AP Chemistry Unit 5 Kinetics: Building and Using Rate Laws

Reaction Rates

What a reaction rate is

A reaction rate describes how fast reactants are being consumed or products are being formed. In chemistry you rarely care only that a reaction can happen—you care whether it happens on a useful time scale. A reaction that is thermodynamically favorable might still be useless if it is so slow that nothing noticeable happens.

Because rates are based on changing amounts, we usually measure rate using concentration (molarity) and time. Conceptually:

• If a reactant’s concentration drops quickly, the reaction is fast.
• If it barely changes over minutes or hours, the reaction is slow.

A common way to define rate for a reactant $A$ is the rate of disappearance:

$\text{rate} = -\frac{\Delta[A]}{\Delta t}$

The negative sign matters: $\Delta[A]$ is negative for a reactant (it decreases), but we want “rate” reported as a positive quantity.

For a product $P$, the rate of appearance is:

$\text{rate} = \frac{\Delta[P]}{\Delta t}$

Why the definition matters (and what AP likes to test)

Those signs and definitions aren’t just bookkeeping. On exams and in labs, you often infer rate from a graph or a table of concentrations. If you don’t handle signs correctly, you can get a negative “rate” and think you did something wrong—or worse, you might keep the negative and carry it through calculations.

Average rate vs instantaneous rate

Rates can change over time as reactants are used up. That means you must be clear whether you’re calculating an average rate over a time interval or the instantaneous rate at a particular moment.

• Average rate uses a finite interval (two data points):

$\text{average rate} = -\frac{\Delta[A]}{\Delta t} = -\frac{[A]_2-[A]_1}{t_2-t_1}$

• Instantaneous rate is the slope of the tangent line to a concentration-time curve at a specific time. Formally it’s a derivative:

$\text{instantaneous rate} = -\frac{d[A]}{dt}$

On AP-style questions, “initial rate” is especially important: it’s the instantaneous rate right at the start (near $t=0$). We like initial rate because concentrations are known accurately at the beginning and reverse reactions or side effects are often minimal.

Units of rate

Rate is “concentration per time,” so common units are:

• $\text{M/s}$
• $\text{M/min}$

Be consistent. If time is in seconds, keep it in seconds unless you intentionally convert.

Stoichiometry links the rates of different species

For a general reaction:

$aA + bB \rightarrow cC + dD$

The species concentrations do not change at the same numerical rate because stoichiometric coefficients tell you how many moles are consumed/formed together. The “single reaction rate” can be defined so that all species give the same value:

$\text{rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$

This matters when you’re given concentration-time data for one species but asked about another.

Worked example 1: converting rates using stoichiometry

For the reaction:

$2NO_2(g) \rightarrow 2NO(g) + O_2(g)$

Suppose $[NO_2]$ decreases at $0.040\ \text{M/s}$ at some moment.

1. Interpret the statement as a rate of disappearance:

$-\frac{d[NO_2]}{dt} = 0.040\ \text{M/s}$

2. Use stoichiometric relationships:

Because coefficients for $NO_2$ and $NO$ are both 2, their concentration rates match in magnitude:

$\frac{d[NO]}{dt} = 0.040\ \text{M/s}$

For $O_2$ the coefficient is 1, so it forms half as fast as $NO_2$ disappears:

$\frac{d[O_2]}{dt} = 0.020\ \text{M/s}$

Common pitfall: forgetting to divide by coefficients and reporting $0.040\ \text{M/s}$ for $O_2$.

How rates are measured in practice

In real experiments you often can’t directly measure concentrations every second. Instead, you measure something that correlates with concentration:

• Gas production (volume or pressure changes)
• Color intensity (spectrophotometry; absorbance relates to concentration)
• Mass change (if gas escapes)
• pH change (for acid/base reactions)

The idea is always the same: track a measurable signal over time and convert that signal into concentration, then compute slopes.

