14.5 Integrated Rate Laws
PRE-HEALTH PREP
Integrated Rate Laws
Zero, First & Second Order
General Chemistry 14.5
Lesson Overview: Integrated rate laws are mathematical equations that describe how reactant concentration changes over time for zero-order, first-order, and second-order reactions. Each reaction order has a distinct half-life behavior — constant for first-order, shortening for zero-order, and lengthening for second-order — with the first-order half-life formula (t½ = 0.693/k) being especially important for radioactive decay. Graphical analysis provides an experimental method for determining reaction order by identifying which plot of concentration versus time produces a straight line.
Table of Contents
Which Rate Law Should You Use?
Before diving into integrated rate laws, it is important to understand when to use them. Kinetics problems fall into two broad categories, and the type of data you are given determines which approach is correct.
Integrated Rate Law vs. Differential Rate Law — Decision Framework
| If the problem gives you… | Then use… |
|---|---|
| Concentration vs. time data — how [A] changes as the reaction proceeds | Integrated rate laws (this lesson) |
| Initial rate data from multiple experiments — how the rate changes when starting concentrations are varied | Differential rate law / method of initial rates (previous lesson) |
Exam strategy: Read the problem carefully and identify the data type first. If you see a table of concentrations measured at different time points during a single experiment, you need an integrated rate law. If you see a table of initial rates from separate experiments with different starting concentrations, you need the differential rate law and the method of initial rates. Choosing the wrong approach is one of the most common mistakes on kinetics exams.
What Are Integrated Rate Laws?
Integrated rate laws are mathematical equations that describe how the concentration of a reactant changes over time. That single sentence captures their entire purpose: they connect concentration to time, allowing you to calculate exactly how much reactant remains after any given interval.
The word "integrated" comes from calculus — specifically from solving a differential rate equation by taking an integral. A general chemistry course does not require you to perform that derivation, so we won't work through it here. What matters is that you know how to use these equations and what each term represents.
Every integrated rate law contains four quantities: an initial concentration [A]₀, a final concentration [A], a rate constant k, and elapsed time t. The 3 integrated rate law equations differ only in how those quantities are arranged, and each applies to a different reaction order in chemical kinetics.
| Reaction Order | Integrated Rate Law | Units of k |
|---|---|---|
| Zero-order | [A] = [A]₀ − kt | M·s−1 |
| First-order | ln[A] = ln[A]₀ − kt | s−1 |
| Second-order | 1/[A] = 1/[A]₀ + kt | M−1·s−1 |
Units of the Rate Constant and Reaction Order
As seen in the table above, the units of the rate constant k are different for each reaction order in chemical kinetics. This is not coincidental — a convenient trick lets you read the reaction order directly from the units of k.
Add the exponents on the molarity unit and the time unit, then take the absolute value. The result equals the reaction order.
Trick: Reading Reaction Order from Rate Constant Units
- Zero-order — units: M·s−1. Exponents: +1 and −1. Sum = 0, |0| = 0.
- First-order — units: s−1. Exponents: 0 and −1. Sum = −1, |−1| = 1.
- Second-order — units: M−1·s−1. Exponents: −1 and −1. Sum = −2, |−2| = 2.
Exam tip: Most instructors will simply tell you the reaction order. In the rarer case where you are given only the rate constant and must determine the order yourself, use the unit trick above. Either way, identifying the correct order is always the first step before you do any integrated rate law calculation.
Part 1: Concentration Calculations Using Integrated Rate Laws
Once you know the reaction order and have the appropriate integrated rate law, the calculations are straightforward substitution. The most important thing to watch is unit consistency: the time units in k and in t must match before you plug values into any integrated rate law equation.
Example: Concentration of A After One Minute
For the reaction A → B, given [A]₀ = 0.500 M and k = 0.001 (in appropriate units), calculate [A] after 1 minute (= 60 seconds) for each reaction order.
