Library / Foundations: Chemistry / FCH-32
Foundations: Chemistry
Concentration over time for a zero-order reaction
Concentration of a reactant that is used up at a constant rate k, independent of how much is left.
- Ref
- FCH-32
- Rev
- v1
- Field
- Foundations: Chemistry
- Result
- CA [mol/dm³]
For a zero-order reaction the rate does not depend on the concentration, so the reactant falls in a straight line: CA = CA0 − kt, with k in concentration per unit time. This happens when the rate is limited by something other than the concentration, such as a saturated catalyst surface or a light source. The half-life is CA0/(2k), and the reactant is used up completely at t = CA0/k; the formula does not apply beyond that time.
Concentration of a reactant that is used up at a constant rate k, independent of how much is left. It applies only until the reactant runs out.
Calculator
- Half-life, CA0/(2k)25 s
- Time to use up all the reactant, CA0/k50 s
- Fraction remaining0.6
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 0.6 mol/dm³ at a steady 0.02 mol/(dm³ s). Half the reactant is used by 25 s and all of it by 50 s.
- Unlike first-order decay, the fall never slows down: the graph of concentration against time is a straight line that reaches zero in finite time.
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 1 mol/(dm³ s) | −19 | 0.998 | 20 | +10% | −6.67 % |
| t Time | 0 to 300 s | −5 | 1 | 6 | +10% | −6.67 % |
| CA0 Initial concentration | 0.01 to 5 mol/dm³ | −0.39 | 4.6 | 4.99 | +10% | +16.7 % |
Sweep of Time
| Time (s) | Concentration at time t (mol/dm³) |
|---|---|
| 0 | 1 |
| 37.5 | 0.25 |
| 75 | −0.5 |
| 112.5 | −1.25 |
| 150 | −2 |
| 187.5 | −2.75 |
| 225 | −3.5 |
| 262.5 | −4.25 |
| 300 | −5 |
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 by the same amount, as the line is shifted up. Raising it by 0.1 (to 1.1) moves the result by +16.7 %, from 0.6 to 0.7 mol/dm³.
Rate constant
The zero-order rate constant, in mol/(dm³ s): how fast the concentration falls. Lowers the concentration in direct proportion to its change, because the line falls faster. Raising it by 0.005 (to 0.025) moves the result by −16.7 %, from 0.6 to 0.5 mol/dm³.
Time
The elapsed time since the start, in s. It must not exceed C₀/k. Lowers the concentration by k per second, a straight-line fall. Raising it by 5 s (to 25 s) moves the result by −16.7 %, from 0.6 to 0.5 mol/dm³.
Worked example with the values above
- Amount used so far: k t = 0.02 × 20 = 0.4 mol/dm³.
- Subtract from the start: 1 − 0.4.
- CA = 0.6 mol/dm³.
Assumptions and limits
- A reaction whose rate is independent of the concentration of the reactant over the time range considered.
- Constant temperature, and a time not beyond CA0/k.
Common mistakes
- Applying the straight-line law after the reactant has run out, which gives a negative concentration.
- Using the first-order half-life formula, which has a different form.
Statistics and uncertainty
Plotting concentration against time gives a straight line of slope −k if the reaction is zero order, so a straight line fit is the test; a curved plot rules zero order out. The slope’s uncertainty from regression is usually the main error in k.
Where it is used
- Catalytic and surface-limited reactions, and slow-release systems.
- Estimating how long a reactant lasts at a steady consumption rate.
Maths you need
- Rate constant: the proportionality between the rate and concentration terms
- Reaction order: the power of concentration in the rate law