Exam Focus

• Typical question patterns
  • Given a concentration-time table, calculate an average rate over a time interval.
  • Given a graph of $[A]$ vs time, estimate the initial rate (slope near $t=0$) or an instantaneous rate at a time.
  • Use stoichiometric coefficients to relate $\frac{d[A]}{dt}$ to $\frac{d[P]}{dt}$.
• Common mistakes
  • Forgetting the negative sign for reactant disappearance (or reporting a negative rate).
  • Mixing up “rate of reaction” with “rate of appearance/disappearance” without accounting for coefficients.
  • Using inconsistent time units (minutes in one step, seconds in the next).

Introduction to Rate Law

What a rate law is

A rate law is an experimentally determined equation that links the reaction rate to reactant concentrations. It answers a key kinetics question: If I change the concentration of a reactant, how does the reaction rate respond?

A general rate law for reactants $A$ and $B$ looks like:

$\text{rate} = k[A]^m[B]^n$

Here:

• $\text{rate}$ is the reaction rate (units like $\text{M/s}$).
• $k$ is the rate constant, a proportionality constant that depends on temperature (and the specific reaction).
• $m$ and $n$ are reaction orders with respect to each reactant.

A critical idea: reaction orders are not taken from the balanced equation (except in special cases you are not expected to assume on AP). They come from experiment.

Why rate laws matter

Rate laws are the bridge between observation and explanation.

• Practically: they let you predict how changing concentrations changes speed.
• Conceptually: the form of the rate law gives clues about what is happening at the molecular level (often through a “rate-determining step” idea, which you may see later in kinetics).

Reaction order and what it means

The order with respect to a reactant is the exponent on that reactant concentration in the rate law.

• If $m=1$, doubling $[A]$ doubles the rate.
• If $m=2$, doubling $[A]$ increases the rate by a factor of 4.
• If $m=0$, changing $[A]$ does not change the rate (rate is independent of $[A]$).

The overall reaction order is the sum of exponents:

$\text{overall order} = m+n$

Overall order is useful for identifying which integrated rate-law model might fit data and for tracking units of $k$.

Units of the rate constant $k$

Because $\text{rate}$ has units of $\text{M/time}$, the units of $k$ depend on the overall order.

Start from:

$\text{rate} = k[A]^m[B]^n$

Rearrange conceptually:

$k = \frac{\text{rate}}{[A]^m[B]^n}$

If overall order is 1, $k$ has units of $1/\text{time}$ (like $\text{s}^{-1}$). If overall order is 2, $k$ has units of $1/(\text{M}\cdot\text{time})$.

A helpful quick table:

Common mistake: treating $k$ as always having units of $\text{s}^{-1}$. It does not.

How rate laws are determined: method of initial rates

Because rate changes over time, experiments often focus on initial rates. You run multiple trials with different starting concentrations and measure the initial rate in each trial.

The strategy:

1. Compare two trials where only one reactant concentration changes.
2. Form a ratio of rates.
3. Use exponent rules to solve for the order.

If $[A]$ changes and $[B]$ stays constant, then:

$\frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^m$

You can then solve for $m$.

Worked example 2: finding a rate law from initial rate data

A reaction has the form $A + B \rightarrow \text{products}$. The data are:

Step 1: Determine order in $A$. Compare trials 1 and 2 (same $[B]$).

$\frac{\text{rate}_2}{\text{rate}_1} = \frac{0.040}{0.020} = 2$

$\frac{[A]_2}{[A]_1} = \frac{0.20}{0.10} = 2$

So:

$2 = 2^m$

Thus:

$m = 1$

Step 2: Determine order in $B$. Compare trials 2 and 3 (same $[A]$).

$\frac{\text{rate}_3}{\text{rate}_2} = \frac{0.160}{0.040} = 4$

$\frac{[B]_3}{[B]_2} = \frac{0.20}{0.10} = 2$

So:

$4 = 2^n$

Thus:

$n = 2$

Step 3: Write the rate law.

$\text{rate} = k[A]^1[B]^2$

Step 4: Solve for $k$ using any trial. Using trial 1:

$0.020 = k(0.10)(0.10)^2$

$0.020 = k(0.10)(0.01)$

$0.020 = k(0.001)$

$k = 20$

Now check units: overall order is $1+2=3$, so $k$ units should be:

$\frac{\text{M/s}}{\text{M}^3} = \text{M}^{-2}\text{s}^{-1}$

So you would report $k = 20\ \text{M}^{-2}\text{s}^{-1}$.

