at which the intermediate B peaks in the irreversible scheme A → B → C starting from pure A. It is when the formation rate of B equals its removal rate. For equal rate constants the peak is at 1/k₁.
Calculator
- Peak concentration of B as a fraction of A₀0.25
- Half-life of A, ln 2/k₁6.931 s
- Half-life of B, ln 2/k₂3.466 s
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The intermediate peaks at 6.931 s, when it reaches 25 % of the initial A.
- Compare the peak time with the half-life of A, 6.931 s: it comes at 1 half-lives, so sample around then to capture the intermediate.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (s) | Local step | Change in result |
|---|---|---|---|---|---|---|
| k2 Rate constant B → C | 1e-4 to 100 1/s | 0.06915 | 69.15 | 69.08 | +10% | −5.21 % |
| k1 Rate constant A → B | 1e-4 to 100 1/s | 0.06227 | 38.02 | 37.96 | +10% | −4.17 % |
Sweep of Rate constant B → C
| Rate constant B → C (1/s) | Time of maximum (s) |
|---|---|
| 1e-4 | 69.15 |
| 5.623e-4 | 52.1 |
| 0.003162 | 35.67 |
| 0.01778 | 21 |
| 0.1 | 10 |
| 0.5623 | 3.735 |
| 3.162 | 1.128 |
| 17.78 | 0.293 |
| 100 | 0.06915 |
Sweep one variable
Variables and how each one contributes
Time of maximum
What the equation gives you. Everything else feeds it.
Rate constant A → B
The rate constant of A → B, in 1/s. A larger k₁ brings the peak earlier. Raising it by 10 % (to 0.11) moves the result by −4.17 %, from 6.931 to 6.643 s.
Rate constant B → C
The rate constant of B → C, in 1/s. A larger k₂ brings the peak earlier. Raising it by 10 % (to 0.22) moves the result by −5.21 %, from 6.931 to 6.57 s.
Worked example with the values above
- Ratio: k₂/k₁ = 0.2 / 0.1 = 2; ln = 0.69315.
- Difference: k₂ − k₁ = 0.1 1/s.
- tmax = 0.69315 / 0.1 = 6.9315 s.
Assumptions and limits
- Irreversible first-order steps for the consecutive scheme, and a single substrate with a steady-state intermediate for the enzyme.
- Constant temperature, pH and volume, with negligible product and substrate depletion for initial rates.
Common mistakes
- Using the formulas after substantial substrate depletion, where the rate no longer is the initial rate.
- Mixing up k₁ and k₂ in the consecutive formulas.
- Using the total inhibitor concentration in place of the free one.
Statistics and uncertainty
Fitted enzyme constants, Km and Vmax, carry uncertainties of 10 % to 30 % from the data scatter, and a wide range of substrate concentrations is needed to define both.
Where it is used
- Following intermediates in sequential reactions and decay chains.
- Describing enzyme saturation and inhibition.
Maths you need
- Integrated rate law: concentration as a function of time
- Steady state: the intermediate’s concentration stays constant
- Michaelis constant: Km, the substrate concentration at half of Vmax