Zero-Order
[A] = 0.500 M − (0.001 M·s−1)(60 s)
[A] = 0.500 − 0.060 = 0.44 M
First-Order
ln[A] = ln(0.500) − (0.001 s−1)(60 s)
ln[A] = −0.693 − 0.060 = −0.753
[A] = e−0.753 = 0.47 M
Because the first-order integrated rate law gives you ln[A], you must exponentiate both sides to recover [A]. Raise e to the power of the result.
Second-Order
1/[A] = 1/0.500 + (0.001 M−1·s−1)(60 s)
1/[A] = 2.00 + 0.06 = 2.06
[A] = 1/2.06 = 0.49 M
Notice that the 3 reaction orders give slightly different answers even when the rate constants share the same numerical value. The higher the order, the slower the consumption at these concentrations — here, the second-order integrated rate law predicts the least reactant depletion in the first minute.
Unit consistency is critical: If the rate constant is expressed in per-second units, time must be converted to seconds before substituting into the integrated rate law. In the example above, 1 minute was converted to 60 seconds. This is the single most common source of arithmetic errors on chemical kinetics problems.
Part 2: Half-Lives and Reaction Order
What Is a Half-Life?
The half-life (t½) of a reaction is the time required for the concentration of a reactant to decrease to half its current value. Starting from 10 M, one half-life brings you to 5 M; another half-life brings you to 2.5 M; and so on. Half-life is one of the most tested concepts in chemical kinetics.
How Half-Life Depends on Reaction Order
The most important distinction among the 3 reaction orders is whether the half-life depends on concentration. The table below summarizes how each integrated rate law predicts a different half-life behavior.
| Order | Half-Life Expression | Depends on [A]₀? | As Reaction Proceeds… |
|---|---|---|---|
| Zero-order | t½ = [A]₀ / 2k | Yes — directly proportional | Half-life gets shorter |
| First-order | t½ = 0.693 / k | No — independent | Half-life stays constant |
| Second-order | t½ = 1 / (k[A]₀) | Yes — inversely proportional | Half-life gets longer |
For zero-order reactions, a higher starting concentration means a longer half-life. As the concentration drops, successive half-lives become shorter. The opposite is true for second-order reactions: the half-life lengthens as concentration decreases, because the inverse relationship amplifies the effect.
First-order stands apart from both. Its half-life is entirely independent of concentration and depends only on the rate constant k. This constant half-life is the hallmark of first-order kinetics and the defining feature of radioactive decay.
Carbon-14 Dating: First-Order Kinetics in Action
One of the most well-known practical applications of first-order kinetics is radiocarbon dating. Carbon-14 (14C) is a radioactive isotope that undergoes first-order decay with a half-life of 5,730 years. While an organism is alive, it continuously exchanges carbon with its environment, maintaining a roughly constant ratio of 14C to stable 12C. Once the organism dies, the 14C begins to decay without being replenished.
By measuring the ratio of 14C to 12C remaining in an organic sample — and comparing it to the ratio found in living organisms — scientists can calculate how many half-lives have elapsed and therefore determine the age of archaeological and geological specimens. Because 14C decay is first-order, its half-life remains constant regardless of how much carbon-14 was originally present, which is precisely what makes this dating technique reliable. The same integrated rate law equations covered in this lesson are the mathematical foundation behind radiocarbon dating.
Half-Life Behavior by Reaction Order
The half life of a reactant is the time it takes until only half of the reactant remains. From the above table it can be seen that the half life for a Zero order or 2nd order reaction depends upon the initial concentration of the reactant. For a Zero order reactant, the half life becomes shorter and shorter over time. For a 2nd order reactant the half life becomes longer and longer over time. It is only for a 1st order reactant that the half life is independent of the initial reactant concentration which means that it will remain constant over time.
Radioactive decay proceeds according to 1st order kinetics and each radioactive nuclide has a characteristic half life. Had radioactive decay been Zero or 2nd order this would not have been the case as the half life of a nuclide would have been dependent upon how much one initially had.
Take for example francium-223 which has a half life of only 22 minutes. If you started with a 100g sample then only 50g of francium-223 would remain after 22 minutes. After an additional 22 minutes only 25g would remain, and after an additional 22 minutes only 12.5g would remain, etc.
