Unit 5: Kinetics

Measuring and Interpreting Reaction Rates

Chemistry is not only about what products form, but also how fast they form. Kinetics is the study of reaction rates and the factors that affect them. You use kinetics to answer practical questions like: Why does food spoil faster when warm? Why does powdered sugar react faster than a sugar cube? Why do catalysts make industrial processes economical?

What a reaction rate means

A reaction rate tells you how quickly reactants are consumed or products are formed. Rates are based on change in concentration over time, so typical units are:

\text{mol L}^{-1}\text{s}^{-1}

(often written as:

\text{M s}^{-1}

).

There are two closely related ideas. An average rate is the change in concentration over a time interval divided by the time elapsed. An instantaneous rate is the slope of the tangent line to a concentration vs. time curve at a specific moment.

Rate of disappearance vs. rate of appearance (and why negatives show up)

If a reactant is being used up, its concentration decreases, so the change in concentration is negative. To keep “rate” as a positive quantity, chemists define:

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

for the rate of disappearance of A, and:

\text{rate} = \frac{\Delta[B]}{\Delta t}

for the rate of appearance of B.

The negative sign is not a “math trick” you can ignore; it encodes a real physical idea: reactants go down.

Relating rates using stoichiometry

For a reaction:

aA + bB \rightarrow cC + dD

the reaction rate is consistent no matter which species you monitor, as long as you account for coefficients:

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

This matters because an experiment might measure only one substance (for example, a colored species in a spectrophotometer or a gas pressure change), and you may need the rate for another substance.

How rates are measured in real experiments

In practice, you rarely “count molecules.” You monitor something that changes as concentration changes:

• Color intensity (spectrophotometry): if a reactant or product is colored, absorbance tracks concentration.
• Gas pressure or volume: for reactions that produce or consume a gas.
• Mass change: if gas escapes from an open container.
• pH or conductivity: if ions are produced or consumed.

A common misconception is thinking rate must be constant. For many reactions, rate decreases over time because reactant concentrations drop, reducing collision frequency.

Worked example 1: average rate from concentration data

A decomposes to products. Concentration drops from 0.80 M to 0.62 M in 50.0 s. Find the average rate of disappearance of A.

Use:

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

Compute the change:

\Delta[A] = 0.62 - 0.80 = -0.18

Then:

\text{rate} = -\frac{-0.18}{50.0} = 0.0036

So the average rate is:

0.0036\ \text{M s}^{-1}

Worked example 2: stoichiometric rate relationship

For:

2NO_2 \rightarrow 2NO + O_2

If the rate of disappearance of NO_2 is:

0.040\ \text{M s}^{-1}

what is the rate of appearance of O_2?

Use:

\text{rate} = -\frac{1}{2}\frac{\Delta[NO_2]}{\Delta t} = \frac{\Delta[O_2]}{\Delta t}

Given:

-\frac{\Delta[NO_2]}{\Delta t} = 0.040

So:

\text{rate} = \frac{1}{2}(0.040) = 0.020

Thus:

\frac{\Delta[O_2]}{\Delta t} = 0.020\ \text{M s}^{-1}

Exam Focus

• Typical question patterns:
  • Compute an average rate from a table of concentration vs. time.
  • Convert a measured disappearance rate to an appearance rate using stoichiometric coefficients.
  • Interpret the slope of a concentration vs. time graph (steeper slope means faster rate).
• Common mistakes:
  • Forgetting the negative sign for disappearance rates (leading to negative “rates”).
  • Ignoring coefficients when relating rates of different species.
  • Using initial concentration instead of the change in concentration over the specified interval.

Rate Laws and Reaction Order (What Controls the Speed)

A rate law is an experimentally determined relationship that connects reaction rate to reactant concentrations. It tells you which reactants matter for speed and how strongly they matter. Determining a rate law requires experimental data, typically using initial concentrations and initial rates.