Common pitfalls here:

• Using the balanced equation coefficients as exponents (not allowed unless told it’s an elementary step).
• Comparing trials where both reactant concentrations changed, which makes solving for a single exponent impossible in one step.

What rate laws do (and do not) tell you

A rate law tells you how rate depends on concentrations under the conditions studied. It does not automatically tell you:

• the full mechanism,
• whether the reaction happens in one step,
• or the value of $k$ at a different temperature.

You should treat the rate law as a model supported by data.

Exam Focus

• Typical question patterns
  • Given initial rate data, determine $m$ and $n$ and then calculate $k$.
  • Ask how the rate changes when a concentration is doubled/halved based on reaction order.
  • Determine units of $k$ from the overall order.
• Common mistakes
  • Pulling exponents from the balanced chemical equation instead of experimental data.
  • Forgetting that “doubling” with second order means multiplying rate by 4 (not 2).
  • Solving for $k$ but reporting incorrect units because overall order was not used.

Concentration Changes over Time

Why concentrations change in a predictable way

If you know how rate depends on concentration (the rate law), you can model how concentrations evolve as the reaction runs. Since reactant concentrations typically decrease, the rate often slows down over time. This is why many concentration-time graphs are curved rather than straight lines.

Chemists use integrated rate laws to connect concentration directly to time. On AP Chemistry, the focus is typically on zero-, first-, and second-order reactions (often for a single reactant $A$).

A key idea: you can often determine the reaction order by which concentration-time plot becomes linear.

Zero-order reactions

What it means

A zero-order reaction has a rate independent of reactant concentration:

$\text{rate} = k$

This can happen when something other than concentration limits the rate—often a catalyst surface is saturated, or a reagent is in such large excess that its concentration is effectively constant.

Integrated rate law (zero order)

For a zero-order reaction in $A$:

$[A]_t = [A]_0 - kt$

• $[A]_0$ is the initial concentration.
• $[A]_t$ is concentration at time $t$.
• The graph of $[A]$ vs $t$ is a straight line with slope $-k$.

Units check: if $[A]$ is M and $t$ is s, then $k$ is $\text{M/s}$.

Worked example 3: using a zero-order integrated law

A zero-order process has $k = 0.0050\ \text{M/s}$ and $[A]_0 = 0.200\ \text{M}$. Find $[A]$ after $30.0\ \text{s}$.

Use:

$[A]_t = [A]_0 - kt$

Substitute:

$[A]_t = 0.200 - (0.0050)(30.0)$

$[A]_t = 0.200 - 0.150$

$[A]_t = 0.050\ \text{M}$

Common mistake: using a logarithm form here (that’s for first-order), or using the wrong sign and making concentration increase.

First-order reactions

What it means

A first-order reaction has rate directly proportional to $[A]$:

$\text{rate} = k[A]$

This is extremely common in processes like radioactive decay and many unimolecular decomposition reactions.

The big consequence: first-order reactions have an exponential concentration decrease over time.

Integrated rate law (first order)

There are equivalent ways to write it; the most used forms are:

$\ln[A]_t = \ln[A]_0 - kt$

and

$\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$

A linear plot test:

• Plot $\ln[A]$ vs $t$.
• If it’s linear, the reaction is first order.
• The slope is $-k$.

Units: for first order, $k$ has units $\text{s}^{-1}$ (or $\text{min}^{-1}$, etc.).

Half-life for first-order reactions

The half-life $t_{1/2}$ is the time for the concentration to fall to half its initial value.

For first-order reactions:

$t_{1/2} = \frac{0.693}{k}$

A distinctive feature: half-life is constant (it does not depend on $[A]_0$) for first-order reactions.

Worked example 4: first-order concentration after time

A first-order reaction has $k = 0.250\ \text{min}^{-1}$ and $[A]_0 = 0.800\ \text{M}$. Find $[A]$ after $6.00\ \text{min}$.