Deriving the First-Order Half-Life Formula
The first-order integrated rate law can be rearranged to:
At the half-life, by definition, [A] = [A]₀/2. Substituting into the first-order integrated rate law:
ln(1/2) = −k·t½
−0.693 = −k·t½
t½ = 0.693 / k
The two negatives cancel cleanly, yielding a positive expression for the half-life. Because [A]₀ cancels out entirely, the result confirms that no concentration term appears in the first-order half-life formula.
Why first-order half-lives matter most: Radioactive decay is universally first-order. Whenever a problem involves nuclear decay, carbon-14 dating, or any radioisotope, you are almost certainly working with a constant half-life and the first-order half-life formula t½ = 0.693/k. The chapter on nuclear reactions later in this course returns to these same integrated rate law equations in that context.
4 Half-Life Calculation Examples
Example 1: Successive Half-Lives Starting from 64 g
A radioactive substance begins at 64 g. Tracking successive half-lives shows exponential decay, the same pattern described by the first-order integrated rate law:
| Half-Lives Elapsed | Mass (g) | Fraction Remaining | Percent Remaining |
|---|---|---|---|
| 0 | 64 | 1 | 100% |
| 1 | 32 | 1/2 | 50% |
| 2 | 16 | 1/4 | 25% |
| 3 | 8 | 1/8 | 12.5% |
| 4 | 4 | 1/16 | 6.25% |
The amount remaining after n half-lives is always the initial amount divided by 2n.
Common misconception — constant half-life does not mean constant rate: During the first half-life, 32 g reacted; during the second, only 16 g; during the third, only 8 g. The reaction is slowing down throughout. A constant half-life reflects the mathematics of exponential decay, not a constant reaction rate. Only zero-order reactions proceed at a truly constant rate.
Example 2: How Much Remains After 30 Minutes? (Half-life = 10 min)
30 minutes ÷ 10 minutes per half-life = 3 half-lives.
Starting from 64 g: 64 → 32 → 16 → 8 g remaining.
Example 3: Finding the Half-Life from the Rate Constant
If k = 0.0693 min−1, use the first-order half-life formula to find t½:
An instructor might withhold the half-life and provide only k, expecting you to derive it before counting half-lives. Always check which quantity you are given.
Example 4: When Time Is Not a Whole Number of Half-Lives
How much of the 64 g sample remains after 27 minutes (half-life = 10 min)?
27 minutes falls between the second half-life (20 min, 16 g) and the third (30 min, 8 g), so simple halving will not give an exact answer. You must return to the first-order integrated rate law directly:
Substitute the initial amount, the rate constant, and t = 27 min to calculate the exact remaining concentration. This is the same procedure as the concentration calculations in Part 1 — the half-life shortcut only works when t is an exact multiple of t½.
Part 3: 3 Graphical Relationships for Determining Reaction Order
Each integrated rate law can be cast into the slope-intercept form of a line, y = mx + b, by choosing the right function of concentration to plot on the y-axis. This is extremely useful in chemical kinetics: if you plot the correct y-variable against time, you get a straight line whose slope equals ±k and whose y-intercept equals the initial value of that variable.
Time is always the independent variable and belongs on the x-axis. The function of concentration is always the dependent variable on the y-axis. Choosing the correct plot is how you determine reaction order experimentally from concentration vs. time data.
Zero-Order: [A] versus t
For a zero-order reaction, plot [A] directly on the y-axis against time on the x-axis. A straight-line concentration vs. time graph is the diagnostic signature of zero-order kinetics.
- y-axis: [A]
- x-axis: t
- Slope: −k (negative — the line slopes downward)
- y-intercept: [A]₀
Only a zero-order reaction produces a straight line when [A] is plotted versus time. A first-order or second-order reaction would produce a curve in this plot.
First-Order: ln[A] versus t
For a first-order reaction, plot ln[A] (the natural logarithm of concentration) on the y-axis. A straight line on this ln concentration vs. time graph confirms first-order kinetics.
