Two ways to find a rate law
- Method of initial rates: run several trials, changing one reactant’s concentration at a time, and compare the initial rates.
- Integrated rate laws (graphing): follow concentration over time in one run and plot [A], ln[A] and 1/[A] against t:
| Straight graph | Order | Slope |
|---|---|---|
| [A] vs t | Zero | −k |
| ln[A] vs t | First | −k |
| 1/[A] vs t | Second | k |
Measuring concentration over time in the lab
- Spectrophotometry: for a colored reactant or product, A is proportional to concentration (4.4.3), so ln A vs t has the same slope as ln[A] vs t.
- Clock reactions: the solution changes color suddenly once a fixed amount of some product has formed. The shorter the time, the faster the rate: rate ∝ 1/t.
- Others: measure the volume or pressure of a gas produced, or take samples at set times, stop the reaction, and titrate.
- Run every trial at the same temperature (in a water bath), because k changes with temperature.
Pseudo-first-order conditions
Rate = k[A]m[B]n. If B is in large excess, [B] barely changes during the reaction, so rate = k′[A]m, where k′ = k[B]n准一级速率常数. This isolates the order in A; repeating with a different [B] and comparing the two values of k′ gives the order in B.
Crystal violet (CV⁺) is purple and reacts with OH⁻ to form a colorless product: CV⁺ + OH⁻ → CVOH. OH⁻ is in large excess, so [OH⁻] barely changes and rate = k′[CV⁺]m. Absorbance A is proportional to [CV⁺]. Choose [OH⁻] and a type of graph:
| [OH⁻] (M) | k′ (s⁻¹) |
|---|---|
| 0.020 | — |
| 0.040 | — |
How does k′ change when [OH⁻] doubles? What is the order with respect to OH⁻?
Example 1: orders from clock-reaction times
For the reaction A + B → products, a clock reaction changes color each time the same amount of product has formed:
| Trial | [A] (M) | [B] (M) | Time (s) |
|---|---|---|---|
| 1 | 0.020 | 0.010 | 50 |
| 2 | 0.040 | 0.010 | 25 |
| 3 | 0.020 | 0.020 | 50 |
Find the order in A and in B, and write the rate law.
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Rate and time. Each trial forms the same amount of product, so rate ∝ 1/t.
-
Order in A (trials 1 and 2). [A] doubles with [B] unchanged, and the time drops from 50 s to 25 s, so the rate doubles:
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Order in B (trials 1 and 3). [B] doubles with [A] unchanged, and the time is the same, so the rate is unchanged:
-
Rate law. .
Example 2: k′ and half-life from a graph of ln A vs t
In the reaction of crystal violet with excess OH⁻, A = 0.800 at t = 0 and A = 0.273 at t = 300 s, and the graph of ln A vs t is straight. Find k′ and the half-life of crystal violet.
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Slope from two points.
-
k′. Slope = −k′, so k′ = 3.58 × 10⁻³ s⁻¹.
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Half-life. A first-order half-life doesn’t depend on concentration:
- The half-life is the same at any concentration
A constant half-life is the mark of first order.
- The rate doesn’t change with concentration
Zero order: the concentration falls at a steady rate.
- Each half-life is twice as long as the one before
A second-order half-life is inversely proportional to concentration.
- Rate = k[A]²
Second order, so 1/[A] vs t is straight.
- Rate = k
Zero order.
- Radioactive decay
Nuclear decay is a first-order process.
- The concentration drops by the same 0.010 M every 20 s
Equal amounts lost, so [A] vs t is straight.
- The concentration halves every 20 s
Equal fractions lost, so ln[A] vs t is straight.
Common mistakes
Mistake 1: “ln A vs t is straight, so the overall order is 1.” With OH⁻ in large excess, the straight line only shows the reaction is first order in crystal violet. The order in OH⁻ needs another run at a different [OH⁻]; the overall order may be 2.
Mistake 2: “The longer a clock reaction takes to change color, the faster it is.” It’s the opposite. Each run forms the same amount of product before the color changes, so a longer time means a slower rate: rate ∝ 1/t.