Use:

$\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$

Substitute:

$\ln\left(\frac{[A]_t}{0.800}\right) = -(0.250)(6.00)$

$\ln\left(\frac{[A]_t}{0.800}\right) = -1.50$

Exponentiate both sides:

$\frac{[A]_t}{0.800} = e^{-1.50}$

$[A]_t = 0.800e^{-1.50}$

Numerically:

$[A]_t \approx 0.800(0.223) \approx 0.178\ \text{M}$

Common mistakes:

• Forgetting that natural log is used (not base-10 log).
• Plugging $[A]_t$ and $[A]_0$ into the fraction backwards, which changes the sign.

Second-order reactions

What it means

A second-order reaction (single-reactant version) has:

$\text{rate} = k[A]^2$

Rates become very sensitive to concentration: if $[A]$ doubles, rate increases by a factor of 4.

Integrated rate law (second order, one reactant)

For a second-order reaction in $A$:

$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$

Linear plot test:

• Plot $\frac{1}{[A]}$ vs $t$.
• If it’s linear, the reaction is second order.
• The slope is $k$ (note it’s positive in this linear form).

Units: second-order $k$ has units $\text{M}^{-1}\text{s}^{-1}$.

Worked example 5: determine $k$ from concentration-time data

A reaction is suspected to be second order in $A$. Data show $[A]_0 = 0.500\ \text{M}$ and after $20.0\ \text{s}$, $[A]_t = 0.250\ \text{M}$.

Assuming second order:

$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$

Substitute:

$\frac{1}{0.250} = \frac{1}{0.500} + k(20.0)$

Compute reciprocals:

$4.00 = 2.00 + 20.0k$

Solve:

$2.00 = 20.0k$

$k = 0.100\ \text{M}^{-1}\text{s}^{-1}$

Common mistake: using the first-order log form and getting a different $k$, then not noticing the units don’t match the assumed order.

Determining reaction order from graphical (linear) tests

A very common AP task is: “Given concentration vs time data, determine the reaction order.” The core logic is that different integrated laws produce different linear relationships.

What “linear” means in practice: points fall roughly on a straight line within experimental uncertainty. You may be given a best-fit line or asked which plot “looks most linear.”

Connecting concentration-time behavior back to rate laws

It’s easy to memorize three integrated forms without understanding them. A deeper way to think about it:

• If rate does not depend on concentration (zero order), concentration drops at a constant rate, so $[A]$ vs $t$ is a straight line.
• If rate depends on $[A]$ (first order), the reaction slows proportionally as $[A]$ decreases, producing exponential decay and a linear $\ln[A]$ vs $t$ plot.
• If rate depends on $[A]^2$ (second order), rate decreases even more dramatically as $[A]$ falls; the reciprocal relationship linearizes it.

Common “gotchas” with concentration changes

• Using the wrong concentration variable: integrated laws for a single-reactant form assume the rate law depends on just $[A]$ (or that other reactants are constant/absorbed into an effective $k$). If multiple reactants change significantly, these simple forms may not apply.
• Treating half-life as always constant: constant half-life is a first-order signature, not a universal rule.
• Slope sign errors: for first-order, the slope of $\ln[A]$ vs $t$ is negative; for second-order, the slope of $\frac{1}{[A]}$ vs $t$ is positive.

Exam Focus

• Typical question patterns
  • Choose which plot ( $[A]$ vs $t$, $\ln[A]$ vs $t$, or $\frac{1}{[A]}$ vs $t$ ) is linear to identify reaction order.
  • Use a linear plot’s slope to calculate $k$ (and report correct units).
  • Use an integrated rate law to solve for $[A]_t$, $t$, or $k$ given the other quantities.
• Common mistakes
  • Mixing the three integrated laws (for example, using $\ln$ for a zero-order situation).
  • Calculating $k$ from a slope but missing the negative sign for first-order plots.
  • Reporting a numerically correct $k$ with incorrect units, signaling the wrong assumed order.