- y-axis: ln[A]
- x-axis: t (lowercase t — not capital T, which denotes temperature)
- Slope: −k (negative)
- y-intercept: ln[A]₀
This plot has direct experimental application. If you know a reaction is first-order, measure [A] at several time points, compute ln[A] for each, and fit a straight line to the scatter plot. The slope of that line equals −k, giving you the rate constant without any additional calculation.
Watch the axis label: Always write the x-axis as lowercase t (time), never capital T (temperature). Some instructors test this distinction directly by including a graph with a capital T as a deliberate wrong-answer choice.
Second-Order: 1/[A] versus t
For a second-order reaction, plot 1/[A] (the reciprocal of concentration) on the y-axis. Only a second-order reaction yields a straight line on this reciprocal concentration vs. time graph.
- y-axis: 1/[A]
- x-axis: t
- Slope: +k (positive — the line slopes upward)
- y-intercept: 1/[A]₀
The positive slope is the key distinguishing feature of the second-order plot. Both zero-order and first-order plots have negative slopes; only the second-order plot has a positive slope.
| Order | y-axis Variable | x-axis Variable | Slope | y-intercept |
|---|---|---|---|---|
| Zero | [A] | t | −k | [A]₀ |
| First | ln[A] | t | −k | ln[A]₀ |
| Second | 1/[A] | t | +k | 1/[A]₀ |
Using Graphs to Determine Reaction Order
Graphical analysis is a powerful experimental method for identifying reaction order from concentration vs. time data. For any given reaction, exactly one of the 3 plots will produce a straight line; the other two will produce curves. The plot that linearizes the data reveals the reaction order.
In practice: collect concentration-versus-time data, then construct all three plots — [A] vs. t, ln[A] vs. t, and 1/[A] vs. t. The plot that yields a straight line identifies the order, and its slope gives ±k directly, allowing you to determine the rate constant for the reaction.
Example: Identifying a Second-Order Reaction from a Graph
Suppose you are given four graphs and asked which one corresponds to a second-order reactant. The correct answer is a plot of 1/[A] versus t producing a straight line with a positive slope. Common wrong-answer traps include:
- A plot of 1/[A] vs. t with a negative slope (wrong sign — the second-order integrated rate law always gives a positive slope here)
- A plot of ln[A] vs. t as a straight line (that indicates first-order kinetics, not second-order)
- A plot of [A] vs. t as a straight line (that indicates zero-order kinetics)
Exam trap — the negative-slope decoy: A common wrong-answer choice for second-order problems is a 1/[A] versus t plot with a negative slope. Remember: the second-order slope is positive (+k). A negative slope on that plot would be physically impossible for a standard second-order reaction.
8 Key Takeaways: Integrated Rate Laws
- Integrated rate laws connect concentration to time. The 3 forms — zero-order ([A] = [A]₀ − kt), first-order (ln[A] = ln[A]₀ − kt), and second-order (1/[A] = 1/[A]₀ + kt) — are each used for calculations once the reaction order is known.
- The units of k reveal the reaction order. Add the exponents on the molarity and time units of k, take the absolute value, and the result equals the order (0, 1, or 2).
- Match time units to the rate constant. Before substituting into any integrated rate law, convert the elapsed time to the same unit used in k (e.g., seconds if k is in s−1).
- The first-order half-life formula is t½ = 0.693/k — constant and independent of concentration. Zero-order half-lives shorten over time; second-order half-lives lengthen.
- Radioactive decay is always first-order, which is why half-lives are the standard way to characterize radioisotopes and why the half-life remains constant throughout radioactive decay.
- When time is not a whole multiple of the half-life, the simple halving method fails. Use the first-order integrated rate law directly to find the exact remaining amount.
- Each reaction order linearizes a different concentration vs. time graph: [A] vs. t (zero-order), ln[A] vs. t (first-order), or 1/[A] vs. t (second-order). The plot that yields a straight line identifies the reaction order, and its slope gives ±k.
- The second-order graph is the only one with a positive slope (+k). Zero- and first-order plots both have negative slopes (−k). This sign difference is a frequent exam question in general chemistry chemical kinetics.