Unit 5: Kinetics
Chemical kinetics asks how fast a reaction proceeds and why that speed changes with conditions. It is distinct from thermochemistry (energy changes and a first look at spontaneity) and from equilibrium (where a reversible process ends up). Balanced equations from Unit 4 supply stoichiometry; kinetics supplies rates, rate laws, and mechanisms.
Collision theory and the molecular picture
Collision theory requires that molecules actually collide, that a collision carries at least the activation energy \(E_a\), and that the partners meet with a geometry that allows the relevant bonds to rearrange. A compact way to think about rate is that it scales with three factors:
\[\text{rate} \propto N \times f_E \times f_O.\]Here \(N\) is roughly the number of collisions per unit time (collision frequency), \(f_E\) is the fraction of collisions with enough energy to surmount the barrier, and \(f_O\) is the fraction with favorable orientation (sometimes folded into a steric factor in \(A\)).
Raising temperature increases \(N\) (faster motion, more frequent collisions) and sharply increases \(f_E\) (the high-energy tail of the speed distribution grows). Increasing concentration in a fixed volume raises \(N\) because more particles occupy the same space. A catalyst mainly increases \(f_E\) for the catalyzed path by offering a lower activation energy; it does not change \(\Delta H\) for the overall conversion. Changing physical state—for example, grinding a solid to increase surface area or improving mixing—raises effective collision frequency or exposes more reactive sites.
Homogeneous catalysis places the catalyst in the same phase as the reactants (solution catalysis is common). Heterogeneous catalysis uses a different phase, often a solid surface where adsorption aligns molecules for reaction. Acid–base catalysis is a major special case: \(\text{H}^+\), \(\text{OH}^-\), or other acids and bases can protonate or deprotonate substrates, stabilizing transition states and speeding steps without being consumed in the net equation.
Factors affecting rate (summary)
Concentration raises collision frequency \(N\) and enters the rate law directly. Temperature raises the amount of collisions and the frequency of effective colisions. Surface area and physical state (powder versus lump, mixing, phase contact) increase the rate of collision. Catalysts (homogeneous, heterogeneous, or acid–base) increases the frequency of effective collisions.
Reaction rate
The reaction rate measures how quickly reactant concentrations fall or product concentrations rise. For a generic reaction
\[a\text{A} + b\text{B} \longrightarrow c\text{C} + d\text{D},\]a common convention ties all species to one rate expression:
\[\text{rate} = -\frac{1}{a}\frac{d[\text{A}]}{dt} = -\frac{1}{b}\frac{d[\text{B}]}{dt} = +\frac{1}{c}\frac{d[\text{C}]}{dt} = +\frac{1}{d}\frac{d[\text{D}]}{dt}.\]The negative sign on reactants makes rate a positive quantity as written. Note that \(d\) represents the derivative, or the measure of the rate of change. If you watch one species by itself, the sign of \(d[\text{X}]/dt\) tells you whether it is being used up (negative for a reactant whose concentration falls) or formed (positive for a product). The stoichiometric factors convert those individual slopes into a single rate for the whole reaction.
In lab data you often measure average rate over an interval (slope of a chord on a concentration–time graph). Instantaneous rate is the slope of the tangent at one time—the limit as \(\Delta t \to 0\) and the quantity that appears in calculus-based rate laws.
Rate laws and the rate constant
Experiment determines how rate depends on concentration. The differential rate law (or simply rate law) has the form
\[\text{rate} = k[\text{A}]^m[\text{B}]^n[\text{C}]^p \cdots\]Here \(k\) is the rate constant (units depend on overall order), and \(m, n, p, \ldots\) are the orders with respect to each reactant. Those exponents are not taken from the balanced equation unless the reaction is a single elementary step (see mechanisms below). Overall reaction order is the sum \(m + n + p + \cdots\).
Important distinction: at fixed temperature (and without changing catalyst), \(k\) does not depend on concentrations or, for gases, on pressure as a separate knob—concentration enters only through the powers \([\text{A}]^m\) etc. Changing pressure in a gas-phase system often changes concentrations and therefore rate, but it does not change \(k\) itself. Temperature and catalysts change \(k\).
