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For a zero-order process, concentration decreases linearly with time at a constant rate.
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Formula
[A] in mol/L · k0 in mol·L−1·s−1 · t in s
| Symbol | Meaning | Unit |
|---|---|---|
| concentration at time t | mol/L | |
| initial concentration | mol/L | |
| zero-order rate constant | mol·L−1·s−1 | |
| time | s |
[A]_t=[A]_0-ktThe relation applies while the rate is independent of reactant concentration and the zero-order regime persists. It cannot predict negative concentrations; after depletion, the mechanism or rate law changes.
The relation applies while the rate is independent of reactant concentration and the zero-order regime persists. It cannot predict negative concentrations; after depletion, the mechanism or rate law changes.
For a zero-order process, concentration decreases linearly with time at a constant rate.
Rearrange for a target quantity
The rearrangement keeps the same model assumptions as the original relation.
Calculate Integrated Zero-Order Rate Law from the stated example data and explain the check.
With [A]0=0.80 mol/L, k0=0.020 mol·L−1·s−1 and t=15 s, [A]=0.50 mol/L.
Check: concentration falls by k0t=0.30 mol/L.
Applying Integrated Zero-Order Rate Law with inconsistent reference states or without checking the stated model assumptions.
Keep the standard state, units, and model assumptions explicit; compare them with the measured system before interpreting the result.
Formula cluster
Integrated Zero-Order Rate Law answers a distinct subquestion. Linked relations complement it when their additional assumptions are met.
With [A]0=0.80 mol/L, k0=0.020 mol·L−1·s−1 and t=15 s, [A]=0.50 mol/L. Check: concentration falls by k0t=0.30 mol/L.
Interpret linear concentration–time data and estimate depletion times.
Here is how to work through a typical Integrated Zero-Order Rate Law ([A]t = [A]0 − kt) task step by step:
Task
Calculate Integrated Zero-Order Rate Law from the stated example data and explain the check.
Solution path
With [A]0=0.80 mol/L, k0=0.020 mol·L−1·s−1 and t=15 s, [A]=0.50 mol/L.
Check: concentration falls by k0t=0.30 mol/L.
For a zero-order process, concentration decreases linearly with time at a constant rate. The relation is useful for Interpret linear concentration–time data and estimate depletion times.. With [A]0=0.80 mol/L, k0=0.020 mol·L−1·s−1 and t=15 s, [A]=0.50 mol/L. Check: concentration falls by k0t=0.30 mol/L. The final check tests a balance, direction or magnitude relevant to this formula. First map each quantity to the actual experiment, then substitute values. The result describes the modelled relationship; it does not automatically establish every mechanism in the real sample.
The symbols mean: [A]_t: concentration at time t (mol/L); [A]_0: initial concentration (mol/L); k: zero-order rate constant (mol·L−1·s−1); t: time (s). Follow [A] in mol/L · k0 in mol·L−1·s−1 · t in s. Concentrations, activities, volumes and times must use compatible scales. A logarithm needs a dimensionless argument, and a balance must include all contributing species. State temperature and standard state whenever they are part of the definition; otherwise the result may refer to a different chemical condition. An amount is not a mass without molar mass, and activity is not an unnormalised concentration. Say whether inputs were measured or taken from a reference.
The relation applies while the rate is independent of reactant concentration and the zero-order regime persists. It cannot predict negative concentrations; after depletion, the mechanism or rate law changes. Check this assumption for the actual sample, temperature and phase. Nonideality, mass transport or side reactions can make a precise calculation systematically wrong. Compare the balanced reaction and the activity or concentration scale with the model, then state the assumption that most affects the result. Near model boundaries, check activities and phase state rather than extending a dilute-solution approximation. Different conditions require a fresh model check.
Recalculate the example: With [A]0=0.80 mol/L, k0=0.020 mol·L−1·s−1 and t=15 s, [A]=0.50 mol/L. Check: concentration falls by k0t=0.30 mol/L. Check [A] in mol/L · k0 in mol·L−1·s−1 · t in s and substitute the value back into the relation. Also ask whether its sign, balance or magnitude is chemically plausible. If a check fails, inspect units, coefficients and reference values before blaming rounding. Substitution checks the algebra; comparing assumptions with the experiment checks whether the model is appropriate. A linear concentration-versus-time plot supports zero-order behavior over that interval; the model stops when concentration reaches zero and cannot predict negative concentration.
Integrated Zero-Order Rate Law is related to Geschwindigkeitsgesetz (Reaktionsordnung), Integrated First-Order Rate Law, Half-Life by Reaction Order, but answers a distinct question: Interpret linear concentration–time data and estimate depletion times.. Do not confuse it with a relation for a neighboring quantity or a different standard state. Name the target quantity and sample conditions first, then choose the equation containing that quantity. Similar-looking equations may use different stoichiometric bases or validity ranges. The linked entries may supply inputs for another step, but they do not replace this distinction. This keeps the interpretation auditable.
Quanta exam set
Curated exam set for "Integrated Zero-Order Rate Law":
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Which formula describes Integrated Zero-Order Rate Law?
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How do you rearrange Integrated Zero-Order Rate Law for Rearrange for a target quantity?
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Which common mistake happens with Integrated Zero-Order Rate Law?
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