What a rate law looks like

For reactants A and B:

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

More generally (for A, B, and C):

\text{rate} = k[A]^x[B]^y[C]^z

• k is the rate constant.
• The exponents are the reaction orders with respect to each reactant.
• The overall order is the sum of the exponents.

A key idea: orders are not taken from the balanced equation (unless the reaction is an elementary step in a mechanism). You are expected to determine orders from data.

What controls the value of the rate constant

The rate constant k depends on the specific reaction and conditions. In particular, it depends strongly on **temperature** and **activation energy**, as described quantitatively by the Arrhenius equation later in this guide. Introducing a catalyst also changes k (by changing the effective activation energy).

Interpreting reaction order

Order tells you how rate responds when concentration changes.

• First order in A: doubling [A] doubles the rate.
• Second order in A: doubling [A] quadruples the rate.
• Zero order in A: changing [A] does not change the rate.

Zero order can happen when something else is limiting, such as saturation of a catalyst surface or a mechanism where the slow step does not involve that reactant.

Units of the rate constant

Because rate has units:

\text{M s}^{-1}

but concentration terms can contribute different powers of M, the units of k depend on overall order. In general:

[k] = \text{M}^{1-\text{overall order}}\text{s}^{-1}

Examples:

• Overall order 0:

[k] = \text{M s}^{-1}

• Overall order 1:

[k] = \text{s}^{-1}

• Overall order 2:

[k] = \text{M}^{-1}\text{s}^{-1}

Determining a rate law: method of initial rates

The method of initial rates compares multiple experiments where initial concentrations are varied and the initial rate is measured. The strategy is:

1. Assume a general form such as:

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

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

This works because if [B] is constant between two trials, its factor cancels when you form a ratio.

Worked example 1: finding orders and k (two-reactant data)

Data for reaction between A and B:

Assume:

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

Find m by comparing Trials 1 and 2 (B constant):

\frac{0.080}{0.020} = 4

\frac{0.20}{0.10} = 2

So:

4 = 2^m

m = 2

Find n by comparing Trials 1 and 3 (A constant):

\frac{0.040}{0.020} = 2

\frac{0.20}{0.10} = 2

So:

2 = 2^n

n = 1

Rate law:

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

Find k using Trial 1:

0.020 = k(0.10)^2(0.10)

(0.10)^2(0.10) = 0.0010

k = \frac{0.020}{0.0010} = 20

Overall order is 3, so:

[k] = \text{M}^{-2}\text{s}^{-1}

Worked example 2: initial concentrations with three reactants

Consider:

A + 2B + C \rightarrow D

Experimental data:

Assume:

\text{rate} = k[A]^x[B]^y[C]^z

Order in A: Compare experiments 3 and 4. Only [A] changes (0.1 to 0.2), while [B] and [C] stay the same. The rate changes from 0.02 to 0.08 (a factor of 4). Doubling [A] causes the rate to quadruple:

2^x = 4

x = 2

So the reaction is second order with respect to A.

Order in B: Compare experiments 1 and 3. Only [B] doubles (0.1 to 0.2), and the rate doubles (0.01 to 0.02):

2^y = 2

y = 1

So the reaction is first order with respect to B.

Order in C: Compare experiments 1 and 2. [C] doubles (0.1 to 0.2) but the rate stays the same, so:

z = 0

Rate law:

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

The overall order is 3.

Find k: Any experiment can be used. Using experiment 3:

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

k = \frac{0.02}{(0.1)^2(0.2)}

k = 10\ \text{M}^{-2}\text{s}^{-1}

Exam Focus

• Typical question patterns:
  • Use initial-rate data to determine reaction orders and write the rate law.
  • Compute k (including correct units) once the rate law is known.
  • Predict how the rate changes when one or more reactant concentrations are changed.
• Common mistakes:
  • Assuming orders match stoichiometric coefficients from the balanced equation.
  • Comparing trials where more than one concentration changes (making exponents impossible to isolate).
  • Forgetting to include units for k or giving units that don’t match overall order.