The method of initial rates compares initial rates while varying one reactant’s initial concentration at a time. Holding other concentrations fixed, if doubling \([\text{A}]\) multiplies the initial rate by \(2^m\), \(m\) is the order of \(A\). More generally, for two trials where only \([\text{A}]\) changes,
\[m = \frac{\ln(\text{rate}_1 / \text{rate}_2)}{\ln([\text{A}]_1 / [\text{A}]_2)},\]with nearest-integer order a common simplification on exams when data are clean. Repeat for each reactant, then substitute any one run’s data to solve for \(k\).
Units of \(k\) follow from \(\text{rate}\) in \(\text{M/s}\) (molarity per second) and the concentration powers. Examples for a single reactant \(\text{A}\):
- Zeroth order (\(\text{rate} = k\)): \(k\) in \(\text{M}\cdot\text{s}^{-1}\)
- First order (\(\text{rate} = k[\text{A}]\)): \(k\) in \(\text{s}^{-1}\)
- Second order (\(\text{rate} = k[\text{A}]^2\)): \(k\) in \(\text{M}^{-1}\cdot\text{s}^{-1}\)
Note that rate always has the units of \(\frac{M}{s}\) and concentration always has the units of \(M\).
Integrated rate laws and graphs
Integrating the differential law links concentration to time. Let \([\text{A}]_0\) be the initial concentration of the species tracked in the simplified one-reactant forms below.
Zeroth order:
\[[\text{A}] - [\text{A}]_0 = -kt\]Equivalently \([\text{A}] = [\text{A}]_0 - kt\). A plot of \([\text{A}]\) versus \(t\) is linear with slope \(-k\).
First order:
\[\ln[\text{A}] - \ln[\text{A}]_0 = -kt \quad\text{or}\quad \ln\!\left(\frac{[\text{A}]}{[\text{A}]_0}\right) = -kt.\]A plot of \(\ln[\text{A}]\) versus \(t\) is linear with slope \(-k\).
Second order:
\[\frac{1}{[\text{A}]} - \frac{1}{[\text{A}]_0} = kt.\]A plot of \(1/[\text{A}]\) versus \(t\) is linear with slope \(k\).
Which graph is linear is a standard way to infer order from concentration–time data. Half-life \(t_{1/2}\) is the time for \([\text{A}]\) to drop to half its initial value:
\[\text{zero:}\quad t_{1/2} = \frac{[\text{A}]_0}{2k}\] \[\text{first:}\quad t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}\] \[\text{second:}\quad t_{1/2} = \frac{1}{k[\text{A}]_0}\]For zero order, half-life shrinks as \([\text{A}]_0\) decreases. For first order (including many nuclear decay kinetics), \(t_{1/2}\) is constant throughout the reaction—independent of \([\text{A}]_0\). For second order, half-life grows as \([\text{A}]_0\) decreases.
Arrhenius equation
The Arrhenius equation relates the rate constant to temperature:
\[k = A e^{-E_a/(RT)}.\]\(A\) is the pre-exponential factor (collision frequency and orientation factors); \(R = 8.314 \text{ J/(mol}\cdot\text{K)}\) when \(E_a\) is in \(\text{J/mol}\); \(T\) is kelvin.
The logarithmic linear form is
\[\ln k = -\frac{E_a}{R}\cdot\frac{1}{T} + \ln A.\]A plot of \(\ln k\) versus \(1/T\) has slope \(-E_a/R\) and intercept \(\ln A\). The two-point relation is
\[\ln\frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right).\]For the same reaction under the same conditions, a larger \(k\) always goes with a higher temperature, because \(e^{-E_a/(RT)}\) increases when \(T\) increases. A modest rise in \(T\) can change \(k\) dramatically because it sits in an exponential.
Reaction mechanisms
Many balanced equations are net processes built from simpler elementary steps. Each elementary step has a molecularity—the number of reactant particles that must collide in that step (unimolecular, bimolecular, termolecular in the rare cases textbooks treat). For an elementary step only, rate-law exponents match stoichiometric coefficients for that step.
A reaction mechanism proposes a sequence of such steps. Species produced in one step and consumed in another, never appearing in the net equation, are intermediates. A catalyst is regenerated after the cycle; an intermediate is made and then used up. Transition states (or activated complexes) are high-energy configurations at barrier maxima along the path; they are not stable intermediates you bottle.