Mistake 3: “k′ is the rate constant k.” k′ = k[OH⁻]n, which includes the concentration of the excess reactant. To get k, divide by [OH⁻]n.
Quick check
Why is OH⁻ used in large excess in the crystal violet lab?
These are pseudo-first-order conditions.
Why can absorbance A stand in for [CV⁺] on a ln graph?
ln A = ln(εb) + ln[CV⁺], so the slope is unchanged.
In a clock reaction, how is the time t related to the rate?
The same amount of product forms each time, so a shorter time means a faster rate.
Why must a set of kinetics trials all be run at the same temperature?
At different temperatures k differs, so rate changes can’t be credited to concentration alone.
A graph of ln[A] vs t has a slope of −0.0050 s⁻¹. The half-life is about:
k = 0.0050 s⁻¹, so t½ = 0.693 ÷ 0.0050 ≈ 139 s.
In the method of initial rates, only one reactant’s concentration is changed at a time so that:
That’s controlling variables.
AP-style practice
A blue food dye reacts with hypochlorite to form colorless products: dye + OCl⁻ → products. OCl⁻ is in large excess ([OCl⁻] = 0.034 M), and the absorbance is measured at 630 nm:
| t (s) | A |
|---|---|
| 0 | 0.600 |
| 60 | 0.450 |
| 120 | 0.338 |
| 180 | 0.253 |
| 240 | 0.190 |
(a) Why can absorbance stand in for the dye concentration when studying the rate?
(b) Use the data to show that the reaction is first order in the dye.
(c) Calculate the pseudo-first-order rate constant k′.
(d) Calculate the half-life of the dye.
(e) Repeating the run with [OCl⁻] = 0.068 M gives k′ = 9.6 × 10⁻³ s⁻¹. Find the order in OCl⁻, write the rate law, and calculate k.
Model answer and scoring
(a) By the Beer-Lambert law, A = εbc; at a fixed wavelength in the same cuvette, A is proportional to the dye concentration, so changes in A track changes in concentration.
(b) Every 60 s, A falls to about 0.75 of its previous value (0.450/0.600 = 0.338/0.450 = 0.253/0.338 = 0.190/0.253 ≈ 0.75). Dropping by the same fraction in equal times means a constant half-life, which is the mark of first order; equivalently, ln A vs t is straight.
(c) k′ = −ln(0.450/0.600) ÷ 60 s = 0.288 ÷ 60 s = 4.79 × 10⁻³ s⁻¹ (the slope of ln A vs t gives the same).
(d) t½ = 0.693 ÷ (4.79 × 10⁻³ s⁻¹) = 145 s.
(e) Doubling [OCl⁻] takes k′ from 4.79 × 10⁻³ to 9.6 × 10⁻³ s⁻¹, about double, so the reaction is first order in OCl⁻. Rate = k[dye][OCl⁻]; k = k′ ÷ [OCl⁻] = 4.79 × 10⁻³ ÷ 0.034 = 0.14 M⁻¹ s⁻¹.
Scoring: (a) A is proportional to concentration; (b) equal fractional drops in equal times (or ln A linear); (c) the value and units of k′; (d) the half-life; (e) first order, the rate law, and the value of k.
Summary
- Method of initial rates: change one concentration at a time and compare rates; in a clock reaction, rate ∝ 1/t.
- Graphing: whichever of [A], ln[A] or 1/[A] vs t is straight gives zero, first or second order; using absorbance in place of concentration leaves the slope unchanged.
- Pseudo-first-order conditions: put one reactant in large excess to find the order in the other; comparing k′ at different excess concentrations gives all the orders, and k = k′ ÷ [B]n.
AP tip: Kinetics lab questions often give a data table and ask you to “show it’s first order.” The fastest way is to point out that the concentration drops by the same fraction in equal time intervals (a constant half-life), then calculate a ratio or two as evidence.
Sources: the data in the simulation, examples and practice are illustrative.