Library / Foundations: Chemistry / FCH-33
Foundations: Chemistry
Concentration over time for a second-order reaction
Concentration of a reactant that disappears at a rate proportional to the square of its concentration.
- Ref
- FCH-33
- Rev
- v1
- Field
- Foundations: Chemistry
- Result
- CA [mol/dm³]
For a second-order reaction with rate = kCA², integration gives 1/CA = 1/CA0 + kt, or CA = CA0/(1 + kCA0t). The half-life is 1/(kCA0), which gets longer as the reaction proceeds because the concentration (and the rate) has fallen. This form applies to one reactant, or to two reactants at equal starting concentrations. The unit of k is dm³/(mol s).
Concentration of a reactant that disappears at a rate proportional to the square of its concentration. The half-life grows as the reactant is used up.
Calculator
- Initial half-life, 1/(kCA0)20 s
- Fraction remaining0.5
- Time for 90 % of the reactant to react, 9/(kCA0)180 s
- Straight-line form, 1/CA2 dm³/mol
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.
- After 20 s the concentration has fallen from 1 to 0.5 mol/dm³, which is 50 % of the start. The first half-life is 20 s.
- Each successive half-life is twice the last (the second takes 40 s), because the rate falls with the square of the concentration: unlike first-order decay, the half-life is not constant.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mol/dm³) | Local step | Change in result |
|---|---|---|---|---|---|---|
| k Rate constant | 1e-4 to 2 dm³/(mol s) | 0.02439 | 0.998 | 0.9736 | +10% | −4.76 % |
| t Time | 0 to 300 s | 0.0625 | 1 | 0.9375 | +10% | −4.76 % |
| CA0 Initial concentration | 0.01 to 5 mol/dm³ | 0.009901 | 0.8333 | 0.8234 | +10% | +4.76 % |
Sweep of Time
| Time (s) | Concentration at time t (mol/dm³) |
|---|---|
| 0 | 1 |
| 37.5 | 0.3478 |
| 75 | 0.2105 |
| 112.5 | 0.1509 |
| 150 | 0.1176 |
| 187.5 | 0.09639 |
| 225 | 0.08163 |
| 262.5 | 0.0708 |
| 300 | 0.0625 |
Sweep one variable
Variables and how each one contributes
Concentration at time t
What the equation gives you. Everything else feeds it.
Initial concentration
The initial concentration of the reactant, in mol/dm³. Raises the concentration at time t, though less than in proportion, because a higher starting concentration also reacts faster. Raising it by 0.1 (to 1.1) moves the result by +4.76 %, from 0.5 to 0.5238 mol/dm³.
Rate constant
The second-order rate constant, in dm³/(mol s). Lowers the concentration, because the reaction is faster. Raising it by 0.01 (to 0.06) moves the result by −9.09 %, from 0.5 to 0.4545 mol/dm³.
Time
The elapsed time since the start, in s. Lowers the concentration, with the fall slowing down as time goes on. Raising it by 5 s (to 25 s) moves the result by −11.1 %, from 0.5 to 0.4444 mol/dm³.
Worked example with the values above
- Product: k CA0 t = 0.05 × 1 × 20 = 1.
- Denominator: 1 + 1 = 2.
- CA = 1 / 2 = 0.5 mol/dm³.
Assumptions and limits
- A single reactant, or two reactants with equal starting concentrations, in a reaction of overall second order.
- Constant temperature and volume.
Common mistakes
- Using the constant half-life of a first-order reaction.
- Using concentration units inconsistent with the units of k.
Statistics and uncertainty
A plot of 1/CA against time is a straight line of slope k for a second-order reaction: this linear plot is the usual test of the order and gives k with a known uncertainty. Concentrations at long times are small, so measurement errors on them have large effects on 1/CA.
Where it is used
- Predicting how long a reaction takes to reach a given conversion.
- Fitting data from reactions such as dimerisation or the decomposition of nitrogen dioxide.
Maths you need
- Rate constant: the proportionality between the rate and concentration terms
- Reaction order: the power of concentration in the rate law