The rate-determining step (RDS) is the slowest step; it has the highest activation energy in the sequence and usually controls the observed rate law. If the first step is slow, its elementary law often appears directly in the experimental rate law (exponents from that step’s stoichiometry). Steps after the RDS do not change the concentration dependence of the rate law (they only consume intermediates as they leak out of the bottleneck).
If the experimental rate law does not match the stoichiometry of the overall equation, you infer multiple steps and a slow step that controls rate.
Pre-equilibrium and deriving rate laws
When the RDS is not the first step, a common pattern is a fast, reversible early step followed by a slow step. The early step sets up pre-equilibrium: forward and reverse rates of that step are large, so an equilibrium constant \(K\) links intermediates to reactant concentrations. You write the rate law from the RDS (using its elementary exponents), then substitute for any intermediate using the equilibrium expression from the fast step so that only overall reactants (and catalysts if present) appear in the final law. A more general steady-state approximation treats \(d[\text{intermediate}]/dt \approx 0\) when the intermediate is consumed as fast as it forms; AP problems often give setups where pre-equilibrium is enough.
Example mechanism:
- \(\text{A} + \text{B} \rightleftharpoons \text{I}\) fast
- \(\text{I} + \text{C} \longrightarrow \text{D}\) slow
The slow elementary step gives
\[\text{rate} = k_2[\text{I}][\text{C}].\]Because the intermediate \(\text{I}\) should not appear in the final experimental rate law, use the fast pre-equilibrium:
\[K = \frac{[\text{I}]}{[\text{A}][\text{B}]} \qquad\Rightarrow\qquad [\text{I}] = K[\text{A}][\text{B}].\]Substitution gives
\[\text{rate} = k_2K[\text{A}][\text{B}][\text{C}] = k_{\text{obs}}[\text{A}][\text{B}][\text{C}].\]Catalytic cycles
A catalytic cycle is a mechanism where the catalyst is consumed in an early elementary step and regenerated in a later step. The catalyst cancels from the net equation, but it can still appear in the rate law because its concentration affects how much of the faster pathway is available.
Simple acid-catalyzed pattern:
- \(\text{S} + \text{H}^+ \rightleftharpoons \text{SH}^+\) fast
- \(\text{SH}^+ \longrightarrow \text{P} + \text{H}^+\) slow or product-forming
Adding the steps cancels \(\text{H}^+\), so acid is not consumed overall. However, increasing \([\text{H}^+]\) can increase \([\text{SH}^+]\) and raise the observed rate.
For heterogeneous catalysts, the same cycle idea happens on a surface:
- Reactants adsorb to active sites.
- Bonds weaken or orient correctly on the surface.
- Products form and desorb, freeing the active site.
Catalyst poisoning occurs when another species binds strongly to active sites and blocks the cycle. Finely divided catalysts usually work faster because they expose more surface area and therefore more active sites.
Reaction coordinate diagrams
A reaction coordinate diagram plots energy versus progress from reactants to products. If reactants lie higher in energy than products, the net reaction is exothermic (\(\Delta H < 0\) in the usual convention); if products lie higher, it is endothermic (\(\Delta H > 0\)). Forward and reverse activation energies \(E_{a,\text{fwd}}\) and \(E_{a,\text{rev}}\) are measured from each side up to the transition state. For a simple one-step profile,
\[E_{a,\text{fwd}} - E_{a,\text{rev}} = \Delta H.\]A catalyst lowers \(E_{a,\text{fwd}}\) and \(E_{a,\text{rev}}\) by about the same amount (same pathway lowering for forward and reverse), so it speeds approach to equilibrium but does not change \(\Delta H\) or the equilibrium constant at fixed \(T\).
Transition state and catalysis (summary)
Along the reaction coordinate, the transition state (or activated complex) sits at an energy maximum separating reactants from products—the peak that must be crossed for the rearrangement to complete.
A catalyst provides an alternative pathway with lower activation energy. It is regenerated by the end of the cycle and does not appear in the net reaction. A catalyst increases \(k\) (both forward and reverse for a reversible path) and speeds approach to equilibrium but does not change \(\Delta G^\circ\) or the equilibrium constant for a given reaction at fixed \(T\).
Kinetics vs equilibrium (preview)
For a reversible elementary reaction, forward and reverse rates balance at equilibrium, linking rate constants to an equilibrium constant in simple cases. The full story belongs with equilibrium; here the essential point is that kinetics answers “how fast” while equilibrium answers “how far” under constant conditions.