Concentration–Time Relationships (Integrated Rate Laws)

A differential rate law tells you how rate depends on concentration at a moment. An integrated rate law tells you how concentration changes over time. AP Chemistry focuses on integrated laws for zero-, first-, and second-order reactions.

Zero-order integrated rate law

Zero-order means the rate does not depend on reactant concentration:

\text{rate} = k

For a reactant A:

[A]_t = [A]_0 - kt

A plot of concentration vs. time is linear.

First-order integrated rate law

First-order means:

\text{rate} = k[A]

Integrated form:

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

Equivalent exponential form:

[A]_t = [A]_0 e^{-kt}

A plot of \ln[A] vs. time is linear, and you can interpret it with a slope-intercept mindset: the slope is -k and the y-intercept is \ln[A]_0.

Second-order integrated rate law

A common second-order case is:

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

Integrated form:

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

A plot of \frac{1}{[A]} vs. time is linear, with slope k and y-intercept \frac{1}{[A]_0}.

Why these forms matter

Each order has a “signature” transformation that creates a straight line when graphed. Also notice how k appears:

• Zero order: concentration changes by k times time.
• First order: the logarithm changes by k times time.
• Second order: the reciprocal changes by k times time.

This is also why the units of k depend on the overall order.

Worked example 1: first-order concentration after time

A first-order reaction has:

k = 0.025\ \text{s}^{-1}

If:

[A]_0 = 0.80\ \text{M}

find [A]_t after 40.0 s.

Use:

[A]_t = [A]_0 e^{-kt}

Compute:

-kt = -(0.025)(40.0) = -1.0

Then:

[A]_t = (0.80)e^{-1.0}

Using:

e^{-1.0} \approx 0.368

So:

[A]_t \approx 0.80 \times 0.368 = 0.294

Thus:

[A]_t \approx 0.29\ \text{M}

Worked example 2: second-order time to reach a concentration

A follows:

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

with:

k = 0.50\ \text{M}^{-1}\text{s}^{-1}

If:

[A]_0 = 0.40\ \text{M}

how long until:

[A]_t = 0.10\ \text{M}

Use:

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

Plug in:

\frac{1}{0.10} = \frac{1}{0.40} + (0.50)t

Compute:

10.0 = 2.5 + 0.50t

Solve:

7.5 = 0.50t

t = 15\ \text{s}

Exam Focus

• Typical question patterns:
  • Use an integrated rate law to compute concentration after a given time.
  • Solve for time needed to reach a certain concentration.
  • Identify which integrated form matches given data (often via linear graph reasoning).
• Common mistakes:
  • Mixing up which expression is linear for which order.
  • Forgetting that the second-order integrated law shown is for \text{rate} = k[A]^2 specifically.
  • Plugging concentrations into logarithms without consistent units.

Graphical Analysis and Half-Life

Integrated rate laws correspond to straight-line graphs once you plot the right transformation. Graphing turns “curvy” concentration-time data into a linear relationship with interpretable slope and intercept.

Linear plots that diagnose reaction order

For a reactant A:

• Zero order: plot [A] vs. t is linear.

  • Slope is -k.
  • Intercept is [A]_0.

• First order: plot \ln[A] vs. t is linear.

  • Slope is -k.
  • Intercept is \ln[A]_0.

• Second order: plot \frac{1}{[A]} vs. t is linear.

  • Slope is k.
  • Intercept is \frac{1}{[A]_0}.

A common pitfall is assuming the raw [A] vs. t plot “looks linear enough,” so it must be zero order. Many reactions are only approximately linear over short time windows.

Half-life: the time to reduce concentration by half

The half-life t_{1/2} is the time required for a reactant concentration to drop to half its current value.

A helpful way to visualize repeated half-lives:

Half-life can also be determined graphically by finding the time it takes for the concentration (or amount) to fall to half its starting value, then repeating that step for successive halves.

