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Integrated Rate Laws & Graphical Analysis

You’ll be able to

One graph tells you the order

Each order integrates to a different equation, and each equation has the form of a straight line, y = mx + b. The trick is that only one transformation of the data — plotting [A], ln[A], or 1/[A] against time — comes out as a straight line, and which one is linear reveals the order. Zero order makes [A] vs t linear, first order makes ln[A] vs t linear, and second order makes 1/[A] vs t linear.

Integrated rate laws as straight lines
Zero: [A] = −kt + [A]₀ (plot [A] vs t) | First: ln[A] = −kt + ln[A]₀ (plot ln[A] vs t) | Second: 1/[A] = kt + 1/[A]₀ (plot 1/[A] vs t)
In each case y = mx + b: the y-intercept is the starting value and the slope is ±k. Zero and first order give slope = −k; second order gives slope = +k.

Extracting k from the slope — mind the sign

Once you know which plot is linear, the slope gives k directly. For the zero-order ([A] vs t) and first-order (ln[A] vs t) plots the reactant is disappearing, so the line slopes downward and slope = −k (take k as the magnitude). For the second-order (1/[A] vs t) plot, 1/[A] grows as [A] falls, so the line slopes upward and slope = +k. The y-intercept is always the t = 0 value: [A]₀, ln[A]₀, or 1/[A]₀.

Half-life is a fingerprint of order

The half-life (t½) — the time for [A] to fall by half — behaves differently for each order, so watching how t½ changes as the reaction proceeds is itself a way to spot the order. First-order t½ is constant (independent of concentration). Zero-order t½ shrinks as the reaction proceeds (it is proportional to [A]₀). Second-order t½ grows as the reaction proceeds (it is inversely proportional to [A]₀).

Half-life by order
Zero: t½ = [A]₀ / (2k) First: t½ = 0.693 / k Second: t½ = 1 / (k[A]₀)
Only first-order t½ is independent of [A]₀ (0.693 = ln 2). If successive half-lives are equal → first order; if they get shorter → zero order; if they get longer → second order.
Worked example

For A → products, concentration is tracked over time: t = 0 s, [A] = 0.100 M t = 20 s, [A] = 0.050 M t = 40 s, [A] = 0.0333 M t = 60 s, [A] = 0.0250 M Determine the reaction order and the value of k.

  1. 1.Test zero order — is [A] vs t linear? The drops are 0.050, then 0.0167, then 0.0083 M over equal 20 s steps. Not constant, so [A] vs t is curved → not zero order.
  2. 2.Test first order — is ln[A] vs t linear? ln[A] = −2.30, −3.00, −3.40, −3.69. The steps are −0.70, −0.40, −0.29. Not constant → not first order.
  3. 3.Test second order — is 1/[A] vs t linear? 1/[A] = 10, 20, 30, 40 M⁻¹. Each 20 s step adds exactly +10, a constant slope → this plot is the straight line, so the reaction is second order.
  4. 4.Extract k from the slope of the 1/[A] vs t line: slope = Δ(1/[A])/Δt = (40 − 10) / (60 − 0) = 30 / 60 = 0.50. For a second-order plot slope = +k.
  5. 5.Attach units so that rate (M·s⁻¹) = k[A]² works out: second-order k carries M⁻¹·s⁻¹.
Answer: Second order in A, with k = 0.50 M⁻¹·s⁻¹ (only the 1/[A] vs t plot is linear).
Tip

Memorize the pairing as "0, 1, 2 → [A], ln[A], 1/[A]." The order that gives a straight line is the order of the reaction. If the problem hands you a graph and says the line is ln[A] vs t, you already know it is first order before doing any arithmetic.

Checkpoint

For A → products, a graph of 1/[A] versus time is a straight line, while graphs of [A] and ln[A] versus time are curved. What is the order, and what does the slope of the straight line equal?

Checkpoint

A reaction gives a straight line when ln[A] is plotted against time, with a slope of −0.0347 min⁻¹. What is the half-life of the reaction?

On the exam

On free-response, justify the order by naming which plot is linear, then read k off the slope with the right sign (−k for zero/first order, +k for second order) and give units by overall order (M·s⁻¹, s⁻¹, M⁻¹·s⁻¹). A bare "second order" without pointing to the linear 1/[A] plot usually loses the justification point.

Answer the 2 checkpoints as you read.

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