Library / Solution chemistry, equilibria and kinetics / SCH-37
Solution chemistry, equilibria and kinetics
Partial reaction order from paired initial rates
Order of a reaction in one reactant from two initial-rate runs that differ only in that reactant’s concentration.
- Ref
- SCH-37
- Rev
- v1
- Field
- Solution chemistry, equilibria and kinetics
- Result
- m
If two initial-rate experiments differ only in the concentration of one reactant, the rate ratio is (C₂/C₁)ᵐ, so m = ln(r₂/r₁)/ln(C₂/C₁). The result is the exponent of the local power-law fit, which can be a fraction or a whole number. Tripling the concentration and getting nine times the rate gives m = 2. Use initial rates so that depletion and products do not interfere.
Order of a reaction in one reactant from two initial-rate runs that differ only in that reactant’s concentration. It is the exponent of the local power-law fit; use positive rates and two distinct concentrations.
Calculator
- Rate ratio r₂/r₁9
- Concentration ratio C₂/C₁3
- Rate ratio for a tenfold concentration increase, 10^m100
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.
- A 3-fold rise in concentration raised the rate 9-fold, so the order in this reactant is 2.
- Orders near whole numbers (0, 1, 2) suggest simple elementary steps, while a fractional value such as 0.5 points to a chain or pre-equilibrium mechanism.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| r2 Initial rate in run 2 | 1e-9 to 10 mol/(L·s) | −12.58 | 8.384 | 20.96 | +10% | +4.34 % |
| r1 Initial rate in run 1 | 1e-9 to 10 mol/(L·s) | −6.384 | 14.58 | 20.96 | +10% | −4.34 % |
| C1 Concentration in run 1 | 0.001 to 0.2 mol/L | 0.3852 | 5.419 | 5.034 | +10% | +9.5 % |
| C2 Concentration in run 2 | 0.25 to 5 mol/L | 0.5617 | 2.398 | 1.836 | +10% | −7.98 % |
Sweep of Initial rate in run 2
| Initial rate in run 2 (mol/(L·s)) | Partial order |
|---|---|
| 1e-9 | −12.58 |
| 1.778e-8 | −9.956 |
| 3.162e-7 | −7.336 |
| 5.623e-6 | −4.716 |
| 1e-4 | −2.096 |
| 0.001778 | 0.524 |
| 0.03162 | 3.144 |
| 0.5623 | 5.764 |
| 10 | 8.384 |
Sweep one variable
Variables and how each one contributes
Partial order
What the equation gives you. Everything else feeds it.
Initial rate in run 2
The initial rate in run 2, in mol/(L·s). A larger rate ratio gives a larger order. Raising it by 10 % (to 0.0099) moves the result by +4.34 %, from 2 to 2.087.
Initial rate in run 1
The initial rate in run 1, in mol/(L·s). A larger starting rate lowers the order. Raising it by 10 % (to 0.0011) moves the result by −4.34 %, from 2 to 1.913.
Concentration in run 2
The concentration of the reactant in run 2. A larger concentration ratio lowers the order for the same rate ratio. Raising it by 10 % (to 0.33) moves the result by −7.98 %, from 2 to 1.84.
Concentration in run 1
The concentration of the reactant in run 1. A larger starting concentration shrinks the ratio, so the order rises. Raising it by 10 % (to 0.11) moves the result by +9.5 %, from 2 to 2.19.
Worked example with the values above
- Rate ratio: r₂/r₁ = 0.009 / 0.001 = 9; ln = 2.1972.
- Concentration ratio: C₂/C₁ = 3; ln = 1.0986.
- m = 2.1972 / 1.0986 = 2.
Assumptions and limits
- Constant volume, a well-mixed closed batch at fixed temperature.
- The simple rate law stated, with the same mechanism over the range studied.
Common mistakes
- Using the wrong sign of the stoichiometric coefficient for reactants and products.
- Forgetting that a pseudo-first-order constant includes the excess reactant’s concentration.
- Using rates from different times in a method of initial rates.
Statistics and uncertainty
Rate constants are normally known to 5 % to 10 % and change by about 7 % per kelvin near room temperature for a typical activation energy, so temperature control matters.
Where it is used
- Finding reaction orders and rate constants from experiments.
- Predicting selectivity and relaxation to equilibrium.
Maths you need
- Rate law: rate = k[A]^m[B]^n
- Stoichiometric coefficient: ν, negative for reactants
- First-order decay: c = c₀ e^(−kt)