First-order half-life (memorize this one)

For first-order reactions, the half-life is constant and independent of [A]_0:

t_{1/2} = \frac{\ln 2}{k}

This is the “half-life equation” that only works in this simple form for first-order kinetics. For example, if after 30 s, 50% of a substance has decayed, then after 30 more seconds, 25% remains; 30 more seconds, 12.5% remains; and so on.

Zero-order half-life

From:

[A]_t = [A]_0 - kt

At half-life:

[A]_t = \frac{1}{2}[A]_0

So:

t_{1/2} = \frac{[A]_0}{2k}

Second-order half-life

From:

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

At half-life:

[A]_t = \frac{1}{2}[A]_0

So:

t_{1/2} = \frac{1}{k[A]_0}

Using “constant half-life” as evidence for first order

If successive half-lives are equal (for example, it takes 10 s to go from 1.0 M to 0.50 M and another 10 s to go from 0.50 M to 0.25 M), that pattern strongly suggests first-order behavior.

Worked example 1: finding k from a half-life

A first-order reaction has:

t_{1/2} = 28.0\ \text{s}

Find k.

Use:

t_{1/2} = \frac{\ln 2}{k}

So:

k = \frac{\ln 2}{t_{1/2}}

k = \frac{0.693}{28.0} = 0.0248\ \text{s}^{-1}

Worked example 2: identify order from graph description

If plotting \ln[A] vs. time produces a straight line with negative slope, while plotting [A] vs. time is curved, that indicates first-order kinetics.

Exam Focus

• Typical question patterns:
  • Choose which plot ( [A] vs t, \ln[A] vs t, or 1/[A] vs t ) is linear for a given order.
  • Use slope of a linearized plot to determine k.
  • Use half-life behavior to infer reaction order or calculate k.
• Common mistakes:
  • Forgetting the sign of the slope (zero and first order have slope -k; second order has slope k).
  • Using the first-order half-life equation for non-first-order reactions.
  • Confusing “linear concentration decay” (zero order) with “linearized plot is linear” (all orders can be linear after transformation).

The Molecular Collision Model (Why Rates Change)

A rate law describes what happens; the molecular collision model (collision theory) helps you understand why rates depend on concentration, temperature, and physical state.

What the collision model says

For molecules to react, they generally must:

1. Collide
2. Collide with enough energy to overcome an energy barrier (activation energy)
3. Collide with the right orientation so bonds can break and form

Most collisions do not lead to reaction because they are not energetic enough or not properly oriented.

Why concentration or pressure increases rate

Increasing concentration (or pressure for gases) increases the number of particles in a given volume, increasing collision frequency. More collisions per second usually means more successful collisions per second, so rate increases. However, the exact concentration dependence must be determined experimentally (the reaction orders).

Why surface area matters (especially for solids)

If a reactant is a solid, only particles at the surface can collide with other reactants. Crushing a solid increases surface area, exposing more particles and increasing collision opportunities. A typical misconception is thinking “surface area changes concentration.” For a pure solid, concentration is not included in the rate law the same way; surface area affects effective contact.

Stirring and mixing

Stirring can sometimes speed up a reaction by improving contact between reactants. If the mixture is heterogeneous (not uniform throughout), stirring mixes it more effectively and can increase rate. If the mixture is homogeneous (uniform throughout), stirring typically does not change the rate because everything is already evenly mixed.

Why temperature increases rate so dramatically

Temperature changes the energy distribution of particles. When temperature rises, average kinetic energy increases and a larger fraction of molecules have energy at or above the activation energy. Temperature also increases the likelihood of sufficiently energetic collisions, which is why rate often increases sharply with temperature.

Worked example (conceptual): explain a rate increase

If a reaction rate doubles when concentration of A doubles (with other conditions constant), that is consistent with first-order dependence on A. Collision-model language would say: doubling [A] roughly doubles the frequency of collisions involving A, so successful collisions happen about twice as often.

