Rates, Mechanisms, and Catalysis

Driving question

What evidence reveals how quickly a reaction proceeds, which molecular events control it, and how a catalyst changes the pathway?

Rate is a measured change

For $\mathrm{A\rightarrow products}$, an average disappearance rate is

$-\frac{\Delta[\mathrm A]}{\Delta t}.$

The negative sign makes the reported rate positive when reactant concentration decreases. For a general balanced equation $a\mathrm A\rightarrow p\mathrm P$, a single reaction rate accounts for coefficients:

$-\frac{1}{a}\frac{d[\mathrm A]}{dt}=\frac{1}{p}\frac{d[\mathrm P]}{dt}.$

Rates depend on collision frequency, collision energy, orientation, and the molecular pathway—not merely whether a balanced equation exists.

Rate laws come from experiments

A common empirical law is

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

Orders $m$ and $n$ are determined from controlled data. They are not generally copied from overall equation coefficients.

Trials 1→2 double $[\mathrm A]$ while holding $[\mathrm B]$ constant; rate quadruples, so $m=2$. Trials 2→3 double $[\mathrm B]$ and rate doubles, so $n=1$. Thus

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

Using trial 1,

$k=\frac{2.00\times10^{-3}\ \mathrm{M\,s^{-1}}}{(0.100\ \mathrm M)^2(0.100\ \mathrm M)}=2.00\ \mathrm{M^{-2}\,s^{-1}}.$

Units of $k$ depend on overall order.

Time data test the model

Linear forms help distinguish common single-reactant laws:

• zero order: $[\mathrm A]$ versus $t$ is linear;
• first order: $\ln[\mathrm A]$ versus $t$ is linear;
• second order: $1/[\mathrm A]$ versus $t$ is linear.

For a first-order process, $t_{1/2}=\ln2/k$ and does not depend on starting concentration. A straight-looking graph alone is insufficient: inspect axes, units, residuals, and range.

A mechanism must pass two tests

A proposed sequence of elementary steps must:

1. sum to the overall reaction after intermediates cancel; and
2. be consistent with the observed rate law.

For an elementary step, molecularity can justify the step's concentration dependence. Do not apply overall coefficients as orders unless the overall reaction is itself an elementary event. An intermediate is produced in one step and consumed in another; a catalyst is consumed and later regenerated.

Catalysts change pathways, not thermodynamics

A catalyst supplies an alternative mechanism with a lower activation-energy barrier. At the same temperature, a larger fraction of collisions can reach the transition region, increasing both forward and reverse rates. A catalyst does not change $\Delta H$, $\Delta G^\circ$, or the equilibrium constant, and it does not change the equilibrium composition; it helps equilibrium be reached faster.

Increasing temperature changes the molecular energy distribution and the rate constant. The Arrhenius relationship is

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

A linear plot of $\ln k$ versus $1/T$ has slope $-E_a/R$ when the model applies.

Retrieval challenge

1. If doubling $[\mathrm A]$ quadruples rate, what is the order in A?
2. Why can overall balanced coefficients fail to predict a rate law?
3. How does an intermediate differ from a catalyst?
4. Does a catalyst increase the equilibrium constant? Defend your answer.

Sources

• College Board AP Chemistry
• AP Chemistry Course and Exam Description
• IUPAC Gold Book
• NIST Chemical Kinetics Database