Library / Solution chemistry, equilibria and kinetics / SCH-43
Solution chemistry, equilibria and kinetics
Time for unequal-concentration bimolecular conversion
Time for x mol/L to react in A + B → products with rate k[A][B], from the starting concentrations A₀ and B₀.
- Ref
- SCH-43
- Rev
- v1
- Field
- Solution chemistry, equilibria and kinetics
- Result
- t [s]
For A + B → products with rate k[A][B], integrating the second-order rate law with unequal starting concentrations gives t = ln[A₀(B₀ − x)/(B₀(A₀ − x))]/[k(B₀ − A₀)], where x is the amount reacted per litre. For equal starting concentrations the limit is t = x/[k A₀(A₀ − x)]. The calculator uses a log1p form to stay accurate when the two concentrations are nearly equal. Require 0 ≤ x < min(A₀, B₀).
Time for x mol/L to react in A + B → products with rate k[A][B], from the starting concentrations A₀ and B₀. For equal starting concentrations the limit x/[k A₀(A₀ − x)] is used. Needs 0 ≤ x < min(A₀, B₀).
Calculator
- Initial rate, k A₀ B₀0.02 mol/(L·s)
- Rate at the end, k(A₀ − x)(B₀ − x)0.0075 mol/(L·s)
- Fraction of the limiting reactant consumed0.5
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.
- Converting 0.05 mol/L takes 4.055 s, 50 % of the limiting reactant, with the rate falling from 0.02 to 0.0075 mol/(L·s).
- Conversion slows as the reactants are used up, and the time diverges as x approaches the smaller initial concentration.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (s) | Local step | Change in result |
|---|---|---|---|---|---|---|
| k Second-order rate constant | 1e-4 to 10,000 L/(mol·s) | 4.055e-4 | 40,550 | 40,550 | +10% | −9.09 % |
| B0 Initial concentration of B | 0.06 to 1 mol/L | 0.7132 | 27.47 | 26.75 | +10% | −10.5 % |
| A0 Initial concentration of A | 0.06 to 1 mol/L | 0.2955 | 10.74 | 10.45 | +10% | −12.7 % |
| x Reacted concentration | 0.001 to 0.05 mol/L | 0.05038 | 4.055 | 4.004 | +10% | +17.6 % |
Sweep of Reacted concentration
| Reacted concentration (mol/L) | Time (s) |
|---|---|
| 0.001 | 0.05038 |
| 0.007125 | 0.3764 |
| 0.01325 | 0.7359 |
| 0.01938 | 1.135 |
| 0.0255 | 1.58 |
| 0.03163 | 2.08 |
| 0.03775 | 2.648 |
| 0.04388 | 3.299 |
| 0.05 | 4.055 |
Sweep one variable
Variables and how each one contributes
Time
What the equation gives you. Everything else feeds it.
Initial concentration of A
The initial concentration of A, in mol/L. A larger A₀ speeds the reaction, so the time falls. Raising it by 10 % (to 0.11) moves the result by −12.7 %, from 4.055 to 3.538 s.
Initial concentration of B
The initial concentration of B, in mol/L. A larger B₀ speeds the reaction, so the time falls. Raising it by 10 % (to 0.22) moves the result by −10.5 %, from 4.055 to 3.628 s.
Second-order rate constant
The second-order rate constant, in L/(mol·s). A larger k shortens the time in inverse proportion. Raising it by 10 % (to 1.1) moves the result by −9.09 %, from 4.055 to 3.686 s.
Reacted concentration
The reacted concentration, in mol/L. A larger conversion takes longer, increasingly so near complete reaction. Raising it by 10 % (to 0.055) moves the result by +17.6 %, from 4.055 to 4.769 s.
Worked example with the values above
- Remaining concentrations: A₀ − x = 0.05, B₀ − x = 0.15.
- Log term: ln[A₀(B₀ − x)/(B₀(A₀ − x))] = ln 1.5 = 0.40547.
- t = 0.40547 / (1 × 0.1) = 4.0547 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