Exam Focus

• Typical question patterns:
  • Explain qualitatively how changing concentration, pressure, surface area, stirring, or temperature affects rate.
  • Use particle-level reasoning (collisions, energy, orientation) to justify observed trends.
• Common mistakes:
  • Claiming that all collisions cause reaction (only successful collisions do).
  • Overgeneralizing collision frequency to exact rate laws (orders must come from data).
  • Ignoring orientation and focusing only on collision energy.

Energy Profiles, Activation Energy, and the Arrhenius Equation

Rate depends on whether particles can overcome an energy barrier. Energy diagrams and the Arrhenius equation connect kinetics to energetics and explain temperature effects quantitatively.

Activation energy and transition state

Activation energy is the minimum energy required for reactants to reach the transition state (activated complex). The transition state is not an isolable substance; it is a high-energy arrangement of atoms at the top of the barrier.

Interpreting a reaction energy (potential energy) diagram

A typical energy diagram plots potential energy vs. reaction progress.

• Reactants start at some energy level.
• The curve rises to a peak (the transition state).
• The curve falls to the products energy level.

Two quantities are often confused:

• Activation energy E_a: energy difference between reactants and the peak.
• Overall enthalpy change \Delta H: energy difference between products and reactants.

You can have an exothermic reaction with negative \Delta H but still a large E_a, meaning it can be slow.

Energy profiles also distinguish forward and reverse activation energies. The activation energy for the reverse reaction is the energy difference between products and the peak; it can be different from the forward activation energy.

Arrhenius equation (temperature dependence of k)

The Arrhenius equation relates the rate constant to temperature:

k = Ae^{-E_a/(RT)}

• A is the frequency factor (collision frequency and orientation probability).
• E_a is activation energy.
• R is the gas constant.
• T is temperature in kelvins.

A useful linear form is:

\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A

So a plot of \ln k vs. \frac{1}{T} is linear with slope:

-\frac{E_a}{R}

A two-temperature form commonly used in problems is:

\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

Worked example: using two temperatures to find activation energy

A reaction has:

k_1 = 2.0\times10^{-3}\ \text{s}^{-1}

at:

T_1 = 300\ \text{K}

and:

k_2 = 8.0\times10^{-3}\ \text{s}^{-1}

at:

T_2 = 320\ \text{K}

Find E_a.

Use:

\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

Compute:

\ln\left(\frac{8.0\times10^{-3}}{2.0\times10^{-3}}\right) = \ln(4) = 1.386

Compute:

\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{320} - \frac{1}{300}

\frac{1}{320} = 0.003125

\frac{1}{300} = 0.003333

0.003125 - 0.003333 = -0.000208

Now solve using:

R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}

1.386 = -\frac{E_a}{8.314}(-0.000208)

1.386 = \frac{E_a}{8.314}(0.000208)

E_a = \frac{1.386\times 8.314}{0.000208}

E_a \approx 5.54\times10^{4}\ \text{J mol}^{-1}

So:

E_a \approx 55\ \text{kJ mol}^{-1}

Connecting Arrhenius to the collision model

It’s true that at higher temperature particles move faster, but the key Arrhenius idea is that higher temperature increases the fraction of collisions with energy exceeding E_a. That exponential dependence is why temperature can have an outsized effect on rate.

Exam Focus

• Typical question patterns:
  • Interpret energy diagrams: identify E_a and compare barriers for catalyzed vs. uncatalyzed pathways.
  • Use the Arrhenius equation (often the two-temperature form) to compute E_a or a new k.
  • Explain why a reaction can be thermodynamically favorable but kinetically slow.
• Common mistakes:
  • Confusing \Delta H with E_a on an energy profile.
  • Using Celsius instead of kelvins in Arrhenius calculations.
  • Algebra sign errors in the temperature-difference term and the negative sign in the equation.

Reaction Mechanisms and Elementary Steps

Many reactions do not occur in a single collision event. Instead, they proceed through a sequence called a reaction mechanism.

Mechanism vocabulary

• A reaction mechanism is a proposed sequence of steps that adds up to the overall balanced reaction.
• An elementary step is one step in the mechanism that occurs in a single event at the molecular level.
• An intermediate is formed in one step and consumed in a later step; it cancels out and does not appear in the overall net equation.

