Integrated Rate Law

Differential vs integrated

The differential rate law expresses rate as a function of concentration:

      −d[A]/dt = k · [A]^n

It is mechanistically informative — the exponent n encodes the molecularity of the rate-determining step — but it doesn't tell you what concentration to expect at, say, 47 minutes. To answer the time question, you need an explicit [A](t). That is the integrated rate law:

      [A] = f(t; k, [A]_0, n)

Getting from the differential to the integrated form is a textbook exercise in separation of variables and integration. The trick is that integrating [A]^−n gives different functions for n = 0, 1, and 2 — which is why the three orders end up with three different-looking integrated forms.

Deriving the zero-order integrated law

Start with the differential rate law for zero-order:

      −d[A]/dt = k

Separate variables (concentration on the left, time on the right):

      d[A] = −k · dt

Integrate both sides — left from [A]₀ at t = 0 to [A] at time t, right from 0 to t:

      ∫_{[A]_0}^{[A]} d[A] = ∫_0^t (−k) dt
      [A] − [A]_0 = −k·t
      [A] = [A]_0 − k·t

A plot of [A] versus t gives a straight line with intercept [A]₀ and slope −k. The reaction depletes [A] linearly until it reaches zero — at which point a new mechanism takes over (e.g., a saturated catalyst desaturates).

Deriving the first-order integrated law

Differential form:

      −d[A]/dt = k · [A]

Separate:

      d[A]/[A] = −k · dt

Integrate (using ∫ dx/x = ln|x|):

      ln[A] − ln[A]_0 = −k·t
      ln([A]/[A]_0) = −k·t
      [A] = [A]_0 · e^(−k·t)

A plot of ln[A] versus t is linear with slope −k. This exponential decay is the same mathematics that governs RC discharging in electronics, viscous damping, and Newton's law of cooling — separation of variables on a first-order ODE always gives an exponential.

Deriving the second-order integrated law

Differential form for [A]² rate dependence:

      −d[A]/dt = k · [A]²

Separate:

      d[A]/[A]² = −k · dt
      [A]^(−2) d[A] = −k · dt

Integrate (∫ x^(−2) dx = −1/x):

      −1/[A] − (−1/[A]_0) = −k·t
      1/[A] − 1/[A]_0 = k·t
      1/[A] = 1/[A]_0 + k·t

A plot of 1/[A] versus t is linear with slope +k (note the sign flip from zero and first order). The reciprocal is the natural variable here because antidifferentiating [A]^−2 produces a reciprocal.

All three orders, side by side

	Zero-order	First-order	Second-order in [A]	Mixed second-order
Differential rate law	−d[A]/dt = k	−d[A]/dt = k[A]	−d[A]/dt = k[A]²	−d[A]/dt = k[A][B]
Integrated rate law	[A] = [A]0 − k·t	ln[A] = ln[A]0 − k·t	1/[A] = 1/[A]0 + k·t	(1/([A]0−[B]0))·ln(([B]0[A])/([A]0[B])) = k·t
Linear plot	[A] vs t	ln[A] vs t	1/[A] vs t	ln([A]/[B]) vs t (after rearrangement)
Slope of plot	−k	−k	+k	complex (depends on [A]0, [B]0)
Half-life	[A]0/(2k)	ln(2)/k = 0.693/k	1/(k·[A]0)	not single-valued
Concentration profile	Linear drop	Exponential decay	Hyperbolic decay	Tracks limiting reagent
Units of k	mol·L⁻¹·s⁻¹	s⁻¹	L·mol⁻¹·s⁻¹	L·mol⁻¹·s⁻¹
Typical use	Saturated enzyme; alcohol metabolism	Radioactive decay; N₂O₅ → NO₂	2NO₂ → 2NO+O₂	Saponification, A+B→P with [A]₀ ≠ [B]₀

Worked example: hydrogen peroxide decomposition

Hydrogen peroxide decomposes catalytically: 2H₂O₂ → 2H₂O + O₂. With Fe³⁺ catalyst at 25 °C, the reaction is first-order in [H₂O₂] with k = 1.06 × 10⁻³ s⁻¹.

Question: starting from [H₂O₂]₀ = 0.30 M, what is the concentration after 10 minutes (600 s)?

      [A](t) = [A]_0 · e^(−k·t)
      [A](600) = 0.30 · e^(−1.06 × 10⁻³ × 600)
              = 0.30 · e^(−0.636)
              = 0.30 · 0.529
              = 0.159 M

About 47 % decomposed in 10 minutes. The half-life:

      t_{1/2} = ln(2) / k = 0.693 / (1.06 × 10⁻³) = 654 s ≈ 10.9 min

Consistent with our 47 % conversion at t = 600 s — slightly less than half a half-life.

Worked example: NO₂ decomposition

2NO₂ → 2NO + O₂ at 300 °C is second-order in NO₂ with k = 0.775 L·mol⁻¹·s⁻¹.

Starting from [NO₂]₀ = 0.0250 M, find [NO₂] after 30 s:

      1/[A] = 1/[A]_0 + k·t
      1/[A] = 1/0.0250 + 0.775 × 30
      1/[A] = 40.0 + 23.25
      1/[A] = 63.25 L/mol
      [A]   = 0.01581 M

About 37 % conversion. The first half-life:

      t_{1/2} = 1/(k·[A]_0) = 1/(0.775 × 0.0250) = 51.6 s

The next half-life — from 0.01250 to 0.00625 M — is twice as long: 1/(0.775 × 0.01250) = 103.2 s. Second-order half-lives lengthen as reaction proceeds, which is exactly why the curve flattens.

