Kinetics Rate Laws: Zero, First, and Second Order Integrated Forms
Rate laws describe how concentration changes over time. Here is exactly how zero, first, and second-order kinetics work — with integrated forms, half-life formulas, and three worked examples.
Learning Objectives
- ✓Distinguish zero, first, and second order rate laws by integrated form and plot.
- ✓Compute half-life and predict concentration over time for each order.
- ✓Apply the Arrhenius equation to determine activation energy from temperature data.
1. Direct Answer: The Three Orders in One Table
For a reaction A → products, the rate of disappearance of A depends on [A] raised to a power that defines the ORDER. ZERO-ORDER: rate = k, independent of [A]; [A]_t = [A]_0 − kt; half-life t_{1/2} = [A]_0 / (2k); plot of [A] vs t is linear. FIRST-ORDER: rate = k[A]; ln[A]_t = ln[A]_0 − kt; t_{1/2} = 0.693/k (concentration-independent — the hallmark of first-order); plot of ln[A] vs t is linear. SECOND-ORDER: rate = k[A]², or rate = k[A][B] for two reactants; 1/[A]_t = 1/[A]_0 + kt; t_{1/2} = 1/(k[A]_0); plot of 1/[A] vs t is linear. The standard exam test for order is to plot [A], ln[A], and 1/[A] vs t and see which yields a straight line. Rate constants k have order-specific units: zero-order M/s, first-order 1/s, second-order 1/(M·s).
Key Points
- •Zero-order: linear in [A] vs t, half-life depends on [A]_0.
- •First-order: linear in ln[A] vs t, half-life is constant (0.693/k).
- •Second-order: linear in 1/[A] vs t, half-life inversely depends on [A]_0.
2. Half-Life as the Order Test
Half-life behavior gives the cleanest qualitative diagnostic for order even without making a plot. ZERO-ORDER half-life is proportional to [A]_0 — doubling the initial concentration doubles the half-life. FIRST-ORDER half-life is INDEPENDENT of [A]_0 — exactly the same time elapses between [A] = 1.0 and [A] = 0.5 as between [A] = 0.25 and [A] = 0.125. This concentration-independence is the diagnostic for first-order kinetics and explains why radioactive decay (always first-order) has a single characteristic half-life regardless of how much you start with. SECOND-ORDER half-life is INVERSELY proportional to [A]_0 — doubling initial concentration HALVES the half-life. Three successive half-lives that shrink toward each other point to second order; three that stay constant point to first; three that grow point to zero.
Key Points
- •Three successive half-lives: shrinking → second order, constant → first, growing → zero.
- •Radioactive decay is canonical first-order.
- •Half-life pattern is often the easiest experimental test.
3. Worked Example 1: First-Order Decomposition
N₂O₅ decomposes to NO₂ and O₂ as a first-order reaction with k = 4.8 × 10⁻⁴ s⁻¹ at 45°C. Starting [N₂O₅]_0 = 0.10 M, find [N₂O₅] after 30 minutes (1800 s). Use ln[A]_t = ln[A]_0 − kt: ln[N₂O₅] = ln(0.10) − (4.8 × 10⁻⁴)(1800) = −2.303 − 0.864 = −3.167. So [N₂O₅] = e^(−3.167) = 0.042 M. Half-life: t_{1/2} = 0.693/k = 0.693/(4.8 × 10⁻⁴) = 1444 s = 24 minutes. After 30 minutes, slightly more than one half-life has passed, and indeed 0.042 M is just under half of 0.10 M — sanity check confirmed.
Key Points
- •ln[A]_t = ln[A]_0 − kt for first-order.
- •t_{1/2} = 0.693/k regardless of starting concentration.
- •Always sanity-check answer against expected half-life count.
4. Worked Example 2: Second-Order Dimerization
2A → A₂ is second-order in A with k = 0.10 M⁻¹s⁻¹. Starting [A]_0 = 0.20 M, find [A] after 50 s. Use 1/[A]_t = 1/[A]_0 + kt: 1/[A] = 1/0.20 + (0.10)(50) = 5.0 + 5.0 = 10.0 M⁻¹, so [A] = 0.10 M. Half-life: t_{1/2} = 1/(k[A]_0) = 1/(0.10 × 0.20) = 50 s. The math confirms the result — we have exactly one half-life elapsed (50 s) and concentration has halved from 0.20 to 0.10. The NEXT half-life will be 1/(k × 0.10) = 100 s — twice as long, because second-order half-lives grow as concentration drops.
Key Points
- •1/[A]_t = 1/[A]_0 + kt for second-order.
- •t_{1/2} = 1/(k[A]_0) grows as [A] decreases.
- •Successive half-lives double in true second-order kinetics.
5. Worked Example 3: Determining Order From Data
Given concentration vs time data for A: t = 0, [A] = 1.00; t = 10s, [A] = 0.80; t = 20s, [A] = 0.64; t = 30s, [A] = 0.51; t = 40s, [A] = 0.41. Test for first-order: ln[A] values are 0, −0.223, −0.446, −0.673, −0.892. Differences are −0.223 (per 10s consistently), so ln[A] decreases linearly with time — FIRST-ORDER confirmed. Slope = −0.0223 s⁻¹, so k = 0.0223 s⁻¹. Sanity check half-life: t_{1/2} = 0.693/0.0223 = 31 s, and indeed [A] drops from 1.00 to about 0.50 by 30 s. If the test had failed (ln[A] not linear), we would test 1/[A] for second-order, then [A] itself for zero-order.
Key Points
- •Test linearity of ln[A] vs t for first order; 1/[A] vs t for second; [A] vs t for zero.
- •Constant differences in ln[A] over equal time = first order.
- •Sanity check by half-life behavior.
6. Arrhenius Equation and Activation Energy
Rate constants grow with temperature according to the Arrhenius equation: k = A × e^(−Ea/RT), where A is the pre-exponential (frequency) factor, Ea is activation energy, R is the gas constant (8.314 J/mol·K), and T is absolute temperature. Taking the natural log: ln k = ln A − Ea/(RT), so a plot of ln k vs 1/T is linear with slope −Ea/R. Two-temperature form: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). Example: k₁ = 2.0 × 10⁻⁵ at 300 K; k₂ = 4.0 × 10⁻⁴ at 350 K. ln(k₂/k₁) = ln(20) = 3.00. (Ea/R)(1/300 − 1/350) = (Ea/8.314)(0.0004762) = 3.00. So Ea = 3.00 × 8.314 / 0.0004762 = 52,300 J/mol = 52 kJ/mol. The temperature dependence captured by Arrhenius is why reaction rates roughly double for every 10 K temperature increase near room temperature.
Key Points
- •k = A × e^(−Ea/RT); ln k vs 1/T is linear, slope = −Ea/R.
- •Two-point Arrhenius: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).
- •Rate roughly doubles per 10 K near room temperature.
7. Mechanisms, Rate-Determining Steps, and Catalysts
Most reactions occur in multiple elementary steps, and the overall rate is governed by the SLOWEST step (rate-determining step). The molecularity of each elementary step (unimolecular, bimolecular, termolecular) directly gives its rate law — A → products is rate = k[A] always for elementary unimolecular. The OBSERVED rate law of an overall reaction is determined experimentally and reflects the mechanism. CATALYSTS provide a lower-energy reaction pathway by lowering Ea; they appear in the rate law if they participate in the rate-determining step. Enzymes are biological catalysts that lower Ea dramatically. STEADY-STATE APPROXIMATION (intermediates have constant low concentration) is the workhorse for deriving mechanism-based rate laws.
Key Points
- •Overall rate equals rate of slowest (rate-determining) elementary step.
- •Elementary step rate law equals its molecularity; observed law reflects mechanism.
- •Catalysts lower Ea; enzymes are biological catalysts.
8. Running Kinetics in ChemistryIQ
Provide concentration-vs-time data or a rate-vs-concentration table and ChemistryIQ tests for zero, first, and second-order behavior, returns the rate constant with units and half-life, and produces the integrated rate law plot. For temperature-dependent data ChemistryIQ runs Arrhenius analysis and returns activation energy and pre-exponential factor with uncertainty. This content is for educational purposes only.
Key Points
- •Automatic order determination from concentration-time data.
- •Rate constant with correct units returned alongside half-life.
- •Arrhenius analysis for activation energy from temperature data.
High-Yield Facts
- ★Zero-order: [A]_t = [A]_0 − kt, t_{1/2} = [A]_0/(2k).
- ★First-order: ln[A]_t = ln[A]_0 − kt, t_{1/2} = 0.693/k (constant).
- ★Second-order: 1/[A]_t = 1/[A]_0 + kt, t_{1/2} = 1/(k[A]_0).
- ★k units: zero = M/s; first = 1/s; second = 1/(M·s).
- ★Arrhenius: k = A·e^(−Ea/RT); ln k vs 1/T linear with slope −Ea/R.
Practice Questions
1. A reaction has half-lives of 100 s, 100 s, and 100 s for three consecutive half-life observations starting from various concentrations. What is the order?
2. For a second-order reaction with k = 0.50 M⁻¹s⁻¹ and [A]_0 = 0.10 M, what is [A] after 20 s?
3. A reaction has k₁ = 0.10 s⁻¹ at 300 K and k₂ = 0.40 s⁻¹ at 320 K. Estimate Ea.
FAQs
Common questions about this topic
Because rate is always concentration/time (M/s) but the rate law multiplies different powers of [A] before reaching that result. Zero-order: rate = k → k must have units of M/s. First-order: rate = k[A] → k = (M/s)/M = 1/s. Second-order: rate = k[A]² → k = (M/s)/M² = 1/(M·s). Always check k units to verify the assumed order is consistent.
The observed rate law of an overall reaction reflects the elementary steps of its mechanism, especially the rate-determining step. A rate law of k[A]²[B] cannot come from a single-step bimolecular reaction; it suggests a mechanism with at least three molecules involved in the rate-determining step (or its leading equivalent). However, the observed rate law alone does NOT uniquely determine the mechanism — multiple mechanisms can produce the same rate law.
When one reactant is in large excess (say A + B → products with [B] ≫ [A]), [B] is effectively constant during the reaction and is absorbed into the rate constant: rate = k[A][B] becomes rate = k\u2032[A] where k\u2032 = k[B]_0. The reaction now LOOKS first-order, even though it is truly second-order. Pseudo-first-order conditions are deliberately created in experiments to simplify analysis.
Because the Arrhenius equation is exponential in 1/T and Ea is typically tens of kJ/mol — large compared to RT. A 10 K increase near 300 K shifts 1/T enough that e^(−Ea/RT) doubles or so for typical activation energies (50–100 kJ/mol). For larger Ea reactions, temperature dependence is even more dramatic, which is why explosives need a precise initiation temperature and biological enzymes operate in tight temperature ranges.
Yes. Provide concentration-vs-time data and ChemistryIQ tests for zero, first, and second-order linearity, returns the rate constant with units and half-life, plots the integrated rate law, and confirms order from half-life behavior. For Arrhenius analysis, provide rate constants at multiple temperatures. This content is for educational purposes only.