Mechanisms matter because they explain rate laws, explain catalysts, and connect kinetics to molecular-level events.

Molecularity of an elementary step

For an elementary step, molecularity is the number of reacting particles in that step:

• Unimolecular: one reactant particle rearranges.
• Bimolecular: two particles collide (even if they are the same species).
• Termolecular: three particles collide simultaneously (rare).

Why elementary steps are special: their rate laws come from their form

For an elementary step, the rate law directly reflects the reactant particles in that step. Examples:

If:

A \rightarrow \text{products}

then:

\text{rate} = k[A]

If:

A + B \rightarrow \text{products}

then:

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

This direct connection is not generally valid for the overall reaction, which is why you cannot write a rate law from the balanced equation unless you are explicitly told the reaction is elementary.

What a rate-determining step means

In a multi-step mechanism, one step is often much slower than the others. The slowest step is the rate-determining step (RDS), and it limits the overall rate (like a bottleneck in an assembly line).

Worked example: intermediates, elementary-step rate laws, and the RDS

Consider the mechanism:

Step I (fast):

A + A \rightleftharpoons X

Step II (slow):

X + B \rightarrow C + Y

Step III (fast):

Y + B \rightarrow D

• X and Y are intermediates because each is created in one step and consumed in a later step.
• Step I is bimolecular (two reactant particles collide, even though they are both A). Step II and Step III are also bimolecular.
• Step II is the rate-determining step, so the initial form of the rate law is:

\text{rate} = k[X][B]

Because X is an intermediate, you often want to express the rate in terms of measurable reactants. In problems that treat Step I as establishing X quickly relative to the slow step, X can be related to reactant concentration; a common AP-style simplification is that [X] is proportional to [A]^2, leading to an overall form consistent with:

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

Multi-step reactions can also be represented on a reaction energy profile with multiple peaks (each peak corresponds to a transition state for an elementary step), and the tallest peak typically corresponds to the slowest step.

Exam Focus

• Typical question patterns:
  • Identify intermediates and catalysts from a proposed mechanism.
  • Determine the overall reaction by adding elementary steps.
  • Decide whether a proposed mechanism is plausible (steps sum to the overall equation; intermediates cancel).
• Common mistakes:
  • Calling a species an intermediate even though it appears in the overall equation.
  • Assuming every mechanism has exactly one slow step (AP problems often do, but real systems can be more complex).
  • Using stoichiometric coefficients from the overall reaction to infer elementary-step molecularity.

Connecting Mechanisms to Rate Laws (and Vice Versa)

Connecting a proposed mechanism to an observed rate law is a core Unit 5 skill. The rate law is a clue about what must be happening in the slow step.

The simplest case: slow step gives the rate law directly

If the slow step is elementary, its reactants determine the rate law.

Example:

Step 1 (slow):

A + B \rightarrow C

Step 2 (fast):

C + B \rightarrow D

Overall reaction:

A + 2B \rightarrow D

Because Step 1 is slow and elementary:

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

The coefficient 2 on B in the overall reaction does not force a squared term in the rate law.

When intermediates appear in the slow-step rate law

If the slow step involves an intermediate, you must eliminate it to match the experimentally measured rate law (which is written using concentrations you can control).

Fast equilibrium followed by slow step (common AP pattern)

Mechanism:

Step 1 (fast equilibrium):

A + B \rightleftharpoons I

Step 2 (slow):

I + B \rightarrow \text{products}

Rate from Step 2:

\text{rate} = k[I][B]

Use a fast-equilibrium relationship:

[I] = K[A][B]

Substitute:

\text{rate} = k(K[A][B])[B]

Combine constants:

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

Problems typically signal when the “fast equilibrium” idea is intended; you shouldn’t invent it without guidance.

Using a rate law to infer mechanistic information

• If rate is independent of [A], A likely is not involved in the slow step.
• If rate is proportional to [B]^2, the slow step might involve two B particles, or it might involve one B and an intermediate whose concentration is proportional to [B].

A rate law does not uniquely determine a mechanism, but it can rule many mechanisms out.

Worked example: check whether a mechanism matches a rate law

Observed rate law:

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

Proposed mechanism:

Step 1 (fast):

B + B \rightleftharpoons I

Step 2 (slow):

A + I \rightarrow \text{products}

Rate from the slow step:

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

From fast equilibrium:

[I] = K[B]^2

Substitute:

\text{rate} = k[A](K[B]^2)

So:

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

This matches, so the mechanism is consistent with the kinetics.

A note on what “consistent” means

Kinetics rarely proves a unique mechanism; it shows consistency. Many different mechanisms can lead to the same rate law.

Exam Focus

• Typical question patterns:
  • Given a mechanism with a slow step, write the rate law and compare to an experimental rate law.
  • Identify the rate-determining step from a mechanism.
  • Determine whether a proposed mechanism is consistent with the observed rate law.
• Common mistakes:
  • Writing the rate law from the overall reaction rather than the slow step.
  • Leaving an intermediate in the final rate law when the question expects a law in terms of measurable reactants.
  • Assuming “fast” means it does not matter at all; fast steps can still shape intermediate relationships.

Catalysts (How They Speed Reactions Without Being Consumed)

A catalyst increases reaction rate without being permanently consumed. Catalysts are central in industry, biology (enzymes), and environmental chemistry.

What catalysts do at the molecular level

Catalysts speed reactions by providing an alternative pathway with a lower activation energy. On an energy diagram, the catalyzed pathway has a lower peak.

• Catalysts change E_a and therefore change k.
• Catalysts do not change the energies of reactants or products.
• Catalysts do not change:

\Delta H

A common misconception is that catalysts make reactions “more exothermic.” They do not; they change kinetics, not thermodynamics.

Catalysts and equilibrium

Catalysts increase the rates of both the forward and reverse reactions by lowering activation energies for both directions. Therefore:

• A catalyst helps the system reach equilibrium faster.
• A catalyst does not change the equilibrium constant.

Homogeneous vs. heterogeneous catalysis

• Homogeneous catalyst: same phase as reactants (often aqueous). The catalyst participates in steps and is regenerated.
• Heterogeneous catalyst: different phase (often a solid). Reactants adsorb onto the surface, react, and desorb.

Because heterogeneous catalysts rely on surface sites, increased surface area usually increases the rate.

Enzymes as biological catalysts

Enzymes (usually proteins) are highly specific catalysts that bind substrates in an active site, orient reactants, and stabilize transition states. They can be sensitive to temperature and pH; denaturation reduces catalytic activity.

Catalysts in mechanisms: they cancel like intermediates, but appear at start and end

In a mechanism, a catalyst is typically present in the beginning, used in an early step, and regenerated in a later step, so it appears on both sides when you add steps and cancels out.

Example overall reaction:

A + B \rightarrow C

Possible catalytic mechanism:

Step I:

A + X \rightarrow Y

Step II:

B + Y \rightarrow C + X

Here, X is the **catalyst** (consumed then regenerated), and Y is an intermediate.

Worked example (conceptual): catalyst effect on rate constant

If temperature is constant but a catalyst is added, the observed rate increases because k increases (the rate law form stays the same, but the numerical value of k changes).

Exam Focus

• Typical question patterns:
  • Compare catalyzed and uncatalyzed energy diagrams (identify lower E_a and unchanged \Delta H).
  • Explain why catalysts do not change equilibrium constants but do change how quickly equilibrium is reached.
  • Describe how heterogeneous catalysts depend on surface area and adsorption.
• Common mistakes:
  • Claiming catalysts increase product yield at equilibrium.
  • Saying catalysts are “used up” (they are regenerated, though they can degrade in real systems).
  • Confusing lowering E_a with lowering reactant energy or changing \Delta H.