ASCII view of the three linear plots

  [A]                    ln[A]                      1/[A]
   │                       │                          │            ●
   │\\\\.                    │\\\\.                       │           ●
   │ \\\\\\\\.                  │ \\\\\\\\.                     │          ●
   │   \\\\\\\\.                │   \\\\\\\\.                   │        ●
   │     \\\\\\\\.              │     \\\\\\\\.                 │      ●
   │       \\\\\\\\.            │       \\\\\\\\.               │   ●
   └────────────► t       └─────────────► t           ●─────────────► t

  zero-order:            first-order:                second-order in [A]:
  slope = −k             slope = −k                  slope = +k
  starts at [A]_0        starts at ln[A]_0           starts at 1/[A]_0

Reversible reactions and equilibrium

The integrated forms above assume the reverse reaction is negligible, which is only valid far from equilibrium. For a reversible first-order reaction A ⇌ B with forward rate constant k_f and reverse k_r, the rate becomes

      −d[A]/dt = k_f·[A] − k_r·[B]

Defining the equilibrium concentration [A]_eq and the deviation Δ[A] = [A] − [A]_eq, the integrated form becomes

      Δ[A](t) = Δ[A]_0 · exp(−(k_f + k_r)·t)

The relaxation rate is the sum of forward and reverse rate constants. This is the basis of Eigen's relaxation methods (temperature-jump, pressure-jump) — small perturbations from equilibrium decay exponentially with a rate constant that gives k_f + k_r, which separates into k_f and k_r once you know K_eq = k_f/k_r.

Real systems analyzed by integrated rate laws

• ¹⁴C dating. First-order radioactive decay with t_1/2 = 5,730 years. The integrated rate law ln(N/N₀) = −k·t plus a known initial activity gives sample age. The Shroud of Turin's 1988 ¹⁴C dating used this with N/N₀ ≈ 0.928, giving t ≈ 700 years (medieval, 13th–14th century).
• Drug pharmacokinetics. First-order plasma elimination C(t) = C₀·e^−k_e·t is the basis of dose-interval calculations. Penicillin V has k_e = 1.39 h⁻¹ (t_1/2 = 0.5 h), so 4-hour dosing is needed to maintain therapeutic level.
• Ozone reaction in stratosphere. O + O₃ → 2O₂ is second-order, rate = k[O][O₃] with k ≈ 8 × 10⁻¹⁵ cm³·molecule⁻¹·s⁻¹ at 220 K. Integrated under pseudo-first-order with [O₃] flooded gives the [O] decay used in atmospheric models.
• Saponification of methyl acetate. CH₃COOCH₃ + OH⁻ → CH₃COO⁻ + CH₃OH, second-order overall, rate = k[ester][OH⁻] with k = 0.155 L·mol⁻¹·s⁻¹ at 25 °C. Run with [ester]₀ = [OH⁻]₀ and use 1/[A] vs t plot to extract k.
• Iodine clock reaction. 2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻, second-order with rate = k[I⁻][S₂O₈²⁻]. Integrated rate law lets you predict the timing of the starch-iodine colour flash to the second.

Identifying the order from data

Given a table of [A] versus t, compute three derived columns: [A] itself, ln[A], and 1/[A]. Plot each against t and fit a line. The plot with the highest correlation coefficient identifies the order. A real-world cleanliness check: R² > 0.995 for the winning fit, while the other two should be visibly curved. If two plots look equally linear, your data range is too narrow — extend the experiment to higher conversion.

For multi-reactant systems, run isolation experiments: hold every concentration in flooded excess except one, and use the first-order integrated law to extract a pseudo-first-order constant. Repeat with different excess values and let the pseudo-first-order constants reveal the true order in each species.

Common mistakes

• Forgetting integration limits. When you integrate d[A]/[A], the limits are [A]₀ and [A], not 0 and 1. Sloppy limits drop or duplicate the [A]₀ term, ruining the integrated form.
• Mixing log bases. ln[A] uses natural log (base e). Using log₁₀ silently changes the slope by a factor of 2.303 and gives the wrong rate constant. Stick to the ln form unless you explicitly multiply by 2.303.
• Applying integrated forms past equilibrium. Integrated forms above assume an irreversible reaction. After 50 % conversion of a reversible system, the apparent k starts shrinking because the reverse rate is no longer negligible — your linear plot bends.
• Linear regression on a tiny range. Fitting [A], ln[A], and 1/[A] over only 5 % conversion makes all three look equally linear (any smooth curve is locally a line). Extend the experiment to at least 60–80 % conversion before deciding.
• Confusing molecularity with order. Molecularity is the count of molecules in an elementary step; order is the empirical concentration exponent. They match only for elementary reactions; they often diverge for stepwise mechanisms.
• Negative t_{1/2} for second-order at low [A]₀. If the second-order plot doesn't intersect [A]/[A]₀ = 0.5 in the experimental window because [A]₀ is too low, your t_{1/2} estimate is unreliable. Run at